17 files changed, 48399 insertions, 57 deletions

diff --git a/šola/ana1/kolokvij2.lyx b/šola/ana1/kolokvij2.lyx
index a057288..4d94e99 100644
--- a/šola/ana1/kolokvij2.lyx
+++ b/šola/ana1/kolokvij2.lyx
@@ -1,5 +1,5 @@
-#LyX 2.3 created this file. For more info see http://www.lyx.org/
-\lyxformat 544
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
 \begin_document
 \begin_header
 \save_transient_properties true
@@ -17,11 +17,11 @@
 enumitem
 theorems-ams
 \end_modules
-\maintain_unincluded_children false
+\maintain_unincluded_children no
 \language slovene
 \language_package default
-\inputencoding auto
-\fontencoding global
+\inputencoding auto-legacy
+\fontencoding auto
 \font_roman "default" "default"
 \font_sans "default" "default"
 \font_typewriter "default" "default"
@@ -29,7 +29,9 @@ theorems-ams
 \font_default_family default
 \use_non_tex_fonts false
 \font_sc false
-\font_osf false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
 \font_sf_scale 100 100
 \font_tt_scale 100 100
 \use_microtype false
@@ -63,7 +65,9 @@ theorems-ams
 \suppress_date false
 \justification false
 \use_refstyle 1
+\use_formatted_ref 0
 \use_minted 0
+\use_lineno 0
 \index Index
 \shortcut idx
 \color #008000
@@ -86,42 +90,20 @@ theorems-ams
 \papercolumns 1
 \papersides 1
 \paperpagestyle default
+\tablestyle default
 \tracking_changes false
 \output_changes false
+\change_bars false
+\postpone_fragile_content false
 \html_math_output 0
 \html_css_as_file 0
 \html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
 \end_header
 
 \begin_body
 
-\begin_layout Title
-List s formulami za 2.
- kolokvij Analize 1
-\end_layout
-
-\begin_layout Author
-
-\noun on
-Anton Luka Šijanec
-\end_layout
-
-\begin_layout Date
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-today
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
 \begin_layout Standard
 \begin_inset ERT
 status open
@@ -161,15 +143,18 @@ begin{multicols}{2}
 \begin_inset Formula $\log_{a}1=0$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\log_{a}a=1$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\log_{a}a^{x}=x$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $a^{\log_{a}x}=x$
 \end_inset
 
@@ -187,7 +172,8 @@ begin{multicols}{2}
 \begin_inset Formula $D=b^{2}-4ac$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}$
 \end_inset
 
@@ -205,7 +191,8 @@ begin{multicols}{2}
 \begin_inset Formula $zw=\left(ac-bd\right)+\left(ad+bc\right)i$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\vert zw\vert=\vert z\vert\vert w\vert$
 \end_inset
 
@@ -241,7 +228,8 @@ begin{multicols}{2}
 \begin_inset Formula $z^{2}=a^{2}+2abi-b^{2}$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $z^{3}=a^{3}-3ab^{2}+\left(3a^{2}b-b^{3}\right)i$
 \end_inset
 
@@ -259,7 +247,8 @@ begin{multicols}{2}
 \begin_inset Formula $z^{n}=r^{3}\left(\cos\left(3\phi\right)+i\sin\left(3\phi\right)\right)$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $\phi=\arctan\frac{\Im z}{\Re z}$
 \end_inset
 
@@ -316,7 +305,8 @@ je konv.
 \end_layout
 
 \begin_layout Standard
-Vrsta je konv., če je konv.
+Vrsta je konv.,
+ če je konv.
  njeno zap.
  delnih vsot.
 \end_layout
@@ -343,7 +333,8 @@ n+1; & q=1
 \series bold
 Primerjalni krit.
 \series default
-: 
+:
+ 
 \begin_inset Formula $\sum_{1}^{\infty}a_{k}$
 \end_inset
 
@@ -389,11 +380,13 @@ majoranta
 \series bold
 Kvocientni
 \series default
-: 
+:
+ 
 \begin_inset Formula $a_{k}>0$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $D_{n}\coloneqq\frac{a_{n}+1}{a_{n}}$
 \end_inset
 
@@ -419,11 +412,13 @@ Kvocientni
 \begin_inset Formula $\exists D\coloneqq\lim_{n\to\infty}D_{n}$
 \end_inset
 
-: 
+:
+ 
 \begin_inset Formula $\vert D\vert<1\Longrightarrow$
 \end_inset
 
-konv., 
+konv.,
+ 
 \begin_inset Formula $\vert D\vert>1\Longrightarrow div.$
 \end_inset
 
@@ -435,7 +430,9 @@ konv.,
 \series bold
 Korenski
 \series default
-: Kot Kvocientni, le da 
+:
+ Kot Kvocientni,
+ le da 
 \begin_inset Formula $D_{n}\coloneqq\sqrt[n]{a_{n}}$
 \end_inset
 
@@ -447,7 +444,8 @@ Korenski
 \series bold
 Leibnizov
 \series default
-: 
+:
+ 
 \begin_inset Formula $a_{n}\to0\Longrightarrow\sum_{1}^{\infty}\left(\left(-1\right)^{k}a_{k}\right)<\infty$
 \end_inset
 
@@ -476,11 +474,13 @@ Pri konv.
 \begin_inset Formula $x$
 \end_inset
 
-, pri enakomerni ni.
+,
+ pri enakomerni ni.
 \end_layout
 
 \begin_layout Standard
-Potenčna vrsta: 
+Potenčna vrsta:
+ 
 \begin_inset Formula $\sum_{j=1}^{\infty}b_{j}x^{j}$
 \end_inset
 
@@ -495,7 +495,8 @@ Potenčna vrsta:
 \end_inset
 
 abs.
- konv., 
+ konv.,
+ 
 \begin_inset Formula $\vert x\vert>R\Longrightarrow$
 \end_inset
 
@@ -719,7 +720,8 @@ divergira
 \end_layout
 
 \begin_layout Standard
-Krožnica: 
+Krožnica:
+ 
 \begin_inset Formula $\left(x-p\right)^{2}+\left(y-q\right)^{2}=r^{2}$
 \end_inset
 
@@ -727,7 +729,8 @@ Krožnica:
 \end_layout
 
 \begin_layout Standard
-Elipsa: 
+Elipsa:
+ 
 \begin_inset Formula $\frac{\left(x-p\right)^{2}}{a^{2}}+\frac{\left(y-q\right)^{2}}{b^{2}}=1$
 \end_inset
 
@@ -800,7 +803,8 @@ Odvod
 \begin_inset Formula $\frac{f'g-fg'}{g^{2}}$
 \end_inset
 
-, 
+,
+ 
 \begin_inset Formula $g\not=0$
 \end_inset
 
@@ -1260,7 +1264,8 @@ Zvezna
 \begin_inset Formula $\sup$
 \end_inset
 
-, je omejena in doseže vse funkcijske vrednosti na 
+,
+ je omejena in doseže vse funkcijske vrednosti na 
 \begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
 \end_inset
 
@@ -1275,7 +1280,8 @@ Zvezna
 \begin_inset Formula $I$
 \end_inset
 
-, če 
+,
+ če 
 \begin_inset Formula $\forall\varepsilon>0\exists\delta_{\left(\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
 \end_inset
 
@@ -1290,7 +1296,8 @@ Zvezna
 \begin_inset Formula $I$
 \end_inset
 
-, če 
+,
+ če 
 \begin_inset Formula $\forall\varepsilon>0\forall x\in I\exists\delta_{\left(x,\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
 \end_inset
 
diff --git a/šola/ana1/prak.lyx b/šola/ana1/prak.lyx
new file mode 100644
index 0000000..b6b21e7
--- /dev/null
+++ b/šola/ana1/prak.lyx
@@ -0,0 +1,1609 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\usepackage{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\DeclareMathOperator{\ctg}{ctg}
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+\end_modules
+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 1cm
+\topmargin 1cm
+\rightmargin 1cm
+\bottommargin 2cm
+\headheight 1cm
+\headsep 1cm
+\footskip 1cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+newcommand
+\backslash
+euler{e}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+setlength{
+\backslash
+columnseprule}{0.2pt}
+\backslash
+begin{multicols}{2}
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\log_{a}1=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}a=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}a^{x}=x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a^{\log_{a}x}=x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\log_{a}x^{n}=n\log_{a}x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $D=b^{2}-4ac$
+\end_inset
+
+,
+ 
+\begin_inset Formula $x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $zw=\left(ac-bd\right)+\left(ad+bc\right)i$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vert zw\vert=\vert z\vert\vert w\vert$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\arg\left(zw\right)=\arg z+\arg w$
+\end_inset
+
+ (kot)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z\overline{z}=a^{2}-\left(bi\right)^{2}=a^{2}+b^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\left(\cos\phi+i\sin\phi\right)$
+\end_inset
+
+
+\begin_inset Formula $\left(\cos\psi+i\sin\psi\right)=\cos\left(\phi+\psi\right)+i\sin\left(\phi+\psi\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z^{2}=a^{2}+2abi-b^{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $z^{3}=a^{3}-3ab^{2}+\left(3a^{2}b-b^{3}\right)i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $(a+b)^{n}=\sum_{k=0}^{n}{n \choose k}ab^{n-k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $z^{n}=r^{n}\left(\cos\left(n\phi\right)+i\sin\left(n\phi\right)\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\phi=\arctan\frac{\Im z}{\Re z}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Odprta množica ne vsebuje robnih točk.
+ Zaprta vsebuje vse.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sin\left(x\pm y\right)=\sin x\cdot\cos y\pm\sin y\cdot\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\cos\left(x\pm y\right)=\cos x\cdot\cos y\mp\sin y\cdot\sin x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\tan\left(x\pm y\right)=\frac{\tan x\pm\tan y}{1\text{\ensuremath{\mp\tan}x\ensuremath{\cdot\tan y}}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $a_{n}$
+\end_inset
+
+je konv.
+ 
+\begin_inset Formula $\Longleftrightarrow$
+\end_inset
+
+ 
+\begin_inset Formula $\forall\varepsilon>0:\exists n_{0}\ni:\forall n,m:n_{0}<n<m\wedge\vert a_{n}-a_{m}\vert<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\euler^{1/k}\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{nk}\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Vrsta je konv.,
+ če je konv.
+ njeno zap.
+ delnih vsot.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $s_{n}=\begin{cases}
+\frac{1-q^{n+1}}{1-q}; & q\not=1\\
+n+1; & q=1
+\end{cases}$
+\end_inset
+
+.
+ Geom.
+ vrsta konv.
+ 
+\begin_inset Formula $\Longleftrightarrow q\in\left(-1,1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Primerjalni krit.
+\series default
+:
+ 
+\begin_inset Formula $\sum_{1}^{\infty}a_{k}$
+\end_inset
+
+ konv.
+ 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ 
+\begin_inset Formula $b_{k}\leq a_{k}$
+\end_inset
+
+za 
+\begin_inset Formula $k>n_{0}$
+\end_inset
+
+ 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ vrsti sta navzdol omejeni 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ 
+\begin_inset Formula $\sum_{1}^{\infty}b_{k}$
+\end_inset
+
+ konv.
+ 
+\begin_inset Formula $\sum_{1}^{\infty}a_{k}$
+\end_inset
+
+ rečemo 
+\shape italic
+majoranta
+\shape default
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Kvocientni
+\series default
+:
+ 
+\begin_inset Formula $a_{k}>0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n}+1}{a_{n}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall n<n_{0}:D_{n}\in\left(0,1\right)\Longrightarrow\sum_{1}^{\infty}a_{k}<\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall n<n_{0}:D_{n}\geq1\Longrightarrow\sum_{1}^{\infty}a_{k}=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\exists D\coloneqq\lim_{n\to\infty}D_{n}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\vert D\vert<1\Longrightarrow$
+\end_inset
+
+konv.,
+ 
+\begin_inset Formula $\vert D\vert>1\Longrightarrow div.$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Korenski
+\series default
+:
+ Kot Kvocientni,
+ le da 
+\begin_inset Formula $D_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Leibnizov
+\series default
+:
+ 
+\begin_inset Formula $a_{n}\to0\Longrightarrow\sum_{1}^{\infty}\left(\left(-1\right)^{k}a_{k}\right)<\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Absolutna konvergenca 
+\begin_inset Formula $\left(\sum_{1}^{\infty}\vert a_{n}\vert<\infty\right)$
+\end_inset
+
+ 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ konvergenca
+\end_layout
+
+\begin_layout Standard
+Pri konv.
+ po točkah je 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni ni.
+\end_layout
+
+\begin_layout Standard
+Potenčna vrsta:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}x^{j}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $R^{-1}=\limsup_{k\to\infty}\sqrt[k]{\vert b_{k}\vert}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\vert x\vert<R\Longrightarrow$
+\end_inset
+
+abs.
+ konv.,
+ 
+\begin_inset Formula $\vert x\vert>R\Longrightarrow$
+\end_inset
+
+divergira
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\lim_{x\to a}\left(\alpha f\left(x\right)\right)=\alpha\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Tabular
+<lyxtabular version="3" rows="4" columns="4">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\sin$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cos$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\tan$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $30^{\circ}=\frac{\pi}{6}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{1}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{\sqrt{3}}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{\sqrt{3}}{3}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $45^{\circ}=\frac{\pi}{4}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{\sqrt{2}}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{\sqrt{2}}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $60^{\circ}=\frac{\pi}{3}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{\sqrt{3}}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{1}{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\sqrt{3}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Krožnica:
+ 
+\begin_inset Formula $\left(x-p\right)^{2}+\left(y-q\right)^{2}=r^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Elipsa:
+ 
+\begin_inset Formula $\frac{\left(x-p\right)^{2}}{a^{2}}+\frac{\left(y-q\right)^{2}}{b^{2}}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Tabular
+<lyxtabular version="3" rows="8" columns="4">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Izraz
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Odvod
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Izraz
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Odvod
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{f}{g}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $g\not=0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f\left(g\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f'\left(g\right)g'$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\tan x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cos^{-2}x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cot x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-sin^{-2}x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $a^{x}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $a^{x}\text{\ensuremath{\ln a}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{x}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{x}\left(1+\ln x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\log_{a}x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{1}{x\ln a}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{-1}\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{1}{f'\left(f^{-1}\left(a\right)\right)}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\arcsin x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\left(1-x^{2}\right)^{-\frac{1}{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\arccos x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-\left(1-x^{2}\right)^{-\frac{1}{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\arctan x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{1}{1+x^{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\text{arccot\,}x$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-\frac{1}{1+x^{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $x^{n}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $nx^{n-1}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(I\right)>0\Leftrightarrow f$
+\end_inset
+
+ konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(I\right)<0\Leftrightarrow f$
+\end_inset
+
+ konkavna na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+ab>0\wedge a<b\Leftrightarrow a^{-1}>b^{-1},\quad ab<0\wedge a<b\Leftrightarrow a^{-1}<b^{-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\sin x}{x}=1\quad\quad\tan\phi=\left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\lim_{x\to0}x\ln x=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+f\text{ zv.+odv.@ }\left[a,b\right]\Rightarrow\exists\xi\in\left[a,b\right]\ni:f\left(b\right)-f\left(a\right)=f'\left(\xi\right)\left(b-a\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+T_{f,a,n}\left(x\right)=\sum_{k=0}^{n}\frac{f^{\left(k\right)}\left(a\right)}{k!}\left(x-a\right)^{k}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f\text{\ensuremath{\in C^{n+1}}}$
+\end_inset
+
+ na odprtem 
+\begin_inset Formula $I\subset\mathbb{R}\Rightarrow\forall a,x\in I\exists c\in\left(\min\left\{ a,x\right\} ,\max\left\{ a,x\right\} \right)\ni:f\left(x\right)-T_{f,a,n}\left(x\right)=R_{f,a,n}\left(x\right)=\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}$
+\end_inset
+
+
+\begin_inset Formula $\left(x-a\right)^{n+1}.\text{ Posledično velja tudi takale ocena:}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\right|\leq M\Rightarrow R_{f,a,n}\left(x\right)=\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+R=\lim_{n\to\infty}\left|\frac{c_{n}}{c_{n+1}}\right|,\quad R=\lim_{n\to\infty}\frac{1}{\sqrt[n]{\left|c_{n}\right|}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zvezna 
+\begin_inset Formula $\text{f}$
+\end_inset
+
+ na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže 
+\begin_inset Formula $\inf$
+\end_inset
+
+ in 
+\begin_inset Formula $\sup$
+\end_inset
+
+,
+ je omejena in doseže vse funkcijske vrednosti na 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je enakomerno zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta_{\left(\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\forall x\in I\exists\delta_{\left(x,\varepsilon\right)}>0\ni:\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zvezna 
+\begin_inset Formula $f$
+\end_inset
+
+ na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+f'\left(x\right)=\lim_{x\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sinh x=\frac{e^{x}-e^{-x}}{2},\quad\cosh x=\frac{e^{x}+e^{-x}}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Uporabne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sin x=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}x^{2n+1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos x=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}x^{2n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{\left(2n+1\right)!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $e^{x}=\sum_{x=0}^{\infty}\frac{x^{n}}{n!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(1+x\right)^{\alpha}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\ln\left(1+x\right)=\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{x^{n}}{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\ln\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+Razcep racionalnih
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)^{3}}=\frac{A}{x-a}+\frac{B}{\left(x-1\right)^{2}}+\frac{C}{\left(x-1\right)^{3}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)\left(x-b\right)^{2}}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{\left(x-b\right)^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{p\left(x\right)}{\left(x-a\right)\left(x^{2}-b\right)}=\frac{A}{x-a}+\frac{Bx-C}{x^{2}-b}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Integrali
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\int\frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\arctan\frac{x}{a}+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{x^{2}-a^{2}}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{a^{2}-x^{2}}dx=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{ax+b}dx=\frac{1}{a}\ln\left|ax+b\right|+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\left(ax+b\right)^{n}dx=\frac{\left(ax+b\right)^{n+1}}{a\left(n+1\right)}+C
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sin^{2}\left(x\right)}dx=-\ctg\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\cos^{2}\left(x\right)}=\tan\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sqrt{a^{2}+x^{2}}}dx=\ln\left|x+\sqrt{x^{2}+a^{2}}\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{1}{\sqrt{x^{2}-a^{2}}}dx=\ln\left|x+\sqrt{x^{2}-a^{2}}\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\sqrt{a^{2}+x^{2}}dx=\frac{1}{2}\left(x\sqrt{a^{2}+x^{2}}+a^{2}\ln\left(\sqrt{a^{2}+x^{2}}+x\right)\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}\left(x\sqrt{a^{2}-x^{2}}+a^{2}\arctan\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{A}{x-a}dx=A\ln\left|x-a\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{A}{\left(x-a\right)^{n}}dx=\frac{-A}{n-1}\cdot\frac{1}{\left(x-a\right)^{n-1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\int\frac{Bx+C}{x^{2}+bx+c}=\frac{B}{2}\ln\left|x^{2}+bx+c\right|+\frac{2C-Bb}{\sqrt{-D}}\arctan\left(\frac{2x+b}{\sqrt{-D}}\right)
+\]
+
+\end_inset
+
+ In velja 
+\begin_inset Formula $D=b^{2}-4c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Pri 
+\begin_inset Formula $\int\sin\left(x\right)^{p}\cos\left(x\right)^{q}dx$
+\end_inset
+
+ lih 
+\begin_inset Formula $q$
+\end_inset
+
+ substituiramo 
+\begin_inset Formula $t=\cos\left(x\right)$
+\end_inset
+
+,
+ lih 
+\begin_inset Formula $p$
+\end_inset
+
+ pa 
+\begin_inset Formula $t=\sin\left(x\right)$
+\end_inset
+
+.
+ Pri sodih nižamo stopnje s formulo dvonega kota.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{multicols}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
new file mode 100644
index 0000000..104ba6c
--- /dev/null
+++ b/šola/ana1/teor.lyx
@@ -0,0 +1,16315 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
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+\use_default_options true
+\begin_modules
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+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
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+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 2cm
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+\end_header
+
+\begin_body
+
+\begin_layout Title
+Teorija Analize 1 —
+ IŠRM 2023/24
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Abstract
+Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića.
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Števila
+\end_layout
+
+\begin_layout Definition*
+Množica je matematični objekt,
+ ki predstavlja skupino elementov.
+ Če element 
+\begin_inset Formula $a$
+\end_inset
+
+ pripada množici 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $a\in A$
+\end_inset
+
+,
+ sicer pa 
+\begin_inset Formula $a\not\in A$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je podmnožica množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $B\subset A$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall b\in B:b\in A$
+\end_inset
+
+.
+ Presek 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $
+\end_inset
+
+.
+ Unijo 
+\begin_inset Formula $B$
+\end_inset
+
+ in 
+\begin_inset Formula $C$
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $
+\end_inset
+
+.
+ Razliko/komplement 
+\begin_inset Quotes gld
+\end_inset
+
+
+\begin_inset Formula $B$
+\end_inset
+
+ manj/brez 
+\begin_inset Formula $C$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ označimo 
+\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Realna števila
+\end_layout
+
+\begin_layout Standard
+Množico realnih števil označimo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ V njej obstajata binarni operaciji seštevanje 
+\begin_inset Formula $a+b$
+\end_inset
+
+ in množenje 
+\begin_inset Formula $a\cdot b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti seštevanja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a+\cdots+z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a+b=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a+c=0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b+0=b+a+c=0+c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in aditivni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $-a$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a+\left(-b\right)$
+\end_inset
+
+ običajno 
+\begin_inset Formula $+$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a-b$
+\end_inset
+
+,
+ čemur pravimo odštevanje 
+\begin_inset Formula $b$
+\end_inset
+
+ od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $b=-a$
+\end_inset
+
+ in 
+\begin_inset Formula $c=-b=-\left(-a\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $b+c$
+\end_inset
+
+ inverz od 
+\begin_inset Formula $\left(-b-c\right)$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $-\left(b+c\right)=-b-c$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Lastnosti množenja
+\end_layout
+
+\begin_layout Axiom
+Komutativnost:
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Asociativnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $a\cdots z$
+\end_inset
+
+ dobro definiran izraz.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj enote:
+ 
+\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Obstoj inverzov:
+ 
+\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Claim*
+Inverz je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $ab=1$
+\end_inset
+
+ in 
+\begin_inset Formula $ac=1$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $b=b1=bac=1c=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Inverz je funkcija in multiplikativni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+.
+ Pri zapisu 
+\begin_inset Formula $a\cdot b^{-1}$
+\end_inset
+
+ lahko 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ izpustimo in pišemo 
+\begin_inset Formula $a/b$
+\end_inset
+
+,
+ čemur pravimo deljenje 
+\begin_inset Formula $a$
+\end_inset
+
+ z 
+\begin_inset Formula $b$
+\end_inset
+
+ za neničeln 
+\begin_inset Formula $b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Skupne lastnosti v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $1\not=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Axiom
+Distributivnost:
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Urejenost 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Realna števila delimo na pozitivna 
+\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $
+\end_inset
+
+,
+ negativna 
+\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $
+\end_inset
+
+ in ničlo 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\geq0$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $x\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Axiom
+Če je 
+\begin_inset Formula $a\not=0$
+\end_inset
+
+,
+ je natanko eno izmed 
+\begin_inset Formula $\left\{ a,-a\right\} $
+\end_inset
+
+ pozitivno,
+ imenujemo ga absolutna vrednost 
+\begin_inset Formula $a$
+\end_inset
+
+ (pišemo 
+\begin_inset Formula $\left|a\right|$
+\end_inset
+
+),
+ in natanko eno negativno,
+ pišemo 
+\begin_inset Formula $-\left|a\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\left|0\right|=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ se 
+\begin_inset Formula $\left|a-b\right|$
+\end_inset
+
+ imenuje razdalja.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je večje od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a>b\Leftrightarrow a-b>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je manjše od 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $a<b\Leftrightarrow a-b<0$
+\end_inset
+
+.
+ Podobno 
+\begin_inset Formula $\leq$
+\end_inset
+
+ in 
+\begin_inset Formula $\geq$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Trikotniška neenakost.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo desni neenačaj.
+ Vemo 
+\begin_inset Formula $ab\leq\left|ab\right|$
+\end_inset
+
+ in 
+\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$
+\end_inset
+
+,
+ korenimo:
+ 
+\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Intervali
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $a<b$
+\end_inset
+
+.
+ Označimo odprti interval 
+\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $
+\end_inset
+
+,
+ zaprti 
+\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $
+\end_inset
+
+,
+ polodprti 
+\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,b)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $
+\end_inset
+
+ in podobno 
+\begin_inset Formula $[a,\infty)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Temeljne številske podmnožice
+\end_layout
+
+\begin_layout Subsubsection
+Naravna števila 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Matematična indukcija
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $A\subseteq\mathbb{N}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $1\in A$
+\end_inset
+
+ (baza) in 
+\begin_inset Formula $a\in A\Rightarrow a+1\in A$
+\end_inset
+
+ (korak),
+ tedaj 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $A=\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $1=\frac{1\cdot2}{2}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Predpostavimo 
+\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ Prištejmo 
+\begin_inset Formula $n+1$
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Cela števila 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Množica 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+ je zaprta za seštevanje in množenje,
+ torej 
+\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$
+\end_inset
+
+,
+ ni pa zaprta za odštevanje,
+ ker recimo 
+\begin_inset Formula $5-3\not\in\mathbb{N}$
+\end_inset
+
+.
+ Zapremo jo za odštevanje in dobimo množico 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Racionalna števila 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najmanjša podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+ in je zaprta za deljenje,
+ je 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja 
+\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Za 
+\begin_inset Formula $a\in\mathbb{Q}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b\not\in\mathbb{Q}$
+\end_inset
+
+ velja 
+\begin_inset Formula $a+b\not\in\mathbb{Q}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $a+b\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a+b-a\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $ab\in\mathbb{Q}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $b\in\mathbb{Q}$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Omejenost-množic"
+
+\end_inset
+
+Omejenost množic
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzgor omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo zgornja meja.
+ Najmanjši zgornji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo supremum ali natančna zgornja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\sup A$
+\end_inset
+
+.
+ Če je zgornja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je maksimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\max A$
+\end_inset
+
+.
+ Če množica ni navzgor omejena,
+ pišemo 
+\begin_inset Formula $\sup A=\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $s=\sup A\in\mathbb{R}$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $\forall a\in A:a\leq s$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$
+\end_inset
+
+,
+ torej za vsak neničeln 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni več natančna zgornja meja za 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je navzdol omejena 
+\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$
+\end_inset
+
+.
+ Takemu 
+\begin_inset Formula $m$
+\end_inset
+
+ pravimo spodnja meja.
+ Največji spodnji meji 
+\begin_inset Formula $A$
+\end_inset
+
+ pravimo infimum ali natančna spodnja meja množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $\inf A$
+\end_inset
+
+.
+ Če je spodnja meja 
+\begin_inset Formula $A$
+\end_inset
+
+ (
+\begin_inset Formula $m$
+\end_inset
+
+) element 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je minimum množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ označimo 
+\begin_inset Formula $m=\min A$
+\end_inset
+
+.
+ Če množica ni navzdol omejena,
+ pišemo 
+\begin_inset Formula $\inf A=-\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ je omejena,
+ če je hkrati navzgor in navzdol omejena.
+\end_layout
+
+\begin_layout Axiom
+\begin_inset CommandInset label
+LatexCommand label
+name "axm:Dedekind.-Vsaka-navzgor"
+
+\end_inset
+
+Dedekind.
+ Vsaka navzgor omejena množica v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natančno zgornjo mejo v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ aksiom 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "axm:Dedekind.-Vsaka-navzgor"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ ne velja.
+ Če 
+\begin_inset Formula $B\subset\mathbb{Q}$
+\end_inset
+
+,
+ se lahko zgodi,
+ da 
+\begin_inset Formula $\sup B\not\in\mathbb{Q}$
+\end_inset
+
+.
+ Primer:
+ 
+\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+
+\end_layout
+
+\begin_layout Subsection
+Decimalni zapis
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $
+\end_inset
+
+,
+ ki število natančno določajo.
+ Pišemo 
+\begin_inset Formula $x=m,d_{1}d_{2}\dots$
+\end_inset
+
+.
+ Natančno določitev mislimo v smislu:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $m\leq x<m+1$
+\end_inset
+
+ —
+ s tem se izognemo dvojnemu zapisu 
+\begin_inset Formula $1=0,\overline{9}$
+\end_inset
+
+ in 
+\begin_inset Formula $1=1,\overline{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $[m,m+1)$
+\end_inset
+
+ razdelimo na 10 enako dolgih polodprtih intervalov 
+\begin_inset Formula $I_{0},\dots,I_{9}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x$
+\end_inset
+
+ leži na natanko enem izmed njih,
+ indeks njega je 
+\begin_inset Formula $d_{1}$
+\end_inset
+
+.
+ Nadaljujemo tako,
+ da 
+\begin_inset Formula $I_{d_{1}}$
+\end_inset
+
+ razdelimo zopet na 10 delov itd.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Števila 
+\begin_inset Formula $x\in\mathbb{R^{-}}$
+\end_inset
+
+ pišemo tako,
+ da zapišemo decimalni zapis števila 
+\begin_inset Formula $-x$
+\end_inset
+
+ in predenj zapišemo 
+\begin_inset Formula $-$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če se decimalke v zaporedju 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ponavljajo,
+ uporabimo periodični zapis,
+ denimo 
+\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Kompleksna števila
+\end_layout
+
+\begin_layout Definition*
+Vpeljimo število 
+\begin_inset Formula $i$
+\end_inset
+
+ z lastnostjo 
+\begin_inset Formula $i^{2}=-1$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $i$
+\end_inset
+
+ rešitev enačbe 
+\begin_inset Formula $x^{2}+1=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $i\not\in\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sicer bi veljajo 
+\begin_inset Formula $i^{2}\geq0$
+\end_inset
+
+,
+ kar po definiciji ne velja.
+\end_layout
+
+\begin_layout Definition*
+Kompleksna števila so 
+\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $
+\end_inset
+
+.
+ 
+\begin_inset Formula $bi$
+\end_inset
+
+ je še nedefinirano,
+ zato za kompleksna števila definirano seštevanje in množenje za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ in 
+\begin_inset Formula $w=c+di$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Definiramo še konjugirano vrednost 
+\begin_inset Formula $z\in\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\overline{z}\coloneqq a-bi$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$
+\end_inset
+
+ za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Velja,
+ da je 
+\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$
+\end_inset
+
+ v smislu identifikacije 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ z množico 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ torej smo 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ razširili v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+,
+ kjer ima vsak polinom vedno rešitev.
+\end_layout
+
+\begin_layout Subsubsection
+Deljenje v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za 
+\begin_inset Formula $w,z\in\mathbb{C},w\not=0$
+\end_inset
+
+ iščemo 
+\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$
+\end_inset
+
+.
+ Ločimo dva primera:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+:
+ definiramo 
+\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $
+\end_inset
+
+ (splošno):
+ 
+\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$
+\end_inset
+
+,
+ z 
+\begin_inset Formula $\left|w\right|^{2}$
+\end_inset
+
+ pa znamo deliti,
+ ker je realen.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ sta komutativni,
+ asociativni,
+ distributivni,
+ 
+\begin_inset Formula $0$
+\end_inset
+
+ je aditivna enota,
+ 
+\begin_inset Formula $1$
+\end_inset
+
+ je multiplikativna.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+ vpeljemo 
+\begin_inset Formula $\Re z=a$
+\end_inset
+
+ in 
+\begin_inset Formula $\Im z=b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Opazimo 
+\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ si lahko predstavljamo kot urejene pare;
+ 
+\begin_inset Formula $a+bi$
+\end_inset
+
+ ustreza paru 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+ Tako 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ enačimo/identificiramo z 
+\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $
+\end_inset
+
+,
+ s čimer dobimo geometrično predstavitev 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ kot vektorje v 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za 
+\begin_inset Formula $z=a+bi$
+\end_inset
+
+,
+ predstavljen z vektorjem s komponentami 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $a=\left|z\right|\cos\varphi$
+\end_inset
+
+ in 
+\begin_inset Formula $v=\left|z\right|\sin\varphi$
+\end_inset
+
+.
+ Kotu 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ pravimo argument kompleksnega števila 
+\begin_inset Formula $z$
+\end_inset
+
+,
+ oznaka 
+\begin_inset Formula $\arg z$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $z=$
+\end_inset
+
+
+\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$
+\end_inset
+
+.
+ Velja
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem).
+ ne razumem.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
+\end_inset
+
+.
+ Množenje kompleksnh števil 
+\begin_inset Formula $z=\left|z\right|e^{i\varphi}$
+\end_inset
+
+ in 
+\begin_inset Formula $w=\left|w\right|e^{i\psi}$
+\end_inset
+
+ vrne število 
+\begin_inset Formula $zw$
+\end_inset
+
+,
+ za katero velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\arg zw=\arg z+\arg w$
+\end_inset
+
+ (do periode 
+\begin_inset Formula $2\pi$
+\end_inset
+
+ natančno)
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Zaporedja
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$
+\end_inset
+
+ se imenuje realno zaporedje,
+ oznaka 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ je funkcijska vrednost pri 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{n}=n$
+\end_inset
+
+:
+ 
+\begin_inset Formula $1,2,3,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $-1,4,-9,16,-25,\dots$
+\end_inset
+
+;
+ 
+\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Zaporedje lahko podamo rekurzivno.
+ Podamo prvi člen ali nekaj prvih členov in pravilo,
+ kako iz prejšnjih členov dobiti naslednje.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$
+\end_inset
+
+.
+ Fibbonacijevo zaporedje:
+ 
+\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$
+\end_inset
+
+ da zaporedje 
+\begin_inset Formula $1,1,2,3,5,8,\dots$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Posebni tipi zaporedij
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je aritmetično,
+ če 
+\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je geometrično,
+ če 
+\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je navzdol oz.
+ navzgor omejeno,
+ če je množica vseh členov tega taporedja navzgor oz.
+ navzdol omejena (glej 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Omejenost-množic"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+ Podobno z množico členov definiramo supremum,
+ infimum,
+ maksimum in infimum zaporedja.
+\end_layout
+
+\begin_layout Definition*
+Zaporedje je naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$
+\end_inset
+
+,
+ padajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$
+\end_inset
+
+,
+ strogo naraščajoče,
+ če 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$
+\end_inset
+
+,
+ strogo padajoče podobno,
+ monotono,
+ če je naraščajoče ali padajoče in strogo monotono,
+ če je strogo naraščajoče ali strogo padajoče.
+\end_layout
+
+\begin_layout Subsection
+Limita zaporedja
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je odprta,
+ če 
+\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $U\subseteq\mathbb{R}$
+\end_inset
+
+ je zaprta,
+ če je 
+\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$
+\end_inset
+
+ odprta.
+\end_layout
+
+\begin_layout Claim*
+Odprt interval je odprta množica.
+\end_layout
+
+\begin_layout Proof
+Za poljubna 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b>a$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $u\in\left(a,b\right)$
+\end_inset
+
+ poljuben.
+ Ustrezen 
+\begin_inset Formula $r$
+\end_inset
+
+ je 
+\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Zaprt interval je zaprt.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ poljubna in 
+\begin_inset Formula $b>a$
+\end_inset
+
+.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprt,
+ torej da je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$
+\end_inset
+
+ odprta množica.
+ Za poljuben 
+\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ velja,
+ da je bodisi 
+\begin_inset Formula $\in\left(-\infty,a\right)$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $\left(b,\infty\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$
+\end_inset
+
+.
+ Po prejšnji trditvi v obeh primerih velja 
+\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$
+\end_inset
+
+ res odprta,
+ torej je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ res zaprta.
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $B$
+\end_inset
+
+ je okolica točke 
+\begin_inset Formula $t\in\mathbb{R}$
+\end_inset
+
+,
+ če vsebuje kakšno odprto množico 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $t$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$
+\end_inset
+
+ 
+\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall V$
+\end_inset
+
+ okolica 
+\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$
+\end_inset
+
+,
+ pravimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$
+\end_inset
+
+ ali drugače 
+\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$
+\end_inset
+
+.
+ Če zaporedje ima limito,
+ pravimo,
+ da je konvergentno,
+ sicer je divergentno.
+\end_layout
+
+\begin_layout Claim*
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ ima natanko eno limito.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $J$
+\end_inset
+
+ in 
+\begin_inset Formula $L$
+\end_inset
+
+ limiti zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$
+\end_inset
+
+.
+ Torej
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ko trdimo,
+ da obstaja 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+,
+ še ne vemo,
+ ali sta za 
+\begin_inset Formula $L$
+\end_inset
+
+ in 
+\begin_inset Formula $J$
+\end_inset
+
+ ta 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ ista.
+ Ampak trditev še vedno velja,
+ ker lahko vzamemo večjega izmed njiju,
+ ako bi bila drugačna.
+\end_layout
+
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$
+\end_inset
+
+.
+ Velja torej 
+\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $J\not=L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left|J-L\right|\not=0$
+\end_inset
+
+,
+ naj bo 
+\begin_inset Formula $\left|J-L\right|=k$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$
+\end_inset
+
+,
+ ustrezen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ je na primer 
+\begin_inset Formula $\frac{\left|J-L\right|}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset CommandInset label
+LatexCommand label
+name "Konvergentno-zaporedje-v-R-je-omejeno"
+
+\end_inset
+
+Konvergentno zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je omejeno.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Znotraj intervala 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ so vsi členi zaporedja razen končno mnogo (
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+).
+ 
+\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$
+\end_inset
+
+ je unija dveh omejenih množic;
+ 
+\begin_inset Formula $\left(L-1,L+1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
+\end_inset
+
+,
+ zato je tudi sama omejena.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pmkdlim}{Naj bosta}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentni zaporedji v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so tudi 
+\begin_inset Formula $\left(a_{n}*b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentna in velja 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}*b_{n}=\lim_{n\to\infty}a_{n}*\lim_{n\to\infty}b_{n}$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot\right\} $
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\lim_{n\to\infty}b_{n}\not=0$
+\end_inset
+
+,
+ isto velja tudi za deljenje.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to A$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}\to B$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{1},n_{2}\ni:\left(n>n_{1}\Rightarrow\left|a_{n}-A\right|<\varepsilon\right)\wedge\left(n>n_{2}\Rightarrow\left|b_{n}-B\right|<\varepsilon\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}=\max\left\{ n_{1},n_{2}\right\} \ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|<\varepsilon\wedge\left|b_{n}-B\right|<\varepsilon$
+\end_inset
+
+.
+ Dokažimo za vse operacije:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $+$
+\end_inset
+
+ Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|+\left|a_{n}-B\right|<2\varepsilon$
+\end_inset
+
+.
+ Oglejmo si sedaj
+\begin_inset Formula 
+\[
+\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|=\left|\left(a_{n}-A\right)+\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in uporabimo še prejšnjo trditev,
+ torej 
+\begin_inset Formula $\forall2\varepsilon\exists n_{0}\ni:\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|\leq2\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-$
+\end_inset
+
+ Oglejmo si
+\begin_inset Formula 
+\[
+\left|\left(a_{n}-b_{n}\right)-\left(A-B\right)\right|=\left|a_{n}-b_{n}-A+B\right|=\left|\left(a_{n}-A\right)+\left(-\left(b_{n}-B\right)\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
+\]
+
+\end_inset
+
+in nato kot zgoraj.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\cdot$
+\end_inset
+
+ Oglejmo si 
+\begin_inset Formula 
+\[
+\left|a_{n}b_{n}-AB\right|=\left|a_{n}b_{n}-Ab_{n}+Ab_{n}-AB\right|=\left|\left(a_{n}-A\right)b_{n}+A\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|.
+\]
+
+\end_inset
+
+Od prej vemo,
+ da sta zaporedji omejeni,
+ ker sta konvergentni,
+ zato 
+\begin_inset Formula $\exists M>0\forall n\in\mathbb{N}:\left|b_{n}\right|\leq M$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ taka,
+ da 
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon}{2M}$
+\end_inset
+
+ in 
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|A\right|}$
+\end_inset
+
+.
+ Tedaj za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|<\frac{\varepsilon}{2M}M+\left|A\right|\frac{\varepsilon}{2\left|A\right|}=\varepsilon$
+\end_inset
+
+,
+ skratka 
+\begin_inset Formula $\left|a_{n}b_{n}-AB\right|<\varepsilon$
+\end_inset
+
+,
+ s čimer dokažemo 
+\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $/$
+\end_inset
+
+ Ker je 
+\begin_inset Formula $B\not=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|b_{n}\right|\geq\frac{\left|B\right|}{2}>0$
+\end_inset
+
+.
+ ZDB vsi členi zaporedja razen končno mnogo so v poljubno majhni okolici 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+.
+ Če torej vzamemo točko na polovici med 0 in 
+\begin_inset Formula $\left|B\right|$
+\end_inset
+
+,
+ to je 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ bo neskončno mnogo absolutnih vrednosti členov večjih od 
+\begin_inset Formula $\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Pri razumevanju pomaga številska premica.
+ Nadalje uporabimo predpostavko z 
+\begin_inset Formula $\varepsilon=\frac{\left|B\right|}{2}$
+\end_inset
+
+,
+ torej je za 
+\begin_inset Formula $n>n_{0}:$
+\end_inset
+
+ 
+\begin_inset Formula $\left|B-b_{n}\right|<\frac{\left|B\right|}{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left|b_{n}\right|=\left|B-\left(B-b_{n}\right)\right|=\left|B+\left(-\left(B-b_{n}\right)\right)\right|\overset{\text{trik. neen.}}{\geq}\left|\left|B\right|-\left|B-b_{n}\right|\right|=\left|B\right|-\left|B-b_{n}\right|\overset{\text{predp.}}{>}\left|B\right|-\frac{\left|B\right|}{2}=\frac{\left|B\right|}{2},
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|b_{n}\right|>\frac{\left|B\right|}{2}$
+\end_inset
+
+.
+ Če spet izpustimo končno začetnih členov,
+ velja
+\begin_inset Formula 
+\[
+\frac{a_{n}}{b_{n}}-\frac{A}{B}=\frac{a_{n}B-Ab_{n}}{b_{n}B}\overset{\text{prištejemo in odštejemo člen}}{=}\frac{\left(a_{n}-A\right)B+A\left(B-b_{n}\right)}{Bb_{n}}=\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A/B}{b_{n}}\left(B-b_{n}\right)
+\]
+
+\end_inset
+
+sedaj uporabimo na obeh straneh absolutno vrednost:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|=\left|\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A}{Bb_{n}}\left(B-b_{n}\right)\right|\leq\frac{1}{\left|b_{n}\right|}\left|a_{n}-A\right|+\frac{\left|A\right|}{\left|B\right|\left|b_{n}\right|}\left|B-b_{n}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|
+\]
+
+\end_inset
+
+skratka 
+\begin_inset Formula $\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|$
+\end_inset
+
+.
+ Opazimo,
+ da 
+\begin_inset Formula $\frac{2}{\left|B\right|}$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{2\left|A\right|}{\left|B\right|^{2}}$
+\end_inset
+
+ nista odvisna od 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Sedaj vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ in 
+\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
+\end_inset
+
+ takšna,
+ da velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon\left|B\right|}{4}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon\left|B\right|^{2}}{4\left|A\right|}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Tedaj iz zgornje ocene sledi za 
+\begin_inset Formula $n\geq\max\left\{ n_{0},n_{1},n_{2}\right\} $
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{\cancel{2}}{\cancel{\left|B\right|}}\cdot\frac{\varepsilon\cancel{\left|B\right|}}{\cancelto{2}{4}}+\frac{\cancel{2}\cancel{\left|A\right|}}{\cancel{\left|B\right|^{2}}}\cdot\frac{\varepsilon\cancel{\left|B\right|^{2}}}{\cancelto{2}{4}\cancel{\left|A\right|}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
+\]
+
+\end_inset
+
+s čimer dokažemo 
+\begin_inset Formula $a_{n}/b_{n}\to A/B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+.
+ Izračunajmo
+\begin_inset Formula 
+\[
+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{\cdots}}}}\eqqcolon\alpha.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\alpha$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $x_{0}=0,x_{1}=\sqrt{a},x_{2}=\sqrt{a+\sqrt{a}},x_{3}=\sqrt{a+\sqrt{a+\sqrt{a}}},\dots,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Iz zadnjega sledi 
+\begin_inset Formula $x_{n+1}^{2}=a+x_{n}$
+\end_inset
+
+.
+ Če torej limita 
+\begin_inset Formula $\alpha\coloneqq\lim x_{n}$
+\end_inset
+
+ obstaja,
+ mora veljati 
+\begin_inset Formula $\alpha^{2}=a+\alpha$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\alpha_{1,2}=\frac{1\pm\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+ Opcija z minusom ni mogoča,
+ ker je zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ očitno pozitivno.
+ Če torej limita obstaja (
+\series bold
+česar še nismo dokazali
+\series default
+),
+ je enaka 
+\begin_inset Formula $\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+,
+ za primer 
+\begin_inset Formula $a=2$
+\end_inset
+
+ je torej 
+\begin_inset Formula $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}=2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Lahko se zgodi,
+ da limita rekurzivno podanega zaporedja ne obstaja,
+ čeprav jo znamo izračunati,
+ če bi obstajala.
+ Na primer 
+\begin_inset Formula $y_{1}\coloneqq1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $y_{n+1}=1-2y_{n}$
+\end_inset
+
+ nam da zaporedje 
+\begin_inset Formula $1,-1,3,-5,11,\dots$
+\end_inset
+
+,
+ kar očitno nima limite.
+ Če bi limita obstajala,
+ bi zanjo veljalo 
+\begin_inset Formula $\beta=1-2\beta$
+\end_inset
+
+ oz.
+ 
+\begin_inset Formula $3\beta=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\beta=\frac{1}{3}$
+\end_inset
+
+.
+ Navedimo torej nekaj zadostnih in potrebnih pogojev za konvergenco zaporedij.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ monotono realno zaporedje.
+ Če narašča,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\sup\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ Če pada,
+ ima limito 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=\inf\left\{ a_{n},n\in\mathbb{N}\right\} $
+\end_inset
+
+.
+ (
+\begin_inset Formula $\sup$
+\end_inset
+
+ in 
+\begin_inset Formula $\inf$
+\end_inset
+
+ imata lahko tudi vrednost 
+\begin_inset Formula $\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $-\infty$
+\end_inset
+
+ —
+ zaporedje s tako limito ni konvergentno v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Denimo,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča.
+ Pišimo 
+\begin_inset Formula $s\coloneqq\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $s-\varepsilon$
+\end_inset
+
+ ni zgornja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s-\varepsilon<a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+naraščajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\geq a_{n_{0}}>s-\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\leq s$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $s$
+\end_inset
+
+ zgornja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in(s-\varepsilon,s]\subset\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+,
+ s čimer dokažemo konvergenco.
+\end_layout
+
+\begin_layout Proof
+Denimo sedaj,
+ da 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada.
+ Dokaz je povsem analogen.
+ Pišimo 
+\begin_inset Formula $m\coloneqq\inf_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $m+\varepsilon$
+\end_inset
+
+ ni spodnja meja za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:m+\varepsilon>a_{n_{0}}$
+\end_inset
+
+.
+ Ker pa je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\leq a_{n_{0}}<m+\varepsilon$
+\end_inset
+
+.
+ Hkrati je 
+\begin_inset Formula $a_{n}\geq m$
+\end_inset
+
+,
+ saj je 
+\begin_inset Formula $m$
+\end_inset
+
+ spodnja meja.
+ Torej 
+\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in[m,m+\varepsilon)\subset\left(m-\varepsilon,m+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Za monotono zaporedje velja,
+ da je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno natanko tedaj,
+ ko je omejeno.
+\end_layout
+
+\begin_layout Example*
+Naj bo,
+ kot prej,
+ 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $x_{0}=0,x_{n+1}=\sqrt{a+x_{n}}$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $\left(x_{n}\right)_{n}$
+\end_inset
+
+ konvergentno.
+ Dovolj je pokazati,
+ da je naraščajoče in navzgor omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}>x_{1}=\sqrt{a}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(x_{n+1}-x_{n}\right)\left(x_{n+1}+x_{n}\right)=x_{n+1}^{2}-x_{n}^{2}=\left(a+x_{n}\right)-\left(a+x_{n-1}\right)=x_{n}-x_{n-1}
+\]
+
+\end_inset
+
+Ker je zaporedje pozitivno,
+ je 
+\begin_inset Formula $x_{n+1}+x_{n}>0$
+\end_inset
+
+.
+ Desna stran je po I.
+ P.
+ pozitivna,
+ torej tudi 
+\begin_inset Formula $x_{n+1}-x_{n}>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Če je zaporedje res omejeno,
+ je po zgornjem tudi konvergentno in je 
+\begin_inset Formula $\sup_{n\in\mathbb{N}}x_{n}=\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}\leq\frac{1+\sqrt{1+4a+4a^{2}}}{2}=\frac{1+\sqrt{\left(2a+1\right)^{2}}}{}=1+a$
+\end_inset
+
+.
+ Uganili smo neko zgornjo mejo.
+ Domnevamo,
+ da 
+\begin_inset Formula $\forall n\in\mathbb{N}:x_{n}\leq1+a$
+\end_inset
+
+.
+ Dokažimo to z indukcijo:
+ Baza:
+ 
+\begin_inset Formula $0=x_{0}<1+a$
+\end_inset
+
+.
+ Po I.
+ P.
+ 
+\begin_inset Formula $x_{n}>1+a$
+\end_inset
+
+.
+ Korak:
+\begin_inset Formula 
+\[
+x_{n+1}=\sqrt{x_{n}+a}\leq\sqrt{1+a+a}=\sqrt{1+2a}<\sqrt{1+2a+2a^{2}}=1+a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+S tem smo dokazali,
+ da 
+\begin_inset Formula $\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+To lahko dokažemo tudi na alternativen način.
+ Vidimo,
+ da je edini kandidat za limito,
+ če obstaja 
+\begin_inset Formula $L=\frac{1+\sqrt{1+4a}}{2}$
+\end_inset
+
+ in da torej velja 
+\begin_inset Formula $L^{2}=a+L$
+\end_inset
+
+.
+ Preverimo,
+ da je 
+\begin_inset Formula $L$
+\end_inset
+
+ res limita:
+ 
+\begin_inset Formula 
+\[
+x_{n+1}-L=\sqrt{a+x_{n}}-L=\frac{\left(\sqrt{a+x_{n}}-L\right)\left(\sqrt{a+x_{n}}+L\right)}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-L^{2}}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-\left(a+L\right)}{\sqrt{a+x_{n}}+L}=\frac{x_{n}-L}{\sqrt{a+x_{n}}+L}.
+\]
+
+\end_inset
+
+Vpeljimo sedaj 
+\begin_inset Formula $y_{n}\coloneqq x_{n}-L$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|y_{n+1}\right|\leq\frac{\left|y_{n}\right|}{\sqrt{a+x_{n}}+L}\leq\frac{\left|y_{n}\right|}{L}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left|y_{0}\right|=L$
+\end_inset
+
+,
+ dobimo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za razumevanje si oglej nekaj členov rekurzivnega zaporedje 
+\begin_inset Formula $y_{0}=L,y_{n}=\frac{\left|y_{n+1}\right|}{L}$
+\end_inset
+
+.
+ Začnemo z 1 in nato vsakič delimo z 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ oceno 
+\begin_inset Formula $\left|y_{n}\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\left|x_{n}-L\right|\leq\frac{1}{L^{n-1}}$
+\end_inset
+
+.
+ Ker iz definicije 
+\begin_inset Formula $L$
+\end_inset
+
+ sledi 
+\begin_inset Formula $L>1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $L^{n}\to\infty$
+\end_inset
+
+ za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+,
+ torej smo dokazali,
+ da 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ eksponentno pada proti 0 za 
+\begin_inset Formula $n\to\infty$
+\end_inset
+
+.
+ Eksponentno padanje 
+\begin_inset Formula $\left|x_{n}-L\right|$
+\end_inset
+
+ proti 0 je dovolj,
+ da rečemo,
+ da zaporedje konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+a res,
+ vprašaj koga.
+ ne razumem.
+ zakaj.
+ TODO.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\lim_{n\to\infty}\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\cos n$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Proof
+Pišimo 
+\begin_inset Formula $a_{n}=\sin n$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=\cos n$
+\end_inset
+
+.
+ Iz adicijskih izrekov dobimo 
+\begin_inset Formula $a_{n+1}=\sin\left(n+1\right)=\sin n\cos1+\cos n\sin1=a_{n}\cos1+b_{n}\sin1$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $b_{n}=\frac{a_{n+1}-a_{n}\cos1}{\sin1}$
+\end_inset
+
+.
+ Torej če 
+\begin_inset Formula $\exists a\coloneqq\lim_{n\to\infty}a_{n},a\in\mathbb{R}$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists b\coloneqq\lim_{n\to\infty}b_{n},b\in\mathbb{R}$
+\end_inset
+
+.
+ Podobno iz adicijske formule za 
+\begin_inset Formula $\cos\left(n+1\right)$
+\end_inset
+
+ sledi 
+\begin_inset Formula $a_{n}=\frac{b_{n}\cos1-b_{n+1}}{\sin1}$
+\end_inset
+
+,
+ torej če 
+\begin_inset Formula $\exists b$
+\end_inset
+
+,
+ potem tudi 
+\begin_inset Formula $\exists a$
+\end_inset
+
+.
+ Iz obojega sledi,
+ da 
+\begin_inset Formula $\exists a\Leftrightarrow\exists b$
+\end_inset
+
+.
+ Posledično,
+ če 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ obstajata,
+ iz zgornjih obrazcev za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}$
+\end_inset
+
+ sledi,
+ da za
+\begin_inset Formula 
+\[
+\lambda=\frac{1-\cos1}{\sin1}\in\left(0,1\right)
+\]
+
+\end_inset
+
+velja 
+\begin_inset Formula $b=\lambda a$
+\end_inset
+
+ in 
+\begin_inset Formula $a=-\lambda b$
+\end_inset
+
+ in zato 
+\begin_inset Formula $b=\lambda\left(-\lambda b\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $-1=\lambda^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda=i$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\lambda\in\left(0,1\right)$
+\end_inset
+
+.
+ Podobno za 
+\begin_inset Formula $a=-\lambda\left(\lambda a\right)$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $1=-\lambda^{2}$
+\end_inset
+
+,
+ kar je zopet 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Edina druga opcija je,
+ da je 
+\begin_inset Formula $a=b=0$
+\end_inset
+
+.
+ Hkrati pa vemo,
+ da 
+\begin_inset Formula $a_{n}^{2}+b_{n}^{2}=1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $a+b=1$
+\end_inset
+
+,
+ kar ni mogoče za ničelna 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $b$
+\end_inset
+
+ ne obstajata.
+\end_layout
+
+\begin_layout Subsection
+Eulerjevo število
+\end_layout
+
+\begin_layout Theorem*
+Bernoullijeva neenakost.
+ 
+\begin_inset Formula $\forall\alpha\leq1,n\in\mathbb{N}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Z indukcijo na 
+\begin_inset Formula $n$
+\end_inset
+
+ ob fiksnem 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{1}=1-1\alpha$
+\end_inset
+
+.
+ Velja celo enakost.
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ Velja 
+\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\left(1-\alpha\right)^{n+1}=\left(1-\alpha\right)\left(1-\alpha\right)^{n}\geq\left(1-\alpha\right)$
+\end_inset
+
+
+\begin_inset Formula $\left(1-n\alpha\right)=1-n\alpha-\alpha-n\alpha^{2}=1-\left(n+1\right)\alpha-n\alpha^{2}\geq1-\left(n+1\right)\alpha$
+\end_inset
+
+,
+ torej ocena velja tudi za 
+\begin_inset Formula $n+1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Vpeljimo oznaki:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ označimo 
+\begin_inset Formula $n!=1\cdot2\cdot3\cdot\cdots\cdot n$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+faktoriala oziroma 
+\begin_inset Formula $n-$
+\end_inset
+
+fakulteta).
+ Ker velja 
+\begin_inset Formula $n!=n\cdot\left(n-1\right)!$
+\end_inset
+
+ za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+,
+ je smiselno definirati še 
+\begin_inset Formula $0!=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Za 
+\begin_inset Formula $n,k\in\mathbb{N}$
+\end_inset
+
+ označimo še binomski simbol:
+ 
+\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n$
+\end_inset
+
+ nad 
+\begin_inset Formula $k$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Itemize
+Če je 
+\begin_inset Formula $\left(a_{k}\right)_{k}$
+\end_inset
+
+ neko zaporedje (lahko tudi končno),
+ lahko pišemo 
+\begin_inset Formula $\sum_{k=1}^{n}a_{k}\coloneqq a_{1}+a_{2}+\cdots+a_{n}$
+\end_inset
+
+ (pravimo summa) in 
+\begin_inset Formula $\prod_{k=1}^{n}a_{k}\coloneqq a_{1}\cdot\cdots\cdot a_{n}$
+\end_inset
+
+ (pravimo produkt).
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\prod_{k=1}^{n}k=1\cdot2\cdot3\cdot\cdots\cdot n=n!$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Binomska formula.
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{R},n\in\mathbb{N}:\left(a+b\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Indukcija po 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza 
+\begin_inset Formula $n=1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\sum_{k=0}^{1}\binom{n}{k}a^{k}b^{n-k}=\binom{1}{0}a^{0}b^{1-0}+\binom{1}{1}a^{1}b^{1-1}=a+b=\left(a+b\right)^{1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.
+ 
+\begin_inset Formula $\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\left(a+b\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula 
+\[
+\left(a+b\right)^{n+1}=\left(a+b\right)\left(a+b\right)^{n}=\left(a+b\right)\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=
+\]
+
+\end_inset
+
+sedaj naj bo 
+\begin_inset Formula $m=k+1$
+\end_inset
+
+ v levem členu:
+\begin_inset Formula 
+\[
+=\sum_{m=1}^{n+1}\binom{n}{m-1}a^{m}b^{n-\left(m-1\right)}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+Sedaj obravnavajmo le izraz v oglatih oklepajih:
+\begin_inset Formula 
+\[
+\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}+\frac{n!}{k!\left(n-k\right)!}=\frac{kn!}{k!\left(n-k+1\right)!}+\frac{n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=\frac{kn!+n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{n!\left(\cancel{k+}n\cancel{-k}+1\right)}{k!\left(n+1-k\right)!}=\frac{n!\left(n+1\right)}{k!\left(n+1-k\right)!}=\frac{\left(n+1\right)!}{k!\left(n+1-k\right)}=\binom{n+1}{k}
+\]
+
+\end_inset
+
+in skratka dobimo 
+\begin_inset Formula $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$
+\end_inset
+
+.
+ Vstavimo to zopet v naš zgornji račun:
+\begin_inset Formula 
+\[
+\cdots=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{k}b^{n-k+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Bernoulli.
+ Zaporedje 
+\begin_inset Formula $a_{n}\coloneqq\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ je konvergentno.
+\end_layout
+
+\begin_layout Proof
+Dokazali bomo,
+ da je naraščajoče in omejeno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naraščanje:
+ Dokazujemo,
+ da za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+ velja 
+\begin_inset Formula $a_{n}\geq a_{n-1}$
+\end_inset
+
+ oziroma 
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n}{n-1}\right)^{n-1}=\left(\frac{n-1}{n}\right)^{1-n}=\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1-\frac{1}{n}\right)^{1-n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}=\left(\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right)\right)^{n}=\left(1+\frac{1}{n}\right)^{n}\left(1-\frac{1}{n}\right)^{n}\overset{?}{\geq}1-\frac{1}{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(1+\frac{1}{n^{2}}\right)^{n}\overset{?}{\geq}1-\frac{1}{n},
+\]
+
+\end_inset
+
+kar je poseben primer Bernoullijeve neenakosti za 
+\begin_inset Formula $\alpha=\frac{1}{n^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Omejenost:
+ Po binomski formuli je 
+\begin_inset Formula 
+\[
+a_{n}=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{n}\right)^{k}=\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=1+1+\sum_{k=2}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{n^{k}}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\cdots\cdot\frac{n-k+1}{n}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\cancel{\left(1-0\right)}\cdot\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdot\cdots\cdot\left(1-\frac{k-1}{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<2+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}^{n}\frac{1}{2^{k-1}}=\cdots
+\]
+
+\end_inset
+
+Opomnimo,
+ da je 
+\begin_inset Formula $1-\frac{j}{n}<1$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<1$
+\end_inset
+
+ (prvi neenačaj) ter 
+\begin_inset Formula $k!=1\cdot2\cdot3\cdot\cdots\cdot k\geq1\cdot2\cdot2\cdot\cdots\cdot2=2^{k-1}$
+\end_inset
+
+ (drugi).
+ Sedaj si z indukcijo dokažimo 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $n=2$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\frac{1}{2^{2-1}}=1-\frac{1}{2^{2-1}}=1-\frac{1}{2}=\frac{1}{2}$
+\end_inset
+
+.
+ Velja!
+\end_layout
+
+\begin_layout Itemize
+I.
+ P.:
+ 
+\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ 
+\begin_inset Formula $\sum_{k=2}^{n+1}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}+\frac{1}{2^{n+1-1}}=1-2\cdot2^{-n}+2^{-n}=1+2^{-n}\left(1-2\right)=1+2^{-n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In nadaljujmo z računanjem:
+\begin_inset Formula 
+\[
+\cdots=2+1-\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}},
+\]
+
+\end_inset
+
+s čimer dobimo zgornjo mejo 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}<3$
+\end_inset
+
+.
+ Ker je očitno 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}>0$
+\end_inset
+
+,
+ je torej zaporedje omejeno in ker je tudi monotono po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{prejšnjem izreku}
+\end_layout
+
+\end_inset
+
+ konvergira.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Definition*
+Označimo število 
+\begin_inset Formula $e\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$
+\end_inset
+
+ in ga imenujemo Eulerjevo število.
+ Velja 
+\begin_inset Formula $e\approx2,71828$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+V dokazu vidimo moč izreka 
+\begin_inset Quotes gld
+\end_inset
+
+omejenost in monotonost 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca
+\begin_inset Quotes grd
+\end_inset
+
+,
+ saj nam omogoča dokazati konvergentnost zaporedja brez kandidata za limito.
+ Jasno je,
+ da ne bi mogli vnaprej uganiti,
+ da je limita ravno 
+\shape italic
+transcendentno število
+\shape default
+ 
+\begin_inset Formula $e$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Podzaporedje zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ je poljubno zaporedje oblike 
+\begin_inset Formula $\left(a_{\varphi\left(n\right)}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ strogo naraščajoča funkcija.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $L$
+\end_inset
+
+ tudi limita vsakega podzaporedja.
+\end_layout
+
+\begin_layout Proof
+Po predpostavki velja 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}:\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po predpostavki obstaja 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+,
+ da bodo vsi členi zaporedja po 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tem v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+ Iz definicijskega območja 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ vzemimo poljuben element 
+\begin_inset Formula $n_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+.
+ Gotovo obstaja,
+ ker je definicijsko območje števno neskončne moči in s pogojem 
+\begin_inset Formula $n_{1}\geq n_{0}$
+\end_inset
+
+ onemogočimo izbiro le končno mnogo elementov.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Če slednji ne obstaja,
+ je v 
+\begin_inset Formula $D_{\varphi}$
+\end_inset
+
+ končno mnogo elementov,
+ tedaj vzamemo 
+\begin_inset Formula $n_{1}\coloneqq\max D_{\varphi}+1$
+\end_inset
+
+ in je pogoj za limito izpolnjen na prazno.
+ Sicer pa v
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:n>n_{1}\Rightarrow\left|a_{\varphi n}-L\right|<\varepsilon$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ strogo naraščajoča in izbiramo le elemente podzaporedja,
+ ki so v izvornem zaporedju za 
+\begin_inset Formula $n_{0}-$
+\end_inset
+
+tim členom in zato v 
+\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{n\to\infty}\frac{1}{2n+3}=\lim_{n\to\infty}\frac{1}{n}=0$
+\end_inset
+
+ za zaporedje 
+\begin_inset Formula $a_{n}=\frac{1}{n}$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $a_{\varphi n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi\left(n\right)=2n+3$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Karakterizacija limite s podzaporedji.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje in 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Leftrightarrow$
+\end_inset
+
+ za vsako podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+ zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+ obstaja njegovo podzaporedje 
+\begin_inset Formula $\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Dokazano poprej.
+ Limita se pri prehodu na podzaporedje ohranja.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0$
+\end_inset
+
+ in podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}\ni:\forall k\in\mathbb{N}:\left|a_{n_{k}}-K\right|>\varepsilon$
+\end_inset
+
+ (*)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+tu je na 
+\begin_inset Quotes gld
+\end_inset
+
+Zaporedja 2
+\begin_inset Quotes grd
+\end_inset
+
+ napaka,
+ neenačaj obrne v drugo smer
+\end_layout
+
+\end_inset
+
+.
+ Po predpostavki sedaj 
+\begin_inset Formula $\exists\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}\ni:\lim_{l\to\infty}a_{n_{k_{l}}}=L$
+\end_inset
+
+.
+ To pa je v protislovju z (*),
+ torej je začetna predpostavka 
+\begin_inset Formula $a_{n}\not\to L$
+\end_inset
+
+ napačna,
+ torej 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Stekališča
+\end_layout
+
+\begin_layout Definition*
+Točka 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ je stekališče zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subset\mathbb{R}$
+\end_inset
+
+,
+ če v vsaki okolici te točke leži neskončno členov zaporedja.
+\end_layout
+
+\begin_layout Remark*
+Pri limiti zahtevamo več;
+ da izven vsake okolice limite leži le končno mnogo členov.
+\end_layout
+
+\begin_layout Example*
+Primeri stekališč.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Rightarrow L$
+\end_inset
+
+ je stekališče za 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $0,1,0,1,\dots$
+\end_inset
+
+ stekališči sta 
+\begin_inset Formula $\left\{ 0,1\right\} $
+\end_inset
+
+ in zaporedje nima limite (ni konvergentno)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1,1,2,1,2,3,1,2,3,4,\dots$
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $b_{n}=n-$
+\end_inset
+
+to racionalno število
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Racionalnih števil je števno mnogo,
+ zato jih lahko linearno uredimo in oštevilčimo.
+\end_layout
+
+\end_inset
+
+ ima neskončno stekališč,
+ 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Limita je stakališče,
+ stekališče pa ni nujno limita.
+ Poleg tega,
+ če se spomnimo,
+ velja,
+ da vsota konvergentnih zaporedij konvergira k vsoti njunih limit,
+ ni pa nujno res,
+ da so stekališča vsote dveh zaporedij paroma vsote stekališč teh dveh zaporedij.
+ Primer:
+ 
+\begin_inset Formula $a_{n}=\left(-1\right)^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{n}=-\left(-1\right)^{n}$
+\end_inset
+
+.
+ Njuni stekališči sta 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+,
+ toda 
+\begin_inset Formula $a_{n}+b_{n}=0$
+\end_inset
+
+ ima le stekališče 
+\begin_inset Formula $\left\{ 0\right\} $
+\end_inset
+
+,
+ ne pa tudi 
+\begin_inset Formula $\left\{ 1,-1,2,-2\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset Formula $S$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
+\end_inset
+
+ je limita nekega podzaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Očitno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Definirajmo 
+\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče,
+ 
+\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
+\end_inset
+
+.
+ Podzaporedje 
+\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
+\end_inset
+
+ konvergira k 
+\begin_inset Formula $S$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $L$
+\end_inset
+
+ limita nekega zaporedja,
+ je 
+\begin_inset Formula $L$
+\end_inset
+
+ edino njegovo stekališče.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $S$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Po izreku zgoraj je 
+\begin_inset Formula $S$
+\end_inset
+
+ limita nekega podzaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Toda limita vsakega podzaporedja je enaka limiti zaporedja,
+ iz katerega to podzaporedje izhaja,
+ če ta limita obstaja.
+ Potemtakem je 
+\begin_inset Formula $S=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{bw}{Bolzano-Weierstraß}
+\end_layout
+
+\end_inset
+
+.
+ Eksistenčni izrek.
+ Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
+\end_inset
+
+.
+ Očitno je 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
+\end_inset
+
+.
+ Interval 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ razdelimo na dve polovici:
+ 
+\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
+\end_inset
+
+.
+ Izberemo polovico (vsaj ena obstaja),
+ v kateri leži neskončno mnogo členov,
+ in jo označimo z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+.
+ Spet jo razdelimo na pol in z 
+\begin_inset Formula $I_{2}$
+\end_inset
+
+ označimo tisto polovico,
+ v kateri leži neskončno mnogo členov.
+ Postopek ponavljamo in dobimo zaporedje zaprtih intervalov 
+\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo sedaj 
+\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
+\end_inset
+
+.
+ Iz konstrukcije je očitno,
+ da 
+\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ narašča in 
+\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada ter da sta obe zaporedji omejeni.
+ Posledično 
+\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
+\end_inset
+
+ sledi ocena 
+\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ kar konvergira k 0.
+ Posledično 
+\begin_inset Formula $d=l$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Treba je pokazati še,
+ da je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Torej za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja 
+\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $I_{n_{0}}$
+\end_inset
+
+ po konstrukciji vsebuje neskončno mnogo elementov,
+ jih torej tudi 
+\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
+\end_inset
+
+ oziroma poljubno majhna okolica 
+\begin_inset Formula $d=l$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $d=l$
+\end_inset
+
+ stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ edino stekališče omejenega zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $a_{n}\not\to s$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\varepsilon>0\ni:$
+\end_inset
+
+ izven 
+\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+ se nahaja neskončno mnogo členov zaporedja.
+ Ti členi sami zase tvorijo omejeno zaporedje,
+ ki ima po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{B.-W.}
+\end_layout
+
+\end_inset
+
+ izreku stekališče.
+ Slednje gotovo ne more biti enako 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ torej imamo vsaj dve stekališči,
+ kar je v je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s predpostavko.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da ima realno zaporedje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+stekališče v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje neskončno mnogo členov zapopredja
+\end_layout
+
+\begin_layout Itemize
+limito v 
+\begin_inset Formula $\infty$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
+\end_inset
+
+ vsebuje vse člene zaporedja od nekega indeksa dalje
+\end_layout
+
+\begin_layout Standard
+in podobno za 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Povezava s pojmom realnega stekališča/limite:
+ okolice 
+\begin_inset Quotes gld
+\end_inset
+
+točke
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\infty$
+\end_inset
+
+ so intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+ To je smiselno,
+ saj biti 
+\begin_inset Quotes gld
+\end_inset
+
+blizu 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ pomeni bizi zelo velik,
+ kar je ravno biti v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+za poljubno velik 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+Okolica točke 
+\begin_inset Formula $\infty$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+ so torej vsi intervali oblike 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Limes superior in limes inferior
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+realno zaporedje.
+ Tvorimo novo zaporedje 
+\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
+\end_inset
+
+.
+ Očitno je padajoče (
+\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
+\end_inset
+
+),
+ ker je supremum množice vsaj supremum njene stroge podmnožice.
+ Zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ima limito,
+ ki ji rečemo limes superior oziroma zgornja limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ in velja,
+ da leži v 
+\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+.
+ Podobno definiramo tudi limes inferior oz.
+ spodnjo limito zaporedja:
+ 
+\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za razliko od običajne limite,
+ ki lahko ne obstaja,
+ 
+\begin_inset Formula $\limsup$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ vedno obstajata.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+ je največje stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\liminf_{n\to\infty}$
+\end_inset
+
+ najmanjše.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\liminf$
+\end_inset
+
+ je dokaz analogen in ga ne bomo pisali.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče in 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Ker
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
+\end_inset
+
+.
+ Torej imamo 
+\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
+\end_inset
+
+ (zadnji neenačaj za 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+),
+ skratka 
+\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v poljubni okolici 
+\begin_inset Formula $s$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Treba je dokazati še,
+ da 
+\begin_inset Formula $\forall t>s:t$
+\end_inset
+
+ ni stekališče.
+ Naj bo 
+\begin_inset Formula $t>s$
+\end_inset
+
+.
+ Označimo 
+\begin_inset Formula $\delta\coloneqq t-s>0$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ zato v poljubno majhni okolici obstaja tak 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s_{n_{1}}$
+\end_inset
+
+ torej tu najdemo v 
+\begin_inset Formula $[s,s+\frac{\delta}{2})$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Po definiciji
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ je supremum členov od vključno 
+\begin_inset Formula $n$
+\end_inset
+
+ dalje
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
+\end_inset
+
+.
+ Za takšne 
+\begin_inset Formula $n$
+\end_inset
+
+ je 
+\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
+\end_inset
+
+.
+ Torej v 
+\begin_inset Formula $\frac{\delta}{2}-$
+\end_inset
+
+okolici točke 
+\begin_inset Formula $t$
+\end_inset
+
+ leži kvečjemu končno mnogo členov zaporedja oziroma členi 
+\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $t$
+\end_inset
+
+ ni stekališče za 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $M>0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}=\infty$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $N\left(n\right)\geq n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
+\end_inset
+
+ neskončna množica,
+ torej je neskončno mnogo členov v 
+\begin_inset Formula $\left(M,\infty\right)$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $M$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $s=\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Večjih stekališč od 
+\begin_inset Formula $\infty$
+\end_inset
+
+ očitno ni.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $m<0$
+\end_inset
+
+ poljuben.
+ Ker je 
+\begin_inset Formula $s=\inf s_{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
+\end_inset
+
+ Ker 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ pada proti 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
+\end_inset
+
+.
+ Ker je za poljuben 
+\begin_inset Formula $m$
+\end_inset
+
+ neskončno mnogo členov v 
+\begin_inset Formula $\left(-\infty,m\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s=-\infty$
+\end_inset
+
+ res stekališče.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Cauchyjev pogoj
+\end_layout
+
+\begin_layout Definition*
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ ustreza Cauchyjevemu pogoju (oz.
+ je Cauchyjevo),
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ ZDB Dovolj pozni členi so si poljubno blizu.
+\end_layout
+
+\begin_layout Claim*
+Zaporedje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je konvergentno 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je Cauchyjevo.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Če 
+\begin_inset Formula $a_{n}\to L$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
+\end_inset
+
+.
+ Cauchyjev pogoj sledi iz definicije limite za 
+\begin_inset Formula $\frac{\varepsilon}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Če je zaporedje Cauchyjevo,
+ je omejeno:
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
+\end_inset
+
+.
+ V posebnem,
+ 
+\begin_inset Formula $m=n_{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
+\end_inset
+
+.
+ Preostali členi tvorijo končno veliko množico,
+ ki ima 
+\begin_inset Formula $\min$
+\end_inset
+
+ in 
+\begin_inset Formula $\max$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
+\end_inset
+
+ tudi omejena.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{bw}{izreku od prej}
+\end_layout
+
+\end_inset
+
+ sledi,
+ da ima zaporedje stekališče 
+\begin_inset Formula $s$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ Cauchyjevo,
+ 
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $s$
+\end_inset
+
+ 
+\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Moč izreka je v tem,
+ da lahko konvergenco preverjamo tudi tedaj,
+ ko nimamo kandidatov za limito.
+\end_layout
+
+\begin_layout Section
+Številske vrste
+\end_layout
+
+\begin_layout Standard
+Kako sešteti neskončno mnogo števil?
+ Nadgradimo pristop končnih vsot na neskončne vsote!
+\end_layout
+
+\begin_layout Definition*
+Imejmo zaporedje 
+\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
+\end_inset
+
+.
+ Izraz 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ se imenuje vrsta s členi 
+\begin_inset Formula $a_{j}$
+\end_inset
+
+.
+ Pomen izraza opredelimo na naslednjo način:
+\end_layout
+
+\begin_layout Definition*
+Tvorimo novo zaporedje,
+ pravimo mu zaporedje delnih vsot vrste:
+ 
+\begin_inset Formula $s_{1}=a_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{2}=a_{1}+a_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+ —
+ številu 
+\begin_inset Formula $s_{n}$
+\end_inset
+
+ pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ta delna vsota.
+\end_layout
+
+\begin_layout Definition*
+Vrsta je konvergentna,
+ če je v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ konvergentno zaporedje 
+\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Številu 
+\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
+\end_inset
+
+ tedaj pravimo vsota vrste in pišemo 
+\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+.
+ Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
+ Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
+\end_layout
+
+\begin_layout Definition*
+Če vrsta ni konvergentna,
+ rečemo,
+ da je divergentna.
+ Enako,
+ če je 
+\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
+\end_inset
+
+,
+ torej zaporedje 
+\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
+\end_inset
+
+.
+ Ali se sešteje v 1?
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
+\end_inset
+
+.
+ Pišimo 
+\begin_inset Formula $q=\frac{1}{2}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $a_{n}=q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula 
+\[
+s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
+\]
+
+\end_inset
+
+Izračunajmo 
+\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
+\end_inset
+
+ (velja,
+ ker 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+),
+ torej je 
+\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Geometrijska vrsta (splošno).
+ Naj bo 
+\begin_inset Formula $q\in\mathbb{R}$
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
+\end_inset
+
+ se imenuje geometrijska vrsta.
+ Velja 
+\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $q=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $s_{n}=n+1$
+\end_inset
+
+,
+ sicer množimo izraz z 
+\begin_inset Formula $\left(1-q\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
+\end_inset
+
+ in vrsta konvergira 
+\begin_inset Formula $\Leftrightarrow q\not=1$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ To pa se zgodi natanko za 
+\begin_inset Formula $q\in\left(-1,1\right)$
+\end_inset
+
+,
+ takrat je 
+\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Harmonična vrsta.
+ Je vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ toda vrsta divergira.
+ Dokaz sledi kmalu malce spodaj.
+\end_layout
+
+\end_deeper
+\begin_layout Question*
+Kako lahko enostavno določimo,
+ ali dana vrsta konvergira?
+\end_layout
+
+\begin_layout Subsection
+Konvergenčni kriteriji
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{cauchyvrste}{Cauchyjev pogoj}
+\end_layout
+
+\end_inset
+
+.
+ Vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ je konvergentna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ delne vrste ustrezajo Cauchyjevemu pogoju;
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
+\end_inset
+
+ konvergira 
+\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo izrek zgoraj za 
+\begin_inset Formula $n=m-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Vrsti 
+\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
+\end_inset
+
+ divergirata,
+ saj smo videli,
+ da členi ne ene ne druge ne konvergirajo nikamor,
+ torej tudi ne proti 0,
+ kar je potreben pogoj za konvergenco vrste.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta divergira.
+ Protiprimer Cauchyjevega pogoja:
+ Naj bo 
+\begin_inset Formula $\varepsilon=\frac{1}{4}$
+\end_inset
+
+.
+ Tedaj ne glede na izbiro 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ najdemo:
+\begin_inset Formula 
+\[
+s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
+\]
+
+\end_inset
+
+Dokaz divergence brez Cauchyjevega pogoja:
+ 
+\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Geometrični argument za divergenco:
+ TODO XXX FIXME DODAJ
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{pk}{Primerjalni kriterij}
+\end_layout
+
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ vrsti z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
+\end_inset
+
+ (od nekod naprej) —
+ pravimo,
+ da je 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ majoranta za 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ od nekod naprej.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
+\end_inset
+
+ konvergira,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergira.
+\end_layout
+
+\begin_layout Itemize
+Če 
+\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Videli smo,
+ da 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
+\end_inset
+
+ divergira.
+ Kaj pa 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+?
+ Preverimo naslednje in uporabimo primerjalni kriterij:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+?
+ Računajmo 
+\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
+\end_inset
+
+.
+ Velja,
+ ker 
+\begin_inset Formula $k\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
+\end_inset
+
+ konvergira?
+ Opazimo 
+\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+.
+ Za delne vsote vrste 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
+\end_inset
+
+ velja:
+\begin_inset Formula 
+\[
+\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
+\]
+
+\end_inset
+
+torej 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
+\end_inset
+
+.
+ Posledično po primerjalnem kriteriju tudi 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
+\end_inset
+
+ konvergira.
+ Izkaže se 
+\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Kvocientni oz.
+ d'Alembertov kriterij.
+ Za vrsto s pozitivnimi členi 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:kvocientni3a"
+
+\end_inset
+
+
+\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Razlaga.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
+\end_inset
+
+.
+ Vrsto smo majorizirali z geometrijsko vrsto,
+ ki ob 
+\begin_inset Formula $q\in\left(0,1\right)$
+\end_inset
+
+ konvergira po primerjalnem kriteriju,
+ zato tudi naša vrsta konvergira.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $a_{n+2}\geq a_{n+1}$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $a_{n+2}\geq a_{n}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
+\end_inset
+
+.
+ Naša vrsta torej majorizira konstantno vrsto,
+ ki očitno divergira;
+ 
+\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
+\end_inset
+
+.
+ Potemtakem tudi naša vrsta divergira.
+ Poleg tega niti ne velja 
+\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
+\end_inset
+
+,
+ torej vrsta gotovo divergira.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Za 
+\begin_inset Formula $x>0$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+ Vrsta res konvergira po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:kvocientni3a"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Korenski oz.
+ Cauchyjev kriterij.
+ Naj bo 
+\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
+\end_inset
+
+ vrsta z nenegativnimi členi.
+ Naj bo 
+\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
+\end_inset
+
+ (vrsta konvergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
+\end_inset
+
+ (vrsta divergira)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c=1\Longrightarrow$
+\end_inset
+
+ s tem kriterijem ne moremo določiti konvergence.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Skica dokazov.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\leq q^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{n+1}\leq q^{n+1}$
+\end_inset
+
+,
+ torej je vrsta majorizirana z geometrijsko vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Velja 
+\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
+\end_inset
+
+.
+ To pomeni 
+\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n}\geq1$
+\end_inset
+
+,
+ torej je vrsta majorizirana s konstantno in zato divergentno vrsto 
+\begin_inset Formula $\sum_{n=1}^{\infty}1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Enako kot 1 in 2.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Alternirajoče vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta je alternirajoča,
+ če je predznak naslednjega člena nasproten predznaku tega člena.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\sgn a=\begin{cases}
+-1 & ;a<0\\
+1 & ;a>0\\
+0 & ;a=0
+\end{cases}$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Leibnizov konvergenčni kriterij.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ padajoče zaporedje in 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
+\end_inset
+
+.
+ Tedaj vrsta 
+\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ konvergira.
+ Če je 
+\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skica dokaza.
+ Vidimo,
+ da delne vsote 
+\begin_inset Formula $s_{2n}$
+\end_inset
+
+ padajo k 
+\begin_inset Formula $s''$
+\end_inset
+
+ in delne vsote 
+\begin_inset Formula $s_{2n-1}$
+\end_inset
+
+ naraščajo k 
+\begin_inset Formula $s'$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $s'=s''$
+\end_inset
+
+.
+ Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
+ torej 
+\begin_inset Formula $s'=s''=s$
+\end_inset
+
+.
+ 
+\begin_inset Formula $s$
+\end_inset
+
+ je supremum lihih in infimum sodih vsot.
+ 
+\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Harmonična vrsta 
+\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
+\end_inset
+
+,
+ toda alternirajoča harmonična vrsta 
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Absolutno konvergentne vrste
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je absolutno konvergentna,
+ če je 
+\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
+\end_inset
+
+ konvergentna.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ konvergenca.
+\end_layout
+
+\begin_layout Proof
+Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
+\end_layout
+
+\end_inset
+
+ in trikotniško neenakost.
+\begin_inset Formula 
+\[
+\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
+\]
+
+\end_inset
+
+za 
+\begin_inset Formula $m,n\geq n_{0}$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Obrat ne velja,
+ protiprimer je alternirajoča harmonična vrsta.
+\end_layout
+
+\begin_layout Subsection
+Pogojno konvergentne vrste
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
+\end_inset
+
+,
+ temveč 
+\begin_inset Formula $\infty-\infty=$
+\end_inset
+
+ nedefinirano.
+\end_layout
+
+\begin_layout Question*
+Ross-Littlewoodov paradoks.
+ Ali smemo zamenjati vrstni red seštevanja,
+ če imamo neskončno mnogo sumandov?
+\end_layout
+
+\begin_layout Standard
+Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
+\end_inset
+
+.
+ Permutacija 
+\begin_inset Formula $\mathcal{M}$
+\end_inset
+
+ je vsaka bijektivna preslikava 
+\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
+\end_inset
+
+ končna množica,
+ tedaj 
+\begin_inset Formula $\pi$
+\end_inset
+
+ označimo s tabelo:
+\begin_inset Formula 
+\[
+\left(\begin{array}{ccc}
+a_{1} & \cdots & a_{n}\\
+\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula 
+\[
+\pi=\left(\begin{array}{ccccc}
+1 & 2 & 3 & 4 & 5\\
+5 & 3 & 1 & 4 & 2
+\end{array}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ je brezpogojno konvergentna,
+ če za vsako permutacijo 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
+\end_inset
+
+ konvergira in vsota ni odvisna od 
+\begin_inset Formula $\pi$
+\end_inset
+
+.
+ Vrsta je pogojno konvergentna,
+ če je konvergentna,
+ toda ne brezpogojno.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
+\end_inset
+
+ je pogojno konvergentna,
+ ker pri seštevanju z vrstnim redom,
+ pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
+ vrsta ne konvergira.
+\end_layout
+
+\begin_layout Theorem*
+Absolutna konvergenca 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ Brezpogojna konvergenca
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Riemannov sumacijski izrek.
+ Če je vrsta pogojno konvergentna,
+ tedaj 
+\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
+\end_inset
+
+ permutacija 
+\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
+\end_inset
+
+.
+ ZDB Končna vsota je lahko karkoli,
+ če lahko poljubno spremenimo vrstni red seštevanja.
+ Prav tako obstaja taka permutacija 
+\begin_inset Formula $\pi$
+\end_inset
+
+,
+ pri kateri 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
+\end_inset
+
+ nima vsote ZDB delne vsotee ne konvergirajo.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Funkcijske vrste
+\end_layout
+
+\begin_layout Standard
+Tokrat poskušamo seštevati funkcije.
+ V prejšnjem razdelku seštevamo le realna števila.
+ Funkcijska vrsta,
+ če je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedije funkcij 
+\begin_inset Formula $X\to\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+ zunanja konstanta,
+ izgleda takole:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\sum_{n=1}^{\infty}a_{n}\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
+\end_inset
+
+ družina funkcij.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če je 
+\begin_inset Formula $\forall x\in X$
+\end_inset
+
+ zaporedje 
+\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno.
+\end_layout
+
+\begin_layout Definition*
+Označimo limito s 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+.
+ ZDB to pomeni,
+ da 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da funkcije 
+\begin_inset Formula $\varphi_{n}$
+\end_inset
+
+ konvergirajo enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
+\]
+
+\end_inset
+
+ oziroma ZDB 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Poudariti je treba,
+ da je pri konvergenci po točkah 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ lahko odvisen od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pri enakomerni konvergenci pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Očitno enakomerna konvergenca implicira konvergenco po točkah,
+ obratno pa ne velja.
+\end_layout
+
+\begin_layout Example*
+Za 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Tedaj obstaja 
+\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
+0 & ;x\in[0,1)\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+ Torej po definiciji velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ po točkah,
+ toda ne velja 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno.
+ Za poljubno velik pas okoli 
+\begin_inset Formula $\varphi\left(x\right)$
+\end_inset
+
+ bodo še tako pozne funkcijske vrednosti 
+\begin_inset Formula $\varphi_{n}\left(x\right)$
+\end_inset
+
+ od nekega 
+\begin_inset Formula $x$
+\end_inset
+
+ dalje izven tega pasu.
+ Če bi 
+\begin_inset Formula $\varphi_{n}\to\varphi$
+\end_inset
+
+ enakomerno,
+ tedaj bi za poljuben 
+\begin_inset Formula $\varepsilon\in\left(0,1\right)$
+\end_inset
+
+ in dovolj pozne 
+\begin_inset Formula $n$
+\end_inset
+
+ (večje od nekega 
+\begin_inset Formula $n_{0}\in\mathbb{N}$
+\end_inset
+
+) veljalo 
+\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
+\end_inset
+
+.
+ To je ekvivalentno 
+\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
+\end_inset
+
+,
+ zato tak 
+\begin_inset Formula $n$
+\end_inset
+
+ ne obstaja.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neka množica in 
+\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dano zaporedje funkcij.
+ Pravimo,
+ da funkcijska vrsta 
+\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
+\end_inset
+
+ (številska vrsta je konvergentna).
+ ZDB to pomeni,
+ da funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira po točkah na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcijska vrsta 
+\begin_inset Formula $s=\sum_{j=1}^{\infty}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ če funkcijsko zaporedje delnih vsot 
+\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
+\end_inset
+
+ konvergira enakomerno na 
+\begin_inset Formula $X$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija oblike 
+\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
+\end_inset
+
+ se imenuje funkcijska vrsta.
+\end_layout
+
+\begin_layout Exercise*
+Dokaži,
+ da 
+\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
+\end_inset
+
+ ne konvergira enakomerno!
+ Vrsta konvergira po točkah le na intervalu 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+,
+ za druge 
+\begin_inset Formula $x$
+\end_inset
+
+ divergira.
+ Ko fiksiramo zunanjo konstanto,
+ gre za geometrijsko vrsto.
+ Delna vsota 
+\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
+\end_inset
+
+.
+ Sedaj prevedimo,
+ ali 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
+\end_inset
+
+.
+ Za začetekk si oglejmo le 
+\begin_inset Formula $x>0$
+\end_inset
+
+.
+ Ker je tedaj 
+\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
+\end_inset
+
+.
+ Računajmo sedaj 
+\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $n$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ vsota ni enakomerno konvergentna.
+\end_layout
+
+\begin_layout Standard
+Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike 
+\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ torej potence (monomi).
+\end_layout
+
+\begin_layout Definition*
+Potenčna vrsta je funkcijska vrsta oblike 
+\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
+\end_inset
+
+,
+ kjer so a 
+\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
+\end_inset
+
+ dana realna števila.
+\end_layout
+
+\begin_layout Theorem*
+Cauchy-Hadamard.
+ Za vsako potenčno vrsto obstaja konvergenčni radij 
+\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+vrsta absolutno konvergira za 
+\begin_inset Formula $\left|x\right|<R$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Itemize
+vrsta divergira za 
+\begin_inset Formula $\left|x\right|>R$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
+\end_inset
+
+,
+ kjer vzamemo 
+\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Rezultat že poznamo za zelo poseben primer 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+ (geometrijska vrsta).
+ Ideja dokaza je,
+ da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Konvergenca:
+ Za 
+\begin_inset Formula $x=0$
+\end_inset
+
+ vrsta očitno konvergira,
+ zato privzamemo 
+\begin_inset Formula $x\not=0$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $R$
+\end_inset
+
+ s formulo iz definicije (
+\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $x$
+\end_inset
+
+ tak,
+ da 
+\begin_inset Formula $\left|x\right|<R\leq\infty$
+\end_inset
+
+ (sledi 
+\begin_inset Formula $R>0$
+\end_inset
+
+).
+ Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Tedaj po definiciji 
+\begin_inset Formula $R$
+\end_inset
+
+ velja 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
+ Preverimo,
+ da desna stran konvergira.
+ Konvergira,
+ kadar 
+\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pk}{primerjalnem kriteriju}
+\end_layout
+
+\end_inset
+
+ torej naša vrsta absolutno konvergira.
+\end_layout
+
+\begin_layout Itemize
+Divergenca:
+ Vzemimo poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+.
+ Po definciji 
+\begin_inset Formula $R$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
+\end_inset
+
+ za vse dovolj velike 
+\begin_inset Formula $k$
+\end_inset
+
+.
+ Za take 
+\begin_inset Formula $k$
+\end_inset
+
+ sledi
+\begin_inset Formula 
+\[
+\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
+\]
+
+\end_inset
+
+Opazimo,
+ da je desna stran neenačbe člen geometrijske vrste,
+ ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
+ Desna stran divergira,
+ ko 
+\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
+\end_inset
+
+,
+ zato tudi naša vrsta divergira.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primer konvergenčnega radija potenčne vrste od prej:
+ 
+\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
+\end_inset
+
+,
+ torej po zgornjem izreku vrsta konvergira za 
+\begin_inset Formula $x\in\left(-1,1\right)$
+\end_inset
+
+ in divergira za 
+\begin_inset Formula $x\not\in\left[-1,1\right]$
+\end_inset
+
+.
+ Ročno lahko še preverimo,
+ da divergira tudi v 
+\begin_inset Formula $\left\{ -1,1\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Zveznost
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
+ recimo dodaj dokaz zveznosti x^2
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ideja:
+ Izdelati želimo formulacijo,
+ s katero preverimo,
+ če lahko z dovolj majhno spremembo 
+\begin_inset Formula $x$
+\end_inset
+
+ povzročimo majhno spremembo funkcijske vrednosti.
+\end_layout
+
+\begin_layout Example*
+Primer nezvezne funkcije je 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+0 & ;0\leq x<1\\
+1 & ;x=1
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $x\subseteq D$
+\end_inset
+
+,
+ če je zvezna na vsaki točki v 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $D,a,f$
+\end_inset
+
+ kot prej.
+ Velja:
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje na domeni velja,
+ da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ se pravi 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ poljuben.
+ Vsled zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+ velja,
+ da je 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse take 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+ za neko 
+\begin_inset Formula $\delta\in\mathbb{R}$
+\end_inset
+
+.
+ Ker je zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ so vsi členi po nekem 
+\begin_inset Formula $n_{0}$
+\end_inset
+
+ v 
+\begin_inset Formula $\delta-$
+\end_inset
+
+okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej velja pogoj 
+\begin_inset Formula $\left|a_{n}-a\right|<\delta$
+\end_inset
+
+,
+ torej velja 
+\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ za vse 
+\begin_inset Formula $n\geq n_{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Da pridemo do protislovja,
+ moramo dokazati,
+ da 
+\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+,
+ a vendar 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ velja,
+ da 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ Izberimo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ S prvim argumentom konjunkcije smo poskrbeli za to,
+ da je naše konstruiramo zaporedje 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno k 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Konstruirali smo zaporedje,
+ pri katerem so funkcijske vrednosti za vsak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ izven 
+\begin_inset Formula $\varepsilon-$
+\end_inset
+
+okolice 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ torej zaporedje ne konvergira k 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na 
+\begin_inset Formula $D\Leftrightarrow$
+\end_inset
+
+ za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Za funkcijo 
+\begin_inset Formula $f:D\to V$
+\end_inset
+
+ za 
+\begin_inset Formula $X\subseteq V$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da za vsako odprto množico 
+\begin_inset Formula $V\subset\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta množica.
+ Dokazujemo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $a\in D,\varepsilon>0$
+\end_inset
+
+ poljubna.
+ Naj bo 
+\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
+\end_inset
+
+ odprta množica.
+ Po predpostavki sledi,
+ da je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ spet odprta.
+ Ker je 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $a\in V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $V$
+\end_inset
+
+ odprta,
+ 
+\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ to pomeni 
+\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna odprta podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $a\in f^{-1}\left(V\right)$
+\end_inset
+
+ poljuben (torej 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+).
+ Ker je 
+\begin_inset Formula $f\left(a\right)\in V$
+\end_inset
+
+,
+ ki je odprta,
+ 
+\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+,
+ torej je tudi neka odprta okolica 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+.
+ Ker je bil 
+\begin_inset Formula $a$
+\end_inset
+
+ poljuben,
+ je 
+\begin_inset Formula $f^{-1}\left(V\right)$
+\end_inset
+
+ odprta,
+ ker je bila 
+\begin_inset Formula $V$
+\end_inset
+
+ poljubna,
+ je izrek dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g:D\to\mathbb{R}$
+\end_inset
+
+ zvezni v 
+\begin_inset Formula $a\in D$
+\end_inset
+
+.
+ Tedaj so v 
+\begin_inset Formula $a$
+\end_inset
+
+ zvezne tudi funkcije 
+\begin_inset Formula $f+g,f-g,f\cdot g$
+\end_inset
+
+ in 
+\begin_inset Formula $f/g$
+\end_inset
+
+,
+ slednja le,
+ če je 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+ velja za vsako 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
+\end_inset
+
+ tudi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
+\end_inset
+
+.
+ Po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
+\end_layout
+
+\end_inset
+
+ velja,
+ da 
+\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+.
+ Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
+ ki pove,
+ da so tudi 
+\begin_inset Formula $f*g$
+\end_inset
+
+ za 
+\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
+\end_inset
+
+ zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Pri deljenju velja omejitev 
+\begin_inset Formula $f\left(a\right)\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $D,E\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to E$
+\end_inset
+
+ in 
+\begin_inset Formula $g:E\to\mathbb{R}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Hkrati pa,
+ če je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Vzemimo poljubno 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $a_{n}\to a\in D$
+\end_inset
+
+.
+ Zopet uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
+\end_layout
+
+\end_inset
+
+:
+ ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
+\end_inset
+
+ in po istem izreku je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $D$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsi polinomi so zvezni na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Vzemimo 
+\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
+\end_inset
+
+.
+ Uporabimo prejšnji izrek.
+ Polinom je sestavljen iz vsote konstantne funkcije,
+ zmnožene z identiteto,
+ ki je s seboj 
+\begin_inset Formula $n-$
+\end_inset
+
+krat množena.
+ Ker vsota in množenje ohranjata zveznost,
+ je treba dokazati le,
+ da je 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ zvezna in da so 
+\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
+\end_inset
+
+ zvezne.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ Ali velja 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+?
+ Da,
+ velja.
+ Vzamemo lahko katerokoli 
+\begin_inset Formula $\delta\in(0,\varepsilon]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+ poljuben.
+ Tu je 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
+\end_inset
+
+,
+ torej je desna stran implikacije vedno resnična,
+ torej je implikacija vedno resnična.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
+ To so:
+ polinomi,
+ potence,
+ racionalne funkcije,
+ koreni,
+ eksponentne funkcije,
+ logaritmi,
+ trigonometrične,
+ ciklometrične in kombinacije neskončno mnogo naštetih,
+ spojenih s 
+\begin_inset Formula $+,-,\cdot,/,\circ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Tega izreka ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
+\end_inset
+
+ je zvezna povsod,
+ kjer je definirana.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ (zapišemo 
+\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
+\end_inset
+
+),
+ če za vsako zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
+\end_inset
+
+,
+ za katero velja 
+\begin_inset Formula $a_{n}\to a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(a_{n}\right)\to L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+ZDB če za 
+\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
+f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
+L & ;x\in a
+\end{cases}$
+\end_inset
+
+ velja,
+ da je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Vrednost 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+,
+ če sploh obstaja,
+ nima vloge pri vrednosti limite.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $a\in D\subseteq\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Kvadratna funkcija 
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ je zvezna.
+ Vzemimo poljuben 
+\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$
+\end_inset
+
+.
+ Obstajati mora taka 
+\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Podan imamo torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ želimo najti 
+\begin_inset Formula $\delta$
+\end_inset
+
+.
+ Želimo priti do neenakosti,
+ ki ima na manjši strani 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$
+\end_inset
+
+ in na večji strani nek izraz z 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+,
+ da ta 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ nadomestimo z 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in nato večjo stran enačimo z 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da izrazimo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ v odvisnosti od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Računajmo:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$
+\end_inset
+
+.
+ Predelajmo izraz 
+\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$
+\end_inset
+
+.
+ Sedaj nadomestimo 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ z 
+\begin_inset Formula $\delta$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$
+\end_inset
+
+.
+ Iščemo tak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$
+\end_inset
+
+,
+ zato enačimo 
+\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$
+\end_inset
+
+ in dobimo kvadratno enačbo 
+\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$
+\end_inset
+
+,
+ ki jo rešimo z obrazcem za ničle:
+\begin_inset Formula 
+\[
+\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon}
+\]
+
+\end_inset
+
+Toda ker iščemo le pozitivne 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ je edina rešitev
+\begin_inset Formula 
+\[
+\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Število 
+\begin_inset Formula $L_{+}\in\mathbb{R}$
+\end_inset
+
+ je desna limita funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$
+\end_inset
+
+ ZDB če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje s členi desno od 
+\begin_inset Formula $a$
+\end_inset
+
+ velja,
+ da funkcijske vrednosti členov konvergirajo k 
+\begin_inset Formula $L_{+}$
+\end_inset
+
+.
+ Oznaka 
+\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$
+\end_inset
+
+.
+ Podobno definiramo tudi levo limito 
+\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ da velja 
+\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+ V tem primeru velja 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists f\left(a-0\right)$
+\end_inset
+
+,
+ vendar 
+\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+skok
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+\end_inset
+
+ ne obstaja.
+ Zakaj?
+ Izračunajmo levo in desno limito:
+\begin_inset Formula 
+\[
+\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je na intervalu 
+\begin_inset Formula $D$
+\end_inset
+
+ odsekoma zvezna,
+ če je zvezna povsod na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ razen morda v končno mnogo točkah,
+ v katerih ima skok.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
+\end_inset
+
+.
+ Zanima nas,
+ ali obstaja 
+\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
+\end_inset
+
+.
+ Grafični dokaz.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID,
+ glej ZVZ III/ANA1P1120/str.8
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Očitno velja 
+\begin_inset Formula $\triangle ABD\subset$
+\end_inset
+
+ krožni izsek 
+\begin_inset Formula $DAB\subset\triangle ABC$
+\end_inset
+
+,
+ torej za njihove ploščine velja 
+\begin_inset Formula 
+\[
+\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{x}{\sin x}=1
+\]
+
+\end_inset
+
+Da naš sklep res potrdimo,
+ je potreben spodnji izrek.
+\end_layout
+
+\begin_layout Theorem*
+Če za 
+\begin_inset Formula $f,g,h:D\to\mathbb{R}$
+\end_inset
+
+ velja za 
+\begin_inset Formula $a\in D$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
+\end_inset
+
+ in hkrati
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$
+\end_inset
+
+,
+ tedaj tudi 
+\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $
+\end_inset
+
+.
+ Velja
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Posledično 
+\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+ Naj bo sedaj 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Tedaj velja 
+\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $
+\end_inset
+
+ torej velja 
+\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Zvezne funkcije na kompaktnih množicah
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je zaprta in omejena ZDB je unija zaprtih intervalov.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $K\subset\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zfnkm}{omejena in doseže minimum in maksimum}
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ na 
+\begin_inset Formula $I_{1}=(0,1]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f_{1}$
+\end_inset
+
+ je zvezna in 
+\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$
+\end_inset
+
+,
+ torej ni omejena,
+ a 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ ni zaprt.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{2}\left(x\right)=\begin{cases}
+0 & ;x=0\\
+\frac{1}{x} & ;x\in(0,1]
+\end{cases}$
+\end_inset
+
+ ni omejena in je definirana na kompaktni množici,
+ a ni zvezna.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{3}\left(x\right)=x$
+\end_inset
+
+ na 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+.
+ Je omejena,
+ ne doseže maksimuma,
+ a 
+\begin_inset Formula $D_{f_{3}}$
+\end_inset
+
+ ni kompaktna (ni zaprta).
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{4}\left(x\right)=\begin{cases}
+x & ;x\in\left(0,1\right)\\
+\frac{1}{2} & ;x\in\left\{ 0,1\right\} 
+\end{cases}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\sup f_{4}=1$
+\end_inset
+
+,
+ ampak ga ne doseže,
+ a ni zvezna
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Omejenost navzgor:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\begin_layout Itemize
+Omejenost navzdol:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+ 
+\end_layout
+
+\begin_layout Itemize
+Doseže maksimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ravnokar smo dokazali,
+ da 
+\begin_inset Formula $M<\infty$
+\end_inset
+
+.
+ Po definiciji supremuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Doseže minimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ko smo dokazali omejenost,
+ smo dokazali,
+ da 
+\begin_inset Formula $M>-\infty$
+\end_inset
+
+.
+ Po definiciji infimuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $f\left(a\right)f\left(b\right)<0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Interval 
+\begin_inset Formula $I_{0}=\left[a,b\right]$
+\end_inset
+
+ razpolovimo.
+ To pomeni,
+ da pogledamo levo in desno polovico intervala 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\frac{a+b}{2},b\right]$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$
+\end_inset
+
+,
+ smo našli iskano točko 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ sicer z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ označimo katerokoli izmed polovic,
+ ki ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v krajiščih različno predznačene funkcijske vrednosti.
+ Torej 
+\begin_inset Formula $I_{1}=\begin{cases}
+\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\
+\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0
+\end{cases}$
+\end_inset
+
+.
+ S postopkom nadaljujemo.
+ Če v končno mnogo korakih najdemo 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $f\left(\xi\right)=0$
+\end_inset
+
+,
+ fino,
+ sicer pa dobimo zaporedje intervalov 
+\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$
+\end_inset
+
+,
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:različni-predznaki-istoležnih-clenov"
+
+\end_inset
+
+
+\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Ker sta zaporedji 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeni in monotoni,
+ imata po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij}
+\end_layout
+
+\end_inset
+
+ limiti 
+\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha,\beta\in I_{0}$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ zaprt.
+\end_layout
+
+\begin_layout Proof
+Sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:različni-predznaki-istoležnih-clenov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$
+\end_inset
+
+.
+ Ker pa 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I=\left[a,b\right]$
+\end_inset
+
+ omejen zaprt interval 
+\begin_inset Formula $\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj 
+\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ ZDB zvezna funkcija na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže vse funkcijske vrednosti na intervalu 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokaz posledice.
+ Naj bo 
+\begin_inset Formula $y$
+\end_inset
+
+ poljuben.
+ Če je 
+\begin_inset Formula $y=f\left(x_{-}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{-}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $y=f\left(x_{+}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{+}$
+\end_inset
+
+.
+ Sicer pa je 
+\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$
+\end_inset
+
+.
+ Oglejmo si funkcijo 
+\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$
+\end_inset
+
+,
+ torej po prejšnjem izreku 
+\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$
+\end_inset
+
+,
+ kar pomeni ravno 
+\begin_inset Formula $f\left(x\right)=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ poljuben interval med 
+\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ interval med 
+\begin_inset Formula $f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(a-0\right)$
+\end_inset
+
+.
+ Inverzna funkcija 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ je definirana na 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ in zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\coloneqq\arctan$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$
+\end_inset
+
+,
+ zvezna.
+ Naj bo 
+\begin_inset Formula $y\in f\left(I\right)$
+\end_inset
+
+ poljuben.
+ Tedaj 
+\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ in definiramo 
+\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ obstaja in je spet zvezna.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Označimo 
+\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$
+\end_inset
+
+.
+ Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Dokazujemo torej,
+ da 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$
+\end_inset
+
+ je zopet odprta množica 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Torej dokazujemo 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$
+\end_inset
+
+ je spet zopet odprta 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+,
+ kar je ekvivalentno
+\begin_inset Formula 
+\[
+\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right).
+\]
+
+\end_inset
+
+Pišimo 
+\begin_inset Formula $y=f\left(x\right),x\in I\cap V$
+\end_inset
+
+.
+ Privzemimo,
+ da 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča (če pada,
+ ravnamo podobno).
+ Ker jer 
+\begin_inset Formula $ $
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Enakomerna zveznost
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ez}{enakomerno zvezna}
+\end_layout
+
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Primerjajmo to z definicijo 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je (nenujno enakomerno) zvezna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+Pri slednji definiciji je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisna od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ pri enakomerni zveznosti pa le od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ ni enakomerno zvezna,
+ ker je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisen od 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Če pri fiksnem 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ pomaknemo tisto pozitivno točko,
+ v kateri preizkušamo zveznost,
+ bolj v levo,
+ bo na neki točki potreben ožji,
+ manjši 
+\begin_inset Formula $\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Zvezna funkcija na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ kjer je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni enakomerno zvezna.
+ Zanikajmo definicijo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ez}{enakomerne zveznosti}
+\end_layout
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x_{\delta},y_{\delta}\in I:\left|x_{\delta}-y_{\delta}\right|<\delta\wedge\left|f\left(x_{\delta}\right)-f\left(y_{\delta}\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x,y$
+\end_inset
+
+ sta seveda lahko odvisna od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ zato v subskriptu pišemo 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ ki ji pripadata.
+ Ker smo dejali,
+ da to velja,
+ si oglejmo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+ in pripadajoči zaporedji 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(y_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna,
+ ima zaporedje 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ stekališče v 
+\begin_inset Formula $x\in K$
+\end_inset
+
+,
+ torej obstaja podzaporede 
+\begin_inset Formula $\left(x_{1/n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Podobno obstaja podzaporedje 
+\begin_inset Formula $\left(y_{1/n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $y\in K$
+\end_inset
+
+.
+ Pišimo sedaj 
+\begin_inset Formula $x_{l}\coloneqq x_{1/n_{k_{l}}}$
+\end_inset
+
+in 
+\begin_inset Formula $y_{l}\coloneqq y_{1/n_{k_{l}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja torej 
+\begin_inset Formula $x_{l}\to x$
+\end_inset
+
+ in 
+\begin_inset Formula $y_{l}\to y$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|x-y\right|\leq\lim_{l\to\infty}\left(\left|x-x_{l}\right|+\left|x_{l}-y_{l}\right|+\left|y_{l}-y\right|\right)$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja,
+ srednji pa je manjši od 
+\begin_inset Formula $\frac{1}{j}$
+\end_inset
+
+ zaradi naše predpostavke (PDDRAA),
+ potemtakem je 
+\begin_inset Formula $x=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\lim_{l\to\infty}\left(f\left(x_{l}\right)-f\left(y_{l}\right)\right)=\lim_{l\to\infty}\left[\left(f\left(x_{l}\right)-f\left(x\right)\right)+\left(f\left(x\right)-f\left(y\right)\right)+\left(f\left(y\right)-f\left(y_{l}\right)\right)\right]$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja in zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ srednji pa je tudi 0,
+ ker 
+\begin_inset Formula $x=y$
+\end_inset
+
+,
+ potemtakem 
+\begin_inset Formula $f\left(x_{l}\right)-f\left(y_{l}\right)\to0$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\left|f\left(x_{l}\right)-f\left(y_{l}\right)\right|\geq\varepsilon$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall l\in\mathbb{N}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je enakomerno zvezna.
+\end_layout
+
+\begin_layout Corollary*
+En zaprt interval 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ bo enakomerno zvezen,
+ 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ sama po sebi kot 
+\begin_inset Formula $\left(0,\infty\right)\to\mathbb{R}$
+\end_inset
+
+ pa ni definirana na kompaktni množici.
+ Prav tako 
+\begin_inset Formula $\arcsin$
+\end_inset
+
+ in 
+\begin_inset Formula $x\mapsto\sqrt{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Odvod
+\end_layout
+
+\begin_layout Standard
+Najprej razmislek/ideja.
+ Odvod je hitrost/stopnja,
+ s katero se v danem trenutku neka količina spreminja.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID (ali pa —
+ bolje —
+ s čim drugim),
+ glej PS zapiski/ANA1P FMF 2023-12-04.pdf
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Radi bi določili naklon sekante,
+ torej naklon premice,
+ določene z 
+\begin_inset Formula $x$
+\end_inset
+
+ in neko bližnjo točko 
+\begin_inset Formula $x+h$
+\end_inset
+
+ na grafu funkcije,
+ ki je odvisen le od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ ne pa tudi od izbire 
+\begin_inset Formula $h$
+\end_inset
+
+.
+ Bližnjo točko pošljemo proti začetni —
+ 
+\begin_inset Formula $h$
+\end_inset
+
+ pošljemo proti 0.
+ Naklon izračunamo s izrazom 
+\begin_inset Formula $\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $x$
+\end_inset
+
+ označimo 
+\begin_inset Formula $f'\left(x\right)\coloneqq\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Če limita obstaja v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pravimo,
+ da je funkcija odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva na množici 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ če je odvedljiva na vsaki 
+\begin_inset Formula $t\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri odvodov preprostih funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c,c\in\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{\cancelto{c}{f\left(x+h\right)}-\cancelto{c}{f\left(x\right)}}{h}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to0}\frac{2xh+h^{2}}{h}=\lim_{h\to0}2x+h=2x$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{op}{Odvod potence}
+\end_layout
+
+\end_inset
+
+.
+ Za poljuben 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ so funkcije 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+ odvedljive na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{n}-x^{n}=\sum_{k=0}^{n}\binom{n}{k}h^{k}x^{n-k}-x^{n}=\cancel{x^{n}}+nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}\cancel{-x^{n}}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}}{h}=\lim_{h\to0}\frac{\cancel{h}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)}{\cancel{h}}=\lim_{h\to0}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=nx^{n-1}+\cancel{\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}\cancelto{0}{h^{k-1}}x^{n-k}}=nx^{n-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Najprej dokažimo 
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to\infty}\frac{\sin\left(x+h\right)-\sin\left(x\right)=\sin x\cos h+\sin h\cos x-\sin x=\sin x\left(\cos h-1\right)+\sin h\cos x}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\sin x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}+\cos x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{-\sin h}{h}\cdot\frac{\sin h}{\cos h+1}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\cancelto{1}{\frac{\sin h}{h}}\left(\cos x-\cancel{\sin x\frac{\cancelto{0}{\sin h}}{\cos h+1}}\right)=\cos x
+\]
+
+\end_inset
+
+Sedaj dokažimo še 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\cos\left(x+h\right)-\cos\left(x\right)=\cos x\cos h-\sin x\sin h-\cos x=\cos x\left(\cos h-1\right)-\sin x\sin h}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cos x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}-\sin x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{-\sin h}{h}\cdot\frac{\sin^{}h}{\cos h+1}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cancelto{1}{\frac{\sin h}{h}}\left(\cancel{-\cos x\frac{\cancelto{0}{\sin h}}{\cos h+1}}-\sin x\right)\right)=-\sin x
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Od prej vemo 
+\begin_inset Formula $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$
+\end_inset
+
+ (limita zaporedja).
+ Velja tudi 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$
+\end_inset
+
+ (funkcijska limita).
+ Ne bomo dokazali.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oef}{Odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(x\right)=a^{x}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f'\left(x\right)=a^{x}\ln a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=a^{x}a^{h}-a^{x}}{h}=\lim_{h\to0}a^{x}\frac{a^{h}-1}{h}=\cdots
+\]
+
+\end_inset
+
+Sedaj pišimo 
+\begin_inset Formula $\frac{1}{z}\coloneqq a^{h}-1$
+\end_inset
+
+.
+ Ulomek 
+\begin_inset Formula $\frac{a^{h}-1}{h}$
+\end_inset
+
+ namreč ni odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Sedaj
+\begin_inset Formula 
+\[
+a^{h}-1=\frac{1}{z}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a^{h}=\frac{1}{z}+1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\log_{a}\left(\frac{1}{z}+1\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}
+\]
+
+\end_inset
+
+Nadaljujmo s prvotnim računom,
+ ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\searrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\searrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\searrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}}{\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}}=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}\ln a}{\ln\left(\frac{1}{z}+1\right)}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln e}=a^{x}\ln a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\nearrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\nearrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\nearrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\cancelto{e}{\left(\frac{1}{z}+1\right)^{z}}}=a^{x}\ln a
+\]
+
+\end_inset
+
+Kajti 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^{x}=e^{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a\in(0,1]$
+\end_inset
+
+ Podobno kot zgodaj,
+ bodisi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je tam tudi zvezna.
+\end_layout
+
+\begin_layout Proof
+Predpostavimo,
+ da obstaja limita 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Želimo dokazati 
+\begin_inset Formula $f\left(x\right)=\lim_{t\to x}f\left(t\right)$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+f\left(x\right)=\lim_{t\to x}f\left(t\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{t\to x}f\left(t\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}f\left(x+h\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}\left(f\left(x+h\right)-f\left(x\right)\right)=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot h\right)
+\]
+
+\end_inset
+
+Limita obstaja,
+ čim obstajata 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+,
+ ki obstaja po predpostavki,
+ in 
+\begin_inset Formula $\lim_{h\to0}h$
+\end_inset
+
+,
+ ki obstaja in ima vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\left|x\right|=\sqrt{x^{2}}$
+\end_inset
+
+.
+ Je zvezna,
+ ker je kompozitum zveznih funkcij,
+ toda v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\lim_{h\to0}\frac{\left|h\right|-0}{h}=\lim_{h\to0}\sgn h$
+\end_inset
+
+.
+ Limita ne obstaja,
+ ker 
+\begin_inset Formula $-1=\lim_{h\nearrow0}\sgn h\not=\lim_{h\searrow0}\sgn h=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ odvedljivi v 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so 
+\begin_inset Formula $f+g,f-g,f\cdot g,f/g$
+\end_inset
+
+ (slednja le,
+ če 
+\begin_inset Formula $g\left(x\right)\not=0$
+\end_inset
+
+) in velja 
+\begin_inset Formula $\left(f\pm g\right)'=f'\pm g'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(f/g\right)'=\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo vse štiri trditve.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f+g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(f+g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f+g\right)\left(x+h\right)-\left(f+g\right)\left(x\right)=f\left(x+h\right)+g\left(x+h\right)-f\left(x\right)-g\left(x\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}+\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)=f\left(x\right)'+g\left(x\right)'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-f$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $g=-f$
+\end_inset
+
+.
+ 
+\begin_inset Formula $g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{-f\left(x+h\right)+f\left(x\right)}{h}=-\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=-f\left(x\right)'$
+\end_inset
+
+,
+ zato
+\begin_inset Formula 
+\[
+\left(f-g\right)'\left(x\right)=\left(f+\left(-g\right)\right)'\left(x\right)=f'\left(x\right)+\left(-g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\cdot g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(fg\right)'\left(x\right)=\lim_{h\to0}\frac{\left(fg\right)\left(x+h\right)-\left(fg\right)\left(x\right)=f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)+\left[f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x+h\right)\right]}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(x+h\right)\left(f\left(x+h\right)-f\left(x\right)\right)+f\left(x\right)\left(g\left(x+h\right)-g\left(x\right)\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\cancelto{g\left(x\right)}{g\left(x+h\right)}+\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f/g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(f/g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f/g\right)\left(x+h\right)-\left(f/g\right)\left(x\right)=\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x+h\right)g\left(x\right)}{g\left(x+h\right)g\left(x\right)}-\frac{f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}=\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)=f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+\left[f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\frac{g\left(x+h\right)-g\left(x\right)}{h}\cdot\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\left(\frac{1}{g\left(x\right)g\left(x+h\right)}\right)\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}g\left(x\right)-\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)\right)\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{1}{g^{2}\left(x\right)}\left(f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g^{2}\left(x\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\tan'\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)'=\frac{\sin'\left(x\right)\cos\left(x\right)-\sin\left(x\right)\cos'\left(x\right)}{\cos^{2}\left(x\right)}=\frac{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=\cos^{-2}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ok}{Odvod kompozituma}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)$
+\end_inset
+
+ (opomba:
+ 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $a\coloneqq f\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f\left(x+h\right)\coloneqq a+\delta\left(h\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left(g\circ f\right)\left(x+h\right)-\left(g\circ f\right)\left(x\right)=g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(a+\delta_{h}\right)-g\left(a\right)}{h}=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna,
+ zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\varphi\left(x\right)=\sin\left(x^{2}\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=x^{2},g\left(x\right)=\sin x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\varphi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=\sin'\left(x^{2}\right)\left(x^{2}\right)'=2x\cos\left(x^{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\psi\left(x\right)=\sin^{2}\left(x\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=\sin,g\left(x\right)=x^{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\psi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=2\sin x\cos x=\sin2x$
+\end_inset
+
+ (sinus dvojnega kota)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\delta'\left(x\right)=\sin\left(e^{x^{2}}\right)=\sin\left(e^{\left(x^{2}\right)}\right)=\left(g\circ h\circ f\right)\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=e^{x},f\left(x\right)=x^{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\delta'\left(x\right)=\cos\left(e^{x^{2}}\right)e^{x^{2}}2x$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:I\subseteq\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ je zvezno odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če je na 
+\begin_inset Formula $I$
+\end_inset
+
+ odvedljiva in je 
+\begin_inset Formula $f'$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+x^{2}\sin\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}$
+\end_inset
+
+ je na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ odvedljiva,
+ a ne zvezno.
+ Odvedljivost na 
+\begin_inset Formula $\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ je očitna,
+ preverimo še odvedljivost v 
+\begin_inset Formula $0$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\frac{h^{\cancel{2}}\sin\frac{1}{h}}{\cancel{h}}=\lim_{h\to0}h\sin\frac{1}{h}=0,
+\]
+
+\end_inset
+
+ker 
+\begin_inset Formula $h$
+\end_inset
+
+ pada k 0,
+ 
+\begin_inset Formula $\sin\frac{1}{h}$
+\end_inset
+
+ pa je omejen z 1.
+ Velja torej
+\begin_inset Formula 
+\[
+f'\left(x\right)=\begin{cases}
+2x\sin\frac{1}{x}-\cos\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}
+\]
+
+\end_inset
+
+Preverimo nezveznost v 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Spodnja limita ne obstaja.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\left(\cancel{2x\sin\frac{1}{x}}-\cos\frac{1}{x}\right)=-\lim_{x\to0}\cos\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{oi}{Odvod inverza}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ strogo monotona v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva in naj bo 
+\begin_inset Formula $f'\left(a\right)\not=0$
+\end_inset
+
+.
+ Tedaj bo inverzna funkcija,
+ definirana v okolici 
+\begin_inset Formula $b=f\left(a\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $b$
+\end_inset
+
+ odvedljiva in veljalo bo 
+\begin_inset Formula $\left(f^{-1}\right)'\left(b\right)=\frac{1}{f'\left(a\right)}=\frac{1}{f'\left(f^{-1}\left(b\right)\right)}.$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zism}{zvezna in strogo monotona}
+\end_layout
+
+\end_inset
+
+ na okolici 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ inverz na okolici 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+ obstaja in velja 
+\begin_inset Formula $f\left(x\right)=s\Leftrightarrow x=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f^{-1}\left(f\left(x\right)\right)=x$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{odvod kompozituma}
+\end_layout
+
+\end_inset
+
+ in velja 
+\begin_inset Formula 
+\[
+\left(f^{-1}\left(f\left(x\right)\right)\right)'=\left(f^{-1}\right)'\left(f\left(x\right)\right)\cdot f'\left(x\right)=\left(x\right)'=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $x=f^{-1}\left(y\right)$
+\end_inset
+
+ in dobimo za vsak 
+\begin_inset Formula $y$
+\end_inset
+
+ blizu 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\left(f^{-1}\right)'\left(y\right)=\frac{1}{f'\left(f^{-1}\left(y\right)\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov odvodov inverza.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\sqrt[n]{x}=x^{\frac{1}{n}}$
+\end_inset
+
+ za 
+\begin_inset Formula $n\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+ za 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in zgornji izrek.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+ in 
+\begin_inset Formula $f^{-1}=\sqrt[n]{x}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(x\right)=\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}=\frac{1}{f'\left(\sqrt[n]{x}\right)}=\frac{1}{n\sqrt[n]{x}^{n-1}}=\frac{1}{nx^{\frac{n-1}{n}=1-\frac{1}{n}}}=\frac{1}{n}x^{\frac{1}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $h\left(x\right)=\sqrt[n]{x^{m}}=x^{\frac{m}{n}}=g\left(x\right)^{m}$
+\end_inset
+
+ za 
+\begin_inset Formula $n,m\in\mathbb{N},x>0$
+\end_inset
+
+.
+ Uporabimo formulo za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{op}{odvod potence}
+\end_layout
+
+\end_inset
+
+ in 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ok}{kompozituma}
+\end_layout
+
+\end_inset
+
+ in zgornji primer.
+ Velja 
+\begin_inset Formula $g'\left(x\right)=\frac{1}{n}x^{\frac{1}{n}-1}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula 
+\[
+h'\left(x\right)=mg\left(x\right)^{m-1}\cdot g'\left(x\right)=m\left(x^{\frac{1}{n}}\right)^{m-1}\cdot\frac{1}{n}x^{\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m-1}{n}+\frac{1}{n}-1}=\frac{m}{n}x^{\frac{m}{n}-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Izkaže-se,-da"
+
+\end_inset
+
+Izkaže se,
+ da velja celo 
+\begin_inset Formula $\forall x>0,\alpha\in\mathbb{R}:\left(x^{\alpha}\right)'=\alpha x^{\alpha-1}$
+\end_inset
+
+.
+ Mi smo dokazali le za 
+\begin_inset Formula $\alpha\in\mathbb{Q}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Logaritmi,
+ inverz 
+\begin_inset Formula $e^{x}$
+\end_inset
+
+.
+ Gre za 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oef}{odvod eksponentne funkcije}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(a^{x}\right)=a^{x}\ln a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left(e^{x}\right)=e^{x}\ln e=e^{x}$
+\end_inset
+
+.
+ Uporavimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{oi}{odvod inverza}
+\end_layout
+
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left(f^{-1}\right)'\left(x\right)=\frac{1}{f'\left(f^{-1}\left(x\right)\right)}$
+\end_inset
+
+ in za 
+\begin_inset Formula $g\left(x\right)=\log x$
+\end_inset
+
+ uporabimo 
+\begin_inset Formula $g\left(x\right)=f^{-1}\left(x\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\log'\left(x\right)=\left(\left(e^{x}\right)^{-1}\right)'\left(x\right)=\frac{1}{e^{\log x}}=\frac{1}{x}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $g\left(x\right)=\arcsin x$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[-1,1\right]$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $g=f^{-1}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $f=\sin$
+\end_inset
+
+ za 
+\begin_inset Formula $x\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+g'\left(f\left(x\right)\right)=\frac{1}{f'\left(x\right)}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\cos x}
+\]
+
+\end_inset
+
+Ker velja 
+\begin_inset Formula $\sin^{2}x+\cos^{2}x=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\cos^{2}x=1-\sin^{2}x$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\cos x=\sqrt{1-\sin^{2}x}$
+\end_inset
+
+,
+ torej nadaljujemo:
+\begin_inset Formula 
+\[
+g'\left(\sin x\right)=\frac{1}{\sqrt{1-\sin^{2}x}}
+\]
+
+\end_inset
+
+Sedaj zamenjamo 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ s 
+\begin_inset Formula $t$
+\end_inset
+
+ in dobimo:
+\begin_inset Formula 
+\[
+g'\left(t\right)=\frac{1}{\sqrt{1-t^{2}}}=\arcsin^{2}t
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Diferencial
+\end_layout
+
+\begin_layout Standard
+Fiksirajmo funkcijo 
+\begin_inset Formula $f$
+\end_inset
+
+ in točko 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+,
+ v okolici katere je 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana.
+ Želimo oceniti vrednost funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v bližini točke 
+\begin_inset Formula $a$
+\end_inset
+
+ z linearno funkcijo – to je 
+\begin_inset Formula $y\left(x\right)=\lambda x$
+\end_inset
+
+ za neki 
+\begin_inset Formula $\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ ZDB Iščemo najboljši linearni približek,
+ odvisen od 
+\begin_inset Formula $h$
+\end_inset
+
+,
+ za 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Diferencial funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ je linearna preslikava 
+\begin_inset Formula $df\left(a\right):\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ z zahtevo
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left|f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)\right|}{\left|h\right|}=0.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)-df\left(a\right)\left(h\right)}{h}=0=\lim_{h\to0}\left(\frac{f\left(a+h\right)-f\left(a\right)}{h}-\frac{\left(df\left(a\right)\right)\left(h\right)}{h}\right)=
+\]
+
+\end_inset
+
+Upoštevamo linearnost preslikave
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}-df\left(a\right)=f'\left(a\right)-df\left(a\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f'\left(a\right)=df\left(a\right)
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $f\left(a+h\right)-f\left(a\right)\approx df\left(a\right)\left(h\right)$
+\end_inset
+
+ – najboljši linearni približek za 
+\begin_inset Formula $f\left(a+h\right)-f\left(h\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Uporaba diferenciala.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ je točka,
+ v kateri znamo izračunati funkcijsko vrednost,
+ 
+\begin_inset Formula $a+h$
+\end_inset
+
+ pa je točka,
+ v kateri želimo približek funkcijske vrednosti.
+ Izračunajmo približek 
+\begin_inset Formula $\sqrt{2}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sqrt{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a+h=2$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $a=2,25$
+\end_inset
+
+,
+ 
+\begin_inset Formula $h=-0,25$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(a\right)=\sqrt{a}=1,5$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(s\right)=\frac{1}{2\sqrt{x}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f\left(a=2,25\right)=\frac{1}{3}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(2\right)\approx f\left(a\right)+f'\left(2,25\right)\cdot h=1,5-0,25\cdot\frac{1}{3}=\frac{3}{2}-\frac{1}{4}\cdot\frac{1}{3}=\frac{17}{12}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Preizkus:
+ 
+\begin_inset Formula $\left(\frac{17}{12}\right)^{2}=\frac{289}{144}=2+\frac{1}{144}$
+\end_inset
+
+ ...
+ Absolutna napaka 
+\begin_inset Formula $\frac{1}{144}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva povsod na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $a\in I$
+\end_inset
+
+.
+ Če je v 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljiva tudi 
+\begin_inset Formula $f'$
+\end_inset
+
+,
+ pišemo 
+\begin_inset Formula $f''\left(a\right)=\left(f'\left(a\right)\right)'$
+\end_inset
+
+.
+ Podobno pišemo tudi višje odvode:
+ 
+\begin_inset Formula $f^{\left(1\right)}\left(a\right)=f'\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(n+1\right)}=\left(f^{\left(n\right)}\right)'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(0\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f^{\left(2\right)}\left(a\right)=f''\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Pomen besede 
+\begin_inset Quotes gld
+\end_inset
+
+odvod
+\begin_inset Quotes grd
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Odvod v dani točki:
+ 
+\begin_inset Formula $f'\left(a\right)$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ ali
+\end_layout
+
+\begin_layout Itemize
+Funkcija,
+ ki vsaki točki 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+ priredi 
+\begin_inset Formula $f'\left(x\right)$
+\end_inset
+
+ po zgornji definiciji.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+\begin_inset Formula $C^{n}\left(I\right)$
+\end_inset
+
+ je množica funkcije 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $\forall x\in I\exists f'\left(x\right),f''\left(x\right),f^{\left(3\right)},\dots,f^{\left(n\right)}\left(x\right)$
+\end_inset
+
+ in da so 
+\begin_inset Formula $f,f',f'',f^{\left(3\right)},\dots,f^{\left(n\right)}$
+\end_inset
+
+ zvezna funkcije na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ (seveda če obstaja 
+\begin_inset Formula $j-$
+\end_inset
+
+ti odvod,
+ obstaja tudi zvezen 
+\begin_inset Formula $j-1-$
+\end_inset
+
+ti odvod).
+ ZDB je to množica funkcij,
+ ki imajo vse odvode do 
+\begin_inset Formula $n$
+\end_inset
+
+ in so le-ti zvezni.
+ ZDB to so vse 
+\begin_inset Formula $n-$
+\end_inset
+
+krat zvezno odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $C^{\infty}\left(I\right)\coloneqq\bigcap_{n=1}^{\infty}C^{n}\left(I\right)$
+\end_inset
+
+ – to so neskončnokrat odvedljive funkcije na intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Intuitivno
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Baje.
+ Jaz sem itak do vsega skeptičen.
+\end_layout
+
+\end_inset
+
+ velja 
+\begin_inset Formula $C^{1}\left(I\right)\supset C^{2}\left(I\right)\supset C^{3}\left(I\right)\supset C^{4}\left(I\right)\supset\cdots$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Polimomi 
+\begin_inset Formula $\subset C^{\infty}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\left|x\right|^{3}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'\left(x\right)=\begin{cases}
+3x^{2} & ;x\geq0\\
+-3x^{2} & ;x<0
+\end{cases}=3x^{2}\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f''\left(x\right)=\begin{cases}
+6x & ;x\geq0\\
+-6x & ;x<0
+\end{cases}=6x\sgn x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f'''\left(x\right)=\begin{cases}
+6 & ;x>0\\
+-6 & ;x<0
+\end{cases}=6\sgn x$
+\end_inset
+
+ in v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ zato 
+\begin_inset Formula $f\in C^{2}\left(\mathbb{R}\right)$
+\end_inset
+
+ a 
+\begin_inset Formula $f\not\in C^{3}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ ker 
+\begin_inset Formula $\exists f''$
+\end_inset
+
+ in je zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ a 
+\begin_inset Formula $f'''$
+\end_inset
+
+ sicer obstaja,
+ a ni zvezna na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Velja pa 
+\begin_inset Formula $f\in C^{\infty}\left(\mathbb{R}\setminus\left\{ 0\right\} \right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Rolle.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\left(a\right)=f\left(b\right)\Longrightarrow\exists\alpha\in\left(a,b\right)\ni:f'\left(\alpha\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sumimo,
+ da je ustrezna 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ tista,
+ ki je 
+\begin_inset Formula $\max$
+\end_inset
+
+ ali 
+\begin_inset Formula $\min$
+\end_inset
+
+ od 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{zfnkm}{Ker}
+\end_layout
+
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ (kompaktni množici),
+ 
+\begin_inset Formula $\exists\alpha_{1}\in\left[a,b\right],\alpha_{2}\in\left[a,b\right]\ni:f\left(\alpha_{1}\right)=\max f\left(\left[a,b\right]\right)\wedge f\left(\alpha_{2}\right)=\min f\left(\left[a,b\right]\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Če je 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \subseteq\left\{ a,b\right\} $
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f\left(\alpha_{1}\right)=f\left(\alpha_{2}\right)$
+\end_inset
+
+ in je v tem primeru 
+\begin_inset Formula $f$
+\end_inset
+
+ konstanta (
+\begin_inset Formula $\exists!c\in\mathbb{R}\ni:f\left(x\right)=c$
+\end_inset
+
+),
+ ki je odvedljiva in ima povsod odvod nič.
+\end_layout
+
+\begin_layout Proof
+Sicer pa 
+\begin_inset Formula $\left\{ \alpha_{1},\alpha_{2}\right\} \not\subseteq\left\{ a,b\right\} $
+\end_inset
+
+.
+ Tedaj ločimo dva primera:
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{1}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni maksimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem maksimumu odvod 0.
+ Dokaz:
+\begin_inset Formula 
+\[
+f'\left(\alpha_{1}\right)=\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}
+\]
+
+\end_inset
+
+Za 
+\begin_inset Formula $a_{1}$
+\end_inset
+
+ (maksimum) velja 
+\begin_inset Formula $f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)\leq0$
+\end_inset
+
+ (čim se pomaknemo izven točke,
+ v kateri je maksimum,
+ je funkcijska vrednost nižja).
+ Potemtakem velja
+\begin_inset Formula 
+\[
+\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\quad\begin{cases}
+\leq0 & ;h>0\\
+\geq0 & ;h<0
+\end{cases}
+\]
+
+\end_inset
+
+Ker je funkcija odvedljiva na odprtem intervalu,
+ sta leva in desna limita enaki.
+\begin_inset Formula 
+\[
+0\geq\lim_{h\searrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=\lim_{h\nearrow0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}\geq0
+\]
+
+\end_inset
+
+Sledi
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(\alpha_{1}+h\right)-f\left(\alpha_{1}\right)}{h}=f'\left(x\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\alpha_{2}\in\left(a,b\right)$
+\end_inset
+
+ To pomeni,
+ da je globalni minimum na odprtem intervalu.
+ Trdimo,
+ da je v lokalnem minimumu odvod 0.
+ Dokaz je podoben tistemu za lokalni maksimum.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{lagrange}{Lagrange}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in odvedljiva na 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\exists\alpha\in\left(a,b\right)\ni:f\left(b\right)-f\left(a\right)=f'\left(\alpha\right)\left(b-a\right)\sim\frac{f\left(b\right)-f\left(a\right)}{b-a}=f'\left(\alpha\right)
+\]
+
+\end_inset
+
+ZDB na neki točki na grafu funkcije je tangenta na graf funkcije vzporedna premici,
+ ki jo določata točki 
+\begin_inset Formula $\left(a,f\left(a\right)\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b,f\left(b\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za dokaz Lagrangevega uporabimo Rolleov izrek.
+ Splošen primer prevedemo na primer 
+\begin_inset Formula $h\left(a\right)=h\left(b\right)$
+\end_inset
+
+ tako,
+ da od naše splošne funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ odštejemo linearno funkcijo 
+\begin_inset Formula $g$
+\end_inset
+
+,
+ da bo veljalo 
+\begin_inset Formula $\left(f-g\right)\left(a\right)=\left(f-g\right)\left(b\right)$
+\end_inset
+
+.
+ Za funkcijo 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ mora veljati naslednje:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\exists k,n\in\mathbb{R}\ni:f\left(x\right)=kx+n$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(a\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(b\right)=f\left(b\right)-f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Opazimo,
+ da mora biti koeficient funkcije 
+\begin_inset Formula $g$
+\end_inset
+
+ enak 
+\begin_inset Formula $\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ vertikalni odklon pa tolikšen,
+ da ima funkcija 
+\begin_inset Formula $g$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ničlo:
+\begin_inset Formula 
+\[
+\frac{f\left(b\right)-f\left(a\right)}{b-a}a+n=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n=-\frac{f\left(b\right)-f\left(a\right)}{b-a}a
+\]
+
+\end_inset
+
+Našli smo funkcijo 
+\begin_inset Formula $g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $\left(f-g\right)$
+\end_inset
+
+ sedaj ustreza pogojem za Rolleov izrek,
+ torej 
+\begin_inset Formula $\exists\alpha\in\left[a,b\right]\ni:\left(f-g\right)'\left(\alpha\right)=0\Leftrightarrow g'\left(\alpha\right)=f'\left(\alpha\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
+\end_inset
+
+,
+ kar smo želeli dokazati.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ nenujno zaprt niti omejen in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ Lipschitzova.
+ Lipschitzove funkcije so enakomerno zvezne.
+\end_layout
+
+\begin_layout Proof
+Po Lagrangeu velja 
+\begin_inset Formula $\forall x,y\in I\exists\alpha\in\left(x,y\right)\ni:f\left(x\right)-f\left(y\right)=f'\left(\alpha\right)\left(x-y\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\left|f\left(x\right)-f\left(y\right)\right|=\left|f'\left(\alpha\right)\right|\left|x-y\right|\leq\sup_{\beta\in\left(x,y\right)}\left|f'\left(\beta\right)\right|\left|x-y\right|$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists M>0\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|$
+\end_inset
+
+,
+ enakomerno zveznost pa dobimo tako,
+ da 
+\begin_inset Formula $\delta\left(\varepsilon\right)=\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}$
+\end_inset
+
+.
+ Računajmo.
+ Naj bo 
+\begin_inset Formula $M=\sup_{\beta\in I}\left|f'\left(\beta\right)\right|$
+\end_inset
+
+,
+ ki obstaja.
+\begin_inset Formula 
+\[
+\forall x,y:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x,y:\left|x-y\right|<\frac{\varepsilon}{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|<\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}\frac{\varepsilon}{\cancel{\sup_{\beta\in I}\left|f'\left(\beta\right)\right|}}<\varepsilon
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall\varepsilon\exists\delta\left(\varepsilon\right)\forall x,y:\left|x-y\right|<\delta\left(\varepsilon\right)\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Lipschnitzovim funkcijam pravimo tudi Hölderjeve funkcije reda 1.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je Hölderjeva funkcija reda 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ če velja 
+\begin_inset Formula $\exists M>0\forall x,y\in I:\left|f\left(x\right)-f\left(y\right)\right|\leq M\left|x-y\right|^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ odprti interval,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva.
+ Tedaj:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ pada na 
+\begin_inset Formula $I\Leftrightarrow f'\leq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo narašča na 
+\begin_inset Formula $I\Leftarrow f'>0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=x^{3}$
+\end_inset
+
+,
+ ki strogo narašča,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f$
+\end_inset
+
+ strogo pada na 
+\begin_inset Formula $I\Leftarrow f'<0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Protiprimer,
+ da ni 
+\begin_inset Formula $\Leftrightarrow:f\left(x\right)=-x^{3}$
+\end_inset
+
+,
+ ki strogo pada,
+ toda 
+\begin_inset Formula $f'\left(0\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokažimo le 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča na 
+\begin_inset Formula $I\Leftrightarrow f'\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ Drugo točko dokažemo podobno.
+ Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f'\geq0\Rightarrow f$
+\end_inset
+
+ narašča.
+ Vzemimo poljubna 
+\begin_inset Formula $t_{1}<t_{2}\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\alpha\in\left(t_{1},t_{2}\right)\ni:f\left(t_{2}\right)-f\left(t_{1}\right)=f'\left(\alpha\right)\left(t_{2}-t_{1}\right)$
+\end_inset
+
+.
+ Ker je po predpostavki 
+\begin_inset Formula $f'\left(\alpha\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $t_{2}-t_{1}>0$
+\end_inset
+
+,
+ je tudi 
+\begin_inset Formula $f\left(t_{2}\right)-f\left(t_{1}\right)\geq0$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(t_{2}\right)\geq f\left(t_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča 
+\begin_inset Formula $\Rightarrow f'\geq0$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Po predpostavki je 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\geq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h>0$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $f\left(x+h\right)-f\left(x\right)\leq0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $h<0$
+\end_inset
+
+.
+ Torej je ulomek vedno nenegativen.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Konveksnost in konkavnost
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ interval in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall a,b\in I$
+\end_inset
+
+ daljica 
+\begin_inset Formula $\left(a,f\left(a\right)\right),\left(b,f\left(b\right)\right)$
+\end_inset
+
+ leži nad grafom 
+\begin_inset Formula $f$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Enačba premice,
+ ki vsebuje to daljico,
+ se glasi (razmislek je podoben kot pri 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{lagrange}{Lagrangevem izreku}
+\end_layout
+
+\end_inset
+
+)
+\begin_inset Formula 
+\[
+g\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)
+\]
+
+\end_inset
+
+Za konveksno funkcijo torej velja 
+\begin_inset Formula $\forall a,b\in I:\forall x\in\left(a,b\right):f\left(x\right)\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)$
+\end_inset
+
+ oziroma
+\begin_inset Formula 
+\[
+\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}
+\]
+
+\end_inset
+
+Vsak 
+\begin_inset Formula $x$
+\end_inset
+
+ na intervalu lahko zapišemo kot 
+\begin_inset Formula $x=a+t\left(b-a\right)$
+\end_inset
+
+ za nek 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $x-a=t\left(b-a\right)$
+\end_inset
+
+ in konveksnost se glasi
+\begin_inset Formula 
+\[
+\forall a,b\in I:\forall t\in\left(0,1\right):f\left(a+t\left(b-a\right)\right)\leq\frac{f\left(b\right)-f\left(a\right)}{\cancel{b-a}}t\cancel{\left(b-a\right)}+f\left(a\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+f\left(a+t\left(b-a\right)\right)=f\left(a+tb-ta\right)=f\left(\left(1-t\right)a+tb\right)\leq tf\left(b\right)-tf\left(a\right)+f\left(a\right)=\left(1-t\right)f\left(a\right)+tf\left(b\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Konveksna kombinacija izrazov 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je izraz oblike 
+\begin_inset Formula $\left(1-t\right)a+tb$
+\end_inset
+
+ za 
+\begin_inset Formula $t\in\left(0,1\right)$
+\end_inset
+
+.
+ Potemtakem je ZDB definicija konveksnosti 
+\begin_inset Formula $\forall a,b\in I:$
+\end_inset
+
+ funkcijska vrednost konveksne kombinacije 
+\begin_inset Formula $a,b$
+\end_inset
+
+ je kvečjemu konveksna kombinacija funkcijskih vrednosti 
+\begin_inset Formula $a,b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Konkavnost pa je definirana tako,
+ da povsod obrnemo predznake,
+ torej daljica leži pod grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $\forall a,b\in I:f\left(\left(1-t\right)a+tb\right)\geq\left(1-t\right)f\left(a\right)+tf\left(b\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\sin x$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I=\left[-\pi,0\right]$
+\end_inset
+
+.
+ Je konveksna.
+ Se vidi iz grafa.
+ Preveriti analitično bi bilo težko.
+\end_layout
+
+\begin_layout Example*
+Formulirajmo drugačen pogoj za konveksnost.
+ Naj bo spet 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $I$
+\end_inset
+
+ interval.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je konveksna 
+\begin_inset Formula 
+\[
+\Leftrightarrow\forall a,b\in I\forall x\in\left(a,b\right):\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq\frac{f\left(b\right)-f\left(a\right)}{b-a}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Sedaj glejmo le poljuben 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Po prejšnjem pogoju moramo gledati še vse poljubne 
+\begin_inset Formula $b$
+\end_inset
+
+,
+ večje od 
+\begin_inset Formula $a$
+\end_inset
+
+ (ker le tako lahko konstruiramo interval).
+ Za 
+\begin_inset Formula $b$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ mora biti diferenčni kvocient večji od diferenčnega kvocienta 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Ta pogoj pa je ekvivalenten temu,
+ da diferenčni kvocient 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+ s fiksnim 
+\begin_inset Formula $a$
+\end_inset
+
+ in čedalje večjim 
+\begin_inset Formula $x$
+\end_inset
+
+ narašča,
+ torej je pogoj za konveksnost tudi:
+\begin_inset Formula 
+\[
+\forall a\in I\forall x>a:g_{a}\left(x\right)=\frac{f\left(x\right)-f\left(a\right)}{x-a}\text{ je naraščajoča funkcija}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna na odprtem intervalu 
+\begin_inset Formula $I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall a\in I$
+\end_inset
+
+ obstajata funkciji
+\begin_inset Formula 
+\[
+\left(D_{+}f\right)\left(a\right)=\lim_{x\searrow a}g_{a}\left(x\right)=\inf_{x\in I,x>a}g_{a}\left(x\right)\text{ (desni odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(D_{-}f\right)\left(a\right)=\lim_{x\nearrow a}g_{a}\left(x\right)=\sup_{x\in I,x<a}g_{a}\left(x\right)\text{ (levi odvod \ensuremath{f} v \ensuremath{a})}
+\]
+
+\end_inset
+
+in obe sta naraščajoči na 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Obstoj sledi iz monotonosti 
+\begin_inset Formula $g_{a}\left(a\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\lim_{x\searrow a}g_{a}\left(x\right)=\lim_{x\searrow a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$
+\end_inset
+
+ in enako za levo limito.
+ Diferenčni kvocient mora namreč biti naraščajoč.
+ S tem smo dokazali,
+ da je vsaka konveksna funkcija zvezna
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ni pa vsaka konveksna funkcija odvedljiva,
+ protiprimer je 
+\begin_inset Formula $f\left(x\right)=\left|x\right|$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $x_{1},x_{2},x\in I\ni:x_{1}<x_{2}<x$
+\end_inset
+
+.
+ Pomagaj si s skico
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO DORIŠI SKICO ZVZ VII/ANA1UČ/str.
+ 13
+\end_layout
+
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna,
+ sledi 
+\begin_inset Formula $g_{x}\left(x_{1}\right)\leq g_{x}\left(x_{2}\right)$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\forall s,t\in\mathbb{R}:g_{s}\left(t\right)=g_{t}\left(s\right)$
+\end_inset
+
+,
+ lahko našo neenakost zapišemo kot 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+.
+ Sledi (desni neenačaj iz 
+\begin_inset Formula $g_{x_{1}}\left(x\right)\leq g_{x_{2}}\left(x\right)$
+\end_inset
+
+,
+ levi neenačaj pa ker 
+\begin_inset Formula $g$
+\end_inset
+
+ narašča):
+\begin_inset Formula 
+\[
+\left(D_{+}\left(f\right)\right)\left(x_{1}\right)=\inf_{x\in I,x>x_{1}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{1}}\left(x\right)\leq\inf_{x\in I,x>x_{2}}g_{x_{2}}\left(x\right)=\left(D_{+}\left(f\right)\right)\left(x_{2}\right)
+\]
+
+\end_inset
+
+Podobno dokažemo
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+DOPIŠI KAKO!
+ TODO XXX FIXME
+\end_layout
+
+\end_inset
+
+,
+ da 
+\begin_inset Formula $D_{-}$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:I^{\text{odp.}}\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\geq0$
+\end_inset
+
+ in 
+\begin_inset Formula $f$
+\end_inset
+
+ konkavna 
+\begin_inset Formula $\Leftrightarrow\forall x\in I:f''\left(x\right)\leq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco za konveksnost (konkavnost podobno).
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $f$
+\end_inset
+
+ konveksna in dvakrat odvedljiva,
+ torej je odvedljiva in sta levi in desni odvod enaka,
+ po prejšnji posledici pa levi in desni odvod naraščata,
+ torej 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $f''\geq0$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $x,a\in I$
+\end_inset
+
+.
+ Po Lagrangeu 
+\begin_inset Formula $\exists\xi\text{ med \ensuremath{x} in \ensuremath{a}}\ni:f\left(x\right)-f\left(x\right)=f'\left(\xi\right)\left(x-a\right)$
+\end_inset
+
+.
+ Iz predpostavke 
+\begin_inset Formula $f''>0$
+\end_inset
+
+ sledi,
+ da 
+\begin_inset Formula $f'$
+\end_inset
+
+ narašča.
+ Če je 
+\begin_inset Formula $x>\xi>a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\geq f'\left(a\right)$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\geq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $x<\xi<a$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f'\left(\xi\right)\left(x-a\right)\leq f'\left(a\right)\left(x-a\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsection
+Ekstremi funkcij ene spremenljivke
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interal,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\min\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+ Pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni maksimum,
+ če 
+\begin_inset Formula $\exists\delta>0\ni:\max\left\{ f\left(x\right);\forall x\in\left(a-\delta,a+\delta\right)\right\} =f\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ odvedljiva in ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum/maksimum,
+ tedaj je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Glej dokaz Rolleovega izreka.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $f$
+\end_inset
+
+ ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ ekstrem,
+ če ima v 
+\begin_inset Formula $a$
+\end_inset
+
+ lokalni minimum ali lokalni maksimum.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ stacionarno točko.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+ odprt interval,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ dvakrat odvedljiva ter naj bo 
+\begin_inset Formula $f'\left(a\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)>0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni minimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)<0\Rightarrow$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+ ima 
+\begin_inset Formula $f$
+\end_inset
+
+ lokalni maksimum
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f''\left(a\right)=0\Rightarrow$
+\end_inset
+
+ nedoločeno
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Sledi iz 
+\begin_inset Formula $f''>0\Rightarrow$
+\end_inset
+
+ stroga konveksnost in 
+\begin_inset Formula $f''<0\Rightarrow$
+\end_inset
+
+ stroga konkavnost.
+\end_layout
+
+\begin_layout Subsection
+L'Hopitalovo pravilo
+\end_layout
+
+\begin_layout Standard
+Kako izračunati 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Če so funkcije zvezne v 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(a\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če imata funkciji v 
+\begin_inset Formula $a$
+\end_inset
+
+ limito in 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)\not=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\lim_{x\to a}f\left(x\right)}{\lim_{x\to a}g\left(x\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ omejena,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=0$
+\end_inset
+
+ in je na neki okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Formula $g\left(x\right)$
+\end_inset
+
+ navzdol omejena več od nič ali navzgor omejena manj od nič,
+ velja 
+\begin_inset Formula $\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Zanimivi primeri pa so,
+ ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+ ali pa ko 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+,
+ na primer 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}}{x}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{x^{2}}$
+\end_inset
+
+.
+ Tedaj uporabimo L'Hopitalovo pravilo.
+\end_layout
+
+\begin_layout Theorem*
+Če velja hkrati:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Eno-izmed-slednjega:"
+
+\end_inset
+
+Eno izmed slednjega:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=\infty$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=-\infty$
+\end_inset
+
+ in hkrati 
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=-\infty$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f,g$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ odvedljivi
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Potem 
+\begin_inset Formula $\exists L\coloneqq\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\Rightarrow\exists\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}$
+\end_inset
+
+ in ta limita je enaka 
+\begin_inset Formula $L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\end_layout
+
+\begin_layout Example*
+Nekaj primerov uporabe L'Hopitalovega pravila.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula 
+\[
+\lim_{x\to0}x^{x}=\lim_{x\to0}e^{lnx^{x}}=\lim_{x\to0}e^{x\ln x}=e^{\lim_{x\to0}x\ln x}
+\]
+
+\end_inset
+
+Računajmo 
+\begin_inset Formula $\lim_{x\to0}x\ln x$
+\end_inset
+
+ z L'Hopitalom.
+ Potrebujemo ulomek.
+ Ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ tedaj bi dobili 
+\begin_inset Formula $\lim_{x\to0}\frac{x^{2}\ln x}{x}$
+\end_inset
+
+.
+ Toda v tem primeru števec in imenovalec ne ustrezata pogoju 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Eno-izmed-slednjega:"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za L'Hopitalovo pravilo.
+ Druga ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $\left(\ln x\right)^{-1}$
+\end_inset
+
+,
+ tedaj dobimo 
+\begin_inset Formula $\lim_{x\to0}\frac{x}{\left(\ln x\right)^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{1}{\frac{-1}{\log^{2}x}\cdot\frac{1}{x}}=\lim_{x\to0}-x\log^{2}x$
+\end_inset
+
+,
+ kar je precej komplicirano.
+ Tretja ideja:
+ množimo števec in imenovalec z 
+\begin_inset Formula $x^{-1}$
+\end_inset
+
+,
+ tedaj števec in imenovalec divergirata k 
+\begin_inset Formula $-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{\ln x}{x^{-1}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\left(\ln x\right)'}{\left(x^{-1}\right)'}=\lim_{x\to0}\frac{x^{-1}}{-x^{-2}}=\lim_{x\to0}-x=0
+\]
+
+\end_inset
+
+Potemtakem 
+\begin_inset Formula $\lim_{x\to0}x^{x}=e^{0}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\lim_{x\to0}\frac{1-\cos x}{x^{2}}$
+\end_inset
+
+.
+ Obe strani ulomkove črte konvergirata k 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ Prav tako ko enkrat že uporabimo L'H.
+\begin_inset Formula 
+\[
+\lim_{x\to0}\frac{1-\cos x}{x^{2}}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\sin x}{2x}\overset{\text{L'H}}{=}\lim_{x\to0}\frac{\cos x}{2}=\frac{1}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Taylorjev izrek in Taylorjeva formula
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici 
+\begin_inset Formula $a$
+\end_inset
+
+ dovoljkrat odvedljiva.
+ Želimo aproksimirati 
+\begin_inset Formula $f\left(a+h\right)$
+\end_inset
+
+ s polinomi danega reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Iščemo polinome reda 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=0$
+\end_inset
+
+ konstante.
+ 
+\begin_inset Formula $f\left(a+h\right)\approx f\left(a\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=1$
+\end_inset
+
+ linearne funkcije.
+ 
+\begin_inset Formula $f\left(a+h\right)\sim f\left(a\right)+f'\left(a\right)h$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=2$
+\end_inset
+
+ ...
+ Želimo najti 
+\begin_inset Formula $a_{0},a_{1},a_{2}\in\mathbb{R}$
+\end_inset
+
+,
+ odvisne le od 
+\begin_inset Formula $f$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ za katere 
+\begin_inset Formula $f\left(a+b\right)\approx a_{0}+a_{1}h+a_{2}h^{2}$
+\end_inset
+
+.
+ Ko govorimo o aproksimaciji,
+ mislimo take koeficiente,
+ da se približek najbolje prilega dejanski funkcijski vrednosti,
+ v smislu,
+ da
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)}{h^{2}}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula $a_{0}$
+\end_inset
+
+ izvemo takoj,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}f\left(a+h\right)-\left(a_{0}+a_{1}h+a_{2}h^{2}\right)=0=f\left(a\right)-\left(a_{0}+0h+0h^{2}\right)=f\left(a\right)-a_{0}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{0}=f\left(a\right)$
+\end_inset
+
+.
+ Za preostale koeficiente uporabimo L'Hopitalovo pravilo,
+ ki pove,
+ da zadošča,
+ da je
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f'\left(a+h\right)-\left(0+a_{1}+a_{2}h\right)}{2h}=0
+\]
+
+\end_inset
+
+Zopet glejmo števec in vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $f'\left(a\right)-a_{1}=0\Rightarrow f'\left(a\right)=a_{1}$
+\end_inset
+
+.
+ Spet uporabimo L'H:
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f''\left(a+h\right)-\left(0+0+2a_{2}\right)}{2}
+\]
+
+\end_inset
+
+Vstavimo 
+\begin_inset Formula $h=0$
+\end_inset
+
+ v 
+\begin_inset Formula $f''\left(a+h\right)-2a_{2}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $2a_{2}=f''\left(a\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{2}=\frac{f''\left(a\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $n=3$
+\end_inset
+
+ Ugibamo,
+ da je najboljši kubični približek
+\begin_inset Formula 
+\[
+f\left(a+h\right)\approx h\mapsto f\left(a\right)+f'\left(a\right)h+\frac{f''\left(a\right)}{2}h^{2}+\frac{f'''\left(a\right)}{6}h^{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Taylor.
+ Naj bo 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I$
+\end_inset
+
+ interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+krat odvedljiva v točki 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists g_{n}:I-a\to\mathbb{R}\ni:$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $I-a$
+\end_inset
+
+ pomeni interval 
+\begin_inset Formula $I$
+\end_inset
+
+ pomaknjen v levo za 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(a+h\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}h^{j}+g_{n}\left(h\right)h^{n}$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{h\to0}g_{n}\left(h\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Sedaj pišimo 
+\begin_inset Formula $x=a+h$
+\end_inset
+
+.
+ Tedaj se izrek glasi:
+ 
+\begin_inset Formula $\exists\tilde{g_{n}}:I\to\mathbb{R}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}+\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\lim_{x\to a}\tilde{g_{n}}\left(x\right)=0$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Tedaj označimo 
+\begin_inset Formula $T_{n,f,a}\left(x\right)=\sum_{j=0}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(x-a\right)^{j}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom za 
+\begin_inset Formula $f$
+\end_inset
+
+ okrog točke 
+\begin_inset Formula $a$
+\end_inset
+
+) in 
+\begin_inset Formula $R_{n,f,a}\left(x\right)=\tilde{g_{n}}\left(x\right)\left(x-a\right)^{n}$
+\end_inset
+
+ (pravimo ostanek/napaka).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+krat odvedljiva na odprtem intervalu 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\in I$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\forall b\in I\exists\alpha\in I\text{ med \ensuremath{a} in \ensuremath{b}}\ni:R_{n}\left(b\right)=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $T_{n}\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)+\frac{f''\left(a\right)}{2!}\left(x-a\right)^{2}+\cdots+\frac{f^{\left(n\right)}\left(a\right)}{n!}\left(x-a\right)^{n}$
+\end_inset
+
+ torej 
+\begin_inset Formula $n-$
+\end_inset
+
+ti taylorjev polinom in naj bo 
+\begin_inset Formula $K$
+\end_inset
+
+ tako število,
+ da velja 
+\begin_inset Formula $f\left(b\right)-T_{n}\left(b\right)=K\left(b-a\right)^{n+1}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=f\left(x\right)-T_{n}\left(x\right)-K\left(x-a\right)^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{velja}{Velja}
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $T_{n}^{\left(k\right)}\left(a\right)=f\left(a\right)$
+\end_inset
+
+ za 
+\begin_inset Formula $k\leq n$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\frac{d\sum_{j=1}^{n}\frac{f^{\left(j\right)}\left(a\right)}{n!}\left(h\right)^{n}}{dh}=\frac{f^{\left(j\right)}\left(a\right)n!}{n!}\cdot1=f^{\left(j\right)}\left(a\right)$
+\end_inset
+
+.
+ Vsi členi z eksponentom,
+ manjšim od 
+\begin_inset Formula $k$
+\end_inset
+
+,
+ se odvajajo v 0,
+ točno pri eksponentu 
+\begin_inset Formula $k$
+\end_inset
+
+ se člen odvaja v konstanto,
+ pri višjih členih pa ostane potencirana spremenljivka,
+ ki je 
+\begin_inset Formula $0$
+\end_inset
+
+ (tu mislimo odstopanje od 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ označeno s 
+\begin_inset Formula $h$
+\end_inset
+
+),
+ torej se ti členi tudi izničijo.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\forall k\leq n:F^{\left(k\right)}\left(a\right)=0$
+\end_inset
+
+.
+ Nadalje velja 
+\begin_inset Formula $F\left(a\right)=F\left(b\right)=0$
+\end_inset
+
+,
+ ker smo pač tako definirali funkcijo 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ zato obstaja po Rolleovem izreku tak 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'\left(\alpha_{1}\right)=0$
+\end_inset
+
+.
+ Po Rolleovem izreku nadalje obstaja tak 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F''\left(\alpha_{2}\right)=0$
+\end_inset
+
+.
+ Spet po Rolleovem izreku obstaja tak 
+\begin_inset Formula $\alpha_{3}$
+\end_inset
+
+ med 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{2}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F'''\left(\alpha_{3}\right)=0$
+\end_inset
+
+.
+ Postopek lahko ponavljamo in dobimo tak 
+\begin_inset Formula $\alpha=\alpha_{n+1}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $F^{\left(n+1\right)}\left(\alpha\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $\forall x\in I:T_{n}^{\left(n+1\right)}\left(x\right)=0$
+\end_inset
+
+ (očitno,
+ isti argument kot v 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{velja}{drugem odstavku dokaza}
+\end_layout
+
+\end_inset
+
+),
+ to pomeni 
+\begin_inset Formula $f^{\left(n+1\right)}\left(\alpha\right)=\left(K\left(x-a\right)^{n+1}\right)^{\left(n+1\right)}=K\left(n+1\right)!$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $K=\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}$
+\end_inset
+
+ in zato 
+\begin_inset Formula $f\left(b\right)=T_{n}\left(b\right)+\frac{f^{\left(n+1\right)}\left(\alpha\right)}{\left(n+1\right)!}\left(b-a\right)^{\left(n+1\right)}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $\left(n+1\right)-$
+\end_inset
+
+ti odvod omejen na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $\exists M>0\forall x\in I:\left|f^{\left(n+1\right)}\left(x\right)\right|\leq M$
+\end_inset
+
+,
+ lahko ostanek eksplicitno ocenimo,
+ in sicer 
+\begin_inset Formula $\left|R_{n}\left(x\right)\right|\leq\frac{M}{\left(n+1\right)!}\left|x-a\right|^{n+1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kaj pa se zgodi,
+ ko 
+\begin_inset Formula $n$
+\end_inset
+
+ pošljemo v neskončnost?
+ Iskali bi aproksimacije s 
+\begin_inset Quotes gld
+\end_inset
+
+polinomi neskončnega reda
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f\in C^{\infty}$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj definiramo Taylorjevo vrsto 
+\begin_inset Formula $f$
+\end_inset
+
+ v okolici točke 
+\begin_inset Formula $a$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{f,a}\left(x\right)\coloneqq\sum_{j=0}^{\infty}\frac{f^{\left(j\right)}\left(a\right)}{j!}\left(x-a\right)^{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Question*
+Ali Taylorjeva vrsta konvergira oziroma kje konvergira?
+ Kakšna je zveza s 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+?
+ Kakšen je 
+\begin_inset Formula $R_{f,a}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Oglejmo si potenčne vrste (
+\begin_inset Formula $\sum_{j=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+) kot poseben primer funkcijskih vrst (
+\begin_inset Formula $\sum_{j=0}^{\infty}a_{k}\left(x\right)$
+\end_inset
+
+).
+ Vemo,
+ da ima potenčna vrsta konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $x\in\left(-R,R\right)$
+\end_inset
+
+ konvergira,
+ za 
+\begin_inset Formula $x\in\left[-R,R\right]^{C}$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Theorem*
+Naj ima potenčna vrsta 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}b_{k}x^{k}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+ Tedaj ima tudi 
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}kb_{k}x^{k-1}$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\frac{1}{R_{g}}=\limsup_{k\to\infty}\sqrt[k]{\left|ka_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|\left|a_{k}\right|}=\limsup_{k\to\infty}\sqrt[k]{\left|k\right|}\sqrt[k]{\left|a_{k}\right|}=\cdots
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{k\to\infty}\sqrt[k]{\left|k\right|}=\lim_{k\to\infty}k^{1/k}=e^{\lim_{k\to\infty}\frac{1}{k}\ln k}\overset{\text{L'H}}{=}e^{\lim_{k\to\infty}\frac{\frac{1}{k}}{k}}=e^{\lim_{k\to\infty}\cancelto{0}{\frac{1}{k^{2}}}}=e^{0}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots=\limsup_{k\to\infty}1\cdot\sqrt[k]{\left|a_{k}\right|}=\frac{1}{R_{f}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Če ima potenčna vrsta 
+\begin_inset Formula $f$
+\end_inset
+
+ konvergenčni radij 
+\begin_inset Formula $R>0$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f\in C^{\infty}\left(\left(-R,R\right)\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $a_{k}=\frac{f^{\left(k\right)}\left(0\right)}{k!}$
+\end_inset
+
+,
+ potem velja 
+\begin_inset Formula $g=f'$
+\end_inset
+
+ (iz izreka zgoraj).
+ Razlaga:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}a_{k}x^{k}=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{k!}x^{k}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f'\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}$
+\end_inset
+
+ (
+\begin_inset Formula $k$
+\end_inset
+
+ začne z 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ ker se 
+\begin_inset Formula $k=0$
+\end_inset
+
+ člen odvaja v konstanto 
+\begin_inset Formula $0$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)=\sum_{k=1}^{\infty}ka_{k}x^{k-1}=\sum_{k=1}^{\infty}\frac{kf^{\left(k\right)}\left(0\right)}{k!}x^{k-1}=\sum_{k=1}^{\infty}\frac{f^{\left(k\right)}\left(0\right)}{\left(k-1\right)!}x^{k-1}=f'\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ (
+\begin_inset Formula $J$
+\end_inset
+
+ je interval 
+\begin_inset Formula $\subseteq\mathbb{R}$
+\end_inset
+
+) je realno analitična,
+ če se jo da okoli vsake točke 
+\begin_inset Formula $c\in J$
+\end_inset
+
+ razviti v potenčno vrsto,
+ torej če 
+\begin_inset Formula $f\left(x\right)=\sum_{k=0}^{\infty}\frac{f^{\left(k\right)}\left(c\right)}{k!}\left(x-c\right)^{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $x$
+\end_inset
+
+ blizu 
+\begin_inset Formula $c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $f\in C^{\infty}\Rightarrow f$
+\end_inset
+
+ je realno analitična.
+ Protiprimer je 
+\begin_inset Formula $f\left(x\right)=e^{\frac{-1}{\left|x\right|}}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME ZAKAJ?,
+ ne razumem
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri Taylorjevih vrst.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=e^{x}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+ti tayorjev polinom za 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ okoli 
+\begin_inset Formula $0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $T_{n,e^{x},0}\left(x\right)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots+\frac{x^{n}}{n!}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $e^{x}=T_{n,e^{x},0}\left(x\right)+R_{n,e^{x},0}\left(x\right)$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $\lim_{n\to\infty}R_{n,e^{x},0}\left(x\right)=0$
+\end_inset
+
+.
+ Ne bomo dokazali.
+ Sledi 
+\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots+\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente in opazimo učinek odvajanja:
+ 
+\begin_inset Formula $\cos,-\sin,-\cos,\sin,\cos,-\sin,\dots$
+\end_inset
+
+.
+ Členi vrste 
+\begin_inset Formula $\sin x$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ so:
+ 
+\begin_inset Formula $1,0,-1,0,1,0,-1,\dots$
+\end_inset
+
+.
+ Opazimo izpadanje vsakega drugega člena.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots+\left(-1\right)^{k}\frac{x^{2k}}{\left(2k\right)!}$
+\end_inset
+
+.
+ Opazimo sode eksponente.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)$
+\end_inset
+
+.
+ A lahko to funkcijo razvijemo v taylorjevo vrsto okoli točke 0?
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="7" columns="3">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $k$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $f^{\left(k\right)}\left(0\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $0$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{1-x}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-1}{\left(1-x\right)^{2}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-1$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $3$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-2}{\left(1-x\right)^{3}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-2$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\cdots$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $n$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $\frac{-\left(n-1\right)!}{\left(1-x\right)^{n}}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+\begin_inset Formula $-\left(n-1\right)!$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Razvijanje 
+\begin_inset Formula $\log\left(1-x\right)$
+\end_inset
+
+ okoli točke 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Velja 
+\begin_inset Formula $f\left(x\right)=\log\left(1-x\right)=\sum_{k=1}^{\infty}\frac{-\left(k-1\right)!}{k!}x^{k}=-\sum_{k=1}^{\infty}\frac{x^{k}}{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $\left|x\right|<1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Integrali
+\end_layout
+
+\begin_layout Standard
+Radi bi definirali ploščino 
+\begin_inset Formula $P=\left\{ \left(x,t\right)\in\mathbb{R}^{2};x\in\left[a,b\right],t\in\left[0,f\left(x\right)\right]\right\} $
+\end_inset
+
+ za funkcijo 
+\begin_inset Formula $f:\left[a,b\right]\to[0,\infty)$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME skica ANA1P FMF 2024-01-09/str.3
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $P$
+\end_inset
+
+ aproksimiramo s pravokotniki,
+ katerih ploščino smo predhodno definirali takole:
+\end_layout
+
+\begin_layout Definition*
+Ploščina pravokotnika s stranicama 
+\begin_inset Formula $c$
+\end_inset
+
+ in 
+\begin_inset Formula $d$
+\end_inset
+
+ je 
+\begin_inset Formula $c\cdot d$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Najprej diskusija.
+ Naj bo 
+\begin_inset Formula $t_{j}$
+\end_inset
+
+ delitev 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
+\end_inset
+
+.
+ Ne zahtevamo ekvidistančne delitve,
+ torej take,
+ pri kateri bi bile razdalje enake.
+ Kako naj definiramo višine pravokotnikov,
+ katerih stranice so delilne točke 
+\begin_inset Formula $t_{n}$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Standard
+Lahko tako,
+ da na vsakem intervalu 
+\begin_inset Formula $\left[t_{i},t_{i+1}\right]$
+\end_inset
+
+ izberemo nek 
+\begin_inset Formula $\xi_{i}$
+\end_inset
+
+,
+ pravokotnicova osnovnica bode 
+\begin_inset Formula $t_{i+1}-t_{i}$
+\end_inset
+
+,
+ njegova višina pa 
+\begin_inset Formula $f\left(\xi_{i}\right)$
+\end_inset
+
+.
+ Ploščina 
+\begin_inset Formula $P$
+\end_inset
+
+ pod grafom funkcije je približno enaka vsoti ploščin teh pravokotnikov,
+ torej 
+\begin_inset Formula $\sum_{k=1}^{n}f\left(\xi_{k}\right)\left(t_{k}-t_{k-1}\right)=R\left(f,\vec{t},\vec{\xi}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ delitev in 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+ izbira točk na intervalih delitve.
+ Temu pravimo Riemannova vsota za 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ ki pripada delitvi 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ in izboru 
+\begin_inset Formula $\vec{\xi}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $D\coloneqq\left\{ \left[t_{j+1},t_{j}\right];j=\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+,
+ definiramo tako oznako 
+\begin_inset Formula $\left|D\right|_{\infty}\coloneqq\max_{j=\left\{ 1..n\right\} }\left(t_{j}-t_{j-1}\right)=\max_{I\in D}\left(\left|I\right|\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če 
+\begin_inset Formula $\exists A\in\mathbb{R}\ni:$
+\end_inset
+
+ za poljubno fine delitve (
+\begin_inset Formula $\left|D\right|_{\infty}=\infty^{-1}$
+\end_inset
+
+) 
+\begin_inset Formula $D$
+\end_inset
+
+ se pripadajoče Riemannove vsote malo razlikujejo od 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ pravimo številu 
+\begin_inset Formula $A$
+\end_inset
+
+ ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sedaj pa še formalna definicija.
+\end_layout
+
+\begin_layout Definition*
+Naj bodo 
+\begin_inset Formula $f,D,\xi$
+\end_inset
+
+ kot prej in 
+\begin_inset Formula $I\in\mathbb{R}$
+\end_inset
+
+ realno število.
+ Če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D\ni:\left|D\right|_{\infty}<\delta$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\forall$
+\end_inset
+
+ nabor 
+\begin_inset Formula $\xi=\xi_{1},\dots,\xi_{n}$
+\end_inset
+
+,
+ pripadajoč delitvi 
+\begin_inset Formula $D$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+velja 
+\begin_inset Formula $\left|R\left(f,D,\xi\right)-I\right|<\varepsilon\Longrightarrow I$
+\end_inset
+
+ je določen integral 
+\begin_inset Formula $f$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in je po definiciji ploščina lika 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če tak 
+\begin_inset Formula $I$
+\end_inset
+
+ obstaja,
+ kar ni 
+\emph on
+a priori
+\emph default
+,
+ pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $I=\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Temu pravimo Riemannov integral funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Darbouxove vsote.
+ Imamo torej delitev 
+\begin_inset Formula $D=\left\{ \left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} ;t_{0}=1,t_{n}=b\right\} $
+\end_inset
+
+ delitev za 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+.
+ Imamo tudi množico izbranih točk 
+\begin_inset Formula $\xi=\left\{ \xi_{j}\in\left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $R\left(f,D,\xi\right)=\sum_{j=1}^{n}f\left(\xi_{j}\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ocenimo 
+\begin_inset Formula $f\left(\xi_{j}\right)$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)\leq f\left(\xi_{j}\right)\leq\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Definirali smo 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ kot limito Riemannovih vsot s kakršnokoli delitvijo in izbiro 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ zato lahko pišemo 
+\begin_inset Formula $\forall j\in\left\{ 1..n\right\} :\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)=f\left(\xi_{j}\right)=\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Zato lahko limito Riemannovih vsot obravnavamo neodvisno od 
+\begin_inset Formula $\xi$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+s\left(f,D\right)\coloneqq\sum_{j=1}^{n}\left(\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)\leq R\left(f,D,\xi\right)\leq\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-f_{j-1}\right)\eqqcolon S\left(f,D\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Definirali smo dva nova pojma,
+ spodnjo Darbouxovo vsoto 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in zgornjo Darbouxovo vsoto 
+\begin_inset Formula $S\left(f,D\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $s\left(f,D\right)\leq R\left(f,D,\xi\right)\leq S\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $D$
+\end_inset
+
+ in 
+\begin_inset Formula $D'$
+\end_inset
+
+ delitvi za interval 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ če je ima 
+\begin_inset Formula $D'$
+\end_inset
+
+ vse delilne točke,
+ ki jih ima 
+\begin_inset Formula $D$
+\end_inset
+
+ in poleg njih še vsaj kakšno.
+ Označimo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+ (
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od 
+\begin_inset Formula $D$
+\end_inset
+
+).
+ Oglejmo si 
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $s\left(f,D'\right)$
+\end_inset
+
+.
+ Tedaj velja 
+\begin_inset Formula $s\left(f,D\right)\leq s\left(f,D'\right)$
+\end_inset
+
+,
+ ker je infimum po manjši množici lahko le večji —
+ s finejšo delitvijo smo vsaj neko množico (delitveni interval) razdelili na dva dela.
+ Za zgornjo Darbouxovo vsoto velja obratno,
+ torej 
+\begin_inset Formula $S\left(f,D\right)\geq S\left(f,D'\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Za poljubni različni delitvi 
+\begin_inset Formula $D_{1},D_{2}$
+\end_inset
+
+ intervala 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $s\left(f,D_{1}\right)\leq S\left(f,D_{2}\right)$
+\end_inset
+
+ ZDB Katerakoli spodnja Darbouxova vsota je kvečjemu tolikšna kot katerakoli zgornja.
+\end_layout
+
+\begin_layout Proof
+Označimo z 
+\begin_inset Formula $D_{1}\cup D_{2}$
+\end_inset
+
+ delitev,
+ ki vsebuje vse delilne točke tako 
+\begin_inset Formula $D_{1}$
+\end_inset
+
+ kot tudi 
+\begin_inset Formula $D_{2}$
+\end_inset
+
+.
+ Očitno velja,
+ da sta 
+\begin_inset Formula $D_{1}\subset D_{1}\cup D_{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $D_{2}\subset D_{1}\cup D_{2}$
+\end_inset
+
+.
+ Po prejšnjem izreku veljata leva in desna neenakost,
+ srednja pa iz definicije (očitno).
+\begin_inset Formula 
+\[
+s\left(f,D_{1}\right)\leq s\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{2}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ omejena.
+ Označimo 
+\begin_inset Formula $s\left(f\right)\coloneqq\sup_{\text{vse možne delitve }D}s\left(f,D\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $S\left(f\right)\coloneqq\inf_{\text{vse možne delitve }D}S\left(f,D\right)$
+\end_inset
+
+.
+ Funkcija 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ je Riemannovo,
+ če 
+\begin_inset Formula $s\left(f\right)=S\left(f\right)$
+\end_inset
+
+ oziroma če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D$
+\end_inset
+
+ na 
+\begin_inset Formula $J\ni:S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Integrabilnost 
+\begin_inset Formula $f$
+\end_inset
+
+ ne pomeni,
+ da 
+\begin_inset Formula $\exists D\ni:s\left(f,D\right)=S\left(f,D\right)$
+\end_inset
+
+.
+ Ni namreč nujno,
+ da množica vsebuje svoj supremum.
+ Primer:
+ za 
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall D:S\left(f,D\right)>s\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsaka zvezna funkcija je integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Po definiciji 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)=\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ je na zaprtem 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ enakomerno zvezna,
+ torej 
+\begin_inset Formula $\exists\delta>0\forall x_{1},x_{2}\in J:\left|x_{1}-x_{2}\right|<\delta\Rightarrow\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\frac{\varepsilon}{b-a}$
+\end_inset
+
+.
+ Izberimo tako delitev 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\forall j\in\left\{ 1..\left|D\right|\right\} :t_{j}-t_{j-1}<\delta$
+\end_inset
+
+.
+ Tedaj bo veljalo 
+\begin_inset Formula $\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)<\sum_{j=1}^{n}\frac{\varepsilon}{b-a}\left(t_{j}-t_{j-1}\right)=\frac{\varepsilon\left(b-a\right)}{b-a}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skratka dokazali smo 
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ torej je funkcija Riemannovo integrabilna po zgornji definiciji.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ družina intervalov 
+\begin_inset Formula $I_{j}\ni:A\subset\bigcup I_{j}\wedge\sum\left|I_{j}\right|<\varepsilon$
+\end_inset
+
+.
+ Primer:
+ vse števne in končne množice.
+\end_layout
+
+\begin_layout Theorem*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na intervalu 
+\begin_inset Formula $J\Leftrightarrow\left\{ x\in J;f\text{ ni zvezna v }x\right\} $
+\end_inset
+
+ ima mero 
+\begin_inset Formula $0$
+\end_inset
+
+.
+ ZDB če ima množica točk z definicijskega območja 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ v katerih 
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ mero 
+\begin_inset Formula $0$
+\end_inset
+
+ (recimo če je teh točk končno mnogo),
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna.
+\end_layout
+
+\begin_layout Fact*
+Označimo z 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ množico vseh integrabilnih funkcij na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ 
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ je vektorski prostor za množenje s skalarji iz 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $f,g\in I\left(J\right),\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ Velja aditivnost 
+\begin_inset Formula $f\left(x\right)+g\left(x\right)\in J\left(I\right)$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\int_{a}^{b}\left(f\left(x\right)+g\left(x\right)\right)dx=\int_{a}^{b}\left(f\left(x\right)\right)dx+\int_{a}^{b}\left(g\left(x\right)\right)dx$
+\end_inset
+
+ in homogenost 
+\begin_inset Formula $\int_{a}^{b}\lambda f\left(x\right)dx=\lambda\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $c\in J$
+\end_inset
+
+,
+ tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[c,b\right]$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=\int_{a}^{c}f\left(x\right)dx+\int_{c}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če sta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ integrabilni funkciji in če je 
+\begin_inset Formula $\forall x\in J:f\left(x\right)\leq g\left(x\right)$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx\leq\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Posledično velja ob isti predpostavki 
+\begin_inset Formula $\left|\int_{a}^{b}f\left(x\right)dx\right|\leq\int_{a}^{b}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+,
+ definiramo povprečje 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $J$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula 
+\[
+\left\langle f\right\rangle _{J}\coloneqq\frac{\int_{a}^{b}f\left(x\right)dx}{b-a}\in\mathbb{R}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Velja 
+\begin_inset Formula $\inf_{x\in J}f\left(x\right)\leq\left\langle f\right\rangle _{J}\leq\sup_{x\in J}f\left(x\right)$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ 
+\begin_inset Formula $\exists\xi\in J\ni:f\left(\xi\right)=\left\langle f\right\rangle _{J}$
+\end_inset
+
+ (izrek o vmesni vrednosti).
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ dana funkcija.
+ Nedoločeni integral 
+\begin_inset Formula $f$
+\end_inset
+
+ je takšna funkcija 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ če obstaja,
+ 
+\begin_inset Formula $\ni:F'=f\sim\forall x\in J:F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+ Pišemo tudi 
+\begin_inset Formula $Pf$
+\end_inset
+
+ ali 
+\begin_inset Formula $\mathbb{P}f$
+\end_inset
+
+ in pravimo,
+ da je 
+\begin_inset Formula $F=Pf$
+\end_inset
+
+ primitivna funkcija za 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $P\left(f+g\right)=Pf+Pg$
+\end_inset
+
+ (aditivnost odvoda) in 
+\begin_inset Formula $P\left(\lambda f\right)=\lambda Pf$
+\end_inset
+
+ (homogenost odvoda).
+\end_layout
+
+\begin_layout Definition*
+Nedoločeni integral je na intervalu določen do aditivne konstante natančno.
+ Če je 
+\begin_inset Formula $F'_{1}=f=F_{2}'$
+\end_inset
+
+ na intervalu 
+\begin_inset Formula $J$
+\end_inset
+
+ oziroma če na 
+\begin_inset Formula $J$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left(F_{1}-F_{2}\right)'=0$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $F_{1}-F_{2}=c$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $F_{1}=F_{2}+c$
+\end_inset
+
+ za neko konstanto 
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $F\left(x\right)=Pf\left(x\right)=\int f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Integracija po delih 
+\begin_inset Formula $\sim$
+\end_inset
+
+ per partes.
+ Velja 
+\begin_inset Formula $\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izhaja iz odvoda produkza 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|F\left(x_{1}\right)-F\left(x_{2}\right)\right|=$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left|\int_{a}^{x_{1}}f\left(t\right)dt-\int_{a}^{x_{2}}f\left(t\right)dt\right|=\left|\int_{a}^{x_{1}}f\left(t\right)dt+\int_{x_{2}}^{a}f\left(t\right)dt\right|=\left|\int_{x_{2}}^{x_{1}}f\left(t\right)dt\right|=\left|\int_{x_{1}}^{x_{2}}f\left(t\right)dt\right|\leq\int_{x_{1}}^{x_{2}}f\left(t\right)dt
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Osnovni izrek analize/fundamental theorem of calcusus.
+ Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $F$
+\end_inset
+
+ odvedljiva na 
+\begin_inset Formula $J$
+\end_inset
+
+ in velja 
+\begin_inset Formula $F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+F\left(x+h\right)-F\left(x\right)=\int_{x}^{x+h}f\left(t\right)dt\quad\quad\quad\quad/:h
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{F\left(x+h\right)-F\left(x\right)}{h}=\frac{\int_{x}^{x+h}f\left(t\right)dt}{h}=\left\langle f\right\rangle _{\left[x,x+h\right]}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+F'\left(x\right)=\lim_{h\to0}\left\langle f\right\rangle _{x,x+h}=f\left(x\right).
+\]
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+glej ANA1P FMF 2024-01-15.pdf/str.
+ 5 za dokaz,
+ ki ga ne razumem,
+ zakaj je 
+\begin_inset Formula $\lim_{h\to0}\left\langle f\right\rangle _{\left[x,x+h\right]}-f\left(x\right)=0$
+\end_inset
+
+...
+ ampak sej to je nekak očitno
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $G=Pf$
+\end_inset
+
+ (
+\begin_inset Formula $G'=f$
+\end_inset
+
+).
+ Tedaj je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $F'=f=G'$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\left(F-G\right)'=0\Rightarrow F-G=c\in\mathbb{R}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $G\left(x\right)=F\left(x\right)+c$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $G\left(a\right)=F\left(a\right)=0$
+\end_inset
+
+ po definiciji 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $G\left(a\right)=c$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $F\left(x\right)=G\left(x\right)-G\left(a\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $F\left(b\right)=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $F\left(b\right)=\int_{a}^{b}f\left(t\right)dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Iskanje primitivne funkcije
+\end_layout
+
+\begin_layout Itemize
+Uganemo jo
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(x^{n}\right)=\frac{x^{n+1}}{n+1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(e^{x}\right)=e^{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\sin x\right)=-\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\ln x\right)=x\left(\ln x-1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Substitucija/uvedba nove spremenljivke
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+ne razumem.
+ mogoče bom v naslednjem življenju.
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $F\left(x\right)$
+\end_inset
+
+ nedoločeni integral funkcije 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ ter 
+\begin_inset Formula $\phi\left(x\right)$
+\end_inset
+
+ odvedljiva funkcija.
+ Potem velja
+\begin_inset Formula 
+\[
+F\left(\phi\left(t\right)\right)=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Formula je posledica odvoda kompozituma:
+\begin_inset Formula 
+\[
+\left(F\left(\phi\left(t\right)\right)\right)'=F'\left(\phi\left(t\right)\right)\phi'\left(t\right)=f\left(\phi\left(t\right)\right)\phi'\left(t\right)
+\]
+
+\end_inset
+
+integrirajmo levo in desno stran:
+\begin_inset Formula 
+\[
+\int\left(F\left(\phi\left(t\right)\right)\right)'dt=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dt.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Izlimitirani integrali
+\end_layout
+
+\begin_layout Standard
+Doslej smo računali določene integrale omejene funkcije na omejenem intervalu,
+ torej 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Kaj pa neomejen interval,
+ torej 
+\begin_inset Formula $\lim_{b\to\infty}\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,\infty)\to\mathbb{R}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $\forall m>a:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,-m\right]$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+,
+ pracimo,
+ da integral 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ konvergira,
+ sicer pa divergira.
+ Označimo 
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx\coloneqq\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+.
+ Podobno definiramo 
+\begin_inset Formula $\int_{-\infty}^{a}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Pomemben primer.
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{\alpha}dx=?$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\int_{1}^{M}x^{\alpha}dx=\frac{M^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}=\frac{M^{\alpha+1}-1}{\alpha+1}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\exists\lim_{M\to\infty}\int_{1}^{M}x^{\alpha}dx\Leftrightarrow\alpha\not=-1$
+\end_inset
+
+.
+ Poglejmo,
+ kaj se zgodi v 
+\begin_inset Formula $\alpha=-1$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx=\ln M-\ln1=\ln M$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\lim_{n\to\infty}\ln M=\infty$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ je absolutno konvergenten,
+ če je 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Velja 
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<0\Rightarrow\int_{a}^{\infty}f\left(x\right)dx<\infty$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left|\int_{a}^{\infty}f\left(x\right)dx\right|\leq\int_{a}^{\infty}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ali je predpostavka,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena,
+ sploh potrebna?
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:[a,b)\to\mathbb{R}\ni:\forall c<b:f$
+\end_inset
+
+ integrabilna na 
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+.
+ V točki 
+\begin_inset Formula $b$
+\end_inset
+
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ lahko neomejena.
+ Če 
+\begin_inset Formula $\exists$
+\end_inset
+
+ končna limita 
+\begin_inset Formula $\lim_{c\to b}\int_{a}^{c}f\left(x\right)dx$
+\end_inset
+
+,
+ je integral 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ konvergenten,
+ sicer je divergenten.
+ Podobno definiramo,
+ če je funkcija definirana na intervalu 
+\begin_inset Formula $(a,b]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\alpha<0$
+\end_inset
+
+ ima graf 
+\begin_inset Formula $x^{\alpha}$
+\end_inset
+
+ v 
+\begin_inset Formula $x=0$
+\end_inset
+
+ pol.
+ Računajmo
+\begin_inset Formula 
+\[
+\lim_{\varepsilon\to0}\int_{\varepsilon}^{1}x^{\alpha}dx=\lim_{\varepsilon\to0}\frac{x^{\alpha+1}}{\alpha+1}\vert_{\varepsilon}^{1}=\lim_{\varepsilon\to0}\left(\frac{1}{\alpha+1}-\frac{\varepsilon^{\alpha+1}}{\alpha+1}\right)=\lim_{\varepsilon\to0}\frac{1-\varepsilon^{\alpha+1}}{\alpha+1}=\lim_{\varepsilon\to0}\frac{1-\cancelto{0}{e^{\left(\alpha+1\right)\ln\varepsilon}}}{\alpha+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Pridobimo pogoj 
+\begin_inset Formula $\alpha\not=-1$
+\end_inset
+
+ (imenovalec) in 
+\begin_inset Formula $\alpha+1>0$
+\end_inset
+
+ (da bo 
+\begin_inset Formula $\left(\alpha+1\right)\ln\varepsilon\to-\infty$
+\end_inset
+
+),
+ torej skupaj s predpostavko 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Torej 
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx=\frac{1}{\alpha+1}$
+\end_inset
+
+ za 
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Uporaba integrala
+\end_layout
+
+\begin_layout Itemize
+Ploščine:
+ 
+\begin_inset Formula $f\geq0$
+\end_inset
+
+ na 
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je 
+\begin_inset Formula $f\in I\left(J\right)$
+\end_inset
+
+,
+ je ploščina lika med 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in grafom 
+\begin_inset Formula $f$
+\end_inset
+
+ definirana kot 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $f$
+\end_inset
+
+ ni pozitivna,
+ pa je 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=pl\left(L_{1}\right)-pl\left(L_{2}\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $L_{1}$
+\end_inset
+
+ lik nad 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in 
+\begin_inset Formula $L_{2}$
+\end_inset
+
+ lik pod 
+\begin_inset Formula $x$
+\end_inset
+
+ osjo.
+\end_layout
+
+\begin_layout Example*
+Ploščina kroga:
+ Enačba krožnice je 
+\begin_inset Formula $x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ za 
+\begin_inset Formula $r>0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $y=\sqrt{r^{2}-x^{2}}$
+\end_inset
+
+.
+ Ploščina kroga z radijem 
+\begin_inset Formula $r$
+\end_inset
+
+ je torej 
+\begin_inset Formula $2\int_{-r}^{r}\sqrt{r^{2}-x^{2}}dx=\cdots=\pi r^{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/ana1/teor3.lyx b/šola/ana1/teor3.lyx
new file mode 100644
index 0000000..97befd1
--- /dev/null
+++ b/šola/ana1/teor3.lyx
@@ -0,0 +1,1238 @@
+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\use_default_options true
+\maintain_unincluded_children false
+\language slovene
+\language_package default
+\inputencoding utf8
+\fontencoding global
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+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
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+\cite_engine basic
+\cite_engine_type default
+\use_bibtopic false
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+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
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+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Rešen tretji izpit teorije Analize 1 — IŠRM 2023/24
+\end_layout
+
+\begin_layout Abstract
+Izpit je potekal v petek, 30.
+ avgusta 2024 od desete
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Avtor tega besedila je na izpit zamudil poldrugo uro.
+\end_layout
+
+\end_inset
+
+ do dvanajste ure.
+ Nosilec predmeta je 
+\noun on
+Oliver Dragičević
+\noun default
+.
+ Naloge in rešitve sem po spominu spisal 
+\noun on
+Anton Luka Šijanec
+\noun default
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left[15\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Podaj natančne definicije naslednjih pojmov:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+limita zaporedja, stekališče zaporedja
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje in 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $L$
+\end_inset
+
+ je limita 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\sim L=\lim_{n\to\infty}a_{n}\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n>n_{0}:\left|a_{n}-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $L$
+\end_inset
+
+ je stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow\forall\varepsilon>0\exists\mathcal{A}\subseteq\mathbb{N},\left|\mathcal{A}\right|=\left|\mathcal{\mathbb{N}}\right|\ni:\left\{ a_{n};n\in\mathcal{A}\right\} \subseteq\left(L-\varepsilon,L+\varepsilon\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+vsota (neskončne) konvergentne vrste
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ poljubno zaporedje.
+ 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}\coloneqq\lim_{n\to\infty}\sum_{k=1}^{n}a_{n}$
+\end_inset
+
+.
+ Če limita obstaja, je vrsta 
+\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
+\end_inset
+
+ konvergentna in njena vsota je enaka tej limiti.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Cauchyjev pogoj za zaporedja
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realno zaporedje.
+ Konvergentno je natanko tedaj, ko ustreza Cauchyjevemu pogoju: 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall m,n\geq n_{0}:\left|a_{n}-a_{m}\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+odprte, zaprte, omejene, kompaktne množice v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je odprta, ko 
+\begin_inset Formula $\forall a\in\mathcal{A}\exists\varepsilon>0\ni:\left(a-\varepsilon,a+\varepsilon\right)\subseteq\mathcal{A}$
+\end_inset
+
+, ko za vsako točko množice obstaja neka njena okolica, ki je podmnožica
+ množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je zaprta, ko je 
+\begin_inset Formula $\mathcal{A}^{\mathcal{C}}\coloneqq\mathbb{R}\setminus\mathcal{A}$
+\end_inset
+
+ odprta.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je omejena, ko 
+\begin_inset Formula $\exists m,M\in\mathbb{R}\forall a\in\mathcal{A}:a\leq M\wedge a\geq m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Množica 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow\mathcal{A}$
+\end_inset
+
+ zaprta 
+\begin_inset Formula $\wedge$
+\end_inset
+
+ 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+ omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+limita funkcije v dani točki
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathcal{D}$
+\end_inset
+
+ okolica 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\setminus\left\{ a\right\} \to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ je limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a\sim L=\lim_{x\to a}f\left(x\right)\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}\setminus\left\{ a\right\} :\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+zveznost funkcije
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $a\in\mathcal{D}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+ .
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna na množici 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+, če je zvezna na vsaki točki množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+odvedljivost funkcije
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je odvedljiva v 
+\begin_inset Formula $a\text{\ensuremath{\Leftrightarrow\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}}}\in\mathbb{R}$
+\end_inset
+
+, ZDB ko obstaja slednja limita.
+ Tedaj definiramo 
+\begin_inset Quotes eld
+\end_inset
+
+odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+
+\begin_inset Quotes erd
+\end_inset
+
+: 
+\begin_inset Formula $f'\left(a\right)=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je odvedljiva na množici 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+, če je odvedljiva na vsaki točki množice 
+\begin_inset Formula $\mathcal{A}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+določen integral realne funkcije na zaprtem omejenem intervalu.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Naj bodo 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+\end_layout
+
+\begin_layout Enumerate
+Definirajmo pojem delitve 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+.
+ Delitev so točke 
+\begin_inset Formula $t_{0},\dots,t_{n}$
+\end_inset
+
+, da velja 
+\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
+\end_inset
+
+ za nek 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Točke identificiramo z delilnimi intervali takole: 
+\begin_inset Formula $D_{n}=\left[t_{n-1},t_{n}\right]$
+\end_inset
+
+.
+ Delitev torej identificiramo z množico teh dedlilnih intervalov: 
+\begin_inset Formula $D=\left\{ D_{k};\forall k\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+.
+ Definiramo tudi velikost delitve: 
+\begin_inset Formula $\left|D_{\infty}\right|=\max_{k\in\left\{ 1..n\right\} }\left|D_{k}\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Definirajmo pojem izbire za dano delitev.
+ Naj bo 
+\begin_inset Formula $D$
+\end_inset
+
+ delitev.
+ Pripadajoča izbira so take izbirne točke 
+\begin_inset Formula $\xi_{1},\dots,\xi_{n}$
+\end_inset
+
+, da velja 
+\begin_inset Formula $\forall k\in\left\{ 1..n\right\} :\xi_{k}\in D_{k}$
+\end_inset
+
+.
+ Množico teh izbirnih točk označimo z 
+\begin_inset Formula $\xi\coloneqq\left\{ \xi_{k};\forall k\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+, če 
+\begin_inset Formula $\exists I\in\mathbb{R}\forall\varepsilon>0\exists\delta>0\forall$
+\end_inset
+
+ delitev 
+\begin_inset Formula $D\forall$
+\end_inset
+
+ izbiro 
+\begin_inset Formula $\xi$
+\end_inset
+
+, pripadajočo delitvi 
+\begin_inset Formula $D:\left|D_{\infty}\right|<\delta\Rightarrow\left|\sum_{k=1}^{n}\left|D_{k}\right|f\left(\xi\right)-I\right|<\varepsilon$
+\end_inset
+
+.
+ Tedaj pravimo, da je 
+\begin_inset Formula $I$
+\end_inset
+
+ določen integral 
+\begin_inset Formula $f$
+\end_inset
+
+ na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $I\eqqcolon\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[15\right]$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Pojasni princip matematične indukcije.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(P_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedje logičnih vrednosti/izjav/izrazov.
+ Če velja
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $P_{1}$
+\end_inset
+
+ drži in hkrati
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
+\end_inset
+
+ drži 
+\begin_inset Formula $\Rightarrow P_{n+1}$
+\end_inset
+
+ drži,
+\end_layout
+
+\begin_layout Standard
+potem velja 
+\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
+\end_inset
+
+ drži.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Z matematično indukcijo dokaži
+\begin_inset Formula 
+\[
+\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Baza 
+\begin_inset Formula $n=1$
+\end_inset
+
+: 
+\begin_inset Formula $1=\frac{1\left(1+1\right)}{2}$
+\end_inset
+
+ Velja.
+\end_layout
+
+\begin_layout Enumerate
+Indukcijska predpostavka: 
+\begin_inset Formula $1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Korak 
+\begin_inset Formula $n\to n+1$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+1+2+\cdots+n+\cancel{n+1}\overset{?}{=}\frac{\left(n+1\right)\left(n+1+1\right)}{2}=\frac{n^{2}+2n+n+2}{2}=\frac{n\left(n+1\right)}{2}+\cancel{n+1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+1+2+\cdots+n\overset{\text{I.P.}}{=}\frac{n\left(n+1\right)}{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Sklep: 
+\begin_inset Formula $\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[25\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Naj bosta 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ realni konvergentni zaporedji.
+ Dokaži, da je 
+\begin_inset Formula $c_{n}\coloneqq a_{n}b_{n}$
+\end_inset
+
+ prav tako konvergentno zaporedje.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Označimo 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}\eqqcolon A$
+\end_inset
+
+ in 
+\begin_inset Formula $\lim_{n\to\infty}b_{n}\eqqcolon B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Uganemo, da je 
+\begin_inset Formula $\lim_{n\to\infty}a_{n}b_{n}=AB$
+\end_inset
+
+.
+ To moramo sedaj dokazati.
+\end_layout
+
+\begin_layout Itemize
+Dokazujemo, da 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}b_{n}-AB\right|<\varepsilon\sim\left|a_{n}b_{n}+a_{n}B-a_{n}B-AB\right|=\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|<\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Ker po trikotniški neenakosti velja 
+\begin_inset Formula $\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|\leq\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|$
+\end_inset
+
+, je dovolj za poljuben 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ dokazati
+\begin_inset Formula 
+\[
+\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\varepsilon
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno, 
+\begin_inset Formula $\exists n_{1}\in\mathbb{N}\forall n\geq n_{1}:\left|a_{n}-A\right|<\frac{\varepsilon}{2\left|a\right|}$
+\end_inset
+
+, kjer je 
+\begin_inset Formula $a$
+\end_inset
+
+ zgornja meja zaporedja 
+\begin_inset Formula $a_{n}$
+\end_inset
+
+.
+ Slednje je omejeno, ker je konvergentno.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ konvergentno, 
+\begin_inset Formula $\exists n_{2}\in\mathbb{N}\forall n\geq n_{1}:\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|B\right|}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Tedaj za 
+\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
+\end_inset
+
+ velja
+\begin_inset Formula 
+\[
+\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\frac{\varepsilon\left|a\right|}{2\left|a_{n}\right|}+\frac{\varepsilon}{2}\leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
+\]
+
+\end_inset
+
+in izrek je dokazan.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[?\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Dokaži, da je zvezna realna funkcija na zaprtem intervalu omejena.
+ Natančno navedi vse izreke, ki jih pri tem dokazu uporabiš.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bodo 
+\begin_inset Formula $a,b\in\mathbb{R}$
+\end_inset
+
+ in zvezna 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ poljubne.
+\end_layout
+
+\begin_layout Itemize
+Dokaz, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzgor.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ na zaprti množici, je omejeno zaporedje, torej ima stekališče.
+ Recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprta, je 
+\begin_inset Formula $s\in\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in s tem v 
+\begin_inset Formula $s$
+\end_inset
+
+, velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Po konstrukciji 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=\infty$
+\end_inset
+
+, torej 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+, kar ni mogoče, saj 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+ po predpostavki.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Predpostavka 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena
+\begin_inset Quotes erd
+\end_inset
+
+ ne velja, torej smo dokazali, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Dokaz, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzdol.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{\mathbb{N}}}$
+\end_inset
+
+ na zaprti množici, je omejeno zaporedje, torej ima stekališče.
+ Recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ zaprta, je 
+\begin_inset Formula $s\in\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ in s tem v 
+\begin_inset Formula $s$
+\end_inset
+
+, velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Po konstrukciji 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=-\infty$
+\end_inset
+
+, torej 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+, kar ni mogoče, saj 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+ po predpostavki.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Predpostavka 
+\begin_inset Quotes eld
+\end_inset
+
+
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena
+\begin_inset Quotes erd
+\end_inset
+
+ ne velja, torej smo dokazali, da je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzdol omejena.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena navzgor in navzdol, je omejena.
+\end_layout
+
+\begin_layout Itemize
+Uporabljeni izreki.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Zaporedje 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ s členi na kompaktni množici je omejeno.
+\end_layout
+
+\begin_layout Itemize
+Omejeno zaporedje ima stekališče.
+\end_layout
+
+\begin_layout Itemize
+Če je 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+ stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+, obstaja konvergentno podzaporedje 
+\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+, da je 
+\begin_inset Formula $s$
+\end_inset
+
+ njegova limita.
+\end_layout
+
+\begin_layout Itemize
+Množica je kompaktna natanko tedaj, ko vsebuje limite vseh konvergentnih
+ zaporedij s členi v njej.
+\end_layout
+
+\begin_layout Itemize
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $s$
+\end_inset
+
+, če za vsako k 
+\begin_inset Formula $s$
+\end_inset
+
+ konvergentno zaporedje velja, da njegovi s 
+\begin_inset Formula $f$
+\end_inset
+
+ preslikani členi konvergirajo v 
+\begin_inset Formula $f\left(s\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[?\right]$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+Za realno funkcijo ene spremenljivke dokaži verižno pravilo.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naj bodo 
+\begin_inset Formula $\mathcal{D},\mathcal{E},\mathcal{F}\subseteq\mathbb{R}$
+\end_inset
+
+, 
+\begin_inset Formula $x\in\mathcal{D}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:\mathcal{D}\to\mathcal{E}$
+\end_inset
+
+, 
+\begin_inset Formula $g:\mathcal{E}\to\mathcal{F}$
+\end_inset
+
+ poljubne.
+ Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Dokažimo, da je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in da velja 
+\begin_inset Formula 
+\[
+\left(g\circ f\right)'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right).
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Označimo 
+\begin_inset Formula $a\coloneqq f\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $f\left(x+h\right)=\delta_{h}+a$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(g\circ f\right)'\left(x\right)=\lim_{h\to0}\frac{g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(\delta_{h}+a\right)-g\left(a\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $x$
+\end_inset
+
+ odvedljiva, je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna, zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_body
+\end_document
diff --git a/šola/ana2/teor.lyx b/šola/ana2/teor.lyx
new file mode 100644
index 0000000..e1ee169
--- /dev/null
+++ b/šola/ana2/teor.lyx
@@ -0,0 +1,864 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\usepackage{hyperref}
+\usepackage{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\usepackage{amsmath}
+\usepackage{tikz}
+\newcommand{\udensdash}[1]{%
+    \tikz[baseline=(todotted.base)]{
+        \node[inner sep=1pt,outer sep=0pt] (todotted) {#1};
+        \draw[densely dashed] (todotted.south west) -- (todotted.south east);
+    }%
+}%
+\DeclareMathOperator{\Lin}{Lin}
+\DeclareMathOperator{\rang}{rang}
+\DeclareMathOperator{\sled}{sled}
+\DeclareMathOperator{\Aut}{Aut}
+\DeclareMathOperator{\red}{red}
+\DeclareMathOperator{\karakteristika}{char}
+\usepackage{algorithm,algpseudocode}
+\providecommand{\corollaryname}{Posledica}
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+\end_modules
+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 2cm
+\topmargin 2cm
+\rightmargin 2cm
+\bottommargin 2cm
+\headheight 2cm
+\headsep 2cm
+\footskip 1cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Title
+ANA2 IŠRM 2023/24
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Množice v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Razdalja točk v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ je norma njune razlike.
+ 
+\begin_inset Formula $\varepsilon-$
+\end_inset
+
+okolica točke 
+\begin_inset Formula $a\in\mathbb{R}^{n}$
+\end_inset
+
+ so take točke,
+ ki so od 
+\begin_inset Formula $a$
+\end_inset
+
+ oddaljene manj od 
+\begin_inset Formula $\varepsilon\in\mathbb{R}$
+\end_inset
+
+.
+ Robna točka množice 
+\begin_inset Formula $A$
+\end_inset
+
+ je taka točka,
+ katere poljubno majhna okolica vsebuje tako točke iz 
+\begin_inset Formula $A$
+\end_inset
+
+ kot tudi točke,
+ ki niso iz 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Odprta množica ne vsebuje robnih točk.
+ Zaprta množica je komplement neke odprte množice.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ zaprta 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ za vsako zaporedje s členi v 
+\begin_inset Formula $A$
+\end_inset
+
+ velja,
+ da so vsa njegova stekališča,
+ čim obstajajo,
+ v 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco
+\end_layout
+
+\begin_deeper
+\begin_layout Description
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $s$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a_{n}\in A$
+\end_inset
+
+ in 
+\begin_inset Formula $s\not\in A$
+\end_inset
+
+ (RAAPDD).
+ Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ zaprta,
+ je 
+\begin_inset Formula $\mathbb{R}\setminus A$
+\end_inset
+
+ odprta,
+ zato 
+\begin_inset Formula $\exists\varepsilon>0\ni:\left(s-\varepsilon,s+\varepsilon\right)\subset\mathbb{R}\setminus A$
+\end_inset
+
+,
+ torej v 
+\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
+\end_inset
+
+ ni nobenega člena zaporedja,
+ torej 
+\begin_inset Formula $s$
+\end_inset
+
+ ni stekališče 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Dokazujemo,
+ da je 
+\begin_inset Formula $A$
+\end_inset
+
+ zaprta,
+ torej,
+ da je 
+\begin_inset Formula $B=\mathbb{R}\setminus A$
+\end_inset
+
+ odprta.
+ PDDRAA 
+\begin_inset Formula $B$
+\end_inset
+
+ ni odprta 
+\begin_inset Formula $\Rightarrow\exists x\in B\ni:\forall n\in\mathbb{N}:$
+\end_inset
+
+ 
+\begin_inset Formula $n^{-1}-$
+\end_inset
+
+okolica 
+\begin_inset Formula $x$
+\end_inset
+
+ vsebuje nek element 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Našli smo torej zaporedje v 
+\begin_inset Formula $A$
+\end_inset
+
+ s stekališčem v 
+\begin_inset Formula $B$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Stroga podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je kompaktna,
+ če ima vsako zaporedje s členi v njej v njej tudi stekališče.
+ Množica je omejena,
+ če je podmnožica neke okolice izhodišča.
+\end_layout
+
+\begin_layout Theorem*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ kompaktna 
+\begin_inset Formula $\Leftrightarrow A$
+\end_inset
+
+ zaprta in omejena.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ zaporedje v 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ omejena,
+ je zaporedje omejeno,
+ torej premore stekališča.
+ Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ zaprta,
+ vsebuje vsa ta stekališča.
+ Torej je 
+\begin_inset Formula $A$
+\end_inset
+
+ kompaktna.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je omejeno,
+ sicer bi našli zaporedje,
+ da velja 
+\begin_inset Formula $a_{i}\geq i$
+\end_inset
+
+,
+ ki nima stekališča.
+ Treba je dokazati še,
+ da je 
+\begin_inset Formula $A$
+\end_inset
+
+ zaprta.
+ Vsa stekališča zaporedij s členi v 
+\begin_inset Formula $A$
+\end_inset
+
+ imajo v 
+\begin_inset Formula $A$
+\end_inset
+
+ stekališče (kompaktnost).
+ Torej za vsako stekališče zaporedja s členi v 
+\begin_inset Formula $A$
+\end_inset
+
+ velja,
+ da ima v 
+\begin_inset Formula $A$
+\end_inset
+
+ stekališče,
+ torej je 
+\begin_inset Formula $A$
+\end_inset
+
+ zaprta.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Vsako zaporedje v kompaktni množici ima stekališče,
+ kar za zaprto množico ni rečeno.
+ Zaprta množica lahko vsebuje zaporedja brez stekališč.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+razloži normo,
+ trikotniško neenakost,
+ itd.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Točka 
+\begin_inset Formula $a\in A\subseteq\mathbb{R}^{n}$
+\end_inset
+
+ je notranja,
+ če obstaja neka njena okolica,
+ ki je podmnožica 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Točka 
+\begin_inset Formula $a\in\mathbb{R}^{n}$
+\end_inset
+
+ je stekališče množice 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če vsaka njena okolica seka
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+t.
+ j.
+ ima neprazen presek z
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $A\setminus\left\{ a\right\} $
+\end_inset
+
+.
+ Točka 
+\begin_inset Formula $a\in A$
+\end_inset
+
+,
+ ki ni stekališče 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je izolirana točka 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Zaporedje s členi v 
+\begin_inset Formula $\mathbb{R}^{k}$
+\end_inset
+
+ je funkcija 
+\begin_inset Formula $\mathbb{N}\to\mathbb{R}^{k}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n\mapsto a_{n}=\left(a_{n}^{\left(1\right)},\dots,a_{n}^{\left(k\right)}\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $a\in\mathbb{R}^{k}$
+\end_inset
+
+ je limita zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+ s členi v 
+\begin_inset Formula $\mathbb{R}^{k}$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n>n_{0}:\left|a-a_{n}\right|<\varepsilon$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $a=\lim_{n\to\infty}a_{n}$
+\end_inset
+
+.
+ Če zaporedje ima limito,
+ je konvergentno,
+ sicer je divergentno.
+ Točka 
+\begin_inset Formula $s\in\mathbb{R}^{k}$
+\end_inset
+
+ je stekališče zaporedja 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+ s členi v 
+\begin_inset Formula $\mathbb{R}^{k}$
+\end_inset
+
+,
+ če je v vsaki okolici 
+\begin_inset Formula $s$
+\end_inset
+
+ neskončno členov 
+\begin_inset Formula $\left(a_{n}\right)_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Vsako konvergentno zaporedje je omejeno in ima natanko eno limito,
+ ki je njegovo edino stekališče.
+\end_layout
+
+\begin_layout Itemize
+Vsako omejeno zaporedje ima stekališče.
+\end_layout
+
+\begin_layout Itemize
+Stekališče zaporedja je limita nekega podzaporedja in obratno.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $A\subset\mathbb{R}^{k}$
+\end_inset
+
+ je zaprta 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ vsako stekališče zaporedja s členi v 
+\begin_inset Formula $A$
+\end_inset
+
+ leži v 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Funkcije več spremenljivk
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R}^{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+ preslikava.
+ Če je 
+\begin_inset Formula $k\geq2$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ funkcija več spremenljivk.
+ 
+\begin_inset Formula $\Gamma_{f}=\left\{ \left(x,fx\right);x\in D\right\} \subset\mathbb{R}^{k}\times\mathbb{R}$
+\end_inset
+
+ je graf funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $a\in\mathbb{R}^{k}$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $D$
+\end_inset
+
+ je 
+\begin_inset Formula $L\in\mathbb{R}$
+\end_inset
+
+ limita 
+\begin_inset Formula $f$
+\end_inset
+
+ v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta=\delta\left(a,\varepsilon\right)>0\forall x\in D,x\not=a:\left|\left|x-a\right|\right|<\delta\Rightarrow\left|fx-L\right|<\varepsilon$
+\end_inset
+
+ in pišemo 
+\begin_inset Formula $\lim_{x\to a}fx=L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Medtem ko imamo pri funkcijah ene spremenljivke levo in desno limito,
+ je tu obnašanje bolj zapleteno,
+ saj obstaja veliko različnih načinov približevanja k 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subseteq\mathbb{R}^{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+ funkcija in 
+\begin_inset Formula $a\in D$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\varepsilon>0\exists\delta=\delta\left(a,\varepsilon\right)>0\forall x\in D:\left|\left|x-a\right|\right|<\delta\Rightarrow\left|fx-fa\right|<\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je zvezna,
+ če je zvezna na vsaki točki svojega definicijskega območja.
+\end_layout
+
+\begin_layout Remark*
+Če je 
+\begin_inset Formula $a$
+\end_inset
+
+ stekališče 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}fx=fa$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če je 
+\begin_inset Formula $a$
+\end_inset
+
+ izolirana točka 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $f:D\subseteq\mathbb{R}^{k}\to\mathbb{R}$
+\end_inset
+
+ funkcija,
+ 
+\begin_inset Formula $Z=fD$
+\end_inset
+
+ njena zaloga vrednosti in 
+\begin_inset Formula $g:Z\to\mathbb{R}$
+\end_inset
+
+ funkcija.
+ Kompozitum ali sestavljena funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ je funkcija 
+\begin_inset Formula $k$
+\end_inset
+
+ spremenljivk 
+\begin_inset Formula $g\circ f:D\to\mathbb{R}$
+\end_inset
+
+,
+ definirana s predpisom 
+\begin_inset Formula $\left(g\circ f\right)x=gfx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ funkcija 
+\begin_inset Formula $k$
+\end_inset
+
+ spremenljivk,
+ zvezna v 
+\begin_inset Formula $a\in\mathbb{R}^{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ funkcija ene spremenljivke,
+ zvezna v 
+\begin_inset Formula $fa\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ zvezna v 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izberimo poljuben 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/aps1/dn/osvetlitev/Makefile b/šola/aps1/dn/osvetlitev/Makefile
new file mode 100644
index 0000000..2d78386
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/Makefile
@@ -0,0 +1,4 @@
+program: resitev.cpp
+	g++ -Wall -Wextra -pedantic -Wformat-security -std=c++20 -o$@ $<
+clean:
+	rm program
diff --git a/šola/aps1/dn/osvetlitev/in.txt b/šola/aps1/dn/osvetlitev/in.txt
new file mode 100644
index 0000000..76b90fe
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/in.txt
@@ -0,0 +1,9 @@
+30
+7
+10 2
+23 2
+14 1
+4 1
+14 4
+11 5
+1 2
diff --git a/šola/aps1/dn/osvetlitev/resitev.cpp b/šola/aps1/dn/osvetlitev/resitev.cpp
new file mode 100644
index 0000000..0a17f31
--- /dev/null
+++ b/šola/aps1/dn/osvetlitev/resitev.cpp
@@ -0,0 +1,46 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+struct event {
+	int pos;
+	bool tip; // true za začetek, false za konec
+};
+int compar_events (const void * a, const void * b) {
+	if (((struct event *) a)->pos == ((struct event *) b)->pos)
+		return 0;
+	if (((struct event *) a)->pos < ((struct event *) b)->pos)
+		return -1;
+	return 1;
+}
+int main (void) {
+	struct event events[20000];
+	int M, N, x, d;
+	scanf("%d %d", &M, &N);
+	for (int i = 0; i < N; i++) {
+		scanf("%d %d", &x, &d);
+		events[2*i].pos = x-d >= 0 ? x-d : 0;
+		events[2*i].tip = true;
+		events[2*i+1].pos = x+d <= M ? x+d : M;
+		events[2*i+1].tip = false;
+	}
+	qsort(events, 2*N, sizeof events[0], compar_events);
+	int osv = 0;
+	int depth = 0;
+	int start;
+	for (int i = 0; i < 2*N; i++) {
+		// fprintf(stderr, "pos=%d\ttip=%d\n", events[i].pos, events[i].tip);
+		if (events[i].tip == true) {
+			if (depth == 0)
+				start = events[i].pos;
+			depth++;
+		}
+		if (events[i].tip == false) {
+			depth--;
+			if (depth == 0)
+				osv += events[i].pos - start;
+		}
+	}
+	if (depth != 0)
+		fprintf(stderr, "depth == %d\n", depth);
+	printf("%d\n", M-osv);
+}
diff --git a/šola/aps1/dn/zlivanje/in.txt b/šola/aps1/dn/zlivanje/in.txt
new file mode 100644
index 0000000..eaca3bf
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/in.txt
@@ -0,0 +1,26 @@
+25 3 2
+13
+18
+7
+8
+17
+3
+16
+9
+10
+11
+11
+0
+2
+19
+14
+5
+6
+15
+4
+5
+12
+3
+18
+1
+3
diff --git a/šola/aps1/dn/zlivanje/out.txt b/šola/aps1/dn/zlivanje/out.txt
new file mode 100644
index 0000000..6def2f9
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/out.txt
@@ -0,0 +1 @@
+0 2 3 3 4 5 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 18 19 1 3
diff --git a/šola/aps1/dn/zlivanje/resitev.cpp b/šola/aps1/dn/zlivanje/resitev.cpp
new file mode 100644
index 0000000..8926019
--- /dev/null
+++ b/šola/aps1/dn/zlivanje/resitev.cpp
@@ -0,0 +1,39 @@
+#include <sys/param.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+int compar_long (const void * a, const void * b) {
+	if (*(long *)a < *(long *)b)
+		return -1;
+	return *(long *)a > *(long *) b;
+}
+int main (void) {
+	long N, K, A;
+	scanf("%ld %ld %ld", &N, &K, &A);
+	long * d = (long *) malloc(N*sizeof *d);
+	long čet = 0;
+	long lastidx = 0;
+	long long četkončno = 1;
+	for (long i = 0; i < A && četkončno <= 2000000; i++)	// pravzaprav četkončno := K**A,
+		četkončno *= K;		// toda C nima int potence
+	// fprintf(stderr, "aaaaaaaa %ld\n", četkončno);
+	for (long i = 0; i < N; i++) {
+		scanf("%ld", d+i);
+		if (i && d[i-1] > d[i])
+			if (++čet >= četkončno) {
+				qsort(d+lastidx, i-lastidx, sizeof d[0], compar_long);
+				čet = 0;
+				lastidx = i;
+			}
+	}
+	qsort(d+lastidx, N-lastidx, sizeof d[0], compar_long);
+	bool devica = true;
+	for (long i = 0; i < N; i++) {
+		if (devica)
+			devica = false;
+		else
+			printf(" ");
+		printf("%ld", d[i]);
+	}
+	printf("\n");
+}
diff --git a/šola/citati.bib b/šola/citati.bib
new file mode 100644
index 0000000..f553435
--- /dev/null
+++ b/šola/citati.bib
@@ -0,0 +1,242 @@
+Osebna bibtex knjižnica citatov.
+Canonical: ~/projects/r/šola/citati.bib od 2024-07-28
+Stara različica je v ~/projects/sola-gimb-4/citati.bib
+
+@online{hpstrings,
+	author = {Nave, Carl Rod},
+	title = {{HyperPhysics Concepts: Standing Waves on a String}},
+	year = {2016},
+	url = {http://hyperphysics.phy-astr.gsu.edu./hbase/Waves/string.html},
+	urldate = {2016-11-09}
+}
+@online{cohen01,
+	author = {Cohen, Bram},
+	title = {BitTorrent - a new P2P app},
+	year = {2001},
+	url = {http://finance.groups.yahoo.com/group/decentralization/message/3160},
+	urldate = {2007-04-15},
+	note = "Internet Archive"
+}
+@online{harrison07,
+	author = {Harrison, David},
+	url = {http://www.bittorrent.org/beps/bep_0000.html},
+	urldate = {2023-02-28},
+	year = {2008},
+	title = {Index of BitTorrent Enhancement Proposals}
+}
+@online{norberg08,
+	author = {Loewenstern, Andrew and Norberg, Arvid},
+	url = {https://www.bittorrent.org/beps/bep_0005.html},
+	urldate = {2023-02-28},
+	year = {2020},
+	title = {DHT Protocol}
+}
+@online{jones15,
+	author = {Jones, Ben},
+	url = {https://torrentfreak.com/bittorrents-dht-turns-10-years-old-150607/},
+	urldate = {2023-02-28},
+	year = {2015},
+	title = {BitTorrent’s DHT Turns 10 Years Old}
+}
+@online{muo11,
+	author = {Mukherjee, Abhijeet},
+	url = {https://www.makeuseof.com/tag/btdigg-trackerless-torrent/},
+	urldate = {2023-02-28},
+	year = {2011},
+	title = {BTDigg: A Trackerless Torrent Search Engine}
+}
+@online{evseenko11,
+	author = {Evseenko, Nina},
+	url = {http://btdig.com},
+	urldate = {2023-02-28},
+	year = {2011},
+	title = {Btdigg BitTorrent DHT search engine}
+
+}
+@online{griffin17,
+	author = {Griffin, Andrew},
+	url = {https://www.independent.co.uk/tech/torrent-website-download-safe-legal-privacy-i-know-what-you-friends-spying-a7504266.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {'I Know What You Download': Website claims to let people see everything their friends have torrented}
+}
+@online{ikwyd,
+	url = {http://iknowwhatyoudownload.com},
+	urldate = {2023-02-28},
+	title = {I know what you download}
+}
+@online{cohen08,
+	author = {Choen, Bram},
+	url = {https://www.bittorrent.org/beps/bep_0003.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {The BitTorrent Protocol Specification}
+}
+@online{norberg09,
+	author = {Norberg, Arvid},
+	url = {https://www.bittorrent.org/beps/bep_0029.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {uTorrent transport protocol}
+}
+@online{hazel08,
+	author = {Hazel, Greg and Norberg, Arvid},
+	url = {https://www.bittorrent.org/beps/bep_0009.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {Extension for Peers to Send Metadata Files}
+}
+@online{strigeus08,
+	author = {Norberg, Arvid and Strigeus, Ludvig and Hazel, Greg},
+	url = {https://www.bittorrent.org/beps/bep_0010.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {Extension Protocol}
+}
+@online{v2,
+	author = {Cohen, Bram},
+	url = {https://www.bittorrent.org/beps/bep_0052.html},
+	urldate = {2023-02-28},
+	year = {2017},
+	title = {The BitTorrent Protocol Specification v2}
+}
+@incollection{maymounkov2002kademlia,
+	title={Kademlia: A peer-to-peer information system based on the xor metric},
+	author={Maymounkov, Petar and Mazieres, David},
+	booktitle={Peer-to-Peer Systems: First InternationalWorkshop, IPTPS 2002 Cambridge, MA, USA, March 7--8, 2002 Revised Papers},
+	pages={53--65},
+	year={2002},
+	publisher={Springer}
+}
+@inproceedings{pezoa2016foundations,
+	title={Foundations of JSON schema},
+	author={Pezoa, Felipe and Reutter, Juan L and Suarez, Fernando and Ugarte, Mart{\'\i}n and Vrgo{\v{c}}, Domagoj},
+	booktitle={Proceedings of the 25th International Conference on World Wide Web},
+	pages={263--273},
+	year={2016},
+	organization={International World Wide Web Conferences Steering Committee}
+}
+@online{harrison08,
+	author = {Harrison, David},
+	title = {Private Torrents},
+	year = {2008},
+	url = {https://www.bittorrent.org/beps/bep_0027.html},
+	urldate = {2023-02-28},
+}
+@techreport{Eastlake2001,
+	added-at = {2011-05-12T13:55:38.000+0200},
+	author = {Eastlake, D. and Jones, P.},
+	biburl = {https://www.bibsonomy.org/bibtex/2d2dcad25e57d68db18495f062e955cae/voj},
+	interhash = {276256be69453291def2a6fdbed1389d},
+	intrahash = {d2dcad25e57d68db18495f062e955cae},
+	keywords = {hash identifier sha1},
+	month = {9},
+	number = 3174,
+	organization = {IETF},
+	publisher = {IETF},
+	series = {RFC},
+	timestamp = {2013-01-06T13:13:36.000+0100},
+	title = {{US Secure Hash Algorithm 1 (SHA1)}},
+	type = {RFC},
+	year = 2001
+}
+@online{dis,
+	author = {Gams, Matjaž and others},
+	year = {2008},
+	title = {DIS Slovarček},
+	url = {https://dis-slovarcek.ijs.si/},
+	urldate = {2023-02-28}
+}
+@online{rhilip20,
+	author = {Rhilip},
+	year = {2020},
+	urldate = {2023-02-28},
+	url = {https://github.com/Rhilip/Bencode},
+	title = {A pure and simple PHP library for encoding and decoding Bencode data}
+}
+@conference{Kluyver2016jupyter,
+	Title = {Jupyter Notebooks -- a publishing format for reproducible computational workflows},
+	Author = {Thomas Kluyver and Benjamin Ragan-Kelley and Fernando P{\'e}rez and Brian Granger and Matthias Bussonnier and Jonathan Frederic and Kyle Kelley and Jessica Hamrick and Jason Grout and Sylvain Corlay and Paul Ivanov and Dami{\'a}n Avila and Safia Abdalla and Carol Willing},
+	Booktitle = {Positioning and Power in Academic Publishing: Players, Agents and Agendas},
+	Editor = {F. Loizides and B. Schmidt},
+	Organization = {IOS Press},
+	Pages = {87 - 90},
+	Year = {2016}
+}
+@online{sampleih,
+	author = {The 8472},
+	url = {https://www.bittorrent.org/beps/bep_0051.html},
+	title = {DHT Infohash Indexing},
+	urldate = {2023-02-28},
+	year = {2017}
+}
+@Article{Hunter:2007,
+	Author    = {Hunter, J. D.},
+	Title     = {Matplotlib: A 2D graphics environment},
+	Journal   = {Computing in Science \& Engineering},
+	Volume    = {9},
+	Number    = {3},
+	Pages     = {90--95},
+	abstract  = {Matplotlib is a 2D graphics package used for Python for
+	application development, interactive scripting, and publication-quality
+	image generation across user interfaces and operating systems.},
+	publisher = {IEEE COMPUTER SOC},
+	doi       = {10.1109/MCSE.2007.55},
+	year      = 2007
+}
+@online{norberg14,
+	author = {Norberg, Arvid},
+	title = {DHT Security extension},
+	url = {https://www.bittorrent.org/beps/bep_0042.html},
+	urldate = {2023-02-28},
+	year = {2016}
+}
+@InProceedings{10.1007/3-540-45748-8_24,
+	author="Douceur, John R.",
+	editor="Druschel, Peter
+		and Kaashoek, Frans
+		and Rowstron, Antony",
+	title="The Sybil Attack",
+	booktitle="Peer-to-Peer Systems",
+	year="2002",
+	publisher="Springer Berlin Heidelberg",
+	address="Berlin, Heidelberg",
+	pages="251--260",
+	abstract="Large-scale peer-to-peer systems face security threats from faulty or hostile remote computing elements. To resist these threats, many such systems employ redundancy. However, if a single faulty entity can present multiple identities, it can control a substantial fraction of the system, thereby undermining this redundancy. One approach to preventing these ``Sybil attacks'' is to have a trusted agency certify identities. This paper shows that, without a logically centralized authority, Sybil attacks are always possible except under extreme and unrealistic assumptions of resource parity and coordination among entities.",
+	isbn="978-3-540-45748-0"
+}
+@book{cedilnik12,
+	place={Radovljica},
+	title={Matematični Priročnik},
+	publisher={Didakta},
+	author={Cedilnik, Anton and Razpet, Nada and Razpet, Marko},
+	year={2012}
+}
+@online{wikihashuniformity,
+	author="{Wikipedia contributors}",
+	title={Hash function},
+	url={https://w.wiki/An3L},
+	urldate={2024-07-29},
+	year={2024}
+}
+@online{maxmindgeoip2,
+	author={MaxMind},
+	title={GeoLite2 Free Geolocation Data},
+	url={https://dev.maxmind.com/geoip/geolite2-free-geolocation-data},
+	urldate={2024-07-31},
+	year={2024}
+}
+@misc{rfc4086,
+	series = {Request for Comments},
+	number = 4086,
+	howpublished = {RFC 4086},
+	publisher = {RFC Editor},
+	doi = {10.17487/RFC4086},
+	url = {https://www.rfc-editor.org/info/rfc4086},
+	author = {Donald E. Eastlake 3rd and Steve Crocker and Jeffrey I. Schiller},
+	title = {{Randomness Requirements for Security}},
+	pagetotal = 48,
+	year = 2005,
+	month = jun,
+	abstract = {Security systems are built on strong cryptographic algorithms that foil pattern analysis attempts. However, the security of these systems is dependent on generating secret quantities for passwords, cryptographic keys, and similar quantities. The use of pseudo-random processes to generate secret quantities can result in pseudo-security. A sophisticated attacker may find it easier to reproduce the environment that produced the secret quantities and to search the resulting small set of possibilities than to locate the quantities in the whole of the potential number space. Choosing random quantities to foil a resourceful and motivated adversary is surprisingly difficult. This document points out many pitfalls in using poor entropy sources or traditional pseudo-random number generation techniques for generating such quantities. It recommends the use of truly random hardware techniques and shows that the existing hardware on many systems can be used for this purpose. It provides suggestions to ameliorate the problem when a hardware solution is not available, and it gives examples of how large such quantities need to be for some applications. This document specifies an Internet Best Current Practices for the Internet Community, and requests discussion and suggestions for improvements.},
+}
diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx
new file mode 100644
index 0000000..fcb46f9
--- /dev/null
+++ b/šola/la/teor.lyx
@@ -0,0 +1,25345 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\usepackage{hyperref}
+\usepackage{siunitx}
+\usepackage{pgfplots}
+\usepackage{listings}
+\usepackage{multicol}
+\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}}
+\usepackage{amsmath}
+\usepackage{tikz}
+\newcommand{\udensdash}[1]{%
+    \tikz[baseline=(todotted.base)]{
+        \node[inner sep=1pt,outer sep=0pt] (todotted) {#1};
+        \draw[densely dashed] (todotted.south west) -- (todotted.south east);
+    }%
+}%
+\DeclareMathOperator{\Lin}{\mathcal Lin}
+\DeclareMathOperator{\rang}{rang}
+\DeclareMathOperator{\sled}{sled}
+\DeclareMathOperator{\Aut}{Aut}
+\DeclareMathOperator{\red}{red}
+\DeclareMathOperator{\karakteristika}{char}
+\DeclareMathOperator{\Ker}{Ker}
+\DeclareMathOperator{\Slika}{Ker}
+\DeclareMathOperator{\sgn}{sgn}
+\DeclareMathOperator{\End}{End}
+\DeclareMathOperator{\n}{n}
+\DeclareMathOperator{\Col}{Col}
+\usepackage{algorithm,algpseudocode}
+\providecommand{\corollaryname}{Posledica}
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+\end_modules
+\maintain_unincluded_children no
+\language slovene
+\language_package default
+\inputencoding auto-legacy
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\float_placement class
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification false
+\use_refstyle 1
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 2cm
+\topmargin 2cm
+\rightmargin 2cm
+\bottommargin 2cm
+\headheight 2cm
+\headsep 2cm
+\footskip 1cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Teorija linearne algebre za ustni izpit —
+ IŠRM 2023/24
+\end_layout
+
+\begin_layout Author
+
+\noun on
+Anton Luka Šijanec
+\end_layout
+
+\begin_layout Date
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+today
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Abstract
+Povzeto po zapiskih s predavanj prof.
+ Cimpriča.
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset toc
+LatexCommand tableofcontents
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Part
+Teorija
+\end_layout
+
+\begin_layout Section
+Prvi semester
+\end_layout
+
+\begin_layout Subsection
+Vektorji v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Identificaramo 
+\begin_inset Formula $n-$
+\end_inset
+
+terice realnih števil,
+ točke v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+,
+ množice paroma enakih geometrijskih vektorjev.
+\end_layout
+
+\begin_layout Standard
+Osnovne operacije z vektorji:
+ Vsota (po komponentah) in množenje s skalarjem (po komponentah),
+ kjer je skalar realno število.
+\end_layout
+
+\begin_layout Standard
+Lastnosti teh računskih operacij:
+ asociativnost in komutativnost vsote,
+ aditivna enota,
+ 
+\begin_inset Formula $-\vec{a}=\left(-1\right)\cdot\vec{a}$
+\end_inset
+
+,
+ leva in desna distributivnost,
+ homogenost,
+ multiplikativna enota.
+\end_layout
+
+\begin_layout Subsubsection
+Linearna kombinacija vektorjev
+\end_layout
+
+\begin_layout Definition*
+Linearna kombinacija vektorjev 
+\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
+\end_inset
+
+ je izraz oblike 
+\begin_inset Formula $\alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}}$
+\end_inset
+
+,
+ kjer so 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
+\end_inset
+
+ skalarji.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množico vseh linearnih kombinacij vektorjev 
+\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} $
+\end_inset
+
+ in ji pravimo linearna ogrinjača (angl.
+ span).
+ 
+\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} =\left\{ \alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}};\forall\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Linearna neodvisnost vektorjev
+\end_layout
+
+\begin_layout Paragraph*
+Ideja
+\end_layout
+
+\begin_layout Standard
+En vektor je linearno neodvisen,
+ če ni enak 
+\begin_inset Formula $\vec{0}$
+\end_inset
+
+.
+ Dva,
+ če ne ležita na isti premici.
+ Trije,
+ če ne ležijo na isti ravnini.
+\end_layout
+
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:odvisni"
+
+\end_inset
+
+Vektorji 
+\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
+\end_inset
+
+ so linearno odvisni,
+ če se da enega izmed njih izraziti z linearno kombinacijo preostalih 
+\begin_inset Formula $n-1$
+\end_inset
+
+ vektorjev.
+ Vektorji so linearno neodvisni,
+ če niso linearno odvisni (in obratno).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:vsi0"
+
+\end_inset
+
+Vektorji 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ so linearno neodvisni,
+ če za vsake skalarje,
+ ki zadoščajo 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
+\end_inset
+
+.
+ ZDB poleg 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
+\end_inset
+
+ ne obstajajo nobeni drugi 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
+\end_inset
+
+,
+ kjer bi veljalo 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:kvečjemu1"
+
+\end_inset
+
+
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ so linearno neodvisni,
+ če se da vsak vektor na kvečjemu en način izraziti kot linearno kombinacijo 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Te tri definicije so ekvivalentne.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\ref{def:odvisni}\Rightarrow\ref{def:vsi0}\right)$
+\end_inset
+
+ Recimo,
+ da so 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ linearno odvisni v smislu 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:odvisni"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+ Dokažimo,
+ da so tedaj linearno odvisni tudi v smislu 
+\begin_inset Formula $\ref{def:vsi0}$
+\end_inset
+
+.
+ Obstaja tak 
+\begin_inset Formula $i$
+\end_inset
+
+,
+ da lahko 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ izrazimo z linearno kombinacijo preostalih,
+ torej 
+\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+ za neke 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $0=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\left(-1\right)v_{i}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+,
+ kar pomeni,
+ da obstaja linearna kombinacija,
+ ki je enaka 0,
+ toda niso vsi koeficienti 0 (že koeficient pred 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ je 
+\begin_inset Formula $-1$
+\end_inset
+
+),
+ tedaj so vektorji po definiciji 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:vsi0"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ linearno odvisni.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\ref{def:vsi0}\Rightarrow\ref{def:odvisni}\right)$
+\end_inset
+
+ Recimo,
+ da so 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ linearno odvisno v smislu 
+\begin_inset Formula $\ref{def:vsi0}$
+\end_inset
+
+.
+ Tedaj obstajajo 
+\begin_inset Formula $\alpha$
+\end_inset
+
+,
+ ki niso vse 0,
+ da velja 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists i\ni:\alpha_{i}\not=0$
+\end_inset
+
+ in velja
+\begin_inset Formula 
+\[
+\alpha_{i}v_{i}=-\alpha_{1}v_{1}-\cdots-\alpha_{i-1}v_{i-1}-\alpha_{i+1}v_{i+1}-\cdots-\alpha_{n}v_{n}\quad\quad\quad\quad/:\alpha_{i}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{i}=-\frac{\alpha_{1}}{\alpha_{i}}v_{i}-\cdots-\frac{\alpha_{i-1}}{\alpha_{i}}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_{i}}v_{i+1}-\cdots-\frac{\alpha_{n}}{\alpha_{i}}v_{n}\text{,}
+\]
+
+\end_inset
+
+s čimer smo 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ izrazili kot linearno kombinacijo preostalih vektorjev.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\ref{def:vsi0}\Leftrightarrow\ref{def:kvečjemu1}\right)$
+\end_inset
+
+ Naj bodo 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ LN.
+ Recimo,
+ da obstaja 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ ki se ga da na dva načina izraziti kot linearno kombinacijo 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{n}v_{n}$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $0=\left(\alpha_{1}-\beta_{1}\right)v_{1}+\cdots+\left(\alpha_{n}-\beta_{n}\right)v_{n}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:vsi0"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall i:\alpha_{i}-\beta_{i}=0\Leftrightarrow\alpha_{i}=\beta_{i}$
+\end_inset
+
+,
+ torej sta načina,
+ s katerima izrazimo 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ enaka,
+ torej lahko 
+\begin_inset Formula $v$
+\end_inset
+
+ izrazimo na kvečjemu en način z 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+,
+ kar ustreza definiciji 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:kvečjemu1"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Ogrodje in baza
+\end_layout
+
+\begin_layout Definition*
+Vektorji 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ so ogrodje (angl.
+ span),
+ če 
+\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} =\mathbb{R}^{n}\Leftrightarrow\forall v\in\mathbb{R}^{n}\exists\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Vektorji 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ so baza,
+ če so LN in ogrodje 
+\begin_inset Formula $\Leftrightarrow\forall v\in\mathbb{R}^{n}:\exists!\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+ ZDB vsak vektor 
+\begin_inset Formula $\in\mathbb{R}^{n}$
+\end_inset
+
+ se da na natanko en način izraziti kot LK 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primer baze je standardna baza 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\{ \left(1,0,0,\dots,0\right),\left(0,1,0,\dots,0\right),\left(0,0,1,\dots,0\right),\left(0,0,0,\dots,1\right)\right\} $
+\end_inset
+
+.
+ To pa ni edina baza.
+ Primer nestandardne baze v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+ je 
+\begin_inset Formula $\left\{ \left(1,1,1\right),\left(0,1,1\right),\left(0,0,1\right)\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Norma in skalarni produkt
+\end_layout
+
+\begin_layout Definition*
+Norma vektorja 
+\begin_inset Formula $v=\left(\alpha_{1},\dots,\alpha_{n}\right)$
+\end_inset
+
+ je definirana z 
+\begin_inset Formula $\left|\left|v\right|\right|=\sqrt{\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}}$
+\end_inset
+
+.
+ Geometrijski pomen norme je dolžina krajevnega vektorja z glavo v 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Osnovne lastnosti norme:
+ 
+\begin_inset Formula $\left|\left|v\right|\right|\geq0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|\left|v\right|\right|=0\Rightarrow v=\vec{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|\left|\alpha v\right|\right|=\left|\alpha\right|\cdot\left|\left|v\right|\right|$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
+\end_inset
+
+ (trikotniška neenakost)
+\end_layout
+
+\begin_layout Definition*
+Skalarni produkt 
+\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right),v=\left(\beta_{1},\dots,\beta_{n}\right)$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $\left\langle u,v\right\rangle \coloneqq\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
+\end_inset
+
+.
+ Obstaja tudi druga oznaka in pripadajoča drugačna definicija 
+\begin_inset Formula $u\cdot v\coloneqq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ kot med 
+\begin_inset Formula $u,v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Velja 
+\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Uporabimo kosinusni izrek,
+ ki pravi,
+ da v trikotniku s stranicami dolžin 
+\begin_inset Formula $a,b,c$
+\end_inset
+
+ velja 
+\begin_inset Formula $c^{2}=a^{2}+b^{2}-2ab\cos\varphi$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ kot med 
+\begin_inset Formula $b$
+\end_inset
+
+ in 
+\begin_inset Formula $c$
+\end_inset
+
+.
+ Za vektorja 
+\begin_inset Formula $v$
+\end_inset
+
+ in 
+\begin_inset Formula $u$
+\end_inset
+
+ z vmesnim kotom 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ torej velja 
+\begin_inset Formula 
+\[
+\left|\left|u-v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
+\]
+
+\end_inset
+
+Obenem velja 
+\begin_inset Formula $\left|\left|u\right|\right|^{2}=\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}=\left\langle u,u\right\rangle $
+\end_inset
+
+,
+ torej lahko zgornjo enačbo prepišemo v
+\begin_inset Formula 
+\[
+\left\langle u-v,u-v\right\rangle =\left\langle u,u\right\rangle +\left\langle v,v\right\rangle -2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
+\]
+
+\end_inset
+
+Naj bo 
+\begin_inset Formula $w=u,v$
+\end_inset
+
+.
+ Iz prihodnosti si izposodimo obe linearnosti in simetričnost.
+ 
+\begin_inset Formula 
+\[
+\left\langle u-v,u-v\right\rangle =\left\langle u-v,w\right\rangle =\left\langle u,w\right\rangle -\left\langle v,w\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
+\]
+
+\end_inset
+
+ Prišli smo do enačbe
+\begin_inset Formula 
+\[
+\cancel{\left\langle u,u\right\rangle }-2\left\langle u,v\right\rangle +\cancel{\left\langle v,v\right\rangle }=\cancel{\left\langle u,u\right\rangle }+\cancel{\left\langle v,v\right\rangle }-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\quad\quad\quad\quad/:-2
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle u,v\right\rangle =\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Paralelogramska identiteta.
+ 
+\begin_inset Formula $\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}$
+\end_inset
+
+ ZDB vsota kvadratov dolžin obeh diagonal je enota vsoti kvadratov dolžin vseh štirih stranic.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\left|\left|u+v\right|\right|^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u+v\right\rangle +\left\langle v,u+v\right\rangle =\left\langle u,u\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
+\]
+
+\end_inset
+
+ 
+\begin_inset Formula 
+\[
+\left|\left|u-v\right|\right|^{2}=\left\langle u-v,u-v\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left\langle u,u\right\rangle +2\left\langle v,v\right\rangle =2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Cauchy-Schwarzova neenakost.
+ 
+\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|=\left|\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\right|=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\left|\cos\varphi\right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left|\cos\varphi\right|\in\left[0,1\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Trikotniška neenakost.
+ 
+\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Sledi iz Cauchy-Schwarzove.
+ Velja 
+\begin_inset Formula 
+\[
+-\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/\cdot2
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq2\left|\left\langle u,v\right\rangle \right|\leq2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq\cancel{2\left|\left\langle u,v\right\rangle \right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq}2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
+\]
+
+\end_inset
+
+uporabimo kosinusni izrek na levi strani enačbe,
+ desno pa zložimo v kvadrat:
+\begin_inset Formula 
+\[
+\left|\left|u+v\right|\right|^{2}\leq\left(\left|\left|u\right|\right|+\left|\left|v\right|\right|\right)^{2}\quad\quad\quad\quad/\sqrt{}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Za neničelna vektorja velja 
+\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
+\end_inset
+
+,
+ kar je 0 
+\begin_inset Formula $\Leftrightarrow\varphi=\pi=90°$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Vektorski in mešani produkt
+\end_layout
+
+\begin_layout Standard
+Definirana sta le za vektorje v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $u=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right),v=\left(\beta_{1},\beta_{2},\beta_{3}\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $u\times v=\left(\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2},\alpha_{3}\beta_{1}-\alpha_{1}\beta_{3},\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+Geometrijski pomen
+\end_layout
+
+\begin_layout Standard
+Vektor 
+\begin_inset Formula $u\times v$
+\end_inset
+
+ je pravokoten na 
+\begin_inset Formula $u$
+\end_inset
+
+ in 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ njegova dolžina je 
+\begin_inset Formula $\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\sin\varphi$
+\end_inset
+
+,
+ kar je ploščina paralelograma,
+ ki ga oklepata 
+\begin_inset Formula $u$
+\end_inset
+
+ in 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Pravilo desnega vijaka nam je v pomoč pri doložanju usmeritve vektorskega produkta.
+ Če iztegnjen kazalec desne roke predstavlja 
+\begin_inset Formula $u$
+\end_inset
+
+ in iztegnjen sredinec 
+\begin_inset Formula $v$
+\end_inset
+
+,
+ iztegnjen palec kaže v smeri 
+\begin_inset Formula $u\times v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Lagrangeva identiteta.
+ 
+\begin_inset Formula $\left|\left|u\times v\right|\right|+\left\langle u,v\right\rangle ^{2}=\left|\left|u\right|\right|^{2}\cdot\left|\left|v\right|\right|^{2}$
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+DOKAZ???????
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Mešani produkt vektorjev 
+\begin_inset Formula $u,v,w$
+\end_inset
+
+ je skalar 
+\begin_inset Formula $\left\langle u\times v,w\right\rangle $
+\end_inset
+
+.
+ Oznaka:
+ 
+\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph*
+Geometrijski pomen
+\end_layout
+
+\begin_layout Standard
+Volumen paralelpipeda,
+ ki ga določajo 
+\begin_inset Formula $u,v,w$
+\end_inset
+
+.
+ Razlaga:
+ 
+\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle =\left|\left|u\times v\right|\right|\cdot\left|\left|w\right|\right|\cdot\cos\varphi$
+\end_inset
+
+;
+ 
+\begin_inset Formula $\left|\left|u\times v\right|\right|$
+\end_inset
+
+ je namreč ploščina osnovne ploskve,
+ 
+\begin_inset Formula $\left|\left|w\right|\right|\cdot\cos\varphi$
+\end_inset
+
+ pa je višina paralelpipeda.
+\end_layout
+
+\begin_layout Claim*
+Osnovne lastnosti vektorskega produkta so 
+\begin_inset Formula $u\times u=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u\times v=-\left(v\times u\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\alpha u+\beta v\right)\times w=\alpha\left(u\times w\right)+\beta\left(v\times w\right)$
+\end_inset
+
+ (linearnost)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Osnovne lastnosti mešanega produkta so linearnost v vsakem faktorju,
+ menjava dveh faktorjev spremeni predznak (
+\begin_inset Formula $\left[u,v,w\right]=-\left[v,u,w\right]$
+\end_inset
+
+),
+ cikličen pomik ne spremeni vrednosti (
+\begin_inset Formula $\left[u,v,w\right]=\left[v,w,u\right]=\left[w,u,v\right]$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Subsubsection
+Premica v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Premico lahko podamo z
+\end_layout
+
+\begin_layout Itemize
+dvema različnima točkama
+\end_layout
+
+\begin_layout Itemize
+s točko 
+\begin_inset Formula $\vec{r_{0}}$
+\end_inset
+
+ in neničelnim smernim vektorjem 
+\begin_inset Formula $\vec{p}$
+\end_inset
+
+.
+ Premica je tako množica točk 
+\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+t\vec{p};\forall t\in\mathbb{R}\right\} $
+\end_inset
+
+.
+ Taki enačbi premice rečemo parametrična.
+\end_layout
+
+\begin_layout Itemize
+s točko in normalo (v 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+;
+ v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ potrebujemo točko in 
+\begin_inset Formula $n-1$
+\end_inset
+
+ normal)
+\end_layout
+
+\begin_layout Standard
+Nadaljujmo s parametričnim zapisom 
+\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
+\end_inset
+
+.
+ Če točke zapišemo po komponentah,
+ dobimo parametrično enačbo premice po komponentah:
+ 
+\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+t\left(p_{1},p_{2},p_{3}\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+x=x_{0}+tp_{1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+y=y_{0}+tp_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+z=z_{0}+tp_{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sedaj lahko iz vsake enačbe izrazimo 
+\begin_inset Formula $t$
+\end_inset
+
+ in dobimo normalno enačbo premice v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+t=\frac{x-x_{0}}{p_{1}}=\frac{y-y_{0}}{p_{2}}=\frac{z-z_{0}}{p_{3}}\text{, oziroma v splošnem za premico v \ensuremath{\mathbb{R}^{n}}: }t=\frac{x_{1_{0}}-x_{1}}{p_{1}}=\cdots=\frac{x_{n_{0}}-x_{n}}{p_{n}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Osnovne naloge s premicami so projekcija točke na premico,
+ zrcaljenje točke čez premico in razdalja med točko in premico.
+\end_layout
+
+\begin_layout Paragraph*
+Iskanje projekcije dane točke na dano premico
+\end_layout
+
+\begin_layout Standard
+(skica prepuščena bralcu) 
+\begin_inset Formula $\vec{r_{1}}$
+\end_inset
+
+ projiciramo na 
+\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+ vemo,
+ da leži na premici,
+ torej 
+\begin_inset Formula $\exists t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}$
+\end_inset
+
+.
+ Poleg tega vemo,
+ da je 
+\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
+\end_inset
+
+ pravokoten na premico oz.
+ njen smerni vektor 
+\begin_inset Formula $\vec{p}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0$
+\end_inset
+
+.
+ Ti dve enačbi združimo,
+ da dobimo 
+\begin_inset Formula $t$
+\end_inset
+
+,
+ ki ga nato vstavimo v prvo enačbo:
+\begin_inset Formula 
+\[
+\left\langle \vec{r_{0}}+t\vec{p}-\vec{r_{1},}\vec{p}\right\rangle =0\Longrightarrow\left\langle \vec{r_{0}},\vec{p}\right\rangle +t\left\langle \vec{p},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0\Longrightarrow t=\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}=\vec{r_{0}}+\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }\vec{p}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Spotoma si lahko izpišemo obrazec za oddaljenost točke od premice:
+ 
+\begin_inset Formula $a=\left|\left|\vec{r_{1}'}-\vec{r_{1}}\right|\right|$
+\end_inset
+
+ in obrazec za zrcalno sliko (
+\begin_inset Formula $\vec{r_{1}''}$
+\end_inset
+
+):
+ 
+\begin_inset Formula $\vec{r_{1}'}=\frac{\vec{r_{1}''}+\vec{r_{1}}}{2}\Longrightarrow\vec{r_{1}''}=2\vec{r_{1}'}-\vec{r_{1}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Ravnine v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Ravnino lahko podamo
+\end_layout
+
+\begin_layout Itemize
+s tremi nekolinearnimi točkami
+\end_layout
+
+\begin_layout Itemize
+s točko na ravnini in dvema neničelnima smernima vektorjema,
+ ki sta linarno neodvisna.
+ Ravnina je tako množica točk 
+\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q};\forall s,t\in\mathbb{R}\right\} $
+\end_inset
+
+.
+ Taki enačbi ravnine rečemo parametrična.
+\end_layout
+
+\begin_layout Itemize
+s točko in na ravnini in normalo (v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+;
+ v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ poleg točke potrebujemo 
+\begin_inset Formula $n-2$
+\end_inset
+
+ normal)
+\end_layout
+
+\begin_layout Standard
+Nadaljujmo s parametričnim zapisom 
+\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
+\end_inset
+
+.
+ Če točke zapišemo po komponentah,
+ dobimo parametrično enačbo ravnine po komponentah:
+ 
+\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+s\left(p_{1},p_{2},p_{3}\right)+t\left(q_{1},q_{2},q_{3}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+x=x_{0}+sp_{1}+tq_{1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+y=y_{0}+sp_{2}+tq_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+z=y_{0}+sp_{3}+tq_{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Normalna enačba ravnine v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+(skica prepuščena bralcu) Vemo,
+ da je 
+\begin_inset Formula $\vec{n}$
+\end_inset
+
+ (normala) pravokotna na vse vektorje v ravnini,
+ tudi na 
+\begin_inset Formula $\vec{r}-\vec{r_{0}}$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $\vec{r}$
+\end_inset
+
+ na ravnini.
+ Velja torej normalna enačba ravnine:
+ 
+\begin_inset Formula $\left\langle \vec{r}-\vec{r_{0}},\vec{n}\right\rangle =0$
+\end_inset
+
+.
+ Razpišimo jo po komponentah,
+ da na koncu dobimo normalno enačbo ravnine po komponentah:
+\begin_inset Formula 
+\[
+\left\langle \left(x,y,z\right)-\left(x_{0},y_{0},z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle =0=\left\langle \left(x-x_{0},y-y_{0},z-z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n_{1}\left(x-x_{0}\right)+n_{2}\left(y-y_{0}\right)+n_{3}\left(z-z_{0}\right)=0=n_{1}x-n_{1}x_{0}+n_{2}y-n_{2}y_{0}+n_{3}z-n_{3}z_{0}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+n_{1}x+n_{2}y+n_{3}z=n_{1}x_{0}+n_{2}y_{0}+n_{3}z_{0}=d
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Iskanje pravokotne projekcije dane točke na dano ravnino
+\end_layout
+
+\begin_layout Standard
+(skica prepuščena bralcu) Projicirati želimo 
+\begin_inset Formula $\vec{r_{1}}$
+\end_inset
+
+ v 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+ na ravnini 
+\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
+\end_inset
+
+.
+ Vemo,
+ da 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+ leži na ravnini,
+ zato 
+\begin_inset Formula $\exists s,t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
+\end_inset
+
+.
+ Poleg tega vemo,
+ da je 
+\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
+\end_inset
+
+ pravokoten na ravnino oz.
+ na 
+\begin_inset Formula $\vec{p}$
+\end_inset
+
+ in na 
+\begin_inset Formula $\vec{q}$
+\end_inset
+
+ hkrati,
+ torej 
+\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{q}\right\rangle $
+\end_inset
+
+.
+ Vstavimo 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+ iz prve enačbe v drugo in dobimo
+\begin_inset Formula 
+\[
+\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{q}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle \vec{r_{0}},\vec{p}\right\rangle +s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}},\vec{q}\right\rangle +s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle -\left\langle \vec{r_{1}},\vec{q}\right\rangle 
+\]
+
+\end_inset
+
+dobimo sistem dveh enačb
+\begin_inset Formula 
+\[
+s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{p}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{q}\right\rangle 
+\]
+
+\end_inset
+
+sistem rešimo in dobljena 
+\begin_inset Formula $s,t$
+\end_inset
+
+ vstavimo v prvo enačbo zgoraj,
+ da dobimo 
+\begin_inset Formula $\vec{r_{1}'}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Regresijska premica
+\end_layout
+
+\begin_layout Standard
+Regresijska premica je primer uporabe zgornje naloge.
+ V ravnini je danih 
+\begin_inset Formula $n$
+\end_inset
+
+ točk 
+\begin_inset Formula $\left(x_{1},y_{1}\right),\dots,\left(x_{n},y_{n}\right)$
+\end_inset
+
+.
+ Iščemo tako premico 
+\begin_inset Formula $y=ax+b$
+\end_inset
+
+,
+ ki se najbolj prilega tem točkam.
+ Prileganje premice točkam merimo z metodo najmanjših kvadratov:
+ naj bo 
+\begin_inset Formula $d_{i}$
+\end_inset
+
+ navpična razdalja med 
+\begin_inset Formula $\left(x_{i},y_{i}\right)$
+\end_inset
+
+ in premico 
+\begin_inset Formula $y=ax+b$
+\end_inset
+
+,
+ torej razdalja med točkama 
+\begin_inset Formula $\left(x_{i},y_{i}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(x_{i},ax_{i}+b\right)$
+\end_inset
+
+,
+ kar je 
+\begin_inset Formula $\left|y_{i}-ax_{i}-b\right|$
+\end_inset
+
+.
+ Minimizirati želimo vsoto kvadratov navpičnih razdalj,
+ torej izraz 
+\begin_inset Formula $d_{1}^{2}+\cdots+d_{n}^{2}=\left(y_{1}-ax_{1}-b\right)^{2}+\cdots+\left(y_{n}-ax_{n}-b\right)^{2}=\left|\left|\left(y_{1}-ax_{1}-b,\dots,y_{n}-ax_{n}-b\right)\right|\right|^{2}=\left|\left|\left(y_{1},\dots,y_{n}\right)-a\left(x_{1},\dots,x_{n}\right)-b\left(1,\dots,1\right)\right|\right|^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če je torej 
+\begin_inset Formula $\vec{r}=\vec{0}+a\left(x_{1},\dots,x_{n}\right)+b\left(1,\dots,1\right)$
+\end_inset
+
+ hiperravnina v 
+\begin_inset Formula $n-$
+\end_inset
+
+dimenzionalnem prostoru,
+ bo norma,
+ ki jo želimo minimizirati,
+ najmanjša tedaj,
+ ko 
+\begin_inset Formula $a,b$
+\end_inset
+
+ izberemo tako,
+ da najdemo projekcijo 
+\begin_inset Formula $\left(y_{1},\dots,y_{n}\right)$
+\end_inset
+
+ na to hiperravnino (skica prepuščena bralcu).
+ Rešimo sedaj nalogo projekcije točke na ravnino:
+\end_layout
+
+\begin_layout Standard
+Označimo 
+\begin_inset Formula $\vec{y}\coloneqq\left(y_{1},\dots,y_{n}\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vec{x}\coloneqq\left(x_{1},\dots,x_{n}\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vec{1}=\left(1,\dots,1\right)$
+\end_inset
+
+.
+ Vemo,
+ da 
+\begin_inset Formula $\vec{y}-a\vec{x}-b\vec{1}\perp\vec{x},\vec{1}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{x}\right\rangle =0=\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{1}\right\rangle $
+\end_inset
+
+ in dobimo sistem enačb
+\begin_inset Formula 
+\[
+\left\langle \vec{y},\vec{x}\right\rangle =a\left\langle \vec{x},\vec{x}\right\rangle +b\left\langle \vec{1},\vec{x}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle \vec{y},\vec{1}\right\rangle =a\left\langle \vec{x},\vec{1}\right\rangle +b\left\langle \vec{1},\vec{1}\right\rangle .
+\]
+
+\end_inset
+
+V sistem sedaj vstavimo definicije točk 
+\begin_inset Formula $\left(x_{i},y_{i}\right)$
+\end_inset
+
+ in ga nato delimo s številom točk,
+ da dobimo sistem s povprečji,
+ ki ga nato rešimo (izluščimo 
+\begin_inset Formula $a,b$
+\end_inset
+
+):
+\begin_inset Formula 
+\[
+\sum_{i=1}^{n}y_{i}x_{i}=a\sum_{i=i}^{n}x_{i}^{2}+b\sum_{i=1}^{n}x_{i}\quad\quad\quad\quad/:n
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\sum_{i=1}^{n}y_{i}=a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}1=a\sum_{i=1}^{n}x_{i}+bn\quad\quad\quad\quad/:n
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\overline{yx}=a\overline{x^{2}}+b\overline{x}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\overline{y}=a\overline{x}+b\Longrightarrow\overline{y}-a\overline{x}=b
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\overline{yx}=a\overline{x^{2}}+\left(\overline{y}-a\overline{x}\right)\overline{x}=a\overline{x^{2}}+\overline{y}\cdot\overline{x}-a\overline{x}^{2}\Longrightarrow a\left(\overline{x^{2}}-\overline{x}^{2}\right)=\overline{yx}-\overline{y}\cdot\overline{x}\Longrightarrow a=\frac{\overline{yx}-\overline{y}\cdot\overline{x}}{\overline{x^{2}}-\overline{x}^{2}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Sistemi linearnih enačb
+\end_layout
+
+\begin_layout Standard
+Ta sekcija,
+ z izjemo prve podsekcije,
+ je precej dobesedno povzeta po profesorjevi beamer skripti.
+\end_layout
+
+\begin_layout Subsubsection
+Linearna enačba
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\sim$
+\end_inset
+
+ je enačba oblike 
+\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=b$
+\end_inset
+
+ in vsebuje koeficiente,
+ spremenljivke in desno stran.
+ Množica rešitev so vse 
+\begin_inset Formula $n-$
+\end_inset
+
+terice realnih števil,
+ ki zadoščajo enačbi 
+\begin_inset Formula $R=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n};a_{1}x_{1}+\cdots+a_{n}x_{n}=b\right\} $
+\end_inset
+
+.
+ Če so vsi koeficienti 0,
+ pravimo,
+ da je enačba trivialna,
+ sicer (torej čim je en koeficient neničeln) je netrivialna.
+\end_layout
+
+\begin_layout Remark*
+Za trivialno enačbo velja 
+\begin_inset Formula $R=\begin{cases}
+\emptyset & ;b\not=0\\
+\mathbb{R}^{n} & ;b=0
+\end{cases}$
+\end_inset
+
+.
+ Za netrivialno enačbo pa velja 
+\begin_inset Formula $a_{i}\not=0$
+\end_inset
+
+,
+ torej:
+\begin_inset Formula 
+\[
+a_{1}x_{1}+\cdots+a_{i}x_{i}+\cdots+a_{n}x_{n}=b
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=b-a_{i}x_{i}=-a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i-1},x_{i}-\frac{b}{a_{i}},x_{i+1},\dots,x_{n}\right)\right\rangle =0=\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i},\dots,x_{n}\right)-\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)\right\rangle 
+\]
+
+\end_inset
+
+Tu lahko označimo 
+\begin_inset Formula $\vec{n}\coloneqq\left(a_{i},\dots,a_{n}\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vec{r}=\left(x_{1},\dots,x_{i},\dots,x_{n}\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vec{r_{0}}=\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\left\langle \vec{n},\vec{r}-\vec{r_{0}}\right\rangle $
+\end_inset
+
+,
+ kar je normalna enačba premice v 
+\begin_inset Formula $\mathbb{R}^{2}$
+\end_inset
+
+,
+ normalna enačba ravnine v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+ oziroma normalna enačba hiperravnine v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Sistem linearnih enačb
+\end_layout
+
+\begin_layout Definition*
+Sistem 
+\begin_inset Formula $m$
+\end_inset
+
+ linearnih enačb z 
+\begin_inset Formula $n$
+\end_inset
+
+ spremenljivkami je sistem enačb oblike
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
+\vdots &  &  &  & \vdots &  & \vdots\\
+a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
+\end{array}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Množica rešitev sistema je 
+\begin_inset Formula $\mathbb{R}^{n}\Leftrightarrow\forall i,j:a_{i,j}=b_{i}=0$
+\end_inset
+
+.
+ Sicer je množica rešitev presek hiperravnin v 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ —
+ rešitev posameznih enačb.
+ To vključuje tudi primer prazne množice rešitev,
+ saj je takšna na primer presek dveh vzporednih hiperravnin.
+\end_layout
+
+\begin_layout Example*
+Množica rešitev 
+\begin_inset Formula $2\times2$
+\end_inset
+
+ sistema je lahko
+\end_layout
+
+\begin_layout Itemize
+cela ravnina
+\end_layout
+
+\begin_layout Itemize
+premica v ravnini
+\end_layout
+
+\begin_layout Itemize
+točka v ravnini
+\end_layout
+
+\begin_layout Itemize
+prazna množica
+\end_layout
+
+\begin_layout Remark*
+Enako velja za množico rešitev 
+\begin_inset Formula $3\times2$
+\end_inset
+
+ sistema.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Množica rešitev sistema 
+\begin_inset Formula $2\times3$
+\end_inset
+
+ pa ne more biti točka v prostoru,
+ lahko pa je cel prostor,
+ ravnina v prostoru,
+ premica v prostoru ali prazna množica.
+\end_layout
+
+\begin_layout Paragraph*
+Algebraičen pomen rešitev sistema
+\end_layout
+
+\begin_layout Standard
+Rešitve sistema
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
+\vdots &  &  &  & \vdots &  & \vdots\\
+a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+lahko zapišemo kot linearno kombinacijo stolpecv sistema in spremenljivk:
+\begin_inset Formula 
+\[
+\left(b_{1},\dots,b_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)=x_{1}\left(a_{1,1},\dots,a_{m,1}\right)+\cdots+x_{n}\left(a_{1,n},\dots,a_{m,n}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\vec{b}=x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Klasifikacija sistemov linearnih enačb
+\end_layout
+
+\begin_layout Standard
+Sisteme linearnih enačb delimo glede na velikost na
+\end_layout
+
+\begin_layout Itemize
+kvadratne (toliko enačb kot spremenljivk),
+\end_layout
+
+\begin_layout Itemize
+poddoločene (več spremenljivk kot enačb),
+\end_layout
+
+\begin_layout Itemize
+predoločene (več enačb kot spremenljivk);
+\end_layout
+
+\begin_layout Standard
+glede na rešljivost na
+\end_layout
+
+\begin_layout Itemize
+nerešljive (prazna množica rešitev),
+\end_layout
+
+\begin_layout Itemize
+enolično rešljive (množica rešitev je singleton),
+\end_layout
+
+\begin_layout Itemize
+neenolično rešljive (moč množice rešitev je več kot 1);
+\end_layout
+
+\begin_layout Standard
+glede na obliko desnih strani na
+\end_layout
+
+\begin_layout Itemize
+homogene (vektor desnih stani je ničeln)
+\end_layout
+
+\begin_layout Itemize
+nehomogene (vektor desnih strani je neničen)
+\end_layout
+
+\begin_layout Remark*
+Če sta 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{y}$
+\end_inset
+
+ dve različni rešitvi sistema,
+ je rešitev sistema tudi 
+\begin_inset Formula $\left(1-t\right)\vec{x}+t\vec{y}$
+\end_inset
+
+ za vsak realen 
+\begin_inset Formula $t$
+\end_inset
+
+,
+ torej ima vsak neenolično rešljiv sistem neskončno rešitev.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Pogosto (a nikakor ne vedno) se zgodi,
+ da je kvadraten sistem enolično rešljiv,
+ predoločen sistem nerešljiv,
+ poddoločen sistem pa neenolično rešljiv.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Homogen sistem je vedno rešljiv,
+ saj obstaja trivialna rešitev 
+\begin_inset Formula $\vec{0}$
+\end_inset
+
+.
+ Vprašanje pri homogenih sistemih je torej,
+ kdaj je enolično in kdaj neenolično rešljiv.
+ Dokazali bomo,
+ da je vsak poddoločen homogen sistem linearnih enačb neenolično rešljiv.
+\end_layout
+
+\begin_layout Subsubsection
+Reševanje sistema
+\end_layout
+
+\begin_layout Standard
+Sisteme lahko rešujemo z izločanjem spremenljivk.
+ Iz ene enačbe izrazimo spremenljivko in jo vstavimo v druge enačbe,
+ da izrazimo zopet nove spremenljivke,
+ ki jih spet vstavimo v nove enačbe,
+ iz katerih spremenljivk še nismo izražali in tako naprej,
+ vse dokler ne pridemo do zadnjega možnega izražanja (dodatno branje prepuščeno bralcu).
+\end_layout
+
+\begin_layout Standard
+Sisteme pa lahko rešujemo tudi z Gaussovo metodo.
+ Trdimo,
+ da se rešitev sistema ne spremeni,
+ če na njem uporabimo naslednje elementarne vrstične transformacije:
+\end_layout
+
+\begin_layout Itemize
+menjava vrstnega reda enačb,
+\end_layout
+
+\begin_layout Itemize
+množenje enačbe z neničelno konstanto,
+\end_layout
+
+\begin_layout Itemize
+prištevanje večkratnika ene enačbe k drugi.
+\end_layout
+
+\begin_layout Standard
+Z Gaussovo metodo (dodatno branje prepuščeno bralcu) mrcvarimo razširjeno matriko sistema,
+ dokler ne dobimo reducirane kvadratne stopničaste forme (angl.
+ row echelon),
+ ki izgleda takole (
+\begin_inset Formula $\times$
+\end_inset
+
+ reprezentira poljubno realno številko,
+ 
+\begin_inset Formula $0$
+\end_inset
+
+ ničlo in 
+\begin_inset Formula $1$
+\end_inset
+
+ enico):
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccccccccccccccc|c}
+0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
+\vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
+ &  &  & \vdots & \vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \cdots & \times\\
+ &  &  &  &  &  &  & \vdots & \vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \cdots & \times\\
+ &  &  &  &  &  &  &  &  &  &  & \vdots & \vdots &  & \vdots & 0 & \cdots & \vdots\\
+\vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots &  & \vdots\\
+0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots & \times
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Homogeni sistemi
+\end_layout
+
+\begin_layout Definition*
+Sistem je homogen,
+ če je vektor desnih strani ničeln.
+\end_layout
+
+\begin_layout Standard
+Vedno ima rešitev 
+\begin_inset Formula $\vec{0}$
+\end_inset
+
+.
+ Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev.
+ Splošna rešitev nehomogenega sistema je vsota partikularne rešitve tega nehomogenega sistema in splošne rešitve njemu prirejenega homogenega sistema.
+\end_layout
+
+\begin_layout Remark*
+V tem razdelku nehomogen sistem pomeni nenujno homogen sistem (torej splošen sistem linearnih enačb),
+ torej je vsak homogen sistem nehomogen.
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:Vpoddol-hom-sist-ima-ne0-reš"
+
+\end_inset
+
+Vsak poddoločen homogen sistem ima vsaj eno netrivialno rešitev.
+\end_layout
+
+\begin_layout Proof
+Dokaz z indukcijo po številu enačb.
+\end_layout
+
+\begin_deeper
+\begin_layout Paragraph*
+Baza
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=0$
+\end_inset
+
+ za 
+\begin_inset Formula $n\geq2$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $a_{n}=0$
+\end_inset
+
+,
+ je netrivialna rešitev 
+\begin_inset Formula $\left(0,\dots,0,1\right)$
+\end_inset
+
+,
+ sicer pa 
+\begin_inset Formula $\left(0,\dots,0,-a_{n},a_{n-1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph*
+Korak
+\end_layout
+
+\begin_layout Standard
+Denimo,
+ da velja za vse poddoločene homogene sisteme z 
+\begin_inset Formula $m-1$
+\end_inset
+
+ vrsticami.
+ Vzemimo poljuben homogen sistem z 
+\begin_inset Formula $n>m$
+\end_inset
+
+ stolpci (da je poddoločen).
+ Če je 
+\begin_inset Formula $a_{n}=0$
+\end_inset
+
+,
+ je netriviačna rešitev 
+\begin_inset Formula $\left(0,\dots,0,1\right)$
+\end_inset
+
+,
+ sicer pa iz ene od enačb izrazimo 
+\begin_inset Formula $x_{n}$
+\end_inset
+
+ s preostalimi spremenljivkami.
+ Dobljen izraz vstavimo v preostalih 
+\begin_inset Formula $m-1$
+\end_inset
+
+ enačb z 
+\begin_inset Formula $n-1$
+\end_inset
+
+ spremenljivkami in dobljen sistem uredimo.
+ Po I.
+ P.
+ ima slednji netrivialno rešitev 
+\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1}\right)$
+\end_inset
+
+.
+ To rešitev vstavimo v izraz za 
+\begin_inset Formula $x_{n}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\alpha_{n}$
+\end_inset
+
+ in s tem 
+\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1},\alpha_{n}\right)$
+\end_inset
+
+ kot netrivialno rešitev sistema z 
+\begin_inset Formula $m$
+\end_inset
+
+ vrsticami.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev.
+\end_layout
+
+\begin_layout Proof
+Če sta 
+\begin_inset Formula $\left(s_{1},\dots,s_{n}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(t_{1},\dots,t_{n}\right)$
+\end_inset
+
+ dve rešitvi homogenega sistema,
+ velja za 
+\begin_inset Formula $\vec{s}$
+\end_inset
+
+ 
+\begin_inset Formula $\forall i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\left(s_{1},\dots,s_{n}\right)\right\rangle =a_{i,1}s_{1}+\cdots+a_{i,n}s_{n}=0$
+\end_inset
+
+ in enako za 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $\forall\alpha,\beta\in\mathbb{R},i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =0$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =\left\langle \alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right),\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left\langle \vec{s},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\beta\left\langle \vec{t},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\alpha0+\beta0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Splošna rešitev 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ rešljivega nehomogenega sistema s partikularno rešitvijo 
+\begin_inset Formula $\vec{p}$
+\end_inset
+
+ je 
+\begin_inset Formula $\vec{x}=\vec{p}+\vec{h}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\vec{h}$
+\end_inset
+
+ rešitev temu sistemu prirejenega homogenega sistema (desno stvar smo prepisali z ničlami).
+\end_layout
+
+\begin_layout Remark*
+Trdimo,
+ da je množica rešitev nehomogenega sistema samo množica rešitev prirejenega homogenega sistema,
+ premaknjena za partikularno rešitev nehomogenega sistema.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $\forall i:\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}\wedge\left\langle \vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =0$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $\forall i:\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\left\langle \vec{h}\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}+0=b_{i}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Predoločeni sistemi
+\end_layout
+
+\begin_layout Standard
+Predoločen sistem,
+ torej tak z več enačbami kot spremenljivkami,
+ je običajno,
+ a ne nujno,
+ nerešljiv.
+\end_layout
+
+\begin_layout Definition*
+Posplošena rešitev sistema linearnih enačb je taka 
+\begin_inset Formula $n-$
+\end_inset
+
+terica števil 
+\begin_inset Formula $\left(x_{1},\dots x_{n}\right)$
+\end_inset
+
+,
+ za katero je vektor levih strani 
+\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$
+\end_inset
+
+ najbližje vektorju desnih strani 
+\begin_inset Formula $\left(b_{1},\dots,b_{n}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Če je sistem rešljiv,
+ se njegova rešitev ujema s posplošeno rešitvijo.
+ Po metodi najmanjših kvadratov želimo minimizirati izraz 
+\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n}-b_{1}\right)^{2}+\cdots+\left(a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}-b_{n}\right)^{2}$
+\end_inset
+
+ oziroma kvadrat norme razlike 
+\begin_inset Formula $\left|\left|x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}-\vec{b}\right|\right|^{2}$
+\end_inset
+
+.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Z 
+\begin_inset Formula $\vec{a_{i}}$
+\end_inset
+
+ označujemo stolpične vektorje sistema,
+ torej 
+\begin_inset Formula $\vec{a_{i}}=\left(a_{1,i},\dots,a_{m,i}\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+ Podobno kot pri regresijski premici želimo pravokotno projicirati 
+\begin_inset Formula $\vec{b}$
+\end_inset
+
+ na 
+\begin_inset Formula $\Lin\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $
+\end_inset
+
+.
+ Iščemo torej take skalarje 
+\begin_inset Formula $\left(x_{1},\dots,x_{n}\right)$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b}\perp\vec{a_{1}},\dots,\vec{a_{n}}$
+\end_inset
+
+ (hkrati pravokotna na vse vektorje,
+ ki določajo to linearno ogrinjačo).
+ Preuredimo skalarne produkte in zopet dobimo sistem enačb:
+\begin_inset Formula 
+\[
+\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{1}}\right\rangle =\cdots=\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{n}}\right\rangle =0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+x_{1}\left\langle \vec{a_{1}},\vec{a_{1}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{1}}\right\rangle =\left\langle \vec{b},\vec{a_{1}}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+x_{1}\left\langle \vec{a_{1}},\vec{a_{n}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{n}}\right\rangle =\left\langle \vec{b},\vec{a_{n}}\right\rangle 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Izkaže se,
+ da je zgornji sistem vedno rešljiv.
+ Enolično takrat,
+ ko so 
+\begin_inset Formula $\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $
+\end_inset
+
+ linearno neodvisni.
+ Če je neenolično rešljiv,
+ pa poiščemo njegovo najkrajšo rešitev.
+\end_layout
+
+\begin_layout Subsubsection
+Poddoločeni sistemi
+\end_layout
+
+\begin_layout Claim*
+Poddoločen sistem,
+ torej tak,
+ ki ima več spremenljivk kot enačb,
+ ima neskončno rešitev,
+ čim je rešljiv.
+\end_layout
+
+\begin_layout Proof
+Sledi iz zgornjih dokazov,
+ da ima vsak poddoločen homogen sistem neskončno rešitev in da je 
+\begin_inset Formula $\vec{p}+\vec{h}$
+\end_inset
+
+ splošna rešitev nehomogenega sistema,
+ če je 
+\begin_inset Formula $\vec{p}$
+\end_inset
+
+ partikularna rešitev tega sistema in 
+\begin_inset Formula $\vec{h}$
+\end_inset
+
+ splošna rešitev prirejenega homogenega sistema.
+\end_layout
+
+\begin_layout Remark*
+Seveda je lahko poddoločen sistem nerešljiv.
+ Trivialen primer:
+ 
+\begin_inset Formula $x+y+z=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $x+y+z=2$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kadar ima sistem neskončno rešitev,
+ nas često zanima najkrajša (recimo zadnja opomba v prejšnji sekciji).
+ Geometrijski gledano je najkrajša rešitev pravokotna projekcija izhodišča na presek hiperravnin,
+ ki so množica rešitve sistema.
+ Vsaka enačba določa eno hiperravnino v normalni obliki,
+ torej 
+\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}$
+\end_inset
+
+.
+ Projekcija izhodišča na hiperravnino v normalni obliki je presečišče premice,
+ ki gre skozi izhodišče in je pravokotna na ravnino,
+ torej 
+\begin_inset Formula $\vec{r}=t\vec{n_{i}}$
+\end_inset
+
+,
+ in ravnine same.
+ Vstavimo drugo enačbo v prvo in dobimo 
+\begin_inset Formula $\left\langle t\vec{n_{i}},\vec{n_{i}}\right\rangle =b_{i}$
+\end_inset
+
+ in izrazimo 
+\begin_inset Formula $t=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }$
+\end_inset
+
+,
+ s čimer dobimo 
+\begin_inset Formula $\vec{r}=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }\vec{n_{i}}$
+\end_inset
+
+.
+ Doslej je to le projekcija na eno hiperravnino.
+\end_layout
+
+\begin_layout Standard
+Za pravokotno projekcijo na presek hiperravnin pa najprej določimo ravnino,
+ ki je pravokotna na vse hiperravnine sistema,
+ torej 
+\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$
+\end_inset
+
+,
+ in najdimo presek te ravnine z vsemi hiperravninami.
+ To storimo tako,
+ da enačbo ravnine vstavimo v enačbe hiperravnin in jih uredimo:
+ 
+\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}\sim\left\langle t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}\sim t_{1}\left\langle \vec{n_{1}},\vec{n_{i}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}$
+\end_inset
+
+.
+ To nam da sistem enačb
+\begin_inset Formula 
+\[
+t_{1}\left\langle \vec{n_{1}},\vec{n_{1}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{1}}\right\rangle =b_{1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+t_{1}\left\langle \vec{n_{1}},\vec{n_{m}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{m}}\right\rangle =b_{m}
+\]
+
+\end_inset
+
+Rešimo sistem in dobimo 
+\begin_inset Formula $\left(t_{1},\dots,t_{m}\right)$
+\end_inset
+
+,
+ kar vstavimo v enačbo ravnine 
+\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$
+\end_inset
+
+,
+ da dobimo najkrajšo rešitev.
+\end_layout
+
+\begin_layout Subsection
+Matrike
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matrika je element 
+\begin_inset Formula $\left(\mathbb{R}^{n}\right)^{m}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $A=\left(\left(a_{1,1},\dots,a_{1,n}\right),\dots,\left(a_{m,1},\dots,a_{m,n}\right)\right)$
+\end_inset
+
+.
+ Ima 
+\begin_inset Formula $m$
+\end_inset
+
+ vrstic in 
+\begin_inset Formula $n$
+\end_inset
+
+ stolpcev,
+ zato jo pišemo takole:
+\begin_inset Formula 
+\[
+A=\left[\begin{array}{ccc}
+a_{1,1} & \cdots & a_{1,n}\\
+\vdots &  & \vdots\\
+a_{m,1} & \cdots & a_{m,n}
+\end{array}\right]
+\]
+
+\end_inset
+
+Matrikam velikosti 
+\begin_inset Formula $1\times n$
+\end_inset
+
+ pravimo vrstični vektor,
+ matrikam velikosti 
+\begin_inset Formula $m\times1$
+\end_inset
+
+ pa stolpični vektor.
+ Obe vrsti običajno identificiramo z vektorji.
+ 
+\begin_inset Formula $\left[1\right]$
+\end_inset
+
+ identificiramo z 1.
+ Na preseku 
+\begin_inset Formula $i-$
+\end_inset
+
+te vrstice in 
+\begin_inset Formula $j-$
+\end_inset
+
+tega stolpca matrike se nahaja element 
+\begin_inset Formula $a_{i,j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Seštevanje matrik je definirano le za matrike enakih dimenzij.
+ Vsota matrik 
+\begin_inset Formula $A+B$
+\end_inset
+
+ je matrika
+\begin_inset Formula 
+\[
+A+B=\left[\begin{array}{ccc}
+a_{1,1}+b_{1,1} & \cdots & a_{1,n}+b_{1,n}\\
+\vdots &  & \vdots\\
+a_{m,1}+b_{m.1} & \cdots & a_{m,n}+b_{m,n}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Ničelna matrika 0 je aditivna enota.
+\begin_inset Formula 
+\[
+0=\left[\begin{array}{ccc}
+0 & \cdots & 0\\
+\vdots &  & \vdots\\
+0 & \cdots & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Produkt matrike s skalarjem.
+\begin_inset Formula 
+\[
+A\cdot\alpha=\alpha\cdot A=\left[\begin{array}{ccc}
+\alpha a_{1,1} & \cdots & \alpha a_{1,n}\\
+\vdots &  & \vdots\\
+\alpha a_{m,1} & \cdots & \alpha a_{m,n}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Produkt dveh matrik 
+\begin_inset Formula $A_{m\times n}\cdot B_{n\times p}=C_{m\times p}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $c_{i,j}=\sum_{k=1}^{n}a_{i,k}b_{j,k}$
+\end_inset
+
+.
+ (razmislek prepuščen bralcu)
+\end_layout
+
+\begin_layout Remark*
+Kvadratna matrika identiteta 
+\begin_inset Formula $I$
+\end_inset
+
+ je multiplikativna enota:
+ 
+\begin_inset Formula $i_{ij}=\begin{cases}
+0 & ;i\not=j\\
+1 & ;i=j
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Transponiranje matrike 
+\begin_inset Formula $A_{m\times n}^{T}=B_{n\times m}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $b_{ij}=a_{ji}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Lastnosti transponiranja:
+ 
+\begin_inset Formula $\left(A^{T}\right)^{T}=A$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(A+B\right)^{T}=A^{T}+B^{T}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\alpha A\right)^{T}=\alpha A^{T}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(AB\right)^{T}=B^{T}A^{T}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I^{T}=I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $0^{T}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Matrični zapis sistema linearnih enačb
+\end_layout
+
+\begin_layout Standard
+Matrika koeficientov vsebuje koeficiente,
+ imenujmo jo 
+\begin_inset Formula $A$
+\end_inset
+
+ (ena vrstica matrike je ena enačba v sistemu).
+ Stolpični vektor spremenljivk vsebuje spremenljivke 
+\begin_inset Formula $\vec{x}=\left(x_{1},\dots,x_{n}\right)$
+\end_inset
+
+.
+ Vektor desne strani vsebuje desne strani 
+\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{m}\right)$
+\end_inset
+
+.
+ Sistem torej zapišemo kot 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Tudi Gaussovo metodo lahko zapišemo matrično.
+ Trem elementarnim preoblikovanjem,
+ ki ne spremenijo množice rešitev,
+ priredimo ustrezne t.
+ i.
+ elementarne matrike:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{i,j}\left(\alpha\right)$
+\end_inset
+
+:
+ identiteta,
+ ki ji na 
+\begin_inset Formula $i,j-$
+\end_inset
+
+to mesto prištejemo 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+ Ustreza prištevanju 
+\begin_inset Formula $\alpha-$
+\end_inset
+
+kratnika 
+\begin_inset Formula $j-$
+\end_inset
+
+te vrstice k 
+\begin_inset Formula $i-$
+\end_inset
+
+ti vrstici.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P_{ij}$
+\end_inset
+
+:
+ v 
+\begin_inset Formula $I$
+\end_inset
+
+ zamenjamo 
+\begin_inset Formula $i-$
+\end_inset
+
+to in 
+\begin_inset Formula $j-$
+\end_inset
+
+to vrstico.
+ Ustreza zamenjavi 
+\begin_inset Formula $i-$
+\end_inset
+
+te in 
+\begin_inset Formula $j-$
+\end_inset
+
+te vrstice.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{i}\left(\alpha\right)$
+\end_inset
+
+:
+ v 
+\begin_inset Formula $I$
+\end_inset
+
+ pomnožiš 
+\begin_inset Formula $i-$
+\end_inset
+
+to vrstico z 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+ Ustreza množenju 
+\begin_inset Formula $i-$
+\end_inset
+
+te vrstice s skalarjem 
+\begin_inset Formula $\alpha$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Vsako matriko je moč z levim množenjem z elementarnimi matrikami (Gaussova metoda) prevesti na reducirano vrstično stopničasto formo/obliko.
+ ZDB 
+\begin_inset Formula $\forall A\in M\left(\mathbb{R}\right)\exists E_{1},\dots,E_{k}\ni:R=E_{1}\cdot\cdots\cdot E_{k}\cdot A$
+\end_inset
+
+ je r.
+ v.
+ s.
+ f.
+ Ko rešujemo sistem s temi matrikami množimo levo in desno stran sistema.
+\end_layout
+
+\begin_layout Subsubsection
+Postopek iskanja posplošene rešitve predoločenega sistema
+\end_layout
+
+\begin_layout Enumerate
+Sistem 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ z leve pomnožimo z 
+\begin_inset Formula $A^{T}$
+\end_inset
+
+ in dobimo sistem 
+\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Poiščemo običajno rešitev dobljenega sistema,
+ za katero se izkaže,
+ da vselej obstaja (dokaz v 2.
+ semestru).
+\end_layout
+
+\begin_layout Enumerate
+Dokažemo,
+ da je običajna rešitev 
+\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
+\end_inset
+
+ enaka posplošeni rešitvi 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}$
+\end_inset
+
+ bi radi minimizirali.
+ Naj bo 
+\begin_inset Formula $\vec{x_{0}}$
+\end_inset
+
+ običajna rešitev sistema 
+\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $\vec{u}=A\vec{x}-A\vec{x_{0}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{v}=A\vec{x_{0}}-\vec{b}$
+\end_inset
+
+.
+ Trdimo,
+ da 
+\begin_inset Formula $\vec{u}\perp\vec{v}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left\langle \vec{u},\vec{v}\right\rangle =0$
+\end_inset
+
+.
+ Dokažimo:
+\begin_inset Formula 
+\[
+\left\langle A\vec{x}-A\vec{x_{0}},A\vec{x_{0}}-\vec{b}\right\rangle =\left\langle A\left(\vec{x}-\vec{x_{0}}\right),A\vec{x_{0}}-\vec{b}\right\rangle =\left(A\vec{x_{0}}-\vec{b}\right)^{T}A\left(\vec{x}-\vec{x_{0}}\right)=\left(A\vec{x_{0}}-\vec{b}\right)^{T}\left(A^{T}\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left(A^{T}\left(A\vec{x_{0}}-\vec{b}\right)\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)=\left(A^{T}A\vec{x_{0}}-A^{T}\vec{b}\right)\left(\vec{x}-\vec{x_{0}}\right)\overset{\text{predpostavka o }\vec{x_{0}}}{=}0\left(\vec{x}-\vec{x_{0}}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Ker sedaj vemo,
+ da sta 
+\begin_inset Formula $\vec{u}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+ pravokotna,
+ lahko uporabimo Pitagorov izrek,
+ ki za njiju pravi 
+\begin_inset Formula $\left|\left|\vec{u}+\vec{v}\right|\right|^{2}=\left|\left|\vec{u}\right|\right|^{2}+\left|\left|\vec{v}\right|\right|^{2}$
+\end_inset
+
+.
+ V naslednjih izpeljavah je 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ poljuben,
+ 
+\begin_inset Formula $\vec{x_{0}}$
+\end_inset
+
+ pa kot prej.
+\begin_inset Formula 
+\[
+\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}\right|\right|^{2}+\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2},
+\]
+
+\end_inset
+
+kar pomeni ,
+ da je 
+\begin_inset Formula $\vec{x_{0}}$
+\end_inset
+
+ manjši ali enak kot vsi ostale 
+\begin_inset Formula $n-$
+\end_inset
+
+terice spremenljivk.
+\end_layout
+
+\begin_layout Subsubsection
+Najkrajša rešitev sistema
+\end_layout
+
+\begin_layout Standard
+Ta sekcija je precej dobesedno povzeta po profesorjevi beamer skripti.
+\end_layout
+
+\begin_layout Standard
+Sistem 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ je lahko neenolično rešljiv.
+ Tedaj nas često zanima po normi najkrajša rešitev sistema.
+\end_layout
+
+\begin_layout Claim*
+Najkrajša rešitev sistema 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ je 
+\begin_inset Formula $A^{T}\vec{y_{0}}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\vec{y_{0}}$
+\end_inset
+
+ poljubna rešitev sistema 
+\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $\vec{x_{0}}$
+\end_inset
+
+ poljubna rešitev sistema 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{y_{0}}$
+\end_inset
+
+ poljubna rešitev sistema 
+\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
+\end_inset
+
+.
+ Dokazali bi radi,
+ da velja 
+\begin_inset Formula $\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\leq\left|\left|\vec{x_{0}}\right|\right|^{2}$
+\end_inset
+
+.
+ Podobno,
+ kot v prejšnji sekciji:
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\left|\left|\vec{x_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|u+v\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Dokažimo,
+ da sta 
+\begin_inset Formula $\vec{u}=\vec{x_{0}}-A^{T}\vec{y_{0}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{v}=A^{T}\vec{y_{0}}$
+\end_inset
+
+ pravokotna,
+ da lahko uporabimo pitagorov izrek v drugi vrstici:
+\begin_inset Formula 
+\[
+\left\langle \vec{x_{0}}-A^{T}\vec{y_{0}},A^{T}\vec{y_{0}}\right\rangle =\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)^{T}A^{T}\vec{y_{0}}=\left(A\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)\right)^{T}\vec{y_{0}}=\left(A\vec{x_{0}}-AA^{T}\vec{y_{0}}\right)^{T}\vec{y_{0}}=\left(\vec{b}-\vec{b}\right)^{T}\vec{y_{0}}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|u+v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}\right|\right|^{2}+\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right|
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|\vec{x_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Iz rešljivosti 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ sledi rešljivost 
+\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$
+\end_inset
+
+,
+ toda to znamo dokazati šele v drugem semestru.
+\end_layout
+
+\begin_layout Subsubsection
+Inverzi matrik
+\end_layout
+
+\begin_layout Definition*
+Matrika 
+\begin_inset Formula $B$
+\end_inset
+
+ je inverz matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če velja 
+\begin_inset Formula $AB=I$
+\end_inset
+
+ in 
+\begin_inset Formula $BA=I$
+\end_inset
+
+.
+ Matrika 
+\begin_inset Formula $A$
+\end_inset
+
+ je obrnljiva,
+ če ima inverz,
+ sicer je neobrnljiva.
+\end_layout
+
+\begin_layout Claim*
+Če inverz obstaja,
+ je enoličen.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $B_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $B_{2}$
+\end_inset
+
+ inverza 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $AB_{1}=B_{1}A=AB_{2}=B_{2}A=I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $B_{1}=B_{1}I=B_{1}\left(AB_{2}\right)=\left(B_{1}A\right)B_{2}=IB_{2}=B_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če inverz 
+\begin_inset Formula $A$
+\end_inset
+
+ obstaja,
+ ga označimo z 
+\begin_inset Formula $A^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri obrnljivih matrik:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Identična matrika 
+\begin_inset Formula $I$
+\end_inset
+
+:
+ 
+\begin_inset Formula $I\cdot I=I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I^{-1}=I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Elementarne matrike:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $E_{ij}\left(\alpha\right)\cdot E_{ij}\left(-\alpha\right)=I$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $E_{ij}\left(\alpha\right)^{-1}=E_{ij}\left(-\alpha\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P_{ij}\cdot P_{ij}=I$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $P_{ij}^{-1}=P_{ij}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $E_{i}\left(\alpha\right)\cdot E_{i}\left(\alpha^{-1}\right)=I$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $E_{i}\left(\alpha\right)^{-1}=E_{i}\left(\alpha^{-1}\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Claim*
+Produkt obrnljivih matrik je obrnljiva matrika.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $A_{1},\dots,A_{n}$
+\end_inset
+
+ obrnljive matrike,
+ torej po definiciji velja 
+\begin_inset Formula $A_{1}\cdot\cdots\cdot A_{n}\cdot A_{n}^{-1}\cdot\cdots\cdot A_{1}^{-1}=A_{n}\cdot\cdots\cdot A_{1}\cdot A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}=I$
+\end_inset
+
+.
+ Opazimo,
+ da velja 
+\begin_inset Formula $\left(A_{1}\cdot\cdots\cdot A_{n}\right)^{-1}=A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Vsaka obrnljiva matrika je produkt elementarnih matrik.
+ Dokaz sledi kasneje.
+\end_layout
+
+\begin_layout Example*
+Primeri neobrnljivih matrik:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Ničelna matrika,
+ saj pri množenju s katerokoli matriko pridela ničelno matriko in velja 
+\begin_inset Formula $I\not=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Matrike z ničelnim stolpcem/vrstico.
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Naj ima 
+\begin_inset Formula $A$
+\end_inset
+
+ vrstico samih ničel.
+ Tedaj za vsako 
+\begin_inset Formula $B$
+\end_inset
+
+ velja,
+ da ima 
+\begin_inset Formula $AB$
+\end_inset
+
+ vrstico samih ničel (očitno po definiciji množenja).
+ 
+\begin_inset Formula $AB$
+\end_inset
+
+ zato ne more biti 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ saj 
+\begin_inset Formula $I$
+\end_inset
+
+ ne vsebuje nobene vrstice samih ničel.
+ Podobno za ničelni stolpec.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Nekvadratne matrike
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Naj ima 
+\begin_inset Formula $A_{m\times n}$
+\end_inset
+
+ več vrstic kot stolpcev (
+\begin_inset Formula $m>n$
+\end_inset
+
+).
+ PDDRAA obstaja 
+\begin_inset Formula $B$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $AB=I$
+\end_inset
+
+.
+ Uporabimo Gaussovo metodo na 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Z levim množenjem 
+\begin_inset Formula $A$
+\end_inset
+
+ z nekimi elementarnimi matrikami lahko pridelamo RVSO.
+ 
+\begin_inset Formula $E_{1}\cdots E_{n}A=R$
+\end_inset
+
+.
+ 
+\begin_inset Formula $E_{1}\cdots E_{n}AB=E_{1}\cdots E_{n}I=E_{1}\cdots E_{n}=RB$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $R$
+\end_inset
+
+ ima po konstrukciji ničelno vrstico (je namreč 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna RVSO in a ima več vrstic kot stolpcev).
+ Potemtakem ima tudi 
+\begin_inset Formula $RB$
+\end_inset
+
+ ničelno vrstico,
+ torej je neobrnljiva,
+ toda 
+\begin_inset Formula $RB$
+\end_inset
+
+ je enak produktu elementarnih matrik,
+ torej bi morala biti obrnljiva.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Remark*
+Iz 
+\begin_inset Formula $AB=I$
+\end_inset
+
+ ne sledi nujno 
+\begin_inset Formula $BA=I$
+\end_inset
+
+.
+ Primer:
+ 
+\begin_inset Formula $A=\left[\begin{array}{ccc}
+1 & 0 & 0\\
+0 & 1 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $B=\left[\begin{array}{cc}
+1 & 0\\
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $AB=\left[\begin{array}{cc}
+1 & 0\\
+0 & 1
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $BA=\left[\begin{array}{ccc}
+1 & 0 & 0\\
+0 & 1 & 0\\
+0 & 0 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Velja pa to za kvadratne matrike.
+ Dokaz kasneje.
+\end_layout
+
+\begin_layout Subsubsection
+Karakterizacija obrnljivih matrik
+\end_layout
+
+\begin_layout Theorem*
+Za vsako kvadratno matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ so naslednje trditve ekvivalentne:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ je obrnljiva
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ ima levi inverz (
+\begin_inset Formula $\exists B\ni:BA=I$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ ima desni inverz (
+\begin_inset Formula $\exists B\ni:AB=I$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Enumerate
+stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ so linearno neodvisni
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ so ogrodje
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:VbEx:Ax=b"
+
+\end_inset
+
+
+\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+RVSO 
+\begin_inset Formula $A$
+\end_inset
+
+ je 
+\begin_inset Formula $I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ je produkt elementarnih matrik
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Shema dokaza teh ekvivalenc je zanimiv graf.
+ Bralcu je prepuščena njegova skica.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(1\Rightarrow2\right)$
+\end_inset
+
+ Sledi iz definicije.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(1\Rightarrow3\right)$
+\end_inset
+
+ Sledi iz definicije.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(2\Rightarrow5\right)$
+\end_inset
+
+ Naj 
+\begin_inset Formula $\exists B\ni:BA=I$
+\end_inset
+
+.
+ Dokažimo,
+ da 
+\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
+\end_inset
+
+.
+ Pa dajmo:
+ 
+\begin_inset Formula $A\vec{x}=0\Rightarrow B\left(A\vec{x}\right)=B0=0=\left(BA\right)\vec{x}=I\vec{x}=\vec{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(3\Rightarrow7\right)$
+\end_inset
+
+ Naj 
+\begin_inset Formula $\exists B\ni:AB=I$
+\end_inset
+
+.
+ Dokažimo,
+ da 
+\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $\vec{x}=B\vec{b}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $A\vec{x}=AB\vec{b}=I\vec{b}=\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(5\Rightarrow4\right)$
+\end_inset
+
+ Naj 
+\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$
+\end_inset
+
+.
+ Dokažimo,
+ da so stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ linearno neodvisni.
+ Naj bo 
+\begin_inset Formula $A=\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{mn}
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\vec{x}=\left[\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right]$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $A\vec{x}=\left[\begin{array}{ccc}
+a_{11}x_{1} & \cdots & a_{1n}x_{n}\\
+\vdots &  & \vdots\\
+a_{n1}x_{1} & \cdots & a_{mn}x_{n}
+\end{array}\right]=\left[\begin{array}{c}
+a_{11}\\
+\vdots\\
+a_{m1}
+\end{array}\right]x_{1}+\cdots+\left[\begin{array}{c}
+a_{1n}\\
+\vdots\\
+a_{mn}
+\end{array}\right]x_{n}=\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:vsi0"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za linearno neodvisnost mora veljati 
+\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=0\Rightarrow x_{1}=\cdots=x_{n}=0$
+\end_inset
+
+.
+ Ravno to pa smo predpostavili.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(7\Rightarrow6\right)$
+\end_inset
+
+ Uporabimo iste oznake kot zgoraj.
+ Za poljuben 
+\begin_inset Formula $\vec{b}$
+\end_inset
+
+ iščemo tak 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=A\vec{x}=\vec{b}$
+\end_inset
+
+ (definicija ogrodja).
+ Po predpostavki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:VbEx:Ax=b"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja,
+ da 
+\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$
+\end_inset
+
+.
+ Torej po predpostavki najdemo ustrezen 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ za poljuben 
+\begin_inset Formula $\vec{b}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(4\Rightarrow8\right)$
+\end_inset
+
+ Za dokaz uvedimo nekaj lem,
+ ki dokažejo trditev.
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:kom1"
+
+\end_inset
+
+Če ima 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ LN stolpce in če je 
+\begin_inset Formula $C_{n\times n}$
+\end_inset
+
+ obrnljiva,
+ ima tudi 
+\begin_inset Formula $CA$
+\end_inset
+
+ LN stolpce.
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $a_{1},\dots,a_{n}$
+\end_inset
+
+ stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $Ax=0\Rightarrow x=0$
+\end_inset
+
+.
+ Dokazati želimo,
+ da 
+\begin_inset Formula $CAx=0\Rightarrow x=0$
+\end_inset
+
+.
+ Predpostavimo 
+\begin_inset Formula $CAx=0$
+\end_inset
+
+.
+ Množimo obe strani z 
+\begin_inset Formula $C^{-1}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $C^{-1}CAx=C^{-1}0\sim IAx=0\sim Ax=0\Rightarrow x=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Lemma
+Če ima 
+\begin_inset Formula $A$
+\end_inset
+
+ LN stolpce,
+ ima njena RVSO LN stolpce.
+\end_layout
+
+\begin_layout Proof
+Po Gaussu obstajajo take elementarne 
+\begin_inset Formula $E_{1},\dots,E_{n}$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $E_{n}\cdots E_{1}A=R$
+\end_inset
+
+ RVSO.
+ Po lemi 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "lem:kom1"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ ima 
+\begin_inset Formula $E_{1}A$
+\end_inset
+
+ LN stolpce,
+ prav tako 
+\begin_inset Formula $E_{2}E_{1}A$
+\end_inset
+
+ in tako dalje,
+ vse do 
+\begin_inset Formula $E_{n}\cdots E_{1}A=R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Lemma
+Če ima RVSO 
+\begin_inset Formula $R$
+\end_inset
+
+ LN stolpce,
+ je enaka identiteti.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $R\not=I$
+\end_inset
+
+.
+ Tedaj ima bodisi ničelni stolpec bodisi stopnico,
+ daljšo od 1.
+ Če ima ničelni stolpec,
+ ni LN.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Če ima stopnico,
+ daljšo od 1,
+ kar pomeni,
+ da v vrstici takoj za prvo enico obstajajo neki neničelni 
+\begin_inset Formula $\times-$
+\end_inset
+
+i,
+ pa je stolpec z nekim neničelnim 
+\begin_inset Formula $\times-$
+\end_inset
+
+om linearna kombinacija ostalih stolpcev,
+ torej stolpci 
+\begin_inset Formula $R$
+\end_inset
+
+ niso LN 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(6\Rightarrow8\right)$
+\end_inset
+
+ Predpostavimo,
+ da so stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ ogrodje in dokazujemo,
+ da RVSO 
+\begin_inset Formula $A$
+\end_inset
+
+ je 
+\begin_inset Formula $I$
+\end_inset
+
+.
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:68kom1"
+
+\end_inset
+
+Če so stolpci 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ ogrodje in če je 
+\begin_inset Formula $C_{n\times n}$
+\end_inset
+
+ obrnljica,
+ so tudi stolpci 
+\begin_inset Formula $CA$
+\end_inset
+
+ ogrodje.
+\end_layout
+
+\begin_layout Proof
+Naj bodo stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ ogrodje.
+ Torej 
+\begin_inset Formula $\forall b\exists x\ni:Ax=C^{-1}b$
+\end_inset
+
+.
+ Množimo obe strani z 
+\begin_inset Formula $C^{-1}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall b\exists x\ni:CAx=b$
+\end_inset
+
+ —
+ stolpci 
+\begin_inset Formula $CA$
+\end_inset
+
+ so ogrodje.
+\end_layout
+
+\begin_layout Lemma
+Če so stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+ ogrodje,
+ so stolpci njene RVSO ogrodje.
+\end_layout
+
+\begin_layout Proof
+Po Gaussu obstajajo take elementarne 
+\begin_inset Formula $E_{1},\dots,E_{n}$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $E_{n}\cdots E_{1}A=R$
+\end_inset
+
+ RVSO.
+ Po lemi 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "lem:68kom1"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ so stolpci 
+\begin_inset Formula $E_{1}A$
+\end_inset
+
+ ogrodje in tudi stolpci 
+\begin_inset Formula $E_{2}E_{1}A$
+\end_inset
+
+ so ogrodje in tako dalje vse do 
+\begin_inset Formula $R$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Lemma
+Če so stolpci RVSO 
+\begin_inset Formula $R$
+\end_inset
+
+ ogrodje,
+ je 
+\begin_inset Formula $R=I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $R\not=I$
+\end_inset
+
+.
+ Tedaj ima bodisi ničelni stolpec bodisi stopnico,
+ daljšo od 1.
+ Če ima ničelni stolpec,
+ stolpci niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora).
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ Če ima stopnico,
+ daljšo od 1,
+ pa je stolpec z nekim neničelnim 
+\begin_inset Formula $\times-$
+\end_inset
+
+om linearna kombinacija ostalih stolpcev,
+ torej stolpci 
+\begin_inset Formula $R$
+\end_inset
+
+ niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora) 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+(tegale ne razumem zares dobro,
+ niti med predavanji nismo dokazali) mogoče čim ima stopnico,
+ daljšo od 1,
+ ima ničelno vrstico?
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(8\Rightarrow9\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $R\coloneqq\text{RVSO}\left(A\right)=I$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $A$
+\end_inset
+
+ produkt elementarnih matrik.
+ Po Gaussu obstajajo take elementarne matrike 
+\begin_inset Formula $E_{1},\dots,E_{n}$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $E_{n}\cdots E_{1}A=R$
+\end_inset
+
+.
+ Elementarne matrike so obrnljive,
+ zato množimo z leve najprej z 
+\begin_inset Formula $E_{n}^{-1}$
+\end_inset
+
+,
+ nato z 
+\begin_inset Formula $E_{n-1}^{-1}$
+\end_inset
+
+,
+ vse do 
+\begin_inset Formula $E_{1}^{-1}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}R$
+\end_inset
+
+.
+ Upoštevamo,
+ da je inverz elementarne matrike elementarna matrika in da je 
+\begin_inset Formula $R=I$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+\begin_inset Formula $A$
+\end_inset
+
+ je obrnljiva 
+\begin_inset Formula $\Leftrightarrow A^{T}$
+\end_inset
+
+ obrnljiva.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $AB=I\Leftrightarrow\left(AB\right)^{T}=I^{T}\Leftrightarrow B^{T}A^{T}=I$
+\end_inset
+
+ in 
+\begin_inset Formula $BA=I\Leftrightarrow\left(BA\right)^{T}=I^{T}\Leftrightarrow A^{T}B^{T}=I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+\begin_inset Formula $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$
+\end_inset
+
+ in vrstice so LN in ogrodje.
+\end_layout
+
+\begin_layout Remark*
+Inverz 
+\begin_inset Formula $A$
+\end_inset
+
+ lahko izračunamo po Gaussu.
+ Zapišemo razširjeno matriko 
+\begin_inset Formula $\left[A,I\right]$
+\end_inset
+
+ in na obeh applyamo iste elementarne transformacije,
+ da 
+\begin_inset Formula $A$
+\end_inset
+
+ pretvorimo v RVSO.
+ Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ obrnljiva,
+ dobimo na levi identiteto,
+ na desni pa 
+\begin_inset Formula $A^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Determinante
+\end_layout
+
+\begin_layout Definition*
+Vsaki kvadratni matriki 
+\begin_inset Formula $A$
+\end_inset
+
+ priredimo število 
+\begin_inset Formula $\det A$
+\end_inset
+
+.
+ Definicija za 
+\begin_inset Formula $1\times1$
+\end_inset
+
+ matrike:
+ 
+\begin_inset Formula $\det\left[a\right]\coloneqq a$
+\end_inset
+
+.
+ Rekurzivna definicija za 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matrike:
+\begin_inset Formula 
+\[
+\det\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{nn}
+\end{array}\right]=\sum_{k=1}^{n}\left(-1\right)^{k+1}a_{1k}\det A_{1k},
+\]
+
+\end_inset
+
+kjer 
+\begin_inset Formula $A_{ij}$
+\end_inset
+
+ predstavja 
+\begin_inset Formula $A$
+\end_inset
+
+ brez 
+\begin_inset Formula $i-$
+\end_inset
+
+te vrstice in 
+\begin_inset Formula $j-$
+\end_inset
+
+tega stolpca.
+ Tej formuli razvoja se reče 
+\begin_inset Quotes gld
+\end_inset
+
+razvoj determinante po prvi vrstici
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $2\times2$
+\end_inset
+
+ determinanta.
+ 
+\begin_inset Formula $\det\left[\begin{array}{cc}
+a & b\\
+c & d
+\end{array}\right]=ad-bc$
+\end_inset
+
+.
+ Geometrijski pomen je ploščina paralelograma,
+ ki ga razpenjata 
+\begin_inset Formula $\left(c,d\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(a,b\right)$
+\end_inset
+
+,
+ kajti ploščina bi bila 
+\begin_inset Formula $\left(a+c\right)\left(b+d\right)-2bc-2\frac{cd}{2}-2\frac{ab}{2}=\cancel{ab}+\cancel{cb}+ad+\cancel{cd}-\cancel{2}bc-\cancel{cd}-\cancel{ab}=ad-bc$
+\end_inset
+
+.
+ Če zamenjamo vrstni red vektorjev,
+ pa dobimo za predznak napačen rezultat,
+ torej je ploščina enaka 
+\begin_inset Formula $\left|\det\left[\begin{array}{cc}
+a & b\\
+c & d
+\end{array}\right]\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $3\times3$
+\end_inset
+
+ determinanta.
+\begin_inset Formula 
+\[
+\det\left[\begin{array}{ccc}
+a_{11} & a_{12} & a_{13}\\
+a_{21} & a_{22} & a_{23}\\
+a_{31} & a_{32} & a_{33}
+\end{array}\right]=a_{11}\det\left[\begin{array}{cc}
+a_{22} & a_{23}\\
+a_{32} & a_{33}
+\end{array}\right]-a_{12}\det\left[\begin{array}{cc}
+a_{21} & a_{23}\\
+a_{31} & a_{33}
+\end{array}\right]+a_{13}\left[\begin{array}{cc}
+a_{21} & a_{22}\\
+a_{31} & a_{32}
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a_{11}\left(a_{22}a_{33}-a_{23}a_{32}\right)-a_{12}\left(a_{21}a_{33}-a_{23}a_{31}\right)+a_{13}\left(a_{21}a_{32}-a_{22}a_{31}\right)
+\]
+
+\end_inset
+
+To si lahko zapomnimo s Saurusovim pravilom.
+ Pripišemo na desno stran prva dva stolpca in seštejemo produkte po šestih diagonalah.
+ Naraščajoče diagonale (tiste s pozitivnim koeficientom,
+ če bi jih risali kot premice v ravnini) prej negiramo.
+ Geometrijski 
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Vektorski produkt.
+ Velja:
+\begin_inset Formula 
+\[
+\left\langle \left(x,y,z\right),\left(a_{21},a_{22},a_{23}\right)\times\left(a_{31},a_{32},a_{33}\right)\right\rangle =\det\left[\begin{array}{ccc}
+x & y & z\\
+a_{21} & a_{22} & a_{23}\\
+a_{31} & a_{32} & a_{33}
+\end{array}\right],
+\]
+
+\end_inset
+
+torej je 
+\begin_inset Formula $\left[\left(a_{11},a_{12},a_{13}\right),\left(a_{21},a_{22},a_{23}\right),\left(a_{31},a_{32},a_{33}\right)\right]$
+\end_inset
+
+ (mešani produkt) determinanta matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\left|\det A\right|$
+\end_inset
+
+ ploščina paralelpipeda,
+ ki ga razpenjajo trije vrstični vektorji 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Računanje determinant
+\end_layout
+
+\begin_layout Standard
+Determinante računati po definiciji je precej zahtevno (bojda 
+\begin_inset Formula $O\left(n!\right)$
+\end_inset
+
+) za 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ determinanto.
+ Boljšo računsko zahtevnost dobimo z Gaussovo metodo.
+ Oglejmo si najprej posplošeno definicijo determinante:
+ 
+\begin_inset Quotes gld
+\end_inset
+
+razvoj po poljubni 
+\begin_inset Formula $i-$
+\end_inset
+
+ti vrstici
+\begin_inset Quotes grd
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\det A=\sum_{j=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij}
+\]
+
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+razvoj po poljubnem 
+\begin_inset Formula $j-$
+\end_inset
+
+tem stolpcu
+\begin_inset Quotes grd
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det A=\sum_{i=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij}
+\]
+
+\end_inset
+
+Ti dve formuli sta še vedno nepolinomske zahtevnosti,
+ uporabni pa sta v primerih,
+ ko imamo veliko ničel na kaki vrstici/stolpcu.
+ Determinanta zgornjetrikotne matrike je po tej formuli produkt diagonalcev.
+\end_layout
+
+\begin_layout Standard
+Kako pa se determinanta obnaša pri elementarnih vrstičnih transformacijah iz Gaussove metode?
+\end_layout
+
+\begin_layout Itemize
+menjava vrstic 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ determinanti se spremeni predznak
+\end_layout
+
+\begin_layout Itemize
+množenje vrstice z 
+\begin_inset Formula $\alpha\Longrightarrow$
+\end_inset
+
+ determinanta se pomnoži z 
+\begin_inset Formula $\alpha$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+prištevanje večkratnika ene vrstice k drugi 
+\begin_inset Formula $\Longrightarrow$
+\end_inset
+
+ determinanta se ne spremeni
+\end_layout
+
+\begin_layout Standard
+Časovna zahtevnost Gaussove metode je bojda polinomska 
+\begin_inset Formula $O\left(n^{3}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ideja dokaza veljavnosti Gaussove metode:
+ Indukcija po velikosti matrike.
+\end_layout
+
+\begin_layout Standard
+Baza:
+ 
+\begin_inset Formula $2\times2$
+\end_inset
+
+ matrike
+\end_layout
+
+\begin_layout Standard
+Korak:
+ Razvoj po vrstici,
+ ki je elementarna transformacija ne spremeni,
+ dobiš 
+\begin_inset Formula $n$
+\end_inset
+
+ 
+\begin_inset Formula $\left(n-1\right)\times\left(n-1\right)$
+\end_inset
+
+ determimant,
+ ki so veljavne po I.
+ P.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti determinante
+\end_layout
+
+\begin_layout Claim*
+Velja
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\det\left(AB\right)=\det A\det B$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\det A^{T}=\det A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\det\left[\begin{array}{cc}
+A & B\\
+0 & C
+\end{array}\right]=\det A\det C$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokazujemo tri trditve
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Dokazujemo 
+\begin_inset Formula $\det\left(AB\right)=\det A\det B$
+\end_inset
+
+.
+ Obravnavajmo dva posebna primera:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ je elementarna:
+ obrat pomeni množenje determinante z 
+\begin_inset Formula $-1$
+\end_inset
+
+,
+ množenje vrstice z 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ množi determinanto z 
+\begin_inset Formula $\alpha$
+\end_inset
+
+,
+ prištevanje večkratnika vrstice k drugi vrstici množi determinanto z 
+\begin_inset Formula $1$
+\end_inset
+
+.
+ Očitno torej trditev velja,
+ če je 
+\begin_inset Formula $A$
+\end_inset
+
+ elementarna.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A$
+\end_inset
+
+ ima ničelno vrstico:
+ tedaj ima tudi 
+\begin_inset Formula $AB$
+\end_inset
+
+ ničelno vrstico in je 
+\begin_inset Formula $\det A=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\det AB=0$
+\end_inset
+
+,
+ torej očitno trditev velja,
+ če ima 
+\begin_inset Formula $A$
+\end_inset
+
+ ničelno vrstico.
+\end_layout
+
+\begin_layout Standard
+Obravnavajmo še splošen primer:
+ Po Gaussovi metodi obstajajo take elementarne 
+\begin_inset Formula $E_{1},\dots,E_{n}$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $E_{n}\cdots E_{1}A=R$
+\end_inset
+
+ RVSO.
+ Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ kvadratna,
+ je tudi 
+\begin_inset Formula $R$
+\end_inset
+
+ kvadratna.
+ Ločimo dva primera:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $R=I$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\det\left(E_{n}\cdots E_{1}AB\right)=\det E_{n}\cdots\det E_{1}\det AB=\det\left(RB\right)=\det\left(IB\right)=\det B$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det I=\det R=\det E_{n}\cdots\det E_{1}\det A\quad\quad\quad\quad/\cdot\det B
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det I\det B=\det B=\det E_{n}\cdots\det E_{1}\det A\det B
+\]
+
+\end_inset
+
+
+\begin_inset Formula $\det B$
+\end_inset
+
+ zapišimo na dva načina ne levo in desno stran enačbe.
+ 
+\begin_inset Formula 
+\[
+\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det AB=\det A\det B
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Remark*
+\begin_inset Formula $\exists A^{-1}\Leftrightarrow\det A\not=0$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Predpostavimo 
+\begin_inset Formula $A$
+\end_inset
+
+ je obrnljiva.
+ Tedaj 
+\begin_inset Formula $\exists A^{-1}=B\ni:AB=I\overset{\circ\det}{\Longrightarrow}\det\left(AB\right)=\det I$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $\det A\not=0$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula $\det AB=\det B\det A=0\not=\det I=1$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Predpostavimo sedaj 
+\begin_inset Formula $A$
+\end_inset
+
+ ni obrnljiva.
+ Tedaj 
+\begin_inset Formula $\nexists A^{-1}\Rightarrow\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=R$
+\end_inset
+
+ ima ničelno vrstico.
+ Uporabimo isti razmislek kot spodaj,
+ torej 
+\begin_inset Formula $\det R=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
+\end_inset
+
+.
+ Ker so determinante elementarnih matrik vse neničelne,
+ mora biti 
+\begin_inset Formula $\det A$
+\end_inset
+
+ ničeln,
+ da je produkt ničeln.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $R$
+\end_inset
+
+ ima ničelno vrstico.
+ Tedaj 
+\begin_inset Formula $\det\left(R\right)=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
+\end_inset
+
+.
+ Ker so determinante elementarnih matrik vse neničelne,
+ mora biti 
+\begin_inset Formula $\det A$
+\end_inset
+
+ ničeln,
+ da je produkt ničeln.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Dokazujemo 
+\begin_inset Formula $\det A^{T}=\det A$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ elementarna matrika,
+ to drži:
+ 
+\begin_inset Formula $\det P_{ij}=-1=\det P_{ij}^{T}=\det P_{ij}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\det E_{i}\left(\alpha\right)=\alpha=\det E_{i}\left(\alpha\right)^{T}=\det E_{i}\left(\alpha\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\det E_{ij}\left(\alpha\right)=1=\det E_{ji}\left(\alpha\right)^{T}=\det E_{ij}\left(\alpha\right)^{T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Če ima 
+\begin_inset Formula $A$
+\end_inset
+
+ ničelno vrstico,
+ to drži,
+ saj ima tedaj 
+\begin_inset Formula $A^{T}$
+\end_inset
+
+ ničeln stolpec in 
+\begin_inset Formula $\det A=0=\det A^{T}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Splošen primer:
+ Po Gaussovi metodi 
+\begin_inset Formula $\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=\text{RVSO}\left(A\right)=R$
+\end_inset
+
+.
+ Zopet ločimo dva primera:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $R=I$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\det R=\det R^{T}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $R$
+\end_inset
+
+ ima ničelno vrstico.
+ 
+\begin_inset Formula $\det R=\det R^{T}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sedaj vemo,
+ da 
+\begin_inset Formula $\det R=\det R^{T}$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+\det R=\det R^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left(E_{n}\cdots E_{1}A\right)=\det\left(E_{n}\cdots E_{1}A\right)^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left(E_{n}\cdots E_{1}A\right)=\det\left(A^{T}E_{1}^{T}\cdots E_{n}^{T}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A=\det A^{T}\cancel{\det E_{1}^{T}}\cdots\cancel{\det E_{n}^{T}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det A=\det A^{T}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Enumerate
+Dokazujemo 
+\begin_inset Formula $\det\left[\begin{array}{cc}
+A & B\\
+0 & C
+\end{array}\right]=\det A\det C$
+\end_inset
+
+.
+ Levi izraz v enačbi vsebuje t.
+ i.
+ bločno matriko.
+ Upoštevamo poprej dokazano multiplikativnost determinante in opazimo,
+ da pri bločnem množenju matrik velja
+\begin_inset Formula 
+\[
+\left[\begin{array}{cc}
+A & B\\
+0 & C
+\end{array}\right]=\left[\begin{array}{cc}
+I & 0\\
+0 & C
+\end{array}\right]\left[\begin{array}{cc}
+A & B\\
+0 & I
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left[\begin{array}{cc}
+A & B\\
+0 & C
+\end{array}\right]=\det\left[\begin{array}{cc}
+I & 0\\
+0 & C
+\end{array}\right]\det\left[\begin{array}{cc}
+A & B\\
+0 & I
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left[\begin{array}{cc}
+A & B\\
+0 & C
+\end{array}\right]=\det C\det A
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Pojasnilo:
+ Za 
+\begin_inset Formula $\det C$
+\end_inset
+
+ si razpišemo bločno matriko,
+ za 
+\begin_inset Formula $\det A$
+\end_inset
+
+ si zopet razpišemo bločno matriko in nato z Gaussovimi transformacijami z enicami iz spodnjega desnega bloka izničimo zgornji desni blok (
+\begin_inset Formula $B$
+\end_inset
+
+).
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Cramerjevo pravilo —
+ eksplicitna formula za rešitve kvadratnega sistema linearnih enačb
+\end_layout
+
+\begin_layout Standard
+Radi bi dobili eksplicitne formule za komponente rešitve 
+\begin_inset Formula $x_{i}$
+\end_inset
+
+ kvadratnega sistema linearnih enačb 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+.
+ Izpeljimo torej eksplicitno formulo .
+ Druga/srednja matrika je identična,
+ v kateri smo 
+\begin_inset Formula $i-$
+\end_inset
+
+ti stolpec zamenjali z vektorjem spremenljivk 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ (to označimo z 
+\begin_inset Formula $I_{i}\left(\vec{x}\right)$
+\end_inset
+
+),
+ tretja/desna matrika pa je matrika koeficientov v kateri smo 
+\begin_inset Formula $i-$
+\end_inset
+
+ti stolpec zamenjali z vektorjem desnih strani 
+\begin_inset Formula $b$
+\end_inset
+
+ (to označimo z 
+\begin_inset Formula $A_{i}\left(\vec{x}\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{nn}
+\end{array}\right]\left[\begin{array}{ccccccc}
+1 &  & 0 & x_{1} &  &  & 0\\
+ & \ddots &  & \vdots\\
+ &  & 1 & x_{i-1}\\
+ &  &  & x_{i}\\
+ &  &  & x_{i+1} & 1\\
+ &  &  & \vdots &  & \ddots\\
+0 &  &  & x_{n} & 0 &  & 1
+\end{array}\right]=\left[\begin{array}{ccccc}
+a_{11} & \cdots & b_{1} & \cdots & a_{1n}\\
+\vdots & \ddots & \vdots & \iddots & \vdots\\
+a_{i1} &  & b_{i} &  & a_{in}\\
+\vdots & \iddots & \vdots & \ddots & \vdots\\
+a_{n1} & \cdots & b_{n} & \cdots & a_{nn}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{nn}
+\end{array}\right]\left[\begin{array}{ccccccc}
+1 &  & 0 & x_{1} &  &  & 0\\
+ & \ddots &  & \vdots\\
+ &  & 1 & x_{i-1}\\
+ &  &  & x_{i}\\
+ &  &  & x_{i+1} & 1\\
+ &  &  & \vdots &  & \ddots\\
+0 &  &  & x_{n} & 0 &  & 1
+\end{array}\right]=\left[\begin{array}{ccccc}
+a_{11} & \cdots & a_{11}x_{1}+\cdots+a_{1n}x_{n} & \cdots & a_{1n}\\
+\vdots & \ddots & \vdots & \iddots & \vdots\\
+a_{i1} &  & a_{i1}x_{1}+\cdots+a_{in}x_{n} &  & a_{in}\\
+\vdots & \iddots & \vdots & \ddots & \vdots\\
+a_{n1} & \cdots & a_{n1}x_{1}+\cdots+a_{nn}x_{n} & \cdots & a_{nn}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+AI_{i}\left(\vec{x}\right)=A_{i}\left(\vec{b}\right)\quad\quad\quad\quad/\det
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left(AI_{i}\left(\vec{x}\right)\right)=\det A_{i}\left(\vec{b}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det A\det I_{i}\left(\vec{x}\right)=\det A_{i}\left(\vec{b}\right)
+\]
+
+\end_inset
+
+Izračunamo 
+\begin_inset Formula $\det I_{i}\left(\vec{x}\right)$
+\end_inset
+
+ z razvojem po 
+\begin_inset Formula $i-$
+\end_inset
+
+ti vrstici.
+\begin_inset Formula 
+\[
+\det A\cdot x_{i}=\det A_{i}\left(\vec{b}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Formula-za-inverz-matrike"
+
+\end_inset
+
+Formula za inverz matrike
+\end_layout
+
+\begin_layout Standard
+Za dano obrnljivo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ iščemo eksplicitno formulo za celice 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $AX=I$
+\end_inset
+
+.
+ Ideja:
+ najprej bomo problem prevedli na reševanje sistemov linearnih enačb in uporabili Cramerjevo pravilo ter končno poenostavili formule.
+ Naj bodo 
+\begin_inset Formula $\vec{x_{1}},\dots,\vec{x_{n}}$
+\end_inset
+
+ stolpci 
+\begin_inset Formula $X$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{i_{1}},\dots,\vec{i_{n}}$
+\end_inset
+
+.
+ Potemtakem je 
+\begin_inset Formula $\left[\begin{array}{ccc}
+A\vec{x_{1}} & \cdots & A\vec{x_{n}}\end{array}\right]=A\left[\begin{array}{ccc}
+\vec{x_{1}} & \cdots & \vec{x_{n}}\end{array}\right]=AX=I=\left[\begin{array}{ccc}
+\vec{i_{1}} & \cdots & \vec{i_{n}}\end{array}\right]$
+\end_inset
+
+.
+ Primerjajmo sedaj stolpce na obeh straneh:
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :A\vec{x_{i}}=\vec{i_{i}}$
+\end_inset
+
+.
+ ZDB za vsak stolpec 
+\begin_inset Formula $X$
+\end_inset
+
+ smo dobili sistem 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ linearnih enačb.
+ Te sisteme 
+\begin_inset Formula $A\vec{x_{j}}=\vec{i_{j}}$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Tokrat uporabimo indeks 
+\begin_inset Formula $j$
+\end_inset
+
+,
+ ker z njim reprezentiramo stolpec in ponavadi,
+ ko govorimo o elementu 
+\begin_inset Formula $x_{ij}$
+\end_inset
+
+ matrike 
+\begin_inset Formula $X$
+\end_inset
+
+,
+ z 
+\begin_inset Formula $i$
+\end_inset
+
+ označimo vrstico.
+\end_layout
+
+\end_inset
+
+ bomo rešili s Cramerjevim pravilom.
+\begin_inset Formula 
+\[
+x_{ij}=\left(\vec{x}_{j}\right)_{i}=\frac{\det A_{i}\left(\vec{i_{j}}\right)}{\det A}\overset{\text{razvoj po \ensuremath{j-}ti vrstici}}{=}\frac{\det A_{ji}\cdot\left(-1\right)^{j+i}}{\det A}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+X=A^{-1}=\left[\begin{array}{ccc}
+\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\
+\vdots &  & \vdots\\
+\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
+\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc}
+\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\
+\vdots &  & \vdots\\
+\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n}
+\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T},
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+kjer 
+\begin_inset Formula $\tilde{A}$
+\end_inset
+
+ pravimo kofaktorska matrika.
+\end_layout
+
+\begin_layout Subsection
+Algebrske strukture
+\end_layout
+
+\begin_layout Subsubsection
+Uvod
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $M$
+\end_inset
+
+ neprazna množica.
+ Operacija na 
+\begin_inset Formula $M$
+\end_inset
+
+ pove,
+ kako iz dveh elementov 
+\begin_inset Formula $M$
+\end_inset
+
+ dobimo nov element 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ Na primer,
+ če 
+\begin_inset Formula $a,b\in M$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $a\circ b$
+\end_inset
+
+ nov element 
+\begin_inset Formula $M$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Operacija na 
+\begin_inset Formula $M$
+\end_inset
+
+ je funkcija 
+\begin_inset Formula $\circ:M\times M\to M$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $M\times M$
+\end_inset
+
+ kartezični produkt (urejeni pari).
+ 
+\begin_inset Formula $\left(a,b\right)\mapsto\circ\left(a,b\right)$
+\end_inset
+
+,
+ slednje pa označimo z 
+\begin_inset Formula $\circ\left(a,b\right)=a\circ b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Na isti množici imamo lahko več različno definiranih operacij.
+ Ločimo jih tako,
+ da uvedemo pojem grupoida.
+\end_layout
+
+\begin_layout Definition*
+Grupoid je 
+\begin_inset Formula $\left(\text{neprazna množica},\text{izbrana operacija }\circ:M\times M\to M\right)$
+\end_inset
+
+.
+ Na primer 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Še posebej nas zanimajo operacije z lepimi lastnostmi,
+ denimo asociativnost,
+ komutativnost,
+ obstoj enot,
+ inverzov.
+\end_layout
+
+\begin_layout Definition*
+Grupoid,
+ katerega 
+\begin_inset Formula $\circ$
+\end_inset
+
+ je asociativna 
+\begin_inset Formula $\Leftrightarrow\forall a,b,c\in M:\left(a\circ b\right)\circ c=a\circ\left(b\circ c\right)$
+\end_inset
+
+,
+ je polgrupa.
+ Tedaj skladnja dopušča pisanje brez oklepajev:
+ 
+\begin_inset Formula $a\circ b\circ c\circ d$
+\end_inset
+
+ je nedvoumen/veljaven izraz,
+ ko je 
+\begin_inset Formula $\circ$
+\end_inset
+
+ asociativna.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Komutativnost:
+ 
+\begin_inset Formula $\circ$
+\end_inset
+
+ je komutativna 
+\begin_inset Formula $\Leftrightarrow\forall a,b\in M:a\circ b=b\circ a$
+\end_inset
+
+.
+ Grupoidom s komutativno operacijo pravimo,
+ da so komutativni.
+\end_layout
+
+\begin_layout Example*
+Asociativni in komutativni grupoidi (komutativne polgrupe):
+ 
+\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{Q},\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{N},+\right)$
+\end_inset
+
+ —
+ številske operacije.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Asociativni,
+ a ne komutativni grupoidi (nekomutativne polgrupe):
+ 
+\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
+\end_inset
+
+ —
+ množenje matrik.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Komutativni,
+ a ne asociativni grupoidi:
+ Jordanski produkt matrik:
+ 
+\begin_inset Formula $A\circ B=\frac{1}{2}\left(AB+BA\right)$
+\end_inset
+
+\SpecialChar endofsentence
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Niti komutativni niti asociativni grupoidi:
+ Vektorski produkt v 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left(\mathbb{R}^{3},\times\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $M\not=\emptyset$
+\end_inset
+
+.
+ 
+\begin_inset Formula $F$
+\end_inset
+
+ naj bodo vse funkcije 
+\begin_inset Formula $M\to M$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\circ$
+\end_inset
+
+ pa kompozitum dveh funkcij.
+ Izkaže se,
+ da:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left(F,\circ\right)$
+\end_inset
+
+ je vedno polgrupa.
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Definicija kompozituma:
+ 
+\begin_inset Formula $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(f\circ g\right)\circ h\overset{?}{=}f\circ\left(g\circ h\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x:\left(\left(f\circ g\right)\circ h\right)\left(x\right)=\left(f\circ g\right)\left(h\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall x:\left(f\circ\left(g\circ h\right)\right)\left(x\right)=f\left(\left(g\circ h\right)\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Čim ima 
+\begin_inset Formula $M$
+\end_inset
+
+ vsaj tri elemente,
+ 
+\begin_inset Formula $\left(F,\circ\right)$
+\end_inset
+
+ ni komutativna.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ grupoid.
+ Element 
+\begin_inset Formula $e\in M$
+\end_inset
+
+ je enota,
+ če 
+\begin_inset Formula $\forall a\in M:e\circ a=a\wedge a\circ e=a$
+\end_inset
+
+.
+ Če velja le eno v konjunkciji,
+ je 
+\begin_inset Formula $e$
+\end_inset
+
+ bodisi leva bodisi desna enota (respectively) in v takem primeru 
+\begin_inset Formula $e$
+\end_inset
+
+ ni enota.
+\end_layout
+
+\begin_layout Example*
+Ali spodnji grupoidi imajo enoto in kakšna je?
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{R},+\right)$
+\end_inset
+
+:
+ enota je 0.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
+\end_inset
+
+:
+ enota je 1.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{N},+\right)$
+\end_inset
+
+:
+ ni enote,
+ kajti 
+\begin_inset Formula $0\not\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
+\end_inset
+
+:
+ enota je 
+\begin_inset Formula $I_{n\times n}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Vsak grupoid ima kvečjemu eno enoto.
+ Dve enoti v istem grupoidu sta enaki.
+ Še več:
+ vsaka leva enota je enaka vsaki desni enoti.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $e$
+\end_inset
+
+ leva enota in 
+\begin_inset Formula $f$
+\end_inset
+
+ desna enota,
+ torej 
+\begin_inset Formula $\forall a:e\circ a=a\wedge a\circ f=a$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $e\circ f=f$
+\end_inset
+
+ in 
+\begin_inset Formula $e\circ f=e$
+\end_inset
+
+.
+ Ker je vsaka leva enota vsaki desni,
+ sta poljubni enoti enaki.
+ Enota je,
+ če obstaja,
+ ena sama in je obenem edina leva in edina desna enota.
+\end_layout
+
+\begin_layout Example*
+Lahko se zgodi,
+ da obstaja poljubno različnih levih,
+ a nobene desne enote.
+ Primer so vse matrike oblike 
+\begin_inset Formula $\left[\begin{array}{cc}
+a & b\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Račun 
+\begin_inset Formula $\left[\begin{array}{cc}
+a & b\\
+0 & 0
+\end{array}\right]\cdot\left[\begin{array}{cc}
+c & d\\
+0 & 0
+\end{array}\right]=\left[\begin{array}{cc}
+ac & ad\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ pokaže,
+ da so vsi elementi 
+\begin_inset Formula $\left[\begin{array}{cc}
+1 & \times\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ leve enote.
+ Iz dejstva,
+ da je več (tu celo neskončno) levih enot,
+ sledi dejstvo,
+ da ni desnih.
+\end_layout
+
+\begin_layout Definition*
+Polgrupi z enoto pravimo monoid.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ monoid z enoto 
+\begin_inset Formula $e$
+\end_inset
+
+.
+ Inverz elementa 
+\begin_inset Formula $a\in M$
+\end_inset
+
+ je tak 
+\begin_inset Formula $b\in M\ni:b\circ a=e\wedge a\circ b=e$
+\end_inset
+
+.
+ Elementu,
+ ki zadošča levi strani konjunkcije,
+ pravimo levi inverz 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ elemetu,
+ ki zadošča desni strani konjunkcije,
+ pa desni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ je torej tak element,
+ ki je hkrati levi in desni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Ni nujno,
+ da ima vsak element monoida inverz.
+ Primer je 
+\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
+\end_inset
+
+;
+ niso vse matrike obrnljive.
+\end_layout
+
+\begin_layout Claim*
+Vsak element monoida ima kvečjemu en inverz.
+ Vsak levi inverz je enak vsakemu desnemu.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $b$
+\end_inset
+
+ levi in 
+\begin_inset Formula $c$
+\end_inset
+
+ desni inverz 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $b\circ a=e=a\circ c$
+\end_inset
+
+.
+ Računajmo:
+ 
+\begin_inset Formula $b=b\circ e=b\circ\left(a\circ c\right)=\left(b\circ a\right)\circ c=e\circ c=c$
+\end_inset
+
+.
+ Če obstaja,
+ je torej inverz en sam,
+ in ta je edini levi in edini desni inverz.
+\end_layout
+
+\begin_layout Definition*
+Ker vemo,
+ da je inverz enoličen,
+ lahko vpeljemo oznako 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+ za inverz elementa 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Ali v spodnjih monoidih obstajajo inverzi in kakšni so?
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{Z},+\right)$
+\end_inset
+
+:
+ inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ je 
+\begin_inset Formula $-a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{Z},\cdot\right)$
+\end_inset
+
+:
+ inverz 
+\begin_inset Formula $1$
+\end_inset
+
+ je 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ inverz 
+\begin_inset Formula $-1$
+\end_inset
+
+ je 
+\begin_inset Formula $-1$
+\end_inset
+
+,
+ ostali elementi pa inverza nimajo.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
+\end_inset
+
+:
+ inverz 
+\begin_inset Formula $a$
+\end_inset
+
+ je 
+\begin_inset Formula $\frac{1}{a}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Če desnega inverza ni,
+ je lahko levih inverzov več.
+ Primer:
+ Naj bodo 
+\begin_inset Formula $M$
+\end_inset
+
+ vse funkcije 
+\begin_inset Formula $\mathbb{N}\to\mathbb{N}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $\circ$
+\end_inset
+
+ kompozitum funkcij.
+ Tedaj velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $f\in M$
+\end_inset
+
+ ima levi inverz 
+\begin_inset Formula $\Leftrightarrow f$
+\end_inset
+
+ injektivna.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\in M$
+\end_inset
+
+ ima desni inverz 
+\begin_inset Formula $\Leftrightarrow f$
+\end_inset
+
+ surjektivna.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $f\in M$
+\end_inset
+
+ ima inverz 
+\begin_inset Formula $\Leftrightarrow f$
+\end_inset
+
+ bijektivna.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $f\left(n\right)=n+1$
+\end_inset
+
+ je injektivna,
+ a ne surjektivna.
+ Vsi za komponiranje levi inverzi 
+\begin_inset Formula $f$
+\end_inset
+
+ so funkcije oblike 
+\begin_inset Formula $g\left(x\right)=\begin{cases}
+x-1 & ;x>1\\
+\times & ;x=1
+\end{cases}$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $x$
+\end_inset
+
+ lahko slikajo v karkoli,
+ pa bo 
+\begin_inset Formula $\left(g\circ f\right)$
+\end_inset
+
+ še vedno funkcija identiteta.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+V 
+\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
+\end_inset
+
+ je vsak levi inverz tudi desni inverz.
+ To je res tudi za funkcije na končni množici,
+ toda ni res v splošnem.
+\end_layout
+
+\begin_layout Definition*
+Grupa je tak monoid,
+ v katerem ima vsak element inverz.
+ Daljše:
+ grupa je taka neprazna množica 
+\begin_inset Formula $G$
+\end_inset
+
+ z operacijo 
+\begin_inset Formula $\circ$
+\end_inset
+
+,
+ ki zadošča asociativnosti,
+ obstaja enota in za vsak element obstaja njegov inverz.
+ Grupi s komutativno operacijo pravimo Abelova grupa.
+\end_layout
+
+\begin_layout Example*
+Nekaj abelovih grup:
+ 
+\begin_inset Formula $\left(\mathbb{Z},+\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),+\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{R}^{n},+\right)$
+\end_inset
+
+.
+ Nekaj neabelovih grup:
+ 
+\family roman
+\series medium
+\shape up
+\size normal
+\emph off
+\nospellcheck off
+\bar no
+\strikeout off
+\xout off
+\uuline off
+\uwave off
+\noun off
+\color none
+vse obrnljive matrike fiksne dimenzije
+\family default
+\series default
+\shape default
+\size default
+\emph default
+\nospellcheck default
+\bar default
+\strikeout default
+\xout default
+\uuline default
+\uwave default
+\noun default
+\color inherit
+,
+ 
+\family roman
+\series medium
+\shape up
+\size normal
+\emph off
+\nospellcheck off
+\bar no
+\strikeout off
+\xout off
+\uuline off
+\uwave off
+\noun off
+\color none
+vse permutacije neprazne končne množice
+\family default
+\series default
+\shape default
+\size default
+\emph default
+\nospellcheck default
+\bar default
+\strikeout default
+\xout default
+\uuline default
+\uwave default
+\noun default
+\color inherit
+.
+\end_layout
+
+\begin_layout Subsubsection
+Podstrukture
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ grupoid.
+ Reciumi,
+ da je 
+\begin_inset Formula $N$
+\end_inset
+
+ neprazna podmnožica 
+\begin_inset Formula $M$
+\end_inset
+
+.
+ Pod temi pogoji se lahko zgodi,
+ da 
+\begin_inset Formula $\exists a,b\in N\ni:a\circ b\not\in N$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Oglejmo si grupoid 
+\begin_inset Formula $\left(\mathbb{Z},+\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $N\subseteq\mathbb{Z}$
+\end_inset
+
+ naj bodo liha cela števila.
+ 
+\begin_inset Formula $\forall a,b\in N:a+b\not\in N\Rightarrow\exists a,b\in N\ni:a+b\not\in N$
+\end_inset
+
+,
+ kajti vsota lihih števil je soda.
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da je podmnožica 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+ zaprta za 
+\begin_inset Formula $\circ$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall a,b\in N:a\circ b\in N$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Oglejmo si spet grupoid 
+\begin_inset Formula $\left(\mathbb{Z},+\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $N\subseteq\mathbb{Z}$
+\end_inset
+
+ naj bodo soda cela števila.
+ 
+\begin_inset Formula $N$
+\end_inset
+
+ je zaprta za 
+\begin_inset Formula $+$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Takemu 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ z implicitno podedovano operacijo (
+\begin_inset Formula $a\circ_{N}b=a\circ b$
+\end_inset
+
+) pravimo podgrupoid 
+\begin_inset Formula $\left(N,\circ_{N}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Exercise*
+Pokaži,
+ da je 
+\begin_inset Quotes gld
+\end_inset
+
+general linear
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A\not=0\right\} $
+\end_inset
+
+ grupa za matrično množenje.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Asociativnost je dokazana zgoraj.
+ Enota je 
+\begin_inset Formula $I_{n}$
+\end_inset
+
+.
+ Inverzi obstajajo,
+ ker so determinante neničelne in tudi inverzi imajo neničelne determinante.
+ Preveriti je treba še vsebovanost,
+ torej 
+\begin_inset Formula $\forall A,B\in GL_{n}\left(\mathbb{R}\right):A\cdot B\in GL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+ Vzemimo poljubni 
+\begin_inset Formula $A,B\in GL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\det A\not=0\wedge\det B\not=0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\det\left(AB\right)=\det A\det B=0\Leftrightarrow\det A=0\vee\det B=0$
+\end_inset
+
+,
+ toda ker noben izmed izrazov disjunkcije ne drži,
+ determinanta 
+\begin_inset Formula $AB$
+\end_inset
+
+ nikdar ni 0.
+ Enota 
+\begin_inset Formula $I$
+\end_inset
+
+ je vsebovana v 
+\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ saj 
+\begin_inset Formula $\det I=1\not=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Exercise*
+Ali je 
+\begin_inset Quotes gld
+\end_inset
+
+special linear
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A=1\right\} $
+\end_inset
+
+ grupa za matrično množenje?
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Vse lastnosti (razen vsebovanosti) smo preverili zgoraj.
+ Preveriti je treba vsebovanost,
+ torej ali 
+\begin_inset Formula $\forall A,B\in SL_{n}\left(\mathbb{R}\right):A\cdot B\in SL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+ Vzemimo poljubni 
+\begin_inset Formula $A,B\in SL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\det A=1\wedge\det B=1$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\det\left(AB\right)=\det A\det B=1\cdot1=1$
+\end_inset
+
+.
+ Preveriti je treba še,
+ da so inverzi vsebovani.
+ Za poljubno 
+\begin_inset Formula $A\in SL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+ je 
+\begin_inset Formula $\det A^{-1}=\frac{1}{\det A}=1$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $\det A=1$
+\end_inset
+
+.
+ Enota 
+\begin_inset Formula $I$
+\end_inset
+
+ je vsebovana v 
+\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ saj 
+\begin_inset Formula $\det I=1$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Za podedovano operacijo 
+\begin_inset Formula $\circ_{N}$
+\end_inset
+
+ v podstrukturi se asociativnost in komutativnost podedujeta,
+ ni pa nujno,
+ da če obstaja enota v 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+,
+ obstaja enota tudi v 
+\begin_inset Formula $\left(N,\circ_{N}\right)$
+\end_inset
+
+.
+ Prav tako ni rečeno,
+ da se podeduje obstoj inverzov.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ polgrupa (asociativen grupoid) in 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ pravimo,
+ da je 
+\begin_inset Formula $N$
+\end_inset
+
+ podpolgrupa,
+ če je zaprta za 
+\begin_inset Formula $\circ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:podmonoid"
+
+\end_inset
+
+Če je 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ monoid (polgrupa z enoto) in 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $N$
+\end_inset
+
+ podmonoid,
+ če je zaprt za 
+\begin_inset Formula $\circ$
+\end_inset
+
+ in vsebuje enoto iz 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ (prav tisto enoto,
+ glej primer 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "exa:nxnmonoid"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ spodaj).
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
+\end_inset
+
+ je monoid.
+ Soda števila so podpolgrupa (zaprta so za množenje),
+ niso pa podmonoid,
+ saj ne vsebujejo enice (enote).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example
+\begin_inset CommandInset label
+LatexCommand label
+name "exa:nxnmonoid"
+
+\end_inset
+
+
+\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
+\end_inset
+
+ je monoid za operacijo 
+\begin_inset Formula $\left(a,b\right)\circ\left(c,d\right)=\left(ac,bd\right)$
+\end_inset
+
+,
+ saj je enota 
+\begin_inset Formula $\left(1,1\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left(\mathbb{N}\times\left\{ 0\right\} ,\circ\right)$
+\end_inset
+
+ pa za 
+\begin_inset Formula $\circ$
+\end_inset
+
+ kot prej je sicer podpolgrupa v 
+\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
+\end_inset
+
+ in ima enoto 
+\begin_inset Formula $\left(1,0\right)$
+\end_inset
+
+,
+ vendar,
+ ker 
+\begin_inset Formula $\left(1,0\right)\not=\left(1,1\right)$
+\end_inset
+
+,
+ to ni podmonoid.
+ Enota mora torej biti,
+ kot pravi definicija 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "def:podmonoid"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+,
+ ista kot enota v 
+\begin_inset Quotes gld
+\end_inset
+
+starševski
+\begin_inset Quotes grd
+\end_inset
+
+ strukturi.
+\end_layout
+
+\begin_layout Definition*
+Če je 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ grupa in 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ pravimo,
+ da je 
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupa 
+\begin_inset Formula $\Longleftrightarrow$
+\end_inset
+
+ hkrati velja
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+je zaprta za 
+\begin_inset Formula $\circ$
+\end_inset
+
+,
+\end_layout
+
+\begin_layout Itemize
+vsebuje isto enoto kot 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+vsebuje inverz vsakega svojega elementa;
+ ti inverzi pa so itak po enoličnosti enaki inverzom iz 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+special linear,
+ 
+\begin_inset Formula $SL_{n}$
+\end_inset
+
+,
+ grupa vseh matrik z determinanto enako 1,
+ je podgrupa 
+\begin_inset Quotes gld
+\end_inset
+
+general linear
+\begin_inset Quotes grd
+\end_inset
+
+,
+ 
+\begin_inset Formula $GL_{n}$
+\end_inset
+
+,
+ grupe vseh obrnljivih 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matrik,
+ kajti 
+\begin_inset Formula $\det I=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\det$
+\end_inset
+
+ je multiplikativna (glej vajo zgoraj) in 
+\begin_inset Formula $\det A=1\Leftrightarrow\det A^{-1}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+ortogonalne matrike,
+ 
+\begin_inset Formula $O_{n}$
+\end_inset
+
+,
+ vse 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ ki zadoščajo 
+\begin_inset Formula $A^{T}A=I$
+\end_inset
+
+,
+ je podgrupa 
+\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ kajti:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Je zaprta:
+\begin_inset Formula 
+\[
+A,B\in O_{n}\overset{?}{\Longrightarrow}AB\in O_{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(AB\right)^{T}\left(AB\right)\overset{?}{=}I
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+B^{T}\left(A^{T}A\right)B\overset{?}{=}I
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+I=I
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Vsebuje enoto 
+\begin_inset Formula $I$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+I^{T}I=I
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Vsebuje inverze vseh svojih elementov:
+ Uporabimo 
+\begin_inset Formula $A^{T}A=I\Rightarrow A^{T}=A^{-1}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A\in O_{n}\overset{?}{\Longrightarrow}A^{-1}\in O_{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(A^{-1}\right)^{T}A^{-1}\overset{?}{=}I
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(A^{T}\right)^{T}A^{T}=AA^{T}=I
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+specialna ortogonalna grupa,
+ 
+\begin_inset Formula $SO_{n}\coloneqq O_{n}\cap SL_{n}$
+\end_inset
+
+ je podgrupa 
+\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+ Dokazati je moč še bolj splošno,
+ namreč,
+ da je presek dveh podgrup spet podgrupa.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+DOKAŽI?????
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+ grupa in 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+ neprazna.
+ Tedaj velja 
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupa 
+\begin_inset Formula $\Leftrightarrow\forall a,b\in N:a\circ b^{-1}\in N$
+\end_inset
+
+ (zaprtost za odštevanje —
+ v abelovih grupah namreč običajno operacijo označimo s 
+\begin_inset Formula $+$
+\end_inset
+
+ in označimo 
+\begin_inset Formula $a+b^{-1}=a-b$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupa v 
+\begin_inset Formula $\left(M,\circ\right)$
+\end_inset
+
+.
+ Vzemimo 
+\begin_inset Formula $a,b\in N$
+\end_inset
+
+.
+ Upoštevamo 
+\begin_inset Formula $b\in N\Rightarrow b^{-1}\in N$
+\end_inset
+
+ iz definicije podgrupe.
+ Torej velja 
+\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ b^{-1}\in N$
+\end_inset
+
+,
+ zopet iz definicije podgrupe.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Naj 
+\begin_inset Formula $\forall a,b\in N:a\circ b^{-1}\in N$
+\end_inset
+
+.
+ Preverimo lastnosti iz definicije podgrupe:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Vsebovanost enote:
+ Ker je 
+\begin_inset Formula $N$
+\end_inset
+
+ neprazna,
+ vsebuje nek 
+\begin_inset Formula $a$
+\end_inset
+
+.
+ Po predpostavki je 
+\begin_inset Formula $a\circ a^{-1}\in N$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\circ a^{-1}$
+\end_inset
+
+ pa je po definiciji inverza enota.
+\end_layout
+
+\begin_layout Itemize
+Vsebovanost inverzov:
+ Naj bo 
+\begin_inset Formula $a\in N$
+\end_inset
+
+ poljuben.
+ Od prej vemo,
+ da 
+\begin_inset Formula $e\in N$
+\end_inset
+
+.
+ Po predpostavki,
+ ker 
+\begin_inset Formula $e,a\in N\Rightarrow e\circ a^{-1}\in N$
+\end_inset
+
+,
+ 
+\begin_inset Formula $e\circ a^{-1}$
+\end_inset
+
+ pa je po definiciji enote 
+\begin_inset Formula $a^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Zaprtost:
+ Naj bosta 
+\begin_inset Formula $a,b\in N$
+\end_inset
+
+ poljubna.
+ Od prej vemo,
+ da 
+\begin_inset Formula $b^{-1}\in N$
+\end_inset
+
+.
+ Po predpostavki,
+ ker 
+\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ\left(b^{-1}\right)^{-1}\in N$
+\end_inset
+
+,
+ 
+\begin_inset Formula $a\circ\left(b^{-1}\right)^{-1}$
+\end_inset
+
+ pa je po definiciji inverza 
+\begin_inset Formula $a\circ b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Subsubsection
+Homomorfizmi
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\sim$
+\end_inset
+
+ so operacije,
+ ki 
+\begin_inset Quotes gld
+\end_inset
+
+ohranjajo strukturo
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
+\end_inset
+
+ dva grupoida.
+ Preslikava 
+\begin_inset Formula $f:M_{1}\to M_{2}$
+\end_inset
+
+ je homomorfizem grupoidov,
+ če 
+\begin_inset Formula $\forall a,b\in M_{1}:f\left(a\circ_{1}b\right)=f\left(a\right)\circ_{2}f\left(b\right)$
+\end_inset
+
+.
+ Enaka definicija v polgrupah.
+ Za homomorfizem monoidov zahtevamo še,
+ da 
+\begin_inset Formula $f\left(e_{1}\right)=e_{2}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $e_{1}$
+\end_inset
+
+ enota 
+\begin_inset Formula $M_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $e_{2}$
+\end_inset
+
+ enota 
+\begin_inset Formula $M_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$
+\end_inset
+
+,
+ ki slika 
+\begin_inset Formula $a\mapsto\left(a,0\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\circ_{1}$
+\end_inset
+
+ naj bo množenje,
+ 
+\begin_inset Formula $\circ_{2}$
+\end_inset
+
+ pa 
+\begin_inset Formula $\left(a,b\right)\circ_{2}\left(c,d\right)=\left(ac,bd\right)$
+\end_inset
+
+ (množenje po komponentah).
+ 
+\begin_inset Formula $\left(1,1\right)$
+\end_inset
+
+ je enota v 
+\begin_inset Formula $\mathbb{N\times\mathbb{N}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $1$
+\end_inset
+
+ pa je enota v 
+\begin_inset Formula $\mathbb{N}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je homomorfizem,
+ ker 
+\begin_inset Formula $f\left(a\circ_{1}b\right)=\left(a\cdot b,0\right)=\left(a,0\right)\circ_{2}\left(b,0\right)=f\left(a\right)\circ_{2}f\left(b\right)$
+\end_inset
+
+,
+ ni pa homomorfizem monoidov,
+ saj 
+\begin_inset Formula $f\left(1\right)=\left(1,0\right)\not=\left(1,1\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za homomorfizem grup zahtevamo še,
+ da 
+\begin_inset Formula $f\left(a^{-1}\right)=f\left(a\right)^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Izkaže se,
+ da ohranjanje enote in inverzov pri homomorfizmih grup sledi že iz definicije homomorfizmov grupoidov.
+\end_layout
+
+\begin_layout Claim*
+Naj bosta 
+\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
+\end_inset
+
+ grupi.
+ Naj bo 
+\begin_inset Formula $f:M_{1}\to M_{2}$
+\end_inset
+
+ preslikava,
+ ki je homomorfizem grupoidov.
+ Trdimo,
+ da slika enoto v enoto in inverze v inverze.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $e_{1}$
+\end_inset
+
+ enota za 
+\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $e_{2}$
+\end_inset
+
+ enota za 
+\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
+\end_inset
+
+.
+ Dokažimo,
+ da 
+\begin_inset Formula $f\left(e_{1}\right)\overset{?}{=}e_{2}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+f\left(e_{1}\right)=f\left(e_{1}\circ_{1}e_{1}\right)=f\left(e_{1}\right)\circ_{2}f\left(e_{1}\right)=f\left(e_{1}\right)^{-1}\circ f\left(e_{1}\right)\circ e_{2}=e_{2}\circ e_{2}=e_{2}
+\]
+
+\end_inset
+
+Dokažimo še ohranjanje inverzov,
+ se pravi 
+\begin_inset Formula $b$
+\end_inset
+
+ je inverz 
+\begin_inset Formula $a\overset{?}{\Longrightarrow}f\left(b\right)$
+\end_inset
+
+ je inverz 
+\begin_inset Formula $f\left(a\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+a\circ_{1}b=e_{1}\overset{?}{\Longrightarrow}f\left(a\right)\circ_{2}f\left(b\right)=f\left(a\circ_{1}b\right)=f\left(e_{1}\right)=e_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+b\circ_{1}a=e_{1}\overset{?}{\Longrightarrow}f\left(b\right)\circ_{2}f\left(a\right)=f\left(b\circ_{1}a\right)=f\left(e_{1}\right)=e_{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example
+\begin_inset CommandInset label
+LatexCommand label
+name "exa:primeri-homomorfizmov"
+
+\end_inset
+
+Primeri homomorfizmov.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Determinanta:
+ 
+\begin_inset Formula $M_{n}\left(\mathbb{R}\right)\to\mathbb{R}$
+\end_inset
+
+ je homomorfizem,
+ ker ima multiplikativno lastnost:
+ 
+\begin_inset Formula $\det\left(AB\right)=\det A\det B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:permutacijska-matrika"
+
+\end_inset
+
+
+\begin_inset Formula $S_{n}$
+\end_inset
+
+ so vse permutacije množice 
+\begin_inset Formula $\left\{ 1..n\right\} $
+\end_inset
+
+.
+ Vsaki permutaciji 
+\begin_inset Formula $\sigma\in S_{n}$
+\end_inset
+
+ priredimo permutacijsko matriko 
+\begin_inset Formula $P_{\sigma}\in M_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+ tako,
+ da vsebuje vektorje standardne baze 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ kot stolpce:
+\begin_inset Formula 
+\[
+P_{\sigma}\coloneqq\left[\begin{array}{ccc}
+\vec{e_{\sigma\left(1\right)}} & \cdots & \vec{e_{\sigma\left(n\right)}}\end{array}\right]
+\]
+
+\end_inset
+
+Imamo preslikavo 
+\begin_inset Formula $S\to M_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+,
+ ki slika 
+\begin_inset Formula $\sigma\mapsto P_{\sigma}$
+\end_inset
+
+ in trdimo,
+ da je homomorfizem.
+ Dokažimo,
+ da je 
+\begin_inset Formula $\forall\sigma,\tau\in S_{n}:P_{\sigma\circ\tau}=P_{\sigma}\cdot P_{\tau}$
+\end_inset
+
+.
+ Opazimo,
+ da je 
+\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{i}}=\vec{e_{\sigma\left(i\right)}}$
+\end_inset
+
+ (tu množimo matriko z vektorjem).
+ Če namesto 
+\begin_inset Formula $i$
+\end_inset
+
+ pišemo 
+\begin_inset Formula $\tau\left(i\right)$
+\end_inset
+
+,
+ dobimo 
+\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{\tau\left(i\right)}}=\vec{e_{\left(\sigma\circ\tau\right)\left(i\right)}}$
+\end_inset
+
+.
+ Preverimo sedaj množenje 
+\begin_inset Formula $P_{\sigma}P_{\tau}=P_{\sigma}\left[\begin{array}{ccc}
+\vec{e_{\tau\left(1\right)}} & \cdots & \vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
+P_{\sigma}\vec{e_{\tau\left(1\right)}} & \cdots & P_{\sigma}\vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
+\vec{e_{\left(\sigma\circ\tau\right)\left(1\right)}} & \cdots & \vec{e_{\left(\sigma\circ\tau\right)\left(n\right)}}\end{array}\right]=P_{\sigma\circ\tau}$
+\end_inset
+
+.
+ Preslikava je res homomorfizem.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Kompozitum dveh homomorfizmov je tudi sam zopet homomorfizem.
+\end_layout
+
+\begin_layout Proof
+Imejmo tri grupoide in homomorfizma,
+ ki slikata med njimi takole:
+ 
+\begin_inset Formula $\left(M_{1},\circ_{1}\right)\overset{f}{\longrightarrow}\left(M_{2},\circ_{2}\right)\overset{g}{\longrightarrow}\left(M_{3},\circ_{3}\right)$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ spet homorfizem.
+\begin_inset Formula 
+\[
+\left(g\circ f\right)\left(a\circ_{1}b\right)=g\left(f\left(a\circ_{1}b\right)\right)=g\left(f\left(a\right)\circ_{2}f\left(b\right)\right)=g\left(f\left(a\right)\right)\circ_{3}g\left(f\left(b\right)\right)=\left(g\circ f\right)\left(a\right)\circ_{3}\left(g\circ f\right)\left(b\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $S_{n}\overset{\sigma}{\longrightarrow}M_{n}\left(\mathbb{R}\right)\overset{\det}{\rightarrow}\mathbb{R}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sigma$
+\end_inset
+
+ preslikava iz točke 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:permutacijska-matrika"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ zgleda 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "exa:primeri-homomorfizmov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ zgoraj.
+ 
+\begin_inset Formula $\sgn=\det\circ\sigma$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\sgn$
+\end_inset
+
+ parnost permutacije.
+ Preslikava 
+\begin_inset Formula $\sgn$
+\end_inset
+
+ je homomorfizem,
+ ker je kompozitum dveh homomorfizmov.
+\end_layout
+
+\begin_layout Definition*
+Izomorfizem je preslikava,
+ ki je bijektivna in je homomorfizem.
+ Dve grupi sta izomorfni,
+ kadar med njima obstaja izomorfizem.
+\end_layout
+
+\begin_layout Remark*
+S stališča algebre sta dve izomorfni grupi v abstraktnem smislu enaki,
+ saj je izomorfizem zgolj reverzibilno preimenovanje elementov.
+\end_layout
+
+\begin_layout Subsubsection
+Bigrupoidi,
+ polkolobarji,
+ kolobarji
+\end_layout
+
+\begin_layout Definition*
+Neprazni množici 
+\begin_inset Formula $M$
+\end_inset
+
+ z dvema operacijama 
+\begin_inset Formula $\circ_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $\circ_{2}$
+\end_inset
+
+ pravimo bigrupoid in ga označimo z 
+\begin_inset Formula $\left(M,\circ_{1},\circ_{2}\right)$
+\end_inset
+
+.
+ Običajno operaciji označimo z 
+\begin_inset Formula $+,\cdot$
+\end_inset
+
+,
+ tedaj bigrupoid pišemo kot 
+\begin_inset Formula $\left(M,+,\cdot\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Quotation
+\begin_inset Quotes gld
+\end_inset
+
+Če 
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ ena z drugo nimata nobene zveze,
+ je vseeno,
+ če ju študiramo skupaj ali posebej.
+\begin_inset Quotes grd
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Distributivnost je značilnost bigrupoida 
+\begin_inset Formula $\left(M,+,\cdot\right)$
+\end_inset
+
+.
+ Ločimo levo distributivnost:
+ 
+\begin_inset Formula $\forall a,b,c\in M:a\cdot\left(b+c\right)=a\cdot b+a\cdot c$
+\end_inset
+
+ in desno distributivnost:
+ 
+\begin_inset Formula $\forall a,b,c\in M:\left(a+b\right)\cdot c=a\cdot c+b\cdot c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Bigrupoid,
+ ki zadošča levi in desni distributivnosti,
+ je distributiven.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Distributiven bigrupoid,
+ je polkolobar,
+ če je 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+ komutativna polgrupa.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Distributiven grupoid je kolobar,
+ če je 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+ komutativna grupa.
+\end_layout
+
+\begin_layout Example*
+Primer polkolobarja,
+ ki ni kolobar,
+ je 
+\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
+\end_inset
+
+.
+ Ni enote niti inverza za 
+\begin_inset Formula $+$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{N},+\right)$
+\end_inset
+
+ pa je polgrupa.
+\end_layout
+
+\begin_layout Standard
+Kolobarje delimo glede na lastnosti operacije 
+\begin_inset Formula $\cdot$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Definition*
+Asociativen kolobar je tak,
+ kjer je 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ asociativna operacija 
+\begin_inset Formula $\sim\left(M,\cdot\right)$
+\end_inset
+
+ je polgrupa.
+\end_layout
+
+\begin_layout Example*
+Primer kolobarja,
+ ki ni asociativen,
+ je 
+\begin_inset Formula $\left(\mathbb{R}^{3},+,\times\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\times$
+\end_inset
+
+ vektorski produkt.
+ Primer kolobarja,
+ ki je asociativen,
+ je 
+\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ matrično množenje.
+\end_layout
+
+\begin_layout Definition*
+Asociativen kolobar z enoto je tak,
+ ki ima multiplikativno enoto,
+ torej enoto za drugo operacijo 
+\begin_inset Formula $\sim\left(M,\cdot\right)$
+\end_inset
+
+ je monoid.
+ Tipično se enoto za 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ označi z 1,
+ enoto za 
+\begin_inset Formula $+$
+\end_inset
+
+ pa z 0.
+\end_layout
+
+\begin_layout Example*
+Primer asociativnega kolobarja brez enote je 
+\begin_inset Formula $\left(\text{soda }\mathbb{N},+,\cdot\right)$
+\end_inset
+
+.
+ Primer asociativnega kolobarja z enoto je 
+\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $b$
+\end_inset
+
+ je inverz 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $b\cdot a=e$
+\end_inset
+
+ in 
+\begin_inset Formula $a\cdot b=e$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $e$
+\end_inset
+
+ multiplikativna enota kolobarja.
+\end_layout
+
+\begin_layout Remark*
+Element 0 nima nikoli inverza,
+ ker 
+\begin_inset Formula $\forall a\in M:0\cdot a=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\cancel{0\cdot a}=\left(0+0\right)\cdot a=0\cdot a+\cancel{0\cdot a}$
+\end_inset
+
+ (dokaz velja za kolobarje,
+ ne pa polkolobarje,
+ ker imamo pravilo krajšanja
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Dokaz v mojih Odgovorih na vprašanja za ustni izpit Diskretnih struktur 2 IŠRM
+\end_layout
+
+\end_inset
+
+ le,
+ kadar je 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+ grupa).
+\end_layout
+
+\begin_layout Definition*
+Asociativen kolobar z enoto,
+ v katerem ima vsak neničen element inverz,
+ je obseg.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Kolobar je komutativen,
+ če je 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ komutativna operacija (
+\begin_inset Formula $+$
+\end_inset
+
+ je itak po definiciji že komutativna).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Komutativen obseg je polje.
+\end_layout
+
+\begin_layout Example*
+Primeri polj:
+ 
+\begin_inset Formula $\left(\mathbb{Q},+,\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(F\left[\mathbb{R}\right],+,\cdot\right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $F\left[\mathbb{R}\right]$
+\end_inset
+
+ polje racionalnih funkcij.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primer obsega,
+ ki ni polje:
+ 
+\begin_inset Formula $\left(\mathbb{H},+,\cdot\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Kvaternioni so 
+\begin_inset Formula $M_{2\times2}\left(\mathbb{R}\right)$
+\end_inset
+
+ take oblike:
+ za 
+\begin_inset Formula $\alpha,\beta\in\mathbb{C}$
+\end_inset
+
+ je 
+\begin_inset Formula $\mathbb{H}\coloneqq\left[\begin{array}{cc}
+\alpha & \beta\\
+-\overline{\beta} & \overline{\alpha}
+\end{array}\right]=\left[\begin{array}{cc}
+a+bi & c+di\\
+-c+di & a-bi
+\end{array}\right]=\left[\begin{array}{cc}
+1 & 0\\
+0 & 1
+\end{array}\right]a+\left[\begin{array}{cc}
+i & 0\\
+0 & -i
+\end{array}\right]b+\left[\begin{array}{cc}
+0 & 1\\
+-1 & 0
+\end{array}\right]c+\left[\begin{array}{cc}
+0 & i\\
+i & 0
+\end{array}\right]d=1a+bi+cj+dk$
+\end_inset
+
+ za 
+\begin_inset Formula $a,b,c,d\in\mathbb{R}$
+\end_inset
+
+ in dimenzije 
+\begin_inset Formula $1,i,j,k$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primer kolobarja:
+ Naj bo 
+\begin_inset Formula $X$
+\end_inset
+
+ neprazna množica in 
+\begin_inset Formula $R$
+\end_inset
+
+ kolobar.
+ 
+\begin_inset Formula $R^{X}$
+\end_inset
+
+ so vse funkcije 
+\begin_inset Formula $X\to R$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $f,g\in R^{X}$
+\end_inset
+
+.
+ Definirajmo operaciji:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $+$
+\end_inset
+
+ 
+\begin_inset Formula $f+g\coloneqq\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\cdot$
+\end_inset
+
+ 
+\begin_inset Formula $f\cdot g\coloneqq\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Podkolobarji
+\end_layout
+
+\begin_layout Definition*
+Podbigrupoid od 
+\begin_inset Formula $\left(M,+,\cdot\right)$
+\end_inset
+
+ je taka podmnožica 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ ki je zaprta za 
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+ je podgrupoid v 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(M,\cdot\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Podkolobar kolobarja 
+\begin_inset Formula $\left(M,+,\cdot\right)$
+\end_inset
+
+ je taka podmnožica 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupa v 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupoid v 
+\begin_inset Formula $\left(N,\cdot\right)\Leftrightarrow N$
+\end_inset
+
+ zaprta za 
+\begin_inset Formula $\cdot$
+\end_inset
+
+.
+ Skrajšana definicija je torej,
+ da je 
+\begin_inset Formula $\forall a,b\in N:a+b^{-1}\in N\wedge a\cdot b\in N$
+\end_inset
+
+,
+ torej zaprtost za odštevanje in množenje.
+\end_layout
+
+\begin_layout Example*
+Primeri podkolobarjev
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+v 
+\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+zgornjetrikotne matrike
+\end_layout
+
+\begin_layout Itemize
+diagonalne matrike
+\end_layout
+
+\begin_layout Itemize
+matrike s spodnjo vrstico ničelno
+\end_layout
+
+\begin_layout Itemize
+matrike z ničelnim 
+\begin_inset Formula $i-$
+\end_inset
+
+tim stolpcem
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $M_{n}\left(\mathbb{Z}\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $M_{n}\left(\mathbb{Q}\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+matrike oblike 
+\begin_inset Formula $\left[\begin{array}{cc}
+a & b\\
+b & a
+\end{array}\right]$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+v 
+\begin_inset Formula $\left(\mathbb{R}^{[a,b]},+,\cdot\right)$
+\end_inset
+
+ (vse funkcije 
+\begin_inset Formula $\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ za seštevanje in množenje)
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+vse omejene funkcije
+\end_layout
+
+\begin_layout Itemize
+vse zvezne funkcije
+\end_layout
+
+\begin_layout Itemize
+vse odvedljive funkcije
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Definition*
+Podobseg obsega 
+\begin_inset Formula $\left(M,+,\cdot\right)$
+\end_inset
+
+ je taka 
+\begin_inset Formula $N\subseteq M$
+\end_inset
+
+,
+ da velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $N$
+\end_inset
+
+ podgrupa v 
+\begin_inset Formula $\left(M,+\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $N\setminus\left\{ 0\right\} $
+\end_inset
+
+ podgrupa v 
+\begin_inset Formula $\left(M\setminus\left\{ 0\right\} ,\cdot\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+ZDB:
+ 
+\begin_inset Formula $N$
+\end_inset
+
+ je zaprta za odštevanje (seštevanje z aditivnim inverzom) in za deljenje (množenje z multiplikativnim inverzom) z neničelnimi elementi.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primeri podobsegov:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je podobseg v 
+\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ je podobseg v 
+\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Izkaže se,
+ da je najmanjše podpolje v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+,
+ ki vsebuje 
+\begin_inset Formula $\mathbb{Q}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sqrt{3}$
+\end_inset
+
+ množica 
+\begin_inset Formula $\left\{ a+b\sqrt{3};\forall a,b\in\mathbb{Q}\right\} $
+\end_inset
+
+.
+ Očitno je zaprt za odštevanje.
+ Za deljenje?
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\frac{a+b\sqrt{3}}{c+d\sqrt{3}}=\frac{\left(a+b\sqrt{3}\right)\left(c-d\sqrt{3}\right)}{\left(c+d\sqrt{3}\right)\left(c-d\sqrt{3}\right)}=\frac{ac-ad\sqrt{3}+bc\sqrt{3}-3bd}{c^{2}-3d^{2}}=\frac{ac-3bd+\left(bc-ad\right)\sqrt{3}}{c^{2}-3d^{2}}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{ac-3bd}{c^{2}-3d^{2}}+\frac{bc-ad}{c^{2}-3d^{2}}\sqrt{3}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Homomorfizmi kolobarjev
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $\left(M_{1},+_{1},\cdot_{1}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(M_{2},+_{2},\cdot_{2}\right)$
+\end_inset
+
+ kolobarja.
+ 
+\begin_inset Formula $f:M_{1}\to M_{2}$
+\end_inset
+
+ je homomorfizem kolobarjev 
+\begin_inset Formula $\Leftrightarrow\forall a,b\in M_{1}:f\left(a+_{1}b\right)=f\left(a\right)+_{2}f\left(b\right)\wedge f\left(a\cdot_{1}b\right)=f\left(a\right)\cdot_{2}f\left(b\right)$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $f$
+\end_inset
+
+ mora biti homomorfizem grupoidov 
+\begin_inset Formula $\left(M_{1},+_{1}\right)\to\left(M_{2},+_{2}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(M_{1},\cdot_{1}\right)\to\left(M_{2},\cdot_{2}\right)$
+\end_inset
+
+.
+ Za homomorfizem kolobarjev z enoto zahtevamo še 
+\begin_inset Formula $f\left(1_{1}\right)=1_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f:M_{2}\left(\mathbb{R}\right)\to M_{3}\left(\mathbb{R}\right)$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\left[\begin{array}{cc}
+a & b\\
+c & d
+\end{array}\right]\mapsto\left[\begin{array}{ccc}
+a & b & 0\\
+c & d & 0\\
+0 & 0 & 0
+\end{array}\right]$
+\end_inset
+
+ je homomorfizem kolobarjev,
+ ni pa homomorfizem kolobarjev z enoto,
+ kajti 
+\begin_inset Formula $f\left(\left[\begin{array}{cc}
+1 & 0\\
+0 & 1
+\end{array}\right]\right)=\left[\begin{array}{ccc}
+1 & 0 & 0\\
+0 & 1 & 0\\
+0 & 0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ kar ni enota v 
+\begin_inset Formula $M_{3}\left(\mathbb{R}\right)$
+\end_inset
+
+ (
+\begin_inset Formula $I_{3}$
+\end_inset
+
+) za implicitni operaciji 
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $g:M_{n}\left(\mathbb{R}\right)\to M_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+ ki slika 
+\begin_inset Formula $A\mapsto S^{-1}AS$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $S$
+\end_inset
+
+ neka fiksna obrnljiva matrika v 
+\begin_inset Formula $M_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+.
+ Uporabimo implicitni operaciji 
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ za matrike.
+ Računa 
+\begin_inset Formula $g\left(A+B\right)=S^{-1}\left(A+B\right)S=S^{-1}AS+S^{-1}BS=g\left(A\right)+g\left(B\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(AB\right)=S^{-1}ABS=S^{-1}AIBS=S^{-1}ASS^{-1}BS=g\left(A\right)g\left(B\right)$
+\end_inset
+
+ pokažeta,
+ da je 
+\begin_inset Formula $g$
+\end_inset
+
+ homomorfizem kolobarjev,
+ celo z enoto,
+ kajti 
+\begin_inset Formula $g\left(I\right)=S^{-1}IS=S^{-1}S=I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $h:\mathbb{C}\to M_{n}\left(\mathbb{R}\right)$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\alpha+\beta i\to\left[\begin{array}{cc}
+\alpha & \beta\\
+-\beta & \alpha
+\end{array}\right]$
+\end_inset
+
+ je homomorfizem kolobarjev z enoto.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Kolobar ostankov 
+\begin_inset Formula $\mathbb{Z}_{n}\coloneqq\left\{ 0..\left(n-1\right)\right\} $
+\end_inset
+
+ je asociativni kolobar z enoto.
+ Če je 
+\begin_inset Formula $p$
+\end_inset
+
+ praštevilo,
+ pa je celo 
+\begin_inset Formula $\mathbb{Z}_{p}$
+\end_inset
+
+ polje za implicitni operaciji seštevanje in množenja po modulu.
+\end_layout
+
+\begin_layout Subsection
+Vektorski prostori
+\end_layout
+
+\begin_layout Standard
+Ideja:
+ Vektorski prostor je Abelova grupa z dodatno strukturo —
+ množenje s skalarjem.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(F,+,\cdot\right)$
+\end_inset
+
+ polje.
+ Vektorski prostor z operacijama 
+\begin_inset Formula $V+V\to V$
+\end_inset
+
+ in 
+\begin_inset Formula $F\cdot V\to V$
+\end_inset
+
+ nad 
+\begin_inset Formula $F$
+\end_inset
+
+ je taka 
+\begin_inset Formula $\left(V,+,\cdot\right)$
+\end_inset
+
+,
+ da velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left(V,+\right)$
+\end_inset
+
+ je abelova grupa:
+ komutativnost,
+ asociativnost,
+ enota,
+ aditivni inverzi
+\end_layout
+
+\begin_layout Enumerate
+Lastnosti množenja s skalarjem.
+ 
+\begin_inset Formula $\forall\alpha,\beta\in F,a,b\in V:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\alpha\left(a+b\right)=\alpha a+\alpha b$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\alpha+\beta\right)a=\alpha a+\beta a$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\alpha\cdot\beta\right)\cdot a=\alpha\cdot\left(\beta\cdot a\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $1\cdot a=a$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Alternativna abstraktna formulacija aksiomov množenja s skalarjem se glasi:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\forall\alpha\in F$
+\end_inset
+
+ priredimo preslikavo 
+\begin_inset Formula $\varphi_{\alpha}:V\to V$
+\end_inset
+
+,
+ ki pošlje 
+\begin_inset Formula $v\mapsto\alpha v$
+\end_inset
+
+.
+ Štiri zgornje aksiome množenja s skalarjem sedaj označimo z abstraktnimi formulacijami:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\varphi_{\alpha}\left(a+b\right)\overset{\text{def.}}{=}\alpha\left(a+b\right)=\alpha b+\alpha b\overset{\text{def.}}{=}\varphi_{\alpha}\left(a\right)+\varphi_{\alpha}\left(b\right)$
+\end_inset
+
+ —
+ vidimo,
+ da je 
+\begin_inset Formula $\varphi_{\alpha}$
+\end_inset
+
+ homomorfizem iz 
+\begin_inset Formula $\left(V,+\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $\left(V,+\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\varphi_{\alpha+\beta}\left(a\right)\overset{\text{def.}}{=}\left(\alpha+\beta\right)a=\alpha a+\beta a\overset{\text{def.}}{=}\varphi_{\alpha}a+\varphi_{\beta}a$
+\end_inset
+
+ —
+ torej 
+\begin_inset Formula $\varphi_{\alpha+\beta}=\varphi_{\alpha}+\varphi_{\beta}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\varphi_{\alpha\beta}a\overset{\text{def.}}{=}\left(\alpha\beta\right)a=\alpha\left(\beta a\right)\overset{\text{def.}}{=}\varphi_{\alpha}\left(\varphi_{\beta}\left(a\right)\right)=\left(\varphi_{\alpha}\circ\varphi_{\beta}\right)\left(a\right)$
+\end_inset
+
+ —
+ torej 
+\begin_inset Formula $\varphi_{\alpha\beta}=\varphi_{\alpha}\circ\varphi_{\beta}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\varphi_{1}a\overset{\text{def.}}{=}1a=a$
+\end_inset
+
+ —
+ torej 
+\begin_inset Formula $\varphi_{1}=id$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO ALTERNATIVNA DEFINICIJA VEKTORSKEGA PROSTORA Z GRUPO ENDOMORFIZMOV
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Remark*
+Če v definiciji vektorskega prostora zamenjamo polje 
+\begin_inset Formula $F$
+\end_inset
+
+ s kolobarjem 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ dobimo definicijo 
+\series bold
+modula 
+\series default
+nad 
+\begin_inset Formula $F$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vektorskih prostorov:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+standarden primer:
+ naj bo 
+\begin_inset Formula $F$
+\end_inset
+
+ pojle in 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V=F^{n}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $+$
+\end_inset
+
+ seštevanje po komponentah in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ množenje s skalarjem po komponentah.
+ Pod temi pogoji je 
+\begin_inset Formula $\left(V,+,\cdot\right)$
+\end_inset
+
+ vektorski prostor —
+ ustreza vsem osmim aksiomom.
+\end_layout
+
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $F$
+\end_inset
+
+ polje in 
+\begin_inset Formula $n,m\in\mathbb{N}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $V\coloneqq M_{m,n}\left(\mathbb{F}\right)=m\times n$
+\end_inset
+
+ matrike nad 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ 
+\begin_inset Formula $+$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ definiramo kot pri matrikah.
+\end_layout
+
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $F$
+\end_inset
+
+ polje,
+ 
+\begin_inset Formula $S\not=\emptyset$
+\end_inset
+
+ množica.
+ Naj bo 
+\begin_inset Formula $V\coloneqq F^{S}$
+\end_inset
+
+ (vse funkcije 
+\begin_inset Formula $S\to F$
+\end_inset
+
+).
+ Naj bosta 
+\begin_inset Formula $\varphi,\tau:S\to F$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $\forall s\in S$
+\end_inset
+
+ operaciji 
+\begin_inset Formula $\left(\varphi+\tau\right)\left(s\right)=\varphi\left(s\right)+\tau\left(s\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(\varphi\cdot\tau\right)\left(s\right)=\varphi\left(s\right)\cdot\tau\left(s\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor.
+ Ta definicija je podobna kot definiciji z 
+\begin_inset Formula $n-$
+\end_inset
+
+terico elementov polja,
+ saj lahko 
+\begin_inset Formula $n-$
+\end_inset
+
+terico identificiramo s funkcijo 
+\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} \to F$
+\end_inset
+
+,
+ toda ta primer dovoli neskončno razsežne vektorske prostore,
+ saj 
+\begin_inset Formula $S$
+\end_inset
+
+ ni nujno končna,
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+terica pa nekako implicitno je,
+ saj 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Polinomi.
+ Naj bo 
+\begin_inset Formula $V\coloneqq F\left[x\right]$
+\end_inset
+
+ (polinomi v spremenljivki 
+\begin_inset Formula $x$
+\end_inset
+
+ s koeficienti v 
+\begin_inset Formula $F$
+\end_inset
+
+).
+ Seštevanje definirajmo po komponentah:
+ 
+\begin_inset Formula $\left(\alpha+\beta x+\gamma x^{2}\right)+\left(\pi+\tau x\right)=\left(\alpha+\pi+\left(\beta+\tau\right)x+\gamma x^{2}\right)$
+\end_inset
+
+,
+ množenje s skalarjem pa takole:
+ 
+\begin_inset Formula $\alpha\left(a+bx+cx^{2}\right)=\alpha a+\alpha bx+\alpha cx^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Naj bosta 
+\begin_inset Formula $V_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $V_{2}$
+\end_inset
+
+ dva vektorska prostora nad istim poljem 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Tvorimo nov vektorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ ki mu pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+direktna vsota
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $V_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $V_{2}$
+\end_inset
+
+ in ga označimo z 
+\begin_inset Formula $V_{1}\oplus V_{2}\coloneqq$
+\end_inset
+
+
+\begin_inset Formula $\left\{ \left(v_{1},v_{2}\right);\forall v_{1}\in V_{1},v_{2}\in V:2\right\} $
+\end_inset
+
+.
+ Seštevamo po komponentah:
+ 
+\begin_inset Formula $\left(v_{1},v_{2}\right)+\left(v_{1}',v_{2}'\right)=\left(v_{1}+v_{1}',v_{2}+v_{2}'\right)$
+\end_inset
+
+,
+ s skalarjem pa množimo prvi komponento:
+ 
+\begin_inset Formula $\forall\alpha\in F:\alpha\left(v_{1},v_{2}\right)=\left(\alpha v_{1},v_{2}\right)$
+\end_inset
+
+.
+ Definicijo lahko posplošimo na 
+\begin_inset Formula $n$
+\end_inset
+
+ vektorskih prostorov.
+ Tedaj so elementi prostora urejene 
+\begin_inset Formula $n-$
+\end_inset
+
+terice.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Podprostori vekrorskih prostorov —
+ vektorski podprostori
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $\left(V,+,\cdot\right)$
+\end_inset
+
+ vektorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Vektorski podprostor je taka neprazna podmnožica 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ ki je zaprta za seštevanje in množenje s skalarjem.
+ Natančneje:
+ 
+\begin_inset Formula $\left(W,+,\cdot\right)$
+\end_inset
+
+ je vektorski podprostor 
+\begin_inset Formula $\left(V,+,\cdot\right)\Longleftrightarrow$
+\end_inset
+
+ velja hkrati:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $W\subseteq V$
+\end_inset
+
+ in 
+\begin_inset Formula $W\not=\emptyset$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:zaprtost+"
+
+\end_inset
+
+
+\begin_inset Formula $\forall a,b\in W:a+b\in W$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:zaprtostskalar"
+
+\end_inset
+
+
+\begin_inset Formula $\forall a\in W,\alpha\in F:\alpha a\in W$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Lastnosti 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:zaprtost+"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ in 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:zaprtostskalar"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ je moč združiti v eno:
+ 
+\begin_inset Formula $\forall a_{i},a_{2}\in W,\alpha_{1},\alpha_{2}\in F:\alpha_{1}a_{1}+\alpha_{2}a_{2}\in W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Z drugimi besedami je vektorski podprostor taka podmnožica,
+ ki vsebuje vse linearne kombinacije svojih elementov.
+ Odštevanje 
+\begin_inset Formula $a-b$
+\end_inset
+
+ je poseben primer linearne kombinacije,
+ kajti 
+\begin_inset Formula $a_{1}-a_{2}=1a_{1}+\left(-1\right)a_{2}$
+\end_inset
+
+.
+ Sledi,
+ da mora biti 
+\begin_inset Formula $\left(W,+\right)$
+\end_inset
+
+ podgrupa 
+\begin_inset Formula $\left(V,+\right)$
+\end_inset
+
+,
+ torej taka podmnožica 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ ki je zaprta za odštevanje.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Primeri vektorskih podprostorov:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $V=\mathbb{R}^{2}$
+\end_inset
+
+ (ravnina).
+ Vsi vektorski podprostori 
+\begin_inset Formula $V$
+\end_inset
+
+ so premice,
+ ki gredo skozi izhodišče,
+ izhodišče samo in cela ravnina.
+ Slednja sta t.
+ i.
+ trivialna podprostora.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+\begin_inset Formula $\forall\left(V,+,\cdot\right)$
+\end_inset
+
+ vektorski prostor 
+\begin_inset Formula $:\left\{ 0\right\} ,V$
+\end_inset
+
+ sta vektorska podprostora.
+ Imenujemo ju trivialna vektorska podprostora.
+\end_layout
+
+\begin_layout Claim*
+Vsak podprostor vsebuje aditivno enoto 0.
+\end_layout
+
+\begin_layout Proof
+Po definiciji je vsak vektorski podprostor neprazen,
+ torej 
+\begin_inset Formula $\exists w\in W$
+\end_inset
+
+.
+ Polje gotovo vsebuje aditivno enoto 0,
+ torej po aksiomu 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:zaprtostskalar"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ za podprostore sledi 
+\begin_inset Formula $0\cdot w\in W$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $0\cdot w\overset{?}{=}0$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\cancel{0\cdot w}=\left(0+0\right)\cdot w=0\cdot w+\cancel{0\cdot w}$
+\end_inset
+
+ (pravilo krajšanja v grupi),
+ torej 
+\begin_inset Formula $0=0\cdot w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Množica rešitev homogene (desna stran je 0) linearne enačbe je vselej vektorski podprostor.
+\end_layout
+
+\begin_layout Proof
+Imamo 
+\begin_inset Formula $\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n}=0$
+\end_inset
+
+.
+ Če sta 
+\begin_inset Formula $\vec{a}=\left(a_{1},\dots,a_{n}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{n}\right)$
+\end_inset
+
+ rešitvi,
+ velja 
+\begin_inset Formula $\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}=0$
+\end_inset
+
+.
+ Vzemimo poljubna 
+\begin_inset Formula $\alpha,\beta\in F$
+\end_inset
+
+ in si oglejmo 
+\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\alpha\left(\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}\right)+\beta\left(\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha_{1}\left(\alpha a_{1}+\beta b_{1}\right)+\cdots+\alpha_{n}\left(\alpha a_{n}+\beta b_{n}\right)=0
+\]
+
+\end_inset
+
+Vzemimo koeficiente v oklepajih pred 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+ v enačbi pred to vrstico in jih zložimo v vektor.
+ Tedaj je 
+\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}=\left(\alpha a_{1}+\beta b_{1},\dots,\alpha a_{n}+\beta b_{n}\right)$
+\end_inset
+
+ spet rešitev homogene linearne enačbe.
+ Ker je linearna kombinacija elementov vektorskega podprostora spet element vektorskega podprostora,
+ je po definiciji množica rešitev homogene linearne enačbe res vselej vektorski podprostor.
+\end_layout
+
+\begin_layout Remark*
+Podoben računa velja tudi za množico rešitev sistema linearnih enačb,
+ kar sicer sledi tudi iz naslednje trditve.
+\end_layout
+
+\begin_layout Claim*
+Presek dveh podprostorov je tudi sam spet podprostor.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $W_{1},W_{2}$
+\end_inset
+
+ podprostora v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $W_{1}\cap V_{2}$
+\end_inset
+
+ spet podprostor.
+ Vzemimo poljubna 
+\begin_inset Formula $a,b\in W_{1}\cap W_{2}$
+\end_inset
+
+ in poljubna 
+\begin_inset Formula $\alpha,\beta\in F$
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
+\end_inset
+
+.
+ Vemo,
+ da 
+\begin_inset Formula $a,b\in W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $a,b\in W_{2}$
+\end_inset
+
+.
+ Ker je podprostor po definiciji zaprt za linearne kombinacije svojih elementov,
+ je 
+\begin_inset Formula $\alpha a+\beta b\in W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha a+\beta b\in W_{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
+\end_inset
+
+,
+ torej je presek podprostorov res zaprt za LK svojih elementov in je s tem tudi sam podprostor.
+\end_layout
+
+\begin_deeper
+\begin_layout Remark*
+Slednji dokaz lahko očitno posplošimo na več podprostorov.
+ Presek nikdar ni prazen,
+ saj vsi podprostori vsebujejo aditivno enoto 0 (dokaz za to je malce višje).
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Vsota-podprostorov"
+
+\end_inset
+
+Vsota podprostorov
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $W_{2}$
+\end_inset
+
+ podprostora v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Vsoto podprostorov 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $W_{2}$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $W_{1}+W_{2}=\left\{ w_{1}+w_{w};\forall w_{1}\in W_{1},w_{2}\in W_{2}\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Vsota podprostorov je tudi sama spet podprostor.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $a,b\in W_{1}+W_{2}$
+\end_inset
+
+ poljubna.
+ Tedaj po definiciji 
+\begin_inset Formula $a=a_{1}+a_{2}$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $a_{1}\in W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{2}\in W_{2}$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $b=b_{1}+b_{2}$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $b_{1}\in W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $b_{2}\in W_{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall\alpha,\beta\in F$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\alpha a+\beta b=\alpha\left(a_{1}+a_{2}\right)+\beta\left(b_{1}+b_{2}\right)=\alpha a_{1}+\alpha a_{2}+\beta b_{1}+\beta b_{2}=\left(\alpha a_{1}+\beta b_{1}\right)+\left(\alpha a_{2}+\beta b_{2}\right)\in W_{1}+W_{2},
+\]
+
+\end_inset
+
+kajti 
+\begin_inset Formula $\left(\alpha a_{1}+\beta b_{1}\right)\in W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(\alpha a_{2}+\beta b_{2}\right)\in W_{2}$
+\end_inset
+
+,
+ saj sta to linearni kombinaciji elementov prostorov.
+ Njuna vsota pa je element 
+\begin_inset Formula $W_{1}+W_{2}$
+\end_inset
+
+ po definiciji vsote podprostorov.
+\end_layout
+
+\begin_layout Subsubsection
+Baze
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad poljem 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ je baza,
+ če je LN in če je ogrodje.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ je LN,
+ če za vsake 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
+\end_inset
+
+,
+ ki zadoščajo 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
+\end_inset
+
+ velja 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
+\end_inset
+
+.
+ Ekvivalentni definiciji LN:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ je LN 
+\begin_inset Formula $\Leftrightarrow\forall v\in V$
+\end_inset
+
+ se da kvečjemu na en način izraziti kot linearno kombinacijo 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ je LN 
+\begin_inset Formula $\Leftrightarrow\nexists v\in\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+,
+ da bi se ga dalo izraziti kot LK preostalih elementov.
+\end_layout
+
+\begin_layout Standard
+Dokaz ekvivalentnosti teh definicij je enak tistemu za 
+\begin_inset Formula $V=\mathbb{R}^{n}$
+\end_inset
+
+ višje.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ je ogrodje 
+\begin_inset Formula $\Leftrightarrow\forall v\in V$
+\end_inset
+
+ se da na vsaj en način izraziti kot LK te množice 
+\begin_inset Formula $\Leftrightarrow\Lin\left\{ v_{1},\dots,v_{n}\right\} =V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri baz:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+standardna baza:
+ Naj bo 
+\begin_inset Formula $V=F^{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $v_{1}=\left(1,0,0,\dots,0,0\right)$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v_{2}=\left(0,1,0,\dots,0,0\right)$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $v_{n}=\left(0,0,0,\dots,0,1\right)$
+\end_inset
+
+.
+ Da je 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} \subseteq F^{n}$
+\end_inset
+
+ res baza,
+ preverimo z determinanto (
+\begin_inset Formula $\det A\not=0\Leftrightarrow\exists A^{-1}\Leftrightarrow$
+\end_inset
+
+ stolpci so baza prostora):
+\begin_inset Formula 
+\[
+\det\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]=0\Leftrightarrow\left\{ v_{1},\dots,v_{n}\right\} \text{ \textbf{ni} baza}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+baze v 
+\begin_inset Formula $F\left[x\right]_{<n}$
+\end_inset
+
+ (polinomi stopnje,
+ manjše od 
+\begin_inset Formula $n$
+\end_inset
+
+)
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+standardna baza:
+ 
+\begin_inset Formula $\left\{ 1,x,x^{2},x^{3},\dots,x^{n-1}\right\} $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+vzemimo paroma različne 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
+\end_inset
+
+ in definirajmo 
+\begin_inset Formula $p_{i}\left(x\right)=\left(x-\alpha_{1}\right)\cdots\left(x-\alpha_{i-1}\right)\left(x-\alpha_{i+1}\right)\cdots\left(x-\alpha_{n}\right)$
+\end_inset
+
+ za vsak 
+\begin_inset Formula $i\in\left\{ 1..n\right\} $
+\end_inset
+
+,
+ kar je polimom stopnje 
+\begin_inset Formula $n-1$
+\end_inset
+
+.
+\begin_inset Formula $\left\{ \alpha_{1}p_{1}\left(x\right),\dots,\alpha_{n}p_{n}\left(x\right)\right\} $
+\end_inset
+
+ je baza za 
+\begin_inset Formula $F\left[x\right]_{<n}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Dokazujemo,
+ da so LN in ogrodje:
+\end_layout
+
+\begin_layout Itemize
+LN:
+ 
+\begin_inset Formula $\beta_{1}p_{i}\left(x\right)+\cdots+\beta_{n}p_{n}\left(x\right)=0\overset{?}{\Longrightarrow}\beta_{1}=\cdots=\beta_{n}=0$
+\end_inset
+
+.
+ Opazimo,
+ da 
+\begin_inset Formula $p_{i}\left(\alpha_{j}\right)=0\Leftrightarrow i=j$
+\end_inset
+
+.
+ Torej če za 
+\begin_inset Formula $x$
+\end_inset
+
+ vstavimo katerikoli 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+,
+ bodo vsi členi 0,
+ razen 
+\begin_inset Formula $\beta_{i}p_{i}\left(x\right)$
+\end_inset
+
+.
+ Ker pa 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+ ni ničla 
+\begin_inset Formula $p_{i}\left(x\right)$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\beta_{i}=0$
+\end_inset
+
+,
+ čim je 
+\begin_inset Formula $\beta_{i}p_{i}\left(x\right)=0$
+\end_inset
+
+.
+ Preverjati je treba le 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+,
+ ker je dovolj najti eno vrednost spremenljivke,
+ v kateri se vrednosti polinomov ne ujemajo,
+ da lahko rečemo,
+ da polinomi niso isti.
+\end_layout
+
+\begin_layout Itemize
+ogrodje:
+ Trdimo,
+ da za vsak polimom velja formula 
+\begin_inset Formula $f\left(x\right)=\frac{f\left(\alpha_{1}\right)}{p_{1}\left(\alpha_{1}\right)}p_{1}\left(x\right)+\cdots+\frac{f\left(\alpha_{n}\right)}{p_{n}\left(\alpha_{n}\right)}p_{n}\left(x\right)$
+\end_inset
+
+.
+ Obe strani enačbe imata stopnjo največ 
+\begin_inset Formula $n-1$
+\end_inset
+
+ in se ujemata v 
+\begin_inset Formula $n$
+\end_inset
+
+ različnih točkah.
+ Če za 
+\begin_inset Formula $x$
+\end_inset
+
+ vstavimo 
+\begin_inset Formula $\alpha_{i}$
+\end_inset
+
+ za vsak 
+\begin_inset Formula $i$
+\end_inset
+
+,
+ dobimo 0 v vseh členih,
+ razen v 
+\begin_inset Formula $i-$
+\end_inset
+
+tem,
+ kjer se vrednost 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ ujema z vrednostjo 
+\begin_inset Formula $f\left(\alpha_{1}\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\end_deeper
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Obstoj-baze"
+
+\end_inset
+
+Obstoj baze
+\end_layout
+
+\begin_layout Standard
+Omejimo se na končno razsežne vektorske prostore.
+\end_layout
+
+\begin_layout Definition*
+Vektorski prostor je končno razsežen,
+ če ima končno ogrodje:
+ 
+\begin_inset Formula $\exists n\in\mathbb{N}\exists v_{1},\dots,v_{n}\ni:V=\Lin\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+obstoj baze.
+ Vsak končno razsežen vektorski prostor ima vsaj eno bazo.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVP in naj bo 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ njegovo ogrodje.
+ Ker ogrodje ni nujno LN,
+ naj bo 
+\begin_inset Formula $S$
+\end_inset
+
+ minimalna/najmanjša podmnožica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+,
+ ki je še ogrodje za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Trdimo,
+ da je 
+\begin_inset Formula $S$
+\end_inset
+
+ baza za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Po konstrukciji je ogrodje,
+ dokažimo še,
+ da je LN:
+ PDDRAA 
+\begin_inset Formula $S$
+\end_inset
+
+ je linearno odvisna.
+ Tedaj 
+\begin_inset Formula $\exists v_{i}\in S\ni:v_{i}$
+\end_inset
+
+ je LK 
+\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
+\end_inset
+
+ ogrodje manjše moči,
+ kar bi bilo v protislovju s predpostavko.
+ Tedaj obstajajo koeficienti,
+ da velja 
+\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $v\in V$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $S$
+\end_inset
+
+ ogrodje 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ obstajajo neki koeficienti 
+\begin_inset Formula $\beta_{1},\dots,\beta_{n}$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula 
+\[
+v=\beta_{1}v_{1}+\cdots+\beta_{i}v_{i}+\cdots+\beta_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{i}\left(\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}\right)+\cdots+\beta_{n}v_{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left(\beta_{1}+\beta_{i}\alpha_{1}\right)v_{1}+\cdots+\left(\beta_{i-1}+\beta_{i}\alpha_{i-1}\right)v_{i-1}+\left(\beta_{i+1}+\beta_{i}\alpha_{i+1}\right)v_{i+1}+\cdots+\left(\beta_{n}+\beta_{i}\alpha_{n}\right)v_{n}
+\]
+
+\end_inset
+
+To pa je 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ saj je bilo rečeno,
+ da je 
+\begin_inset Formula $S$
+\end_inset
+
+ najmanjše ogrodje,
+ mi pa smo razvili poljuben 
+\begin_inset Formula $v$
+\end_inset
+
+ po manjšem ogrodju.
+ Torej ima vsak KRVP bazo in vsako ogrodje ima podmnožico,
+ ki je baza.
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "enoličnost-moči-baze."
+
+\end_inset
+
+enoličnost moči baze.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVP z 
+\begin_inset Formula $n-$
+\end_inset
+
+elementno bazo.
+ Tedaj velja vse to:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ LN množica 
+\begin_inset Formula $A$
+\end_inset
+
+ v 
+\begin_inset Formula $V$
+\end_inset
+
+ ima 
+\begin_inset Formula $\leq n$
+\end_inset
+
+ elementov
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ ogrodje v 
+\begin_inset Formula $V$
+\end_inset
+
+ ima 
+\begin_inset Formula $\geq n$
+\end_inset
+
+ elementov
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ baza v 
+\begin_inset Formula $V$
+\end_inset
+
+ ima 
+\begin_inset Formula $n$
+\end_inset
+
+ elementov
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokaz je dolg.
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:Vsak-poddoločen-homogen"
+
+\end_inset
+
+Vsak poddoločen homogen sistem linearnih enačb ima netrivialno rešitev.
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Dokaz se nahaja pod identično trditvijo 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:Vpoddol-hom-sist-ima-ne0-reš"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:ln<=ogr"
+
+\end_inset
+
+Če je 
+\begin_inset Formula $u_{1},\dots,u_{m}$
+\end_inset
+
+ LN množica v 
+\begin_inset Formula $V$
+\end_inset
+
+ in 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ ogrodje za 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $m\leq n$
+\end_inset
+
+.
+ ZDB moč katerekoli LN množice je manjša ali enaka od kateregakoli ogrodja v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+RAAPDD 
+\begin_inset Formula $u_{1},\dots,u_{m}$
+\end_inset
+
+ je LN,
+ 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ je ogrodje in 
+\begin_inset Formula $m>n$
+\end_inset
+
+.
+ Iščemo protislovje.
+ Vsakega od 
+\begin_inset Formula $u_{i}$
+\end_inset
+
+ lahko razvijemo po 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+u_{m} & = & \alpha_{m1}v_{1} & + & \cdots & + & \alpha_{mn}v_{n}
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula $\forall i\in\left\{ 1..m\right\} $
+\end_inset
+
+ pomnožimo 
+\begin_inset Formula $i-$
+\end_inset
+
+to enačbo s skalarjem 
+\begin_inset Formula $x_{i}$
+\end_inset
+
+ in jih seštejmo.
+ 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ so abstraktne spremenljivke.
+ Tedaj:
+\begin_inset Formula 
+\[
+x_{1}u_{1}+\cdots+x_{m}u_{m}=x_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+x_{m}\left(\alpha_{m1}v_{1}+\cdots+\alpha_{mn}v_{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=v_{1}\left(\alpha_{11}x_{1}+\cdots+\alpha_{m1}x_{m}\right)+\cdots+v_{n}\left(\alpha_{1n}x_{1}+\cdots+\alpha_{mn}x_{m}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Izenačimo koeficiente za 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ z 0 in dobimo poddoločen homogen sistem enačb (ima 
+\begin_inset Formula $n$
+\end_inset
+
+ enačb in 
+\begin_inset Formula $m$
+\end_inset
+
+ spremenljivk,
+ po predpostavki pa velja 
+\begin_inset Formula $m>n$
+\end_inset
+
+):
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+\alpha_{11}x_{1} & + & \cdots & + & \alpha_{m1}x_{m} & = & 0\\
+\vdots &  &  &  & \vdots &  & \vdots\\
+\alpha_{1n}x_{1} & + & \cdots & + & \alpha_{mn}x_{m} & = & 0
+\end{array}
+\]
+
+\end_inset
+
+Po lemi 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "lem:Vsak-poddoločen-homogen"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ ima ta sistem netrivialno rešitev,
+ recimo 
+\begin_inset Formula $\left(\mu_{1},\dots,\mu_{m}\right)$
+\end_inset
+
+.
+ Če to rešitev vstavimo v 
+\begin_inset Formula $u_{1}x_{1}+\cdots+u_{m}x_{m}$
+\end_inset
+
+,
+ dobimo 
+\begin_inset Formula $u_{1}\mu_{1}+\cdots+u_{m}\mu_{m}=0$
+\end_inset
+
+.
+ Ker so 
+\begin_inset Formula $u_{1},\dots,u_{m}$
+\end_inset
+
+ LN,
+ so 
+\begin_inset Formula $\mu_{1}=\cdots=\mu_{m}=0$
+\end_inset
+
+,
+ kar je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s predpostavko.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ baza je ogrodje 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ po lemi 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "lem:ln<=ogr"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja,
+ da ima vsaka LN množica manj ali enako elementov kot vsako ogrodje,
+ torej tudi manj ali enako kot 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ baza je LN 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ po lemi 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "lem:ln<=ogr"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja,
+ da ima vsako ogrodje več ali enako elementov kot vsaka LN,
+ torej tudi več ali enako kot 
+\begin_inset Formula $n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Sledi iz zgornjih dveh točk,
+ saj je baza tako ogrodje kot LN hkrati.
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVP.
+ Njegova dimenzija,
+ 
+\begin_inset Formula $\dim V$
+\end_inset
+
+,
+ je moč baze v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\dim F^{n}=n$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\dim M_{m\times n}\left(\mathbb{F}\right)=m\cdot n$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Dopolnitev LN množice do baze
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor z dimenzijo 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Trdimo,
+ da
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+ima vsaka LN množica 
+\begin_inset Formula $\leq n$
+\end_inset
+
+ elementov,
+\end_layout
+
+\begin_layout Enumerate
+je vsaka LN množica v 
+\begin_inset Formula $V$
+\end_inset
+
+ z 
+\begin_inset Formula $n$
+\end_inset
+
+ elementi baza,
+\end_layout
+
+\begin_layout Enumerate
+lahko vsako LN množico v 
+\begin_inset Formula $V$
+\end_inset
+
+ dopolnimo do baze.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokaz je dolg
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:večja-ln"
+
+\end_inset
+
+Če so 
+\begin_inset Formula $v_{1},\dots,v_{m}\in V$
+\end_inset
+
+ LN in če 
+\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+,
+ potem so tudi 
+\begin_inset Formula $v_{1},\dots,v_{m},v_{m+1}$
+\end_inset
+
+ LN.
+\end_layout
+
+\begin_deeper
+\begin_layout Proof
+Naj velja 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0$
+\end_inset
+
+ za nek 
+\begin_inset Formula $\vec{\alpha}\in F^{m+1}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $\vec{a}=\vec{0}$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\alpha_{m+1}=0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m}v_{m}=0$
+\end_inset
+
+,
+ ker pa so po predpostavki 
+\begin_inset Formula $v_{1},\dots,v_{m}$
+\end_inset
+
+ LN,
+ je 
+\begin_inset Formula $\vec{\alpha}=\vec{0}$
+\end_inset
+
+.
+ Sicer pa,
+ če PDDRAA 
+\begin_inset Formula $\alpha_{m+1}\not=0$
+\end_inset
+
+,
+ lahko z 
+\begin_inset Formula $a_{m+1}$
+\end_inset
+
+ delimo:
+\begin_inset Formula 
+\[
+\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha_{m+1}v_{m+1}=-\alpha_{1}v_{1}-\cdots-\alpha_{m}v_{m}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{m+1}=\frac{-\alpha_{1}}{\alpha_{m+1}}v_{1}+\cdots+\frac{-\alpha_{m}}{\alpha_{m+1}}v_{m}
+\]
+
+\end_inset
+
+Tedaj pridemo do 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ saj smo 
+\begin_inset Formula $v_{m+1}$
+\end_inset
+
+ izrazili kot LK 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+,
+ po predpostavki pa je vendar 
+\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+že dokazano z dokazom trditve 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "enoličnost-moči-baze."
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ v razdelku 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Obstoj-baze"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{Vsaka LN množica v 
+\backslash
+ensuremath{V} z 
+\backslash
+ensuremath{n} elementi je baza.}
+\end_layout
+
+\end_inset
+
+ PDDRAA 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ je LN,
+ ki ni baza.
+ Tedaj 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ ni ogrodje.
+ Tedaj 
+\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} \not=V$
+\end_inset
+
+.
+ Zatorej 
+\begin_inset Formula $\exists v_{n+1}\in V\ni:\left\{ v_{1},\dots,v_{n},v_{n+1}\right\} $
+\end_inset
+
+ je LN,
+ kar je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s trditvijo,
+ da ima vsaka 
+\begin_inset Formula $LN$
+\end_inset
+
+ množica v 
+\begin_inset Formula $V$
+\end_inset
+
+ kvečjemu 
+\begin_inset Formula $n$
+\end_inset
+
+ elementov.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{Vsako LN množico v $V$ z $n$ elementi lahko dopolnimo do baze.}
+\end_layout
+
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $v_{1},\dots,v_{m}$
+\end_inset
+
+ LN množica v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Vemo,
+ da je 
+\begin_inset Formula $m\leq n$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $m=n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $v_{1},\dots,v_{m}$
+\end_inset
+
+ baza po zgornji trditvi.
+ Sicer pa je 
+\begin_inset Formula $m<n$
+\end_inset
+
+:
+ Tedaj 
+\begin_inset Formula $v_{1},\dots,v_{m}$
+\end_inset
+
+ ni ogrodje,
+ sicer bi imeli neko LN množico z več elementi kot neko ogrodje,
+ saj ima po popraj dokazanem vsako ogrodje vsaj toliko elementov kot vsaka LN množica.
+ Ker 
+\begin_inset Formula $v_{1},\dots,v_{m}$
+\end_inset
+
+ ni ogrodje,
+ 
+\begin_inset Formula $\exists v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+.
+ Po lemi 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "lem:večja-ln"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ je torej 
+\begin_inset Formula $v_{1},\dots,v_{m+1}$
+\end_inset
+
+ LN množica.
+ Če je 
+\begin_inset Formula $m+1=n$
+\end_inset
+
+,
+ je to že baza,
+ sicer ponavljamo dodajanje elementov,
+ dokler ne dodamo 
+\begin_inset Formula $k$
+\end_inset
+
+ elementov in dosežemo 
+\begin_inset Formula $m+k=n$
+\end_inset
+
+.
+ Tedaj je to baza.
+ Naredili smo 
+\begin_inset Formula $k=m-n$
+\end_inset
+
+ korakov.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Uporabna vrednost tega izreka sta dva nova izreka o dimenzijah podprostorov:
+\end_layout
+
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $V$
+\end_inset
+
+ je KRVP in 
+\begin_inset Formula $W$
+\end_inset
+
+ njegov podprostor,
+ je 
+\begin_inset Formula $\dim W\leq\dim V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+PDDRAA 
+\begin_inset Formula $\dim W>\dim V$
+\end_inset
+
+.
+ Čim ima baza 
+\begin_inset Formula $W$
+\end_inset
+
+ večjo moč kot baza 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ obstaja v 
+\begin_inset Formula $W$
+\end_inset
+
+ LN množica z večjo močjo kot baza 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Toda ker je ta LN množica LN tudi v 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ obstaja v 
+\begin_inset Formula $V$
+\end_inset
+
+ LN množica z več elementi kot baza 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ kar je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s trditvijo 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "enoličnost-moči-baze."
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ v razdelku 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Obstoj-baze"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+dimenzijska formula za podprostore.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVP in 
+\begin_inset Formula $W_{1},W_{2}$
+\end_inset
+
+ podprostora v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}-\dim\left(W_{1}\cap W_{2}\right)$
+\end_inset
+
+.
+ Vsota vektorskih podprostorov je definirana v razdelku 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "subsec:Vsota-podprostorov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izberimo bazo 
+\begin_inset Formula $w_{1},\dots,w_{m}$
+\end_inset
+
+ za 
+\begin_inset Formula $W_{1}\cap W_{2}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $u_{1},\dots,u_{k}$
+\end_inset
+
+ njena dopolnitev do baze 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $v_{1},\dots,v_{l}$
+\end_inset
+
+ njena dopolnitev do baze 
+\begin_inset Formula $W_{2}$
+\end_inset
+
+.
+ Trdimo,
+ da je 
+\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
+\end_inset
+
+ baza za 
+\begin_inset Formula $W_{1}+W_{2}$
+\end_inset
+
+.
+ Tedaj bi namreč veljalo 
+\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=m+k+l$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\dim\left(W_{1}\cap W_{2}\right)=m$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\dim\left(W_{1}\right)=m+k$
+\end_inset
+
+ in 
+\begin_inset Formula $\dim\left(W_{2}\right)=m+l$
+\end_inset
+
+.
+ Treba je dokazati še,
+ da je 
+\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
+\end_inset
+
+ baza za 
+\begin_inset Formula $W_{1}+W_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Je ogrodje?
+ Vzemimo poljuben 
+\begin_inset Formula $v\in W_{1}+W_{2}$
+\end_inset
+
+.
+ Po definiciji 
+\begin_inset Formula $W_{1}+W_{2}\exists z_{1}\in W_{1},z_{2}\in W_{2}\ni:v=z_{1}+z_{2}$
+\end_inset
+
+.
+ Razvijmo 
+\begin_inset Formula $z_{1}$
+\end_inset
+
+ po bazi 
+\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $z_{2}$
+\end_inset
+
+ po bazi 
+\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{k}$
+\end_inset
+
+ za 
+\begin_inset Formula $W_{2}$
+\end_inset
+
+.
+ Takole:
+ 
+\begin_inset Formula $z_{1}=\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}$
+\end_inset
+
+ in 
+\begin_inset Formula $z_{2}=\gamma_{1}w_{1}+\cdots\gamma_{m}w_{m}+\delta_{1}v_{1}+\cdots\delta_{l}v_{l}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $v=z_{1}+z_{2}=\left(\alpha_{1}+\gamma_{1}\right)w_{1}+\cdots+\left(\alpha_{m}+\gamma_{m}\right)w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\delta_{1}v_{1}+\cdots+\delta_{l}v_{l}\in\Lin\left\{ w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}\right\} $
+\end_inset
+
+.
+ Je ogrodje.
+\end_layout
+
+\begin_layout Itemize
+Je LN?
+ Naj bo 
+\begin_inset Formula 
+\[
+\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
+\]
+
+\end_inset
+
+Leva stran enačbe je 
+\begin_inset Formula $\in W_{1}$
+\end_inset
+
+,
+ desna pa 
+\begin_inset Formula $\in W_{2}$
+\end_inset
+
+,
+ zatorej je element,
+ ki ga izraza na obeh straneh enačbe opisujeta,
+ 
+\begin_inset Formula $\in W_{1}\cap W_{2}$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $v_{1},\dots,v_{l}$
+\end_inset
+
+ baza za 
+\begin_inset Formula $W_{1}\cap W_{1}$
+\end_inset
+
+.
+ Toda baza od 
+\begin_inset Formula $W_{1}\cap W_{2}$
+\end_inset
+
+ je tudi 
+\begin_inset Formula $w_{1},\dots,w_{m}$
+\end_inset
+
+,
+ zatorej lahko ta element razpišemo po njej:
+\begin_inset Formula 
+\[
+\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}=\delta_{1}w_{1}+\cdots+\delta_{m}v_{m}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\delta_{1}w_{1}+\cdots+\delta_{m}w_{m}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{l}$
+\end_inset
+
+ je baza za 
+\begin_inset Formula $W_{2}$
+\end_inset
+
+ po naši prejšnji definiciji,
+ torej je LN množica,
+ zato 
+\begin_inset Formula $\delta_{1}=\cdots=\delta_{m}=\gamma_{1}=\cdots=\gamma_{l}=0$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
+\end_inset
+
+,
+ se lahko vrnemo k drugi enačbi te točke in to ugotovitev upoštevamo:
+\begin_inset Formula 
+\[
+\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=0
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
+\end_inset
+
+ je baza za 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ po naši prejšnji definiciji,
+ torej je LN množica,
+ zato 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=0$
+\end_inset
+
+.
+ Torej velja 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=\gamma_{1}=\cdots=\gamma_{l}=0$
+\end_inset
+
+,
+ torej je ta množica res LN.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+Velja torej 
+\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim\left(W_{1}\right)+\dim\left(W_{2}\right)$
+\end_inset
+
+.
+ Enačaj velja 
+\begin_inset Formula $\Leftrightarrow W_{1}\cap W_{2}=\left\{ 0\right\} $
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\dim\left(\left\{ 0\right\} \right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition
+\begin_inset CommandInset label
+LatexCommand label
+name "def:vsota-je-direktna"
+
+\end_inset
+
+Pravimo,
+ da je vsota 
+\begin_inset Formula $W_{1}+W_{2}$
+\end_inset
+
+ direktna,
+ če velja 
+\begin_inset Formula $W_{1}\cap W_{2}=\left\{ 0\right\} $
+\end_inset
+
+ oziroma ekvivalentno če je 
+\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$
+\end_inset
+
+ oziroma ekvivalentno 
+\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Prehod na novo bazo
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor dimenzije 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Recimo,
+ da imamo dve bazi v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ 
+\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ naj bo 
+\begin_inset Quotes gld
+\end_inset
+
+stara baza
+\begin_inset Quotes grd
+\end_inset
+
+,
+ 
+\begin_inset Formula $C=\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ pa naj bo 
+\begin_inset Quotes gld
+\end_inset
+
+nova baza
+\begin_inset Quotes grd
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall v\in V$
+\end_inset
+
+ lahko razvijemo po 
+\begin_inset Formula $B$
+\end_inset
+
+ in po 
+\begin_inset Formula $C$
+\end_inset
+
+.
+ Razvoj po 
+\begin_inset Formula $B$
+\end_inset
+
+:
+ 
+\begin_inset Formula $v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
+\end_inset
+
+,
+ razvoj po 
+\begin_inset Formula $C$
+\end_inset
+
+:
+ 
+\begin_inset Formula $v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}$
+\end_inset
+
+.
+ Kakšna je zveza med 
+\begin_inset Formula $\vec{\beta}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{\gamma}$
+\end_inset
+
+ v obeh razvojih?
+\end_layout
+
+\begin_layout Standard
+Uvedimo oznako 
+\begin_inset Formula $\left[v\right]_{B}$
+\end_inset
+
+,
+ to naj bodo koeficienti vektorja 
+\begin_inset Formula $v$
+\end_inset
+
+ pri razvoju po 
+\begin_inset Formula $B$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left[v\right]_{B}=\left[\begin{array}{c}
+\beta_{1}\\
+\vdots\\
+\beta_{n}
+\end{array}\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Vsak vektor stare baze razvijmo po novi bazi,
+ kjer 
+\begin_inset Formula $\left[u_{i}\right]_{C}=\left[\begin{array}{c}
+\alpha_{i1}\\
+\vdots\\
+\alpha_{in}
+\end{array}\right]$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+u_{n} & = & a_{n1}v_{1} & + & \cdots & + & a_{nn}v_{n}
+\end{array}
+\]
+
+\end_inset
+
+Koeficiente 
+\begin_inset Formula $\alpha$
+\end_inset
+
+ zložimo v tako imenovanj prehodno matriko 
+\begin_inset Formula $P_{C\leftarrow B}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+P_{C\leftarrow B}=\left[\begin{array}{ccc}
+\left[u_{1}\right]_{C} & \cdots & \left[u_{n}\right]_{C}\end{array}\right]=\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{n1}\\
+\vdots &  & \vdots\\
+a_{1n} & \cdots & a_{nn}
+\end{array}\right]
+\]
+
+\end_inset
+
+Sledi 
+\begin_inset Formula 
+\[
+v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}=\beta_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+\beta_{n}\left(\alpha_{n1}v_{1}+\cdots+\alpha_{nn}v_{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=v_{1}\left(\beta_{1}\alpha_{11}+\beta_{2}\alpha_{21}+\cdots+\beta_{n}\alpha_{n1}\right)+\cdots+v_{n}\left(\beta_{1}\alpha_{1n}+\beta_{2}\alpha_{2n}+\cdots+\beta_{n}\alpha_{nn}\right)=
+\]
+
+\end_inset
+
+po drugi strani je 
+\begin_inset Formula $v$
+\end_inset
+
+ tudi lahko razvit po novi bazi:
+\begin_inset Formula 
+\[
+=v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}
+\]
+
+\end_inset
+
+Iz česar,
+ ker je razvoj po bazi enoličen,
+ sledi
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+\gamma_{1} & = & \beta_{1}\alpha_{11} & + & \cdots & + & \beta_{n}\alpha_{n1}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+\gamma_{n} & = & \beta_{1}a_{1n} & + & \cdots & + & \beta_{n}\alpha_{nn}
+\end{array},
+\]
+
+\end_inset
+
+kar v matrični obliki zapišemo
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+\alpha_{11} & \cdots & \alpha_{n1}\\
+\vdots &  & \vdots\\
+\alpha_{1n} & \cdots & \alpha_{nn}
+\end{array}\right]\left[\begin{array}{c}
+\beta_{1}\\
+\vdots\\
+\beta_{n}
+\end{array}\right]=\left[\begin{array}{c}
+\gamma_{1}\\
+\vdots\\
+\gamma_{n}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+P_{C\leftarrow B}\left[v\right]_{B}=\left[v\right]_{C}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $P_{B\leftarrow B}=I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Naj bodo prehodi med bazami takšnile:
+ 
+\begin_inset Formula $B\overset{P_{C\leftarrow B}}{\longrightarrow}C\overset{P_{D\leftarrow C}}{\longrightarrow}D$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $P_{D\leftarrow B}=P_{C\leftarrow B}\cdot P_{C\leftarrow D}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P_{B\leftarrow C}\cdot P_{C\leftarrow B}=I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(P_{B\leftarrow C}\right)^{-1}=P_{C\leftarrow B}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $v\in F^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $S$
+\end_inset
+
+ standardna baza za 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+.
+ Potem 
+\begin_inset Formula $\left[v\right]_{S}=\left[\begin{array}{c}
+\alpha_{1}\\
+\vdots\\
+\alpha_{n}
+\end{array}\right]=v$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $P_{S\leftarrow B}=\left[\begin{array}{ccc}
+\left[u_{1}\right]_{S} & \cdots & \left[u_{n}\right]_{S}\end{array}\right]=\left[\begin{array}{ccc}
+u_{1} & \cdots & u_{n}\end{array}\right]$
+\end_inset
+
+ za 
+\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+.
+ Sledi tudi 
+\begin_inset Formula $P_{S\leftarrow C}=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+,
+ kjer so 
+\begin_inset Formula $v,u,B,C$
+\end_inset
+
+ kot prej (kot definirano na začetku tega razdelka).
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P_{C\leftarrow B}=P_{C\leftarrow S}\cdot P_{S\leftarrow B}$
+\end_inset
+
+ (slednji dve točki veljata samo v 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+,
+ kjer je standardna baza lepa in zapisljiva kot elementi v matriki)
+\end_layout
+
+\end_deeper
+\begin_layout Section
+Drugi semester
+\end_layout
+
+\begin_layout Subsection
+Linearne preslikave
+\end_layout
+
+\begin_layout Standard
+Radi bi definirali homomorfizem vektorskih prostorov.
+ Homomorfizem za abelove grupe smo že definirali,
+ vektorski prostor pa je le abelova grupa z dodatno strukturo (množenje s skalarjem).
+\end_layout
+
+\begin_layout Definition*
+Preslikava 
+\begin_inset Formula $f:V_{1}\to V_{2}$
+\end_inset
+
+ je homomorfizem vektorskih prostorov nad istim poljem oziroma linearna preslikava,
+ če je aditivna (homomorfizem) (
+\begin_inset Formula $\forall u,v\in V_{1}:f\left(u+_{1}v\right)=fu+_{2}fv$
+\end_inset
+
+) in če je homogena:
+ 
+\begin_inset Formula $\forall u\in V_{1},\alpha\in F:f\left(\alpha u\right)=\alpha f\left(u\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Ekvivalentno je preverjati oba pogoja hkrati.
+ Če za 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ velja 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in F,u_{1},u_{2}\in U:L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $L$
+\end_inset
+
+ linearna preslikava.
+\end_layout
+
+\begin_layout Example*
+Vrtež za kot 
+\begin_inset Formula $\tau$
+\end_inset
+
+ v ravnini:
+\begin_inset Formula 
+\[
+\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]\to\left[\begin{array}{cc}
+\cos\tau & -\sin\tau\\
+\sin\tau & \cos\tau
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Linearna funkcija iz analize ni linearna preslikava.
+ Premik za vektor 
+\begin_inset Formula $w$
+\end_inset
+
+ ni linearna preslikava.
+ Odvajanje in integriranje sta linearni preslikavi.
+\end_layout
+
+\begin_layout Fact*
+Vsaka linearna preslikava slika 0 v 0.
+\end_layout
+
+\begin_layout Definition*
+Bijektivni linearni preslikavi pravimo linearni izomorfizem.
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:invLinIzJeLinIz"
+
+\end_inset
+
+Inverz linearnega izomorfizma je zopet linearni izomorfizem.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ bijektivna linearna preslikava med vektorskima prostoroma nad istim poljem 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Dokazati je treba,
+ da je 
+\begin_inset Formula $L^{-1}:V\to U$
+\end_inset
+
+ spet linearna preslikava.
+ Ker je 
+\begin_inset Formula $L$
+\end_inset
+
+ linearna,
+ velja 
+\begin_inset Formula 
+\[
+\forall\alpha_{1},\alpha_{2}\in F,v_{1},v_{2}\in V:L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=\alpha_{1}LL^{-1}v_{1}+\alpha_{2}LL^{-1}v_{2}=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $L$
+\end_inset
+
+ injektivna,
+ iz 
+\begin_inset Formula $L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}=L^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $F^{n}$
+\end_inset
+
+ je linearno izomorfen 
+\begin_inset Formula $n-$
+\end_inset
+
+razsežnem 
+\begin_inset Formula $V$
+\end_inset
+
+ nad 
+\begin_inset Formula $F$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Vsak 
+\begin_inset Formula $n-$
+\end_inset
+
+razsežen vektorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+ je linearno izomorfen 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ 
+\begin_inset Formula $n-$
+\end_inset
+
+razsežen vektorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $B=\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ baza za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Definirajmo preslikavo 
+\begin_inset Formula $\phi_{B}:F^{n}\to V$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $\left(x_{1},\cdots,x_{1}\right)\mapsto x_{1}v_{1}+\cdots+x_{n}v_{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $B$
+\end_inset
+
+ ogrodje,
+ je 
+\begin_inset Formula $\phi_{B}$
+\end_inset
+
+ surjektivna.
+ Ker je 
+\begin_inset Formula $B$
+\end_inset
+
+ linearno neodvisna,
+ je 
+\begin_inset Formula $\phi_{B}$
+\end_inset
+
+ injektivna.
+ Pokažimo še,
+ da je linearna presikava:
+\begin_inset Formula 
+\[
+\phi_{B}\left(\alpha\left(x_{1},\dots,x_{n}\right)+\beta\left(y_{1},\dots,y_{n}\right)\right)=\phi_{B}\left(\alpha x_{1}+\beta y_{1},\dots,\alpha x_{n}+\beta x_{n}\right)=v_{1}\left(\alpha x_{1}+\beta y_{1}\right)+\cdots+v_{n}\left(\alpha x_{n}+\beta x_{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha\left(v_{1}x_{1}+\cdots+v_{n}x_{n}\right)+\cdots+\beta\left(v_{1}y_{1}+\cdots+v_{n}y_{n}\right)=\alpha\phi_{B}\left(x_{1},\dots,x_{n}\right)+\beta\phi_{B}\left(y_{1},\dots y_{n}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Matrika-linearne-preslikave"
+
+\end_inset
+
+Matrika linearne preslikave —
+ linearni izomorfizem 
+\begin_inset Formula $M_{m,n}\left(F\right)\to\mathcal{L}\left(F^{n},F^{m}\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $F$
+\end_inset
+
+ polje in 
+\begin_inset Formula $m,n\in F$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\mathcal{L}\left(F^{n},F^{m}\right)$
+\end_inset
+
+ je vektorski prostor linearnih preslikav iz 
+\begin_inset Formula $F^{n}\to F^{m}$
+\end_inset
+
+.
+ Seštevanje definiramo z 
+\begin_inset Formula $\left(L_{1}+L_{2}\right)u\coloneqq L_{1}u+L_{2}u$
+\end_inset
+
+,
+ množenje s skalarjem pa 
+\begin_inset Formula $\left(\alpha L\right)u=\alpha\left(Lu\right)$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $M_{m,n}\left(F\right)$
+\end_inset
+
+ vektorski prostor vseh 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matrik nad 
+\begin_inset Formula $F$
+\end_inset
+
+ z znanim seštevanjem in množenjem.
+ Obstaja linearni izomorfizem med tema dvema prostoroma.
+\end_layout
+
+\begin_layout Proof
+Oznake kot v trditvi.
+ Za vsako 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matriko 
+\begin_inset Formula $A=\left[a_{i,j}\right]$
+\end_inset
+
+ definirajmo preslikavo 
+\begin_inset Formula $L_{A}$
+\end_inset
+
+ iz 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+ v 
+\begin_inset Formula $F^{m}$
+\end_inset
+
+ takole:
+ 
+\begin_inset Formula $L_{A}\left(x_{1},\dots,x_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$
+\end_inset
+
+.
+ Po definiciji matričnega množenja ta preslikava ustreza 
+\begin_inset Formula $L_{A}\vec{x}=A\vec{x}$
+\end_inset
+
+.
+ Dokažimo,
+ da je linearni izomorfizem.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Linearnost:
+ 
+\begin_inset Formula $L_{\alpha A+\beta B}\vec{x}=\left(\alpha A+\beta B\right)\vec{x}=\alpha A\vec{x}+\beta B\vec{x}=\alpha L_{A}\vec{x}+\beta L_{B}\vec{x}=\left(\alpha L_{A}+\beta L_{B}\right)\vec{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Bijektivnost:
+ Konstruirajmo inverzno preslikavo (iz trditve 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "claim:invLinIzJeLinIz"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ vemo,
+ da bo linearna).
+ Vsaki linearni presikavi 
+\begin_inset Formula $L:F^{n}\to F^{m}$
+\end_inset
+
+ priredimo 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matriko 
+\begin_inset Formula $\left[\begin{array}{ccc}
+Le_{1} & \cdots & Le_{n}\end{array}\right]$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $e_{1},\dots,e_{n}$
+\end_inset
+
+ standardna baza za 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+.
+ Pokažimo,
+ da je ta preslikava res inverz,
+ torej preverimo,
+ da je kompozitum 
+\begin_inset Formula $A\mapsto L_{A}\mapsto\left[\begin{array}{ccc}
+L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]$
+\end_inset
+
+ identiteta in da je 
+\begin_inset Formula $\left[\begin{array}{ccc}
+L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]\mapsto L_{A}\mapsto A$
+\end_inset
+
+ tudi identiteta.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$A
+\backslash
+mapsto L_{A}
+\backslash
+mapsto
+\backslash
+left[
+\backslash
+begin{array}{ccc}Le_{1} & 
+\backslash
+cdots & Le_{n}
+\backslash
+end{array}
+\backslash
+right]
+\backslash
+overset{?}{=}id$}
+\end_layout
+
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]=\left[\begin{array}{ccc}
+Ae_{1} & \cdots & Ae_{n}\end{array}\right]=A\left[\begin{array}{ccc}
+e_{1} & \cdots & e_{n}\end{array}\right]=AI=A
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+left[
+\backslash
+begin{array}{ccc}L_{A}e_{1} & 
+\backslash
+cdots & L_{A}e_{n}
+\backslash
+end{array}
+\backslash
+right]
+\backslash
+mapsto L_{A}
+\backslash
+mapsto A
+\backslash
+overset{?}{=}id$}
+\end_layout
+
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\forall x:L_{\left[\begin{array}{ccc}
+Le_{1} & \cdots & Le_{n}\end{array}\right]}x=\left[\begin{array}{ccc}
+Le_{1} & \cdots & Le_{n}\end{array}\right]\left[\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right]=x_{1}Le_{1}+\cdots+x_{n}Le_{n}=L\left(x_{1}e_{1}+\cdots+x_{n}e_{n}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=L\left(x_{1}\left[\begin{array}{c}
+1\\
+0\\
+\vdots\\
+0
+\end{array}\right]+\cdots+x_{n}\left[\begin{array}{c}
+0\\
+\vdots\\
+0\\
+1
+\end{array}\right]\right)=Lx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Proof
+Vsaki linearni preslikavi med dvema vektorskima prostoroma sedaj lahko priredimo matriko.
+ Prirejanje je odvisno od izbire baz v obeh vektorskih prostorih.
+ Matrika namreč preslika koeficiente iz polja 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ s katerimi je dan vektor,
+ ki ga z leve množimo z matriko,
+ razvit po 
+\begin_inset Quotes gld
+\end_inset
+
+vhodni
+\begin_inset Quotes grd
+\end_inset
+
+ bazi,
+ v koeficiente iz istega polja,
+ s katerimi je rezultantni vektor razvit po 
+\begin_inset Quotes gld
+\end_inset
+
+izhodni
+\begin_inset Quotes grd
+\end_inset
+
+ bazi.
+\end_layout
+
+\begin_layout Proof
+Naj bosta 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorska prostora nad istim poljem 
+\begin_inset Formula $F$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ linearna preslikava.
+ Izberimo bazo 
+\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ za 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+ za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Razvijmo vektorje 
+\begin_inset Formula $Lu_{1},\dots,Lu_{n}$
+\end_inset
+
+ po bazi 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
+\end{array}
+\]
+
+\end_inset
+
+Skalarje 
+\begin_inset Formula $\alpha_{i,j}$
+\end_inset
+
+ sedaj zložimo v spodnjo matriko,
+ ki ji pravimo 
+\series bold
+matrika linearne preslikave 
+\begin_inset Formula $L$
+\end_inset
+
+ glede na bazi 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+
+\series default
+.
+ 
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
+\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[\begin{array}{ccc}
+a_{1,1} & \cdots & \alpha_{n,1}\\
+\vdots &  & \vdots\\
+\alpha_{1,m} & \cdots & \alpha_{n,m}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $L:\mathbb{R}\left[x\right]_{\leq3}\to\mathbb{R}\left[x\right]_{\leq2}$
+\end_inset
+
+ —
+ linearna preslikava iz realnih polinomov stopnje kvečjemu 3 v realne polinome stopnje kvečjemu 2,
+ ki predstavlja odvajanje polinomov.
+ Bazi sta 
+\begin_inset Formula $\mathcal{B}=\left\{ 1,x,x^{2},x^{2}\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ 1,x,x^{2}\right\} $
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+L\left(1\right) & = & 0 & + & 0x & + & 0x^{2}\\
+L\left(x\right) & = & 1 & + & 0x & + & 0x^{2}\\
+L\left(x^{2}\right) & = & 0 & + & 2x & + & 0x^{2}\\
+L\left(x^{3}\right) & = & 0 & + & 0x & + & 3x^{2}
+\end{array}
+\]
+
+\end_inset
+
+Zapišimo matriko te linearne preslikave:
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cccc}
+0 & 1 & 0 & 0\\
+0 & 0 & 2 & 0\\
+0 & 0 & 0 & 3
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Prehodna matrika je poseben primer matrike linearne preslikave.
+ 
+\begin_inset Formula $L:V\to V$
+\end_inset
+
+ z bazama 
+\begin_inset Formula $\mathcal{B}=\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $L=id$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+id\left(u_{1}\right) & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,n}v_{n}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+id\left(u_{n}\right) & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,n}v_{n}
+\end{array}
+\]
+
+\end_inset
+
+Zapišimo matriko te linearne preslikave:
+ 
+\begin_inset Formula $\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}=P_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Lastnosti matrik linearnih preslikav
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem
+\begin_inset CommandInset label
+LatexCommand label
+name "thm:osnovna-formula"
+
+\end_inset
+
+osnovna formula.
+ Posplošitev formule 
+\begin_inset Formula $\left[u\right]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}\cdot\left[u\right]_{\mathcal{B}}$
+\end_inset
+
+ se glasi 
+\begin_inset Formula $\left[Lu\right]_{\mathcal{C}}=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}$
+\end_inset
+
+ za linearno preslikavo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+,
+ 
+\begin_inset Formula $u\in U$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\mathcal{\mathcal{B}}=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ baza za 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+ baza za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po korakih:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Razvijmo 
+\begin_inset Formula $u$
+\end_inset
+
+ po bazi 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $u=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Uporabimo-L-na"
+
+\end_inset
+
+Uporabimo 
+\begin_inset Formula $L$
+\end_inset
+
+ na obeh straneh:
+ 
+\begin_inset Formula $Lu=L\left(\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}\right)=\beta_{1}Lu_{1}+\cdots+\beta_{n}Lu_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Razvijmo bazo 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+,
+ preslikano z 
+\begin_inset Formula $L$
+\end_inset
+
+,
+ po bazi 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
+\end{array}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Razvoj vstavimo v enačbo iz koraka 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Uporabimo-L-na"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ in uredimo:
+\begin_inset Formula 
+\[
+Lu=\beta_{1}\left(\alpha_{1,1}v_{1}+\cdots+\alpha_{1,m}v_{m}\right)+\cdots+\beta_{n}\left(\alpha_{n,1}v_{1}+\cdots+\alpha_{n,m}v_{m}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=v_{1}\left(\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\right)+\cdots+v_{m}\left(\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Odtod sledi:
+\begin_inset Formula 
+\[
+\left[Lu\right]_{\mathcal{C}}=\left[\begin{array}{c}
+\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\\
+\vdots\\
+\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}
+\end{array}\right]=\left[\begin{array}{ccc}
+\alpha_{1,1} & \cdots & \alpha_{n,1}\\
+\vdots &  & \vdots\\
+\alpha_{1,m} & \cdots & \alpha_{n,m}
+\end{array}\right]\left[\begin{array}{c}
+\beta_{1}\\
+\vdots\\
+\beta_{n}
+\end{array}\right]=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Theorem
+\begin_inset CommandInset label
+LatexCommand label
+name "thm:matrika-kompozituma-linearnih"
+
+\end_inset
+
+matrika kompozituma linearnih preslikav.
+ Posplošitev formule 
+\begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+ se glasi 
+\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+.
+ Trdimo,
+ da je kompozitum linearnih preslikav spet linearna preslikava in da enačba velja.
+\end_layout
+
+\begin_layout Proof
+Najprej dokažimo,
+ da je kompozitum linearnih preslikav spet linearna preslikava.
+\begin_inset Formula 
+\[
+\left(K\circ L\right)\left(\alpha u+\beta v\right)=K\left(L\left(\alpha u+\beta v\right)\right)=K\left(\alpha Lu+\beta Lv\right)=\alpha KLu+\beta KLv=\alpha\left(K\circ L\right)u+\beta\left(K\circ L\right)v
+\]
+
+\end_inset
+
+Sedaj pa dokažimo še enačbo 
+\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+.
+ Naj bosta 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ in 
+\begin_inset Formula $K:V\to W$
+\end_inset
+
+ linearni preslikavi,
+ 
+\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ baza 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+ baza 
+\begin_inset Formula $V$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{D}$
+\end_inset
+
+ baza 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Od prej vemo,
+ da:
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
+\left[Lu_{1}\right]_{\mathcal{D}} & \cdots & \left[Lu_{n}\right]_{\mathcal{D}}\end{array}\right],
+\]
+
+\end_inset
+
+zato pišimo
+\begin_inset Formula 
+\[
+\left[K\circ L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
+\left[\left(K\circ L\right)u_{1}\right]_{\mathcal{D}} & \cdots & \left[\left(K\circ L\right)u_{n}\right]_{\mathcal{D}}\end{array}\right]=\left[\begin{array}{ccc}
+\left[KLu_{1}\right]_{\mathcal{D}} & \cdots & \left[KLu_{n}\right]_{\mathcal{D}}\end{array}\right]\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}\left[\begin{array}{ccc}
+\left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[\begin{array}{ccc}
+\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Jedro in slika linearne preslikave
+\end_layout
+
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorska prostora nad istim poljem 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ linearna preslikava.
+ Jedro 
+\begin_inset Formula $L$
+\end_inset
+
+ naj bo 
+\begin_inset Formula $\Ker L\coloneqq\left\{ u\in U;Lu=0\right\} $
+\end_inset
+
+ (angl.
+ kernel/null space) in Slika/zaloga vrednosti 
+\begin_inset Formula $L$
+\end_inset
+
+ naj bo 
+\begin_inset Formula $\Slika L\coloneqq\left\{ Lu;\forall u\in U\right\} $
+\end_inset
+
+ (angl.
+ image/range).
+\end_layout
+
+\begin_layout Claim*
+Trdimo naslednje:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\Ker L$
+\end_inset
+
+ je vektorski podprostor v 
+\begin_inset Formula $U$
+\end_inset
+
+ (če vsebuje 
+\begin_inset Formula $\vec{a}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{b}$
+\end_inset
+
+,
+ vsebuje tudi vse LK 
+\begin_inset Formula $\vec{a}$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{b}$
+\end_inset
+
+)
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\Slika L$
+\end_inset
+
+ je vektorski podprostor v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokazujemo dve trditvi:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall u_{1},u_{2}\in\Ker L,\alpha_{1},\alpha_{2}\in F\overset{?}{\Longrightarrow}\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
+\end_inset
+
+.
+ Po predpostavki velja 
+\begin_inset Formula $Lu_{1}=0$
+\end_inset
+
+ in 
+\begin_inset Formula $Lu_{2}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}=0$
+\end_inset
+
+.
+ Iz linearnosti 
+\begin_inset Formula $L$
+\end_inset
+
+ sledi 
+\begin_inset Formula $L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall v_{1},v_{2}\in\Slika L,\beta_{1},\beta_{2}\in F\overset{?}{\Longrightarrow}\beta_{1}v_{1}+\beta_{2}v_{2}\in\Slika L$
+\end_inset
+
+.
+ Po predpostavki velja 
+\begin_inset Formula $\exists u_{1},u_{2}\in U\ni:v_{1}=Lu_{1}\wedge v_{2}=Lu_{2}$
+\end_inset
+
+.
+ Velja torej 
+\begin_inset Formula $\beta_{1}v_{1}+\beta_{2}v_{2}=\beta_{1}Lu_{1}+\beta_{2}Lu_{2}\overset{\text{linearnost}}{=}L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\beta_{1}u_{1}+\beta_{2}u_{2}\in U$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)\in\Slika L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Ničnost 
+\begin_inset Formula $L$
+\end_inset
+
+ je 
+\begin_inset Formula $\n\left(L\right)\coloneqq\dim\Ker L$
+\end_inset
+
+ (angl.
+ nullity) in rang 
+\begin_inset Formula $L$
+\end_inset
+
+ je 
+\begin_inset Formula $\rang\left(L\right)=\dim\Slika L$
+\end_inset
+
+ (angl.
+ rank).
+\end_layout
+
+\begin_layout Remark*
+Jedro in sliko smo definirali za linearne preslikave,
+ vendar ju lahko definiramo tudi za poljubno matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ nad poljem 
+\begin_inset Formula $F$
+\end_inset
+
+,
+ saj smo v 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Matrika-linearne-preslikave"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ dokazali linearni izomorfizem med 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matrikami nad 
+\begin_inset Formula $F$
+\end_inset
+
+ in linearnimi preslikavami 
+\begin_inset Formula $F^{n}\to F^{m}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+Au=\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{m1} & \cdots & a_{mn}
+\end{array}\right]\left[\begin{array}{c}
+u_{1}\\
+\vdots\\
+u_{n}
+\end{array}\right]=\left[\begin{array}{c}
+a_{11}u_{1}+\cdots+a_{1n}u_{n}\\
+\vdots\\
+a_{m1}u_{1}+\cdots+a_{mn}u_{n}
+\end{array}\right]=\left[\begin{array}{c}
+a_{11}\\
+\vdots\\
+a_{m1}
+\end{array}\right]u_{1}+\cdots+\left[\begin{array}{c}
+a_{1n}\\
+\vdots\\
+a_{mn}
+\end{array}\right]u_{n}
+\]
+
+\end_inset
+
+Iz tega je razvidno,
+ da je 
+\begin_inset Formula $\Slika A$
+\end_inset
+
+ torej linearna ogrinjača stolpcev matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Pravimo tudi,
+ da je 
+\begin_inset Formula $\Slika A$
+\end_inset
+
+ stolpični prostor 
+\begin_inset Formula $A$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\Col A$
+\end_inset
+
+ (angl.
+ column space).
+ 
+\begin_inset Formula $\rang A=\dim\Slika A$
+\end_inset
+
+ je torej največje število linearno neodvisnih stolpcev 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Linearna preslikava 
+\begin_inset Formula $L$
+\end_inset
+
+ je injektivna (
+\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
+\end_inset
+
+) 
+\begin_inset Formula $\Leftrightarrow\Ker L=\left\{ 0\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ da je 
+\begin_inset Formula $L$
+\end_inset
+
+ injektivna,
+ torej 
+\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
+\end_inset
+
+.
+ Vzemimo poljuben 
+\begin_inset Formula $u\in\Ker L$
+\end_inset
+
+.
+ Zanj velja 
+\begin_inset Formula $Lu=0=L0\Rightarrow u=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo,
+ 
+\begin_inset Formula $\Ker L=\left\{ 0\right\} $
+\end_inset
+
+.
+ Računajmo:
+ 
+\begin_inset Formula $Lu_{1}=Lu_{2}\Longrightarrow Lu_{1}-Lu_{2}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(u_{1}-u_{2}\right)=0\Longrightarrow u_{1}-u_{2}=0\Longrightarrow u_{1}=u_{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+osnovna formula.
+ Naj bo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ linearna preslikava.
+ Tedaj je 
+\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim U$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\n L+\rang L=\dim U$
+\end_inset
+
+.
+ Za matrike torej trdimo 
+\begin_inset Formula $\n A+\rang A=\dim F^{n}=n$
+\end_inset
+
+ za 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matriko 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Vemo,
+ da sta jedro in slika podprostora.
+ Naj bo 
+\begin_inset Formula $w_{1},\dots,w_{k}$
+\end_inset
+
+ baza jedra in 
+\begin_inset Formula $u_{1},\dots,u_{l}$
+\end_inset
+
+ njena dopolnitev do baze 
+\begin_inset Formula $U$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\dim U=k+l=\n L+l$
+\end_inset
+
+.
+ Treba je še dokazati,
+ da je 
+\begin_inset Formula $l=\rang A$
+\end_inset
+
+.
+ Konstruirajmo bazo za 
+\begin_inset Formula $\Slika L$
+\end_inset
+
+,
+ ki ima 
+\begin_inset Formula $l$
+\end_inset
+
+ elementov in dokažimo,
+ da so 
+\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
+\end_inset
+
+ baza za 
+\begin_inset Formula $\Slika L$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Je ogrodje?
+ Vzemimo poljuben 
+\begin_inset Formula $v\in\Slika L$
+\end_inset
+
+.
+ Zanj obstaja nek 
+\begin_inset Formula $u\in U\ni:Lu=v$
+\end_inset
+
+,
+ ki ga lahko razvijemo po bazi 
+\begin_inset Formula $U$
+\end_inset
+
+ takole 
+\begin_inset Formula $u=\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}$
+\end_inset
+
+.
+ Sedaj na obeh straneh uporabimo 
+\begin_inset Formula $L$
+\end_inset
+
+ in upoštevamo linearnost:
+\begin_inset Formula 
+\[
+v=Lu=L\left(\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}\right)=\alpha_{1}Lw_{1}+\cdots+\alpha_{k}Lw_{k}+\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
+\]
+
+\end_inset
+
+Ker so 
+\begin_inset Formula $w_{i}$
+\end_inset
+
+ baza 
+\begin_inset Formula $\Ker L$
+\end_inset
+
+,
+ so elementi 
+\begin_inset Formula $\Ker L$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $Lw_{i}=0$
+\end_inset
+
+ za vsak 
+\begin_inset Formula $i$
+\end_inset
+
+.
+ Tako poljuben 
+\begin_inset Formula $v\in\Slika L$
+\end_inset
+
+ razpišemo z bazo velikosti 
+\begin_inset Formula $l$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Je LN?
+ Računajmo:
+ 
+\begin_inset Formula $\gamma_{1}Lu_{1}+\cdots+\gamma_{l}Lu_{l}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\right)=0\Longrightarrow\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\in\Ker L$
+\end_inset
+
+,
+ kar pomeni,
+ da ga je moč razviti po bazi 
+\begin_inset Formula $\Ker L$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}=\delta_{1}w_{1}+\cdots+\delta_{k}w_{k}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}-\delta_{1}w_{1}-\cdots-\delta_{k}w_{k}=0
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $w_{1},\dots,w_{k},u_{1},\dots,u_{l}$
+\end_inset
+
+ baza 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ je LN,
+ zato velja 
+\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=w_{1}=\cdots=w_{k}=0$
+\end_inset
+
+,
+ kar pomeni,
+ da očitno velja 
+\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
+\end_inset
+
+,
+ torej je res LN.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Bralcu prav pride skica s 3.
+ strani zapiskov predavanja 
+\begin_inset Quotes gld
+\end_inset
+
+LA1P FMF 2024-02-28
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph*
+Do preproste matrike preslikave z ustreznimi bazami
+\end_layout
+
+\begin_layout Standard
+Imenujmo sedaj 
+\begin_inset Formula $\mathcal{B}=\left\{ w_{1},\dots,w_{k},u_{1},\dots,u_{l}\right\} $
+\end_inset
+
+ bazo za 
+\begin_inset Formula $U$
+\end_inset
+
+,
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ Lu_{1},\dots,Lu_{l},z_{1},\dots,z_{m}\right\} $
+\end_inset
+
+ baza za 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $z_{1},\dots,z_{m}$
+\end_inset
+
+ dopolnitev 
+\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
+\end_inset
+
+ do baze 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $V$
+\end_inset
+
+ je lahko večji kot samo 
+\begin_inset Formula $\Slika L$
+\end_inset
+
+,
+ in si oglejmo matriko naše preslikave 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+,
+ ki slika iz baze 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ v bazo 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+.
+ Najprej razpišimo preslikane elemente baze 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ po bazi 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\begin{array}{ccccccccccccc}
+Lu_{1} & = & 1\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
+\vdots &  & \vdots &  &  &  & \vdots &  & \vdots &  &  &  & \vdots\\
+Lu_{l} & = & 0\cdot Lu_{1} & + & \cdots & + & 1\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
+Lw_{1} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
+\vdots &  & \vdots &  &  &  & \vdots &  & \vdots &  &  &  & \vdots\\
+Lw_{k} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cc}
+I_{l} & 0\\
+0 & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+S primerno izbiro baz 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $V$
+\end_inset
+
+ je torej matrika preslikave precej preprosta,
+ zgolj bločna matrika z identiteto,
+ veliko 
+\begin_inset Formula $\rang L$
+\end_inset
+
+ in ničlami,
+ ki ustrezajo dimenzijam 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kaj pa,
+ če je 
+\begin_inset Formula $L$
+\end_inset
+
+ matrika?
+ Recimo ji 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $L_{A}=L$
+\end_inset
+
+ od prej.
+ Tedaj 
+\begin_inset Formula $A\in M_{p,n}\left(F\right)$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $P=\left[\begin{array}{cccccc}
+u_{1} & \cdots & u_{l} & w_{1} & \cdots & w_{k}\end{array}\right]$
+\end_inset
+
+ matrika,
+ katere stolpci so baza 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $Q=\left[\begin{array}{cccccc}
+Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]$
+\end_inset
+
+ matrika,
+ katere stolpci so baza 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Po karakterizaciji obrnljivih matrik sta obrnljivi.
+ Tedaj
+\begin_inset Formula 
+\[
+AP=\left[\begin{array}{cccccc}
+Au_{1} & \cdots & Au_{l} & Aw_{1} & \cdots & Aw_{k}\end{array}\right]\overset{\text{jedro}}{=}\left[\begin{array}{cccccc}
+Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Q\left[\begin{array}{cc}
+I_{l} & 0\\
+0 & 0
+\end{array}\right]=\left[\begin{array}{cccccc}
+Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]\left[\begin{array}{cc}
+I_{l} & 0\\
+0 & 0
+\end{array}\right]=\left[\begin{array}{cccccc}
+Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+AP=Q\left[\begin{array}{cc}
+I_{l} & 0\\
+0 & 0
+\end{array}\right]\Longrightarrow Q^{-1}AP=\left[\begin{array}{cc}
+I_{l} & 0\\
+0 & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Ekvivalentnost matrik
+\end_layout
+
+\begin_layout Definition*
+Matriki 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ sta ekvivalentni (oznaka 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Isto oznako uporabljamo tudi za podobne matrike,
+ vendar podobnost ni enako kot ekvivalentnost.
+\end_layout
+
+\end_inset
+
+) 
+\begin_inset Formula $\Leftrightarrow\exists$
+\end_inset
+
+ obrnljivi 
+\begin_inset Formula $P,Q\ni:B=PAQ$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Dokazali smo,
+ da je vsaka matrika 
+\begin_inset Formula $A$
+\end_inset
+
+ ekvivalentna matriki 
+\begin_inset Formula $\left[\begin{array}{cc}
+I_{r} & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $r=\rang A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo,
+ da je relacija 
+\begin_inset Formula $\sim$
+\end_inset
+
+ ekvivalenčna:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+refleksivnost:
+ 
+\begin_inset Formula $A\sim A$
+\end_inset
+
+ velja.
+ Naj bo 
+\begin_inset Formula $A\in M_{m,n}\left(F\right)$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $A=I_{m}AI_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+simetričnost:
+ 
+\begin_inset Formula $A\sim B\Rightarrow B\sim A$
+\end_inset
+
+,
+ kajti če velja 
+\begin_inset Formula $B=PAQ$
+\end_inset
+
+ in sta 
+\begin_inset Formula $P$
+\end_inset
+
+ in 
+\begin_inset Formula $Q$
+\end_inset
+
+ obrnljivi,
+ velja 
+\begin_inset Formula $P^{-1}BQ^{-1}=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+tranzitivnost:
+ 
+\begin_inset Formula $A\sim B\wedge B\sim C\Rightarrow A\sim C$
+\end_inset
+
+,
+ kajti,
+ če velja 
+\begin_inset Formula $B=PAQ$
+\end_inset
+
+ in 
+\begin_inset Formula $C=SBT$
+\end_inset
+
+ in so 
+\begin_inset Formula $P,Q,S,T$
+\end_inset
+
+ obrnljive,
+ velja 
+\begin_inset Formula $C=\left(SP\right)A\left(QT\right)$
+\end_inset
+
+ in produkt obrnljivih matrik je obrnljiva matrika.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Dve matriki sta ekvivalentni natanko tedaj,
+ ko imata enako velikost in enak rang.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Po predpostavki imata 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ enako velikost in enak rang 
+\begin_inset Formula $r$
+\end_inset
+
+.
+ Od prej vemo,
+ da sta obe ekvivalentni 
+\begin_inset Formula $\left[\begin{array}{cc}
+I_{r} & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ ker pa je relacija ekvivalentnosti ekvivalenčna,
+ sta 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki 
+\begin_inset Formula $A\sim B$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\exists P,Q\ni:B=PAQ$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ 
+\begin_inset Formula $m\times n$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $P$
+\end_inset
+
+ 
+\begin_inset Formula $m\times m$
+\end_inset
+
+ in 
+\begin_inset Formula $Q$
+\end_inset
+
+ 
+\begin_inset Formula $n\times n$
+\end_inset
+
+,
+ zatorej je po definiciji matričnega množenja 
+\begin_inset Formula $B$
+\end_inset
+
+ 
+\begin_inset Formula $m\times n$
+\end_inset
+
+.
+ Dokazati je treba še 
+\begin_inset Formula $\rang A=\rang B$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\rang B=\rang PAQ\overset{?}{=}\rang PA=\overset{?}{=}\rang A
+\]
+
+\end_inset
+
+Dokažimo najprej 
+\begin_inset Formula $\rang PAQ=\rang PA$
+\end_inset
+
+ oziroma 
+\begin_inset Formula $\rang CQ=\rang C$
+\end_inset
+
+ za obrnljivo 
+\begin_inset Formula $Q$
+\end_inset
+
+ in poljubno C.
+ Dokažemo lahko celo 
+\begin_inset Formula $\Slika CQ=\Slika C$
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+\forall u:u\in\Slika CQ\Leftrightarrow\exists v\ni:u=\left(CQ\right)v\Leftrightarrow\exists v'\ni:u=Cv'\Leftrightarrow u\in\Slika C.
+\]
+
+\end_inset
+
+Sedaj dokažimo še 
+\begin_inset Formula $\rang\left(PA\right)=\rang\left(A\right)$
+\end_inset
+
+.
+ Zadošča dokazati,
+ da je 
+\begin_inset Formula $\Ker\left(PA\right)=\Ker A$
+\end_inset
+
+,
+ kajti tedaj bi iz enakosti izrazov
+\begin_inset Formula 
+\[
+\dim\Slika A+\dim\Ker A=\dim F^{n}=n
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\dim\Slika PA+\dim\Ker PA=\dim F^{n}=n
+\]
+
+\end_inset
+
+dobili 
+\begin_inset Formula $\dim\Slika PA=\dim\Slika A$
+\end_inset
+
+.
+ Dokažimo torej 
+\begin_inset Formula $\Ker PA=\Ker A$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\forall u:u\in\Ker PA\Leftrightarrow PAu=0\overset{P\text{ obrnljiva}}{\Longleftrightarrow}Au=0\Leftrightarrow u\in\Ker A.
+\]
+
+\end_inset
+
+Torej je res 
+\begin_inset Formula $\Ker PA=\Ker A$
+\end_inset
+
+,
+ torej je res 
+\begin_inset Formula $\rang PA=\rang A$
+\end_inset
+
+,
+ torej je res 
+\begin_inset Formula $\rang A=\rang B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Podobnost matrik
+\end_layout
+
+\begin_layout Definition*
+Kvadratni matriki 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ sta podobni,
+ če 
+\begin_inset Formula $\exists$
+\end_inset
+
+ taka obrnljiva matrika 
+\begin_inset Formula $P\ni:B=PAP^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Podobnost je ekvivalenčna relacija.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo,
+ da je relacija ekvivalenčna,
+ torej:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+refleksivna:
+ 
+\begin_inset Formula $A=IAI^{-1}=IAI=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+simetrična:
+ 
+\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+tranzitivna:
+ 
+\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Remark
+\begin_inset CommandInset label
+LatexCommand label
+name "rem:nista-podobni"
+
+\end_inset
+
+Očitno velja podobnost 
+\begin_inset Formula $\Rightarrow$
+\end_inset
+
+ ekvivalentnost,
+ toda obrat ne velja vedno.
+ Na primer 
+\begin_inset Formula $\left[\begin{array}{cc}
+1 & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ sta ekvivalentni (sta enake velikosti in ranga),
+ toda nista podobni (dokaz kasneje).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Od prej vemo,
+ da je vsaka matrika ekvivalentna matriki 
+\begin_inset Formula $\left[\begin{array}{cc}
+I_{r} & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $r$
+\end_inset
+
+ njen rang.
+ A je vsaka kvadratna matrika podobna kakšni lepi matriki?
+ Ja.
+ Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje).
+ Toda a je vsaka kvadratna matrika podobna diagonalni matriki?
+ Ne.
+\end_layout
+
+\begin_layout Definition*
+Matrika 
+\begin_inset Formula $D$
+\end_inset
+
+ je diagonalna 
+\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kdaj je matrika 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna neki diagonalni matriki?
+ Kdaj 
+\begin_inset Formula $\exists$
+\end_inset
+
+ diagonalna 
+\begin_inset Formula $D$
+\end_inset
+
+ in obrnljiva 
+\begin_inset Formula $P\ni:A=PDP^{-1}$
+\end_inset
+
+?
+ Izpeljimo iz nastavka.
+ 
+\begin_inset Formula $D=\left[\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda n
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $P=\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$
+\end_inset
+
+,
+ kjer sta 
+\begin_inset Formula $D$
+\end_inset
+
+ in 
+\begin_inset Formula $P$
+\end_inset
+
+ neznani.
+ Ker mora biti 
+\begin_inset Formula $P$
+\end_inset
+
+ obrnljiva,
+ so njeni stolpični vektorji LN.
+\begin_inset Formula 
+\[
+A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
+\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda_{n}
+\end{array}\right]\text{ in }P\text{ obrnljiva}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
+\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Porodi se naloga,
+ imenovana 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Iščemo pare 
+\begin_inset Formula $\left(\lambda,\vec{v}\right)$
+\end_inset
+
+,
+ ki zadoščajo enačbi 
+\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ je lastna vrednost matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če obstaja tak 
+\begin_inset Formula $\vec{v}\not=0$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
+\end_inset
+
+.
+ V tem primeru pravimo,
+ da je 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+ lastni vektor,
+ ki pripada lastni vrednosti 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Paru 
+\begin_inset Formula $\left(\lambda,\vec{v}\right)$
+\end_inset
+
+,
+ ki zadošča enačbi,
+ pravimo lastni par matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Nalogo 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+ rešujemo v dveh korakih.
+ Najprej najdemo vse 
+\begin_inset Formula $\lambda$
+\end_inset
+
+,
+ nato za vsako poiščemo pripadajoče 
+\begin_inset Formula $\vec{v}$
+\end_inset
+
+,
+ ki za lastno vrednost obstajajo po definiciji.
+\end_layout
+
+\begin_layout Standard
+Za nek 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ pišimo 
+\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$
+\end_inset
+
+ za nek 
+\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$
+\end_inset
+
+ ni obrnljiva 
+\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Polinom 
+\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$
+\end_inset
+
+ je karakteristični polinom matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Premislek zgoraj nam pove,
+ da so lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ ničle 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Karakteristični polinom lahko nima nobene ničle:
+ 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+-1 & 0
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc}
+-\lambda & 1\\
+-1 & -\lambda
+\end{array}\right]=x^{2}+1$
+\end_inset
+
+,
+ katerega ničli sta 
+\begin_inset Formula $\lambda_{1}=i$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{2}=-i$
+\end_inset
+
+,
+ ki nista realni števili.
+ V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti,
+ saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle.
+\end_layout
+
+\begin_layout Standard
+Kako pa iščemo lastne vektorje za lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+?
+ Spomnimo se na 
+\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$
+\end_inset
+
+.
+ Rešiti moramo homogen sistem linearnih enačb.
+ Po definiciji so lastni vektorji neničelni,
+ zato nas trivialna rešitev ne zanima.
+\end_layout
+
+\begin_layout Definition*
+Množici 
+\begin_inset Formula $\Ker\left(A-\lambda I\right)$
+\end_inset
+
+ pravimo lastni podprostor matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ ki pripada 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Slednji vsebuje 
+\begin_inset Formula $\vec{0}$
+\end_inset
+
+ in množico vektorjev,
+ ki so vsi lastni vektorji 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Exercise*
+Izračunaj lastne vrednosti od 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+-1 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Od prej vemo,
+ da 
+\begin_inset Formula $\lambda_{1}=i$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{2}=-i$
+\end_inset
+
+.
+ Izračunajmo 
+\begin_inset Formula $\Ker\left(A-iI\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\Ker\left(A+iI\right)$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\Ker\left(A-iI\right):\quad\left[\begin{array}{cc}
+-i & 1\\
+-1 & -i
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
+1\\
+i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker\left(A+iI\right):\quad\left[\begin{array}{cc}
+i & 1\\
+-1 & i
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
+1\\
+-i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c}
+1\\
+i
+\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c}
+1\\
+-i
+\end{array}\right]\right\} 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Exercise*
+Vstavimo lastna vektorja v 
+\begin_inset Formula $P$
+\end_inset
+
+ in lastne vrednosti v 
+\begin_inset Formula $D$
+\end_inset
+
+ na pripadajoči mesti.
+ Dobimo obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+ in velja 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+P=\left[\begin{array}{cc}
+1 & 1\\
+i & -i
+\end{array}\right],\quad D=\left[\begin{array}{cc}
+i & 0\\
+0 & -i
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Exercise*
+Temu početju pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+diagonalizacija matrike 
+\begin_inset Formula $A$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primer matrike,
+ ki ni diagonalizabilna:
+ 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc}
+-\lambda & 1\\
+0 & -\lambda
+\end{array}\right]=\lambda^{2}$
+\end_inset
+
+.
+ Ničli/lastni vrednosti sta 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{2}=0$
+\end_inset
+
+.
+ Toda 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c}
+1\\
+0
+\end{array}\right]\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $P=\left[\begin{array}{cc}
+1 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ ni obrnljiva.
+ S tem dokažemo trditev v primeru 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "rem:nista-podobni"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+ 
+\begin_inset Formula $\left[\begin{array}{cc}
+1 & 0\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\begin{array}{cc}
+0 & 1\\
+0 & 0
+\end{array}\right]$
+\end_inset
+
+ nista podobni,
+ ker je prva diagonalna,
+ druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati).
+\end_layout
+
+\begin_layout Standard
+Lastne vrednosti lahko definiramo tudi za linearne preslikave,
+ saj so linearne preslikave linearno izomorfne matrikam.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad 
+\begin_inset Formula $F=\mathbb{C}$
+\end_inset
+
+ in 
+\begin_inset Formula $L:V\to V$
+\end_inset
+
+ linearna preslikava.
+ Število 
+\begin_inset Formula $\lambda\in F$
+\end_inset
+
+ je lastna vrednost 
+\begin_inset Formula $L$
+\end_inset
+
+,
+ le obstaja tak neničelni 
+\begin_inset Formula $v\in V$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $Lv=\lambda v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Kako pa rešujemo 
+\begin_inset Quotes gld
+\end_inset
+
+Lastni problem
+\begin_inset Quotes grd
+\end_inset
+
+ za linearne preslikave?
+ 
+\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$
+\end_inset
+
+.
+ Toda determinante linearne preslikave nismo definirali.
+ Lahko pa determinanto izračunamo na matriki,
+ ki pripada tej linearni preslikavi.
+ Toda dvem različnim bazam pripadata različni matriki linearne preslikave.
+ Dokazati je treba,
+ da sta determinanti dveh matrik,
+ pripadajočih eni linearni preslikavi,
+ enaki,
+ četudi sta matriki v različnih bazah.
+\end_layout
+
+\begin_layout Lemma
+\begin_inset CommandInset label
+LatexCommand label
+name "lem:Podobni-matriki-imata"
+
+\end_inset
+
+Podobni matriki imata isto determinanto.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $B=PAP^{-1}$
+\end_inset
+
+ za neko obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $L:V\to V$
+\end_inset
+
+ naj bo linearna preslikava,
+ 
+\begin_inset Formula $V$
+\end_inset
+
+ prostor nad 
+\begin_inset Formula $F=\mathbb{C}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+ pa bazi 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Priredimo matriki 
+\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$
+\end_inset
+
+ in 
+\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$
+\end_inset
+
+.
+ Spomnimo se izreka 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "thm:matrika-kompozituma-linearnih"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $L=\left[id\circ L\circ id\right]$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$
+\end_inset
+
+ za neko obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ Torej sta matriki 
+\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$
+\end_inset
+
+ podobni,
+ torej imata po lemi 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "lem:Podobni-matriki-imata"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ isto determinanto.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Alternativen dokaz,
+ da imata podobni matriki iste lastne vrednosti:
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna 
+\begin_inset Formula 
+\[
+B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B},
+\]
+
+\end_inset
+
+torej so lastne vrednosti enake.
+ Kaj pa lastni vektorji?
+ Naj bo 
+\begin_inset Formula $v$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv,
+\]
+
+\end_inset
+
+torej za 
+\begin_inset Formula $v$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $A$
+\end_inset
+
+ sledi,
+ da je 
+\begin_inset Formula $Pv$
+\end_inset
+
+ lastni vektor 
+\begin_inset Formula $B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Linearni transformaciji torej priredimo tako matriko,
+ ki ima v začetnem in končnem prostoru isto bazo.
+ Tedaj lahko izračunamo lastne pare na tej matriki.
+\end_layout
+
+\begin_layout Theorem*
+Schurov izrek.
+ Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki.
+\end_layout
+
+\begin_layout Proof
+Indukcija po velikosti matrike.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $A_{1\times1}$
+\end_inset
+
+ je zgornjetrikotna.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Po I.
+ P.
+ trdimo,
+ da je vsaka 
+\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$
+\end_inset
+
+ podobna kaki zgornjetrikotni matriki.
+ Dokažimo še za poljubno 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ pripadajoči lastni vektor ter 
+\begin_inset Formula $v_{2},\dots,v_{n}$
+\end_inset
+
+ dopolnitev 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ do baze 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Potem je matrika 
+\begin_inset Formula $P=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+ obrnljiva.
+\begin_inset Formula 
+\[
+AP=\left[\begin{array}{ccc}
+Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc}
+\lambda & a_{1,2} & \cdots & a_{1,n}\\
+0 & \vdots &  & \vdots\\
+\vdots & \vdots &  & \vdots\\
+0 & a_{m,n} & \cdots & a_{m.n}
+\end{array}\right]=P\left[\begin{array}{cc}
+\lambda & B\\
+0 & C
+\end{array}\right]
+\]
+
+\end_inset
+
+Po I.
+ P.
+ obstaja taka zgornjetrikotna 
+\begin_inset Formula $T$
+\end_inset
+
+ in obrnljiva 
+\begin_inset Formula $Q$
+\end_inset
+
+,
+ da 
+\begin_inset Formula $C=QTQ^{-1}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]^{-1}\left[\begin{array}{cc}
+\lambda & B\\
+0 & C
+\end{array}\right]\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{cc}
+\lambda & C\\
+0 & Q^{-1}B
+\end{array}\right]\left[\begin{array}{cc}
+1 & 0\\
+0 & Q
+\end{array}\right]=\left[\begin{array}{cc}
+\lambda & CQ\\
+0 & B
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula $A$
+\end_inset
+
+ je torej podobna 
+\begin_inset Formula $\left[\begin{array}{cc}
+\lambda & CQ\\
+0 & B
+\end{array}\right]$
+\end_inset
+
+,
+ ki je zgornjetrikotna.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO karakterizacija linearnih preslikav 
+\begin_inset Quotes gld
+\end_inset
+
+LA1V FMF 2024-03-12
+\begin_inset Quotes grd
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Zadosten pogoj za diagonalizabilnost
+\end_layout
+
+\begin_layout Theorem
+\begin_inset CommandInset label
+LatexCommand label
+name "thm:lave-razl-lavr-so-LN"
+
+\end_inset
+
+Lastni vektorji,
+ ki pripadajo različnim lastnim vrednostim,
+ so linearno neodvisni.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika,
+ 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ njene lastne vrednosti in 
+\begin_inset Formula $v_{1},\dots,v_{k}$
+\end_inset
+
+ njim pripadajoči lastni vektorji.
+ Dokazujemo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ paroma različni 
+\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
+\end_inset
+
+ LN.
+ Dokaz z indukcijo po 
+\begin_inset Formula $k$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza 
+\begin_inset Formula $k=1$
+\end_inset
+
+:
+ Elementi 
+\begin_inset Formula $\left\{ \lambda_{1}\right\} $
+\end_inset
+
+ so trivialno paroma različni in 
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ je kot neničen vektor LN.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Dokazujemo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$
+\end_inset
+
+ so paroma različne 
+\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
+\end_inset
+
+ so LN,
+ vedoč I.
+ P.
+ Denimo,
+ da 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+.
+ Množimo z 
+\begin_inset Formula $A$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
+\]
+
+\end_inset
+
+Množimo začetno enačbo z 
+\begin_inset Formula $\lambda_{k+1}$
+\end_inset
+
+ (namesto z 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ kot smo to storili zgoraj):
+\begin_inset Formula 
+\[
+\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
+\]
+
+\end_inset
+
+Odštejmo eno enačbo od druge,
+ dobiti moramo 0,
+ saj odštevamo 0 od 0:
+\begin_inset Formula 
+\[
+\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0
+\]
+
+\end_inset
+
+Ker so lastne vrednosti paroma različne (
+\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$
+\end_inset
+
+),
+ so njihove razlike neničelne.
+ Ker so 
+\begin_inset Formula $v_{1},\dots,v_{k}$
+\end_inset
+
+ po predpostavki LN,
+ sledi 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$
+\end_inset
+
+.
+ Vstavimo te konstante v 
+\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $v_{k+1}$
+\end_inset
+
+ neničeln (je namreč lastni vektor),
+ sledi 
+\begin_inset Formula $\alpha_{k+1}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$
+\end_inset
+
+,
+ zatorej so 
+\begin_inset Formula $v_{1},\dots,v_{k+1}$
+\end_inset
+
+ res LN.
+\end_layout
+
+\end_deeper
+\begin_layout Corollary
+\begin_inset CommandInset label
+LatexCommand label
+name "cor:vsota-lastnih-podpr-direktna"
+
+\end_inset
+
+Vsota vseh lastnih podprostorov matrike je direktna (definicija 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "def:vsota-je-direktna"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Naj bodo 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti matrike 
+\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$
+\end_inset
+
+.
+ Pripadajoči lastni podprostori so torej 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$
+\end_inset
+
+.
+ Trdimo,
+ da je vsota teh podprostorov direktna,
+ torej 
+\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$
+\end_inset
+
+.
+ To sledi iz izreka 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "thm:lave-razl-lavr-so-LN"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Če ima 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matrika 
+\begin_inset Formula $n$
+\end_inset
+
+ paroma različnih lastnih vrednosti,
+ je podobna diagonalni matriki.
+\end_layout
+
+\begin_layout Proof
+Po posledici 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "cor:vsota-lastnih-podpr-direktna"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ je vsota lastnih podprostorov matrike 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ direktna.
+ Če je torej lastnih podprostorov 
+\begin_inset Formula $n$
+\end_inset
+
+,
+ je njihova vsota cel prostor 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Matriko se da diagonalizirati,
+ kadar je vsota vseh lastnih podprostorov enaka podprostoru 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ (tedaj so namreč stolpci matrike 
+\begin_inset Formula $P$
+\end_inset
+
+ linearno neodvisni,
+ zato je 
+\begin_inset Formula $P$
+\end_inset
+
+ obrnljiva).
+\end_layout
+
+\begin_layout Subsubsection
+Algebraične in geometrijske vekčratnosti
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika.
+ 
+\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$
+\end_inset
+
+,
+ kjer so 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Stopnji ničle —
+ 
+\begin_inset Formula $n_{i}$
+\end_inset
+
+ —
+ rečemo algebraična večkratnost lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Geometrijska večkratnost lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ je 
+\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$
+\end_inset
+
+
+\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Algebraično večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ označimo z 
+\begin_inset Formula $n_{i}$
+\end_inset
+
+ in je večkratnost ničle 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ v 
+\begin_inset Formula $p_{A}\left(\lambda\right)$
+\end_inset
+
+ (karakterističnem polinomu).
+ Geometrijsko večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ pa označimo z 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ in je dimenzija lastnega podprostora za 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:geom<=alg"
+
+\end_inset
+
+
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
+\end_inset
+
+ —
+ geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna,
+ kot je algebraična večkratnost te lastne vrednosti.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v_{1},\dots,v_{m_{i}}$
+\end_inset
+
+ baza za lastni podprostor 
+\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$
+\end_inset
+
+ njena dopolnitev do baze 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Tedaj velja:
+ 
+\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $Av_{m_{i}+1}=$
+\end_inset
+
+ linearna kombinacija 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $Av_{n}=$
+\end_inset
+
+ linearna kombinacija 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $P=\left[\begin{array}{cccccc}
+v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+,
+ ki je obrnljiva.
+\begin_inset Formula 
+\[
+P^{-1}AP=\left[\begin{array}{cc}
+\lambda_{i}I_{m_{i}} & B\\
+0 & C
+\end{array}\right]
+\]
+
+\end_inset
+
+Ker je karakteristični polinom neodvisen od izbire baze,
+ velja 
+\begin_inset Formula 
+\[
+\det\left(A-xI_{n}\right)=\det\left(\lambda_{i}I_{m_{i}}-xI\right)\det\left(C-xI_{n-m_{i}}\right)=\left(\lambda_{i}-x\right)^{m_{i}}\det\left(C-xI_{n-m_{i}}\right)
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\left(\lambda-x\right)^{m_{i}}$
+\end_inset
+
+ deli karakteristični polinom,
+ je algebraična večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+ vsaj tolikšna,
+ kot je geometrična.
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:mi=ni=>diag"
+
+\end_inset
+
+Matriko s paroma različnimi lastnimi vrednostmi 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ je moč diagonalizirati 
+\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $V_{i}$
+\end_inset
+
+ lastno podprostor lastne vrednosti 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+ Vemo,
+ da se da 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ diagonalizirati 
+\begin_inset Formula $\Leftrightarrow A$
+\end_inset
+
+ ima 
+\begin_inset Formula $n$
+\end_inset
+
+ LN stolpičnih vektorjev 
+\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$
+\end_inset
+
+ vsota lastnih podprostorov je direktna 
+\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$
+\end_inset
+
+.
+ Toda ker po prejšnjem izreku 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Minimalni polinom matrike
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$
+\end_inset
+
+ polinom in 
+\begin_inset Formula $A$
+\end_inset
+
+ matrika.
+ 
+\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $p\left(A\right)=0$
+\end_inset
+
+ (ničelna matrika),
+ pravimo,
+ da polinom 
+\begin_inset Formula $p$
+\end_inset
+
+ anhilira/uniči matriko 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Izkaže se,
+ da karakteristični polimom anhilira matriko —
+ 
+\begin_inset Formula $p_{A}\left(A\right)=0$
+\end_inset
+
+.
+ Dokaz kasneje.
+\end_layout
+
+\begin_layout Definition*
+Polinom 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ je minimalen polinom 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če velja:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $m\left(A\right)=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $m$
+\end_inset
+
+ ima vodilni koeficient 1
+\end_layout
+
+\begin_layout Enumerate
+med vsemi polinomi,
+ ki zadoščajo prvi in drugi zahtevi,
+ ima 
+\begin_inset Formula $m$
+\end_inset
+
+ najnižjo stopnjo
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+eksistenca minimalnega polinoma —
+ Minimalni polinom obstaja.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ matrika.
+ Očitno je 
+\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ vektorski prostor dimenzije 
+\begin_inset Formula $n^{2}$
+\end_inset
+
+.
+ Matrike 
+\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $
+\end_inset
+
+ so linearno odvisne,
+ ker je moč te množice za 1 večja od moči vektorskega prostora.
+ Torej 
+\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$
+\end_inset
+
+,
+ ki niso vse 0 
+\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$
+\end_inset
+
+.
+ Torej polinom 
+\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Če ta polinom delimo z njegovim vodilnim koeficientom,
+ dobimo polinom,
+ ki ustreza prvima dvema zahevama za minimalni polinom.
+ Če med vsemi takimi izberemo takega z najnižjo stopnjo,
+ le-ta ustreza še tretji zahtevi.
+\end_layout
+
+\begin_layout Theorem*
+Če je 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ minimalni polinom za 
+\begin_inset Formula $A$
+\end_inset
+
+ in če 
+\begin_inset Formula $p\left(x\right)$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$
+\end_inset
+
+ (
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+ deli 
+\begin_inset Formula $p\left(x\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Delimo 
+\begin_inset Formula $p$
+\end_inset
+
+ z 
+\begin_inset Formula $m$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$
+\end_inset
+
+.
+ Vstavimo 
+\begin_inset Formula $A$
+\end_inset
+
+ na obe strani:
+\begin_inset Formula 
+\[
+0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0
+\]
+
+\end_inset
+
+Sledi 
+\begin_inset Formula $r\left(x\right)=0$
+\end_inset
+
+,
+ kajti če 
+\begin_inset Formula $r$
+\end_inset
+
+ ne bi bil ničeln polinom,
+ bi ga lahko delili z vodilnim koeficientom in po predpostavki 
+\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$
+\end_inset
+
+ bi imel manjšo stopnjo kot 
+\begin_inset Formula $m\left(x\right)$
+\end_inset
+
+,
+ torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od 
+\begin_inset Formula $m$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $m$
+\end_inset
+
+ ne bi bil minimalni polinom,
+ kar bi vodilo v protislovje.
+\end_layout
+
+\begin_layout Corollary*
+enoličnost minimalnega polinoma.
+ Naj bosta 
+\begin_inset Formula $m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{2}$
+\end_inset
+
+ minimalna polinoma matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker 
+\begin_inset Formula $m$
+\end_inset
+
+ po definiciji anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ iz prejšnje trditve sledi,
+ če vstavimo 
+\begin_inset Formula $m=m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $p=m_{2}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $m_{1}\vert m_{2}$
+\end_inset
+
+.
+ Toda če vstavimo 
+\begin_inset Formula $m=m_{2}$
+\end_inset
+
+ in 
+\begin_inset Formula $p=m_{1}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $m_{2}\vert m_{1}$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$
+\end_inset
+
+ sledi,
+ da se 
+\begin_inset Formula $m_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{2}$
+\end_inset
+
+ razlikujeta le za konstanten faktor,
+ ki pa je po definiciji minimalnega polinoma 1,
+ torej 
+\begin_inset Formula $m_{1}=m_{2}$
+\end_inset
+
+.
+ Zaradi enoličnosti lahko označimo minimalni polinom 
+\begin_inset Formula $A$
+\end_inset
+
+ z 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Ničle minimalnega polinoma
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+ imata iste ničle 
+\begin_inset Formula $\sim$
+\end_inset
+
+ ničle 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ so lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $p_{A}\left(x\right)$
+\end_inset
+
+ (dokaz kasneje),
+ velja po trditvi v dokazu enoličnosti,
+ da 
+\begin_inset Formula $m_{A}\vert p_{A}$
+\end_inset
+
+,
+ torej je vsaka ničla 
+\begin_inset Formula $m_{A}$
+\end_inset
+
+ tudi ničla 
+\begin_inset Formula $p_{A}$
+\end_inset
+
+.
+ Treba je dokazati še,
+ da je vsaka ničla 
+\begin_inset Formula $p_{A}$
+\end_inset
+
+ tudi ničla 
+\begin_inset Formula $m_{A}$
+\end_inset
+
+,
+ natančneje:
+ Treba je dokazati,
+ da če je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost matrike 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $m_{A}\left(\lambda\right)=0$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ lastni vektor za 
+\begin_inset Formula $\lambda$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+.
+ Potem velja 
+\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$
+\end_inset
+
+ in splošneje 
+\begin_inset Formula $A^{n}v=\lambda^{n}v$
+\end_inset
+
+.
+ Sedaj recimo,
+ da je 
+\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$
+\end_inset
+
+.
+ Potem je,
+ ker minimalni polinom anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ 
+\begin_inset Formula 
+\[
+m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $m_{A}\left(\lambda\right)v=0$
+\end_inset
+
+ in 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ (je namreč lastni vektor),
+ velja 
+\begin_inset Formula $m_{A}\left(\lambda\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph*
+Lastnosti
+\end_layout
+
+\begin_layout Standard
+Ker je 
+\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
+\end_inset
+
+,
+ sledi iz 
+\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$
+\end_inset
+
+.
+ Poleg tega,
+ ker 
+\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$
+\end_inset
+
+.
+ Toda pozor:
+ 
+\series bold
+Ni
+\series default
+ res,
+ da 
+\begin_inset Formula $r_{i}=m_{i}$
+\end_inset
+
+ (stropnja lastnega podprostora).
+\end_layout
+
+\begin_layout Theorem*
+Cayley-Hamilton.
+ 
+\begin_inset Formula $p_{A}\left(A\right)=0$
+\end_inset
+
+ —
+ karakteristični polinom matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ anhilira matriko 
+\begin_inset Formula $A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Spomnimo se eksplicitne formule za celico inverza matrike (razdelek 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "subsec:Formula-za-inverz-matrike"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+),
+ ki pravi 
+\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$
+\end_inset
+
+.
+ Računajmo in naposled vstavimo 
+\begin_inset Formula $B=A-xI$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det B\cdot I=B\tilde{B}^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
+\]
+
+\end_inset
+
+Glede na definicijo 
+\begin_inset Formula $\tilde{A}$
+\end_inset
+
+ je 
+\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$
+\end_inset
+
+ matrika velikosti 
+\begin_inset Formula $n\times n$
+\end_inset
+
+,
+ ki vsebuje polinome stopnje 
+\begin_inset Formula $<n$
+\end_inset
+
+,
+ kajti vsebuje determinante kofaktorskih matrik,
+ torej je takele oblike (
+\begin_inset Formula $\forall i\in\left\{ 1..\left(n-1\right)\right\} :B_{i}\in M_{n}\left(\mathbb{C}\right)$
+\end_inset
+
+):
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+To si predstavljamo tako,
+ da iz vsake celice matrike,
+ ki vsebuje polinom,
+ izpostavimo (homogenost) spremenljivko (torej 
+\begin_inset Formula $x$
+\end_inset
+
+ na fiksno potenco) in nato koeficiente v celicah pred to spremenljivvko zložimo v eno matriko.
+ Slednje ponovimo za vsako potenco in dobimo te matrike 
+\begin_inset Formula $B_{i}$
+\end_inset
+
+.
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$
+\end_inset
+
+.
+ Kot v enačbi množimo to z 
+\begin_inset Formula $I$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$
+\end_inset
+
+.
+ Oglejmo si še desno stran enačbe:
+\begin_inset Formula 
+\[
+\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
+\]
+
+\end_inset
+
+In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe:
+\begin_inset Formula 
+\[
+\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\begin{array}{cccc}
+1: & c_{0}I & = & AB_{0}\\
+x: & c_{1}I & = & AB_{1}-B_{0}\\
+x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\
+\vdots\\
+x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\
+x^{n}: & c_{n}I & = & -B_{n-1}
+\end{array}
+\]
+
+\end_inset
+
+Vstavimo sedaj 
+\begin_inset Formula $A$
+\end_inset
+
+ v enačbo namesto 
+\begin_inset Formula $x$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p_{A}\left(A\right)=0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ se da diagonalizirati 
+\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$
+\end_inset
+
+ ima samo enostavne ničle (nima večkratnih —
+ potence so vse 1).
+ Torej 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$
+\end_inset
+
+ za 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ vse paroma različne lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Po predpostavki je 
+\begin_inset Formula $A$
+\end_inset
+
+ podobna diagonalni matriki —
+ 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+ za diagonalno 
+\begin_inset Formula $D$
+\end_inset
+
+ in obrnljivo 
+\begin_inset Formula $P$
+\end_inset
+
+.
+ BSŠ naj bo 
+\begin_inset Formula $D=\left[\begin{array}{ccc}
+\lambda_{1} & 0 & 0\\
+0 & \cdots & 0\\
+0 & 0 & \lambda_{k}
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$
+\end_inset
+
+.
+ Oglejmo si izraz 
+\begin_inset Formula 
+\[
+\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)=\left[\begin{array}{cccc}
+0 &  &  & 0\\
+ & \lambda_{2}-\lambda_{1}\\
+ &  & \ddots\\
+0 &  &  & \lambda_{k}-\lambda_{1}
+\end{array}\right]\cdots\left[\begin{array}{cccc}
+\lambda_{1}-\lambda_{k} &  &  & 0\\
+ & \ddots\\
+ &  & \lambda_{k-1}-\lambda_{k}\\
+0 &  &  & 0
+\end{array}\right]=0
+\]
+
+\end_inset
+
+Sedaj pa še izraz
+\begin_inset Formula 
+\[
+\left(A-\lambda_{1}I\right)\cdots\left(A-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}I\right)\cdots\left(PDP^{-1}-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}PP^{-1}\right)\cdots\left(PDP^{-1}-\lambda_{k}PP^{-1}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=P\left(D-\lambda_{1}I\right)\cancel{P^{-1}}\cdots\cancel{P}\left(D-\lambda_{k}I\right)P^{-1}=P\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)P^{-1}=P0P^{-1}=0,
+\]
+
+\end_inset
+
+torej ta polinom anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker deli 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+ —
+ vsebuje vse ničle 
+\begin_inset Formula $m_{A}\left(x\right)$
+\end_inset
+
+,
+ je prav to minimalen polinom 
+\begin_inset Formula $A$
+\end_inset
+
+ —
+ ima najmanjšo stopnjo možno.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Potrebujemo nekaj lem:
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Lemma
+Za vse matrike 
+\begin_inset Formula $A,B$
+\end_inset
+
+ velja 
+\begin_inset Formula $\n\left(AB\right)\leq\n\left(A\right)+\n\left(B\right)\sim\dim\Ker\left(AB\right)\leq\dim\Ker\left(A\right)+\dim\Ker\left(B\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Oglejmo si preslikavo 
+\begin_inset Formula $L:\Ker AB\to\Ker A$
+\end_inset
+
+,
+ ki slika 
+\begin_inset Formula $x\mapsto Bx$
+\end_inset
+
+.
+ Je dobro definirana,
+ kajti 
+\begin_inset Formula $x\in\Ker AB\Rightarrow ABx=0\Rightarrow Bx\in\Ker A$
+\end_inset
+
+.
+ Po osnovnem dimenzijskem izreku za preslikavo 
+\begin_inset Formula $L$
+\end_inset
+
+ velja 
+\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim\Ker AB$
+\end_inset
+
+.
+ Ker velja 
+\begin_inset Formula $Lx=0\Rightarrow Bx=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $\Ker L\subseteq\Ker B$
+\end_inset
+
+ in zato 
+\begin_inset Formula $\dim\Ker L\leq\dim\Ker B$
+\end_inset
+
+.
+ Poleg tega iz definicije velja 
+\begin_inset Formula $\Slika L\subseteq\Ker A$
+\end_inset
+
+ in zato 
+\begin_inset Formula $\dim\Slika L\leq\dim\Ker A$
+\end_inset
+
+.
+ Vstavimo te neenakosti v enačbo iz dimenzijskega izreka:
+\begin_inset Formula 
+\[
+\dim\Ker L+\dim\Slika L=\dim\Ker AB
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\dim\Ker B+\dim\Ker A\geq\dim\Ker AB
+\]
+
+\end_inset
+
+Lemo lahko posplošimo na več faktorkev,
+ torej 
+\begin_inset Formula $\n\left(A_{1}\cdots A_{k}\right)\leq\n A_{1}+\cdots+\n A_{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Nadaljujmo z dokazom 
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+.
+ Denimo,
+ da 
+\begin_inset Formula $\left(x-\lambda_{1}\right)\cdots\left(x-\lambda_{k}\right)$
+\end_inset
+
+ anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Upoštevamo 
+\begin_inset Formula 
+\[
+\n\left(\left(A-\lambda_{1}\right)\cdots\left(A-\lambda_{k}\right)\right)\leq\n\left(A-\lambda_{1}\right)+\cdots+\n\left(A-\lambda_{k}\right)
+\]
+
+\end_inset
+
+Členi na desni strani so geometrijske večkratnosti,
+ ker pa 
+\begin_inset Formula $A$
+\end_inset
+
+ anhilira polinom po predpostavki,
+ je ta produkt ničelna preslikava in je dimenzija jedra dimenzija celega prostora.
+\begin_inset Formula 
+\[
+\n\left(0\right)=n=n_{1}+\cdots+n_{k}\leq m_{1}+\cdots+m_{k}
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\forall i:m_{i}\leq n_{i}$
+\end_inset
+
+ (
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:geom<=alg"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+),
+ velja v zgornji neenačbi enakost,
+ torej je matrika po 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:mi=ni=>diag"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ diagonalizabilna.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Subsubsection
+Korenski podprostori
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A\in M_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
+\end_inset
+
+ njen minimalni polinom.
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..k\right\} $
+\end_inset
+
+ označimo z 
+\begin_inset Formula $W_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
+\end_inset
+
+ korenski podprostor matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+.
+ Vpeljimo še oznako 
+\begin_inset Formula $V_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{1}$
+\end_inset
+
+ (tu potenca ni 
+\begin_inset Formula $r_{i}$
+\end_inset
+
+,
+ temveč je 
+\begin_inset Formula $1$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Definition*
+\begin_inset CommandInset counter
+LatexCommand set
+counter "theorem"
+value "0"
+lyxonly "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact
+Očitno je 
+\begin_inset Formula $\Ker\left(A-\lambda_{i}I\right)\subseteq\Ker\left(A-\lambda_{i}\right)^{2}\subseteq\Ker\left(A-\lambda_{i}I\right)^{3}\subseteq\cdots$
+\end_inset
+
+,
+ kajti če 
+\begin_inset Formula $x\in\Ker\left(A-\lambda_{i}I\right)^{m}\Rightarrow\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow\left(A-\lambda_{i}I\right)\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow x\in\Ker\left(A-\lambda_{i}I\right)^{m+1}$
+\end_inset
+
+.
+ Izkaže se,
+ da so vse inkluzije do 
+\begin_inset Formula $r_{i}-te$
+\end_inset
+
+ potence stroge,
+ od 
+\begin_inset Formula $r_{i}-$
+\end_inset
+
+te potence dalje pa so vse inkluzije enačaji,
+ torej za 
+\begin_inset Formula $W_{i}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$
+\end_inset
+
+ velja 
+\begin_inset Formula 
+\[
+\Ker\left(A-\lambda_{i}I\right)\subset\Ker\left(A-\lambda_{i}I\right)^{2}\subset\cdots\subset\Ker\left(A-\lambda_{i}I\right)^{r_{i}}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}+1}=\cdots
+\]
+
+\end_inset
+
+Poleg tega se izkaže,
+ da je 
+\begin_inset Formula $\dim W_{i}=n_{i}$
+\end_inset
+
+ (algebraična večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact
+\begin_inset Formula $\dim V_{i}=\dim\Ker\left(A-\lambda_{i}I\right)=m_{1}$
+\end_inset
+
+ (geometrijska večkratnost 
+\begin_inset Formula $\lambda_{i}$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Claim
+\begin_inset CommandInset label
+LatexCommand label
+name "claim:vsota-kor-podpr-je-vse"
+
+\end_inset
+
+
+\begin_inset Formula $\mathbb{C}^{n}=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k}$
+\end_inset
+
+ —
+ vsota vseh korenskih podprostorov je vse in ta vsota je direktna.
+ Tej vsoti pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+korenski razcep matrike 
+\begin_inset Formula $A$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Dokazujmo:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $V_{1}+\cdots+V_{k}$
+\end_inset
+
+ je tudi direktna,
+ ampak ni nujno enaka 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\mathbb{C}^{n}=V_{1}+\cdots+V_{k}\Leftrightarrow A$
+\end_inset
+
+ se da diagonalizirati (povedano prej).
+ Dokažimo trditev 
+\begin_inset CommandInset ref
+LatexCommand vref
+reference "claim:vsota-kor-podpr-je-vse"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+.
+ Dokažimo najprej,
+ da če 
+\begin_inset Formula $w_{1}\in W_{1},\dots,w_{k}\in W_{k}$
+\end_inset
+
+ zadoščajo 
+\begin_inset Formula $w_{1}+\cdots+w_{k}=0$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $w_{1}=\cdots=w_{k}=0$
+\end_inset
+
+ (direktna).
+ Delajmo indukcijo po številu členov:
+\end_layout
+
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $w_{1}=0\Rightarrow w_{1}=0$
+\end_inset
+
+ je očitno.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Indukcijska predpostavka:
+ 
+\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Naj bodo 
+\begin_inset Formula $w_{1},\dots,w_{i+1}$
+\end_inset
+
+ taki,
+ da 
+\begin_inset Formula 
+\[
+w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+w_{1}'+\cdots+w_{i}'+0=0,
+\]
+
+\end_inset
+
+kajti 
+\begin_inset Formula $w_{i+1}\in\Ker\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
+\end_inset
+
+.
+ Ker je vsak korenski prostor 
+\begin_inset Formula $W_{j}$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$
+\end_inset
+
+,
+ ...
+ Tega dokaza ne najdem,
+ zato 
+\series bold
+tega dokaza ne razumem
+\series default
+.
+ Po definiciji je 
+\begin_inset Formula $V$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ če za vsak 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ velja 
+\begin_inset Formula $Av\in V$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Plain Layout
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+Nadaljuj 
+\begin_inset Quotes gld
+\end_inset
+
+LA1P FMF 2024-03-20.pdf
+\begin_inset Quotes grd
+\end_inset
+
+ na strani 3.
+\end_layout
+
+\end_inset
+
+Če predpostavimo,
+ da je vsota direktna,
+ je lahko dokazati,
+ da je vsota cel prostor.
+ V karakteristični polinom,
+ ki po Cayley-Hamiltonu anhilira 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ vstavimo 
+\begin_inset Formula $A$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $0=\left(-1\right)^{n}\left(A-\lambda_{1}I\right)^{r_{1}}\cdots\left(A-\lambda_{k}I\right)^{r_{k}}=A_{1}\cdots A_{k}$
+\end_inset
+
+.
+ Ker je vsota direktna,
+ velja 
+\begin_inset Formula $\n\left(A_{1}\cdots A_{n}\right)=\n\left(0\right)=n=\n A_{1}\cdots\n A_{k}=\dim\left(W_{1}+\cdots+W_{k}\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $W_{1}+\cdots+W_{k}=\mathbb{C}^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Če predpostavimo,
+ da je 
+\begin_inset Formula $W_{i}\cap W_{j}=\left\{ 0\right\} $
+\end_inset
+
+ za 
+\begin_inset Formula $i\not=j$
+\end_inset
+
+,
+ lahko od tod izpeljemo direktnost vsote korenskih podprostorov.
+ Dokaz z indukcijo:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Baza:
+ 
+\begin_inset Formula $W_{1}$
+\end_inset
+
+ je direktna vsota.
+ Očitno (
+\begin_inset Formula $\forall w_{1}\in W_{1}:w_{1}=0\Rightarrow w_{1}=0$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Itemize
+Indukcijska predpostavka:
+ 
+\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Korak:
+ Naj bodo 
+\begin_inset Formula $w_{1},\dots,w_{i+1}$
+\end_inset
+
+ taki,
+ da 
+\begin_inset Formula 
+\[
+w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{1}+\cdots+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{i}+0=0
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\left(A-\lambda_{h}I\right)^{r_{h}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(A-\lambda_{k}I\right)^{r_{k}}$
+\end_inset
+
+ za vsaka 
+\begin_inset Formula $h,k$
+\end_inset
+
+ komutirata (gre namreč za polinom,
+ v katerega je vstavljen 
+\begin_inset Formula $A$
+\end_inset
+
+),
+ velja za vsak 
+\begin_inset Formula $j$
+\end_inset
+
+ iz 
+\begin_inset Formula $\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}=0=\left(A-\lambda_{i+1}I\right)^{r_{i+1}}\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}$
+\end_inset
+
+ tudi
+\begin_inset Formula 
+\[
+\left(A-\lambda_{j}I\right)^{r_{j}}\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{j}=0
+\]
+
+\end_inset
+
+Ker je po I.
+ P.
+ 
+\begin_inset Formula $W_{1}+\cdots+W_{i}$
+\end_inset
+
+ direktna,
+ velja za vsak 
+\begin_inset Formula $j$
+\end_inset
+
+ 
+\begin_inset Formula $w_{j}\in W_{j}$
+\end_inset
+
+,
+ toda zaradi našega množenja tudi 
+\begin_inset Formula $w_{j}\in W_{i+1}$
+\end_inset
+
+.
+ Zaradi predpostavke 
+\begin_inset Formula $m=n\Rightarrow W_{m}\cup W_{n}=\left\{ 0\right\} $
+\end_inset
+
+ velja za vsak
+\begin_inset Formula $j\in\left\{ 1..i\right\} :$
+\end_inset
+
+ 
+\begin_inset Formula $w_{j}=0$
+\end_inset
+
+.
+ V prvi enačbi ostane le še 
+\begin_inset Formula $w_{i+1}=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Dokazati je treba še 
+\begin_inset Formula $i\not=j\Rightarrow W_{i}\cup W_{j}=\left\{ 0\right\} $
+\end_inset
+
+.
+ Dokažimo,
+ da je 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+ invarianten za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ t.
+ j.
+ 
+\begin_inset Formula $v\in W_{i}\Rightarrow Av\in W_{i}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $v\in W_{i}$
+\end_inset
+
+,
+ tedaj 
+\begin_inset Formula 
+\[
+\left(A-\lambda_{i}I\right)^{r_{i}}v=0\quad\quad\quad\quad/\cdot A
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A\left(A-\lambda_{i}I\right)^{r_{i}}v=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Av\in\Ker\left(A-\lambda_{i}I\right)^{r_{i}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Av\in W_{i}
+\]
+
+\end_inset
+
+Ker so vsi 
+\begin_inset Formula $W_{i}$
+\end_inset
+
+ invariantni za 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ so tudi njihovi preseki invariantni za 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Definirajmo torej linearno preslikavo 
+\begin_inset Formula $L:W_{i}\cap W_{j}\to W_{i}\cap W_{j}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $v\mapsto Av$
+\end_inset
+
+.
+ Vemo,
+ da ima 
+\begin_inset Formula $L$
+\end_inset
+
+ vsaj eno lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ in pripadajoči lastni vektor 
+\begin_inset Formula $w$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $w\in W_{i}\cap W_{j}$
+\end_inset
+
+ in 
+\begin_inset Formula $Lw=\lambda w$
+\end_inset
+
+,
+ toda 
+\begin_inset Formula $Lw=Aw=\lambda w$
+\end_inset
+
+.
+ Ker velja 
+\begin_inset Formula $Av=\lambda v\Rightarrow A^{q}v=\lambda^{q}v\Rightarrow p\left(A\right)v=p\left(\lambda\right)v$
+\end_inset
+
+ za vsak polinom 
+\begin_inset Formula $p$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $p\left(A\right)w=p\left(\lambda\right)w$
+\end_inset
+
+ za vsak polinom 
+\begin_inset Formula $p$
+\end_inset
+
+.
+ Uporabimo polinom 
+\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{i}\right)^{r_{i}}$
+\end_inset
+
+ in dobimo 
+\begin_inset Formula $\left(A-\lambda_{i}\right)^{r_{i}}w=\left(\lambda-\lambda_{i}\right)^{r_{i}}w$
+\end_inset
+
+.
+ Leva stran enačbe je 0,
+ PDDRAA 
+\begin_inset Formula $w$
+\end_inset
+
+ ni 0,
+ torej 
+\begin_inset Formula $\left(\lambda-\lambda_{i}\right)^{r_{i}}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda=\lambda_{i}$
+\end_inset
+
+.
+ Vendar lahko namesto tistega polimoma uporabimo polinom 
+\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{j}\right)^{r_{j}}$
+\end_inset
+
+,
+ kar pokaže 
+\begin_inset Formula $\lambda=\lambda_{j}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda_{j}=\lambda_{i}$
+\end_inset
+
+,
+ kar je v 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+ s tem,
+ da so lastne vrednosti 
+\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
+\end_inset
+
+ paroma različne.
+ Torej 
+\begin_inset Formula $w=0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Jordanska kanonična forma
+\end_layout
+
+\begin_layout Standard
+Vsaka kvadratna matrika je podobna posebni zgornjetrikotni matriki,
+ ki ji pravimo JKF.
+ To je bločno diagonalna matrika,
+ ki ima za diagonalne bloke t.
+ i.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+jordanske kletke
+\begin_inset Quotes grd
+\end_inset
+
+,
+ to so matrike oblike:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+ &  &  &  & \lambda
+\end{array}\right].
+\]
+
+\end_inset
+
+Jordanska matrika je sestavljena iz jordanskih kletk po diagonali (
+\begin_inset Formula $J_{i}$
+\end_inset
+
+ so jordanske kletke):
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+J_{1} &  & 0\\
+ & \ddots\\
+0 &  & J_{m}
+\end{array}\right].
+\]
+
+\end_inset
+
+Običajno zahtevamo še,
+ da so JK,
+ ki imajo isto lastno vrednost,
+ skupaj,
+ ter da so JK padajoče urejene po lastni vrednosti od največje do najmanjše.
+\end_layout
+
+\begin_layout Theorem*
+Za vsako kvadratno kompleksno matriko 
+\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ obstaja taka jordanska matrika 
+\begin_inset Formula $J$
+\end_inset
+
+ in taka obrnljiva matrika 
+\begin_inset Formula $P$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+.
+ ZDB vsaka 
+\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$
+\end_inset
+
+ je podobna neki Jordanski matriki.
+\end_layout
+
+\begin_layout Standard
+Procesu iskanja jordanske matrike pravimo 
+\begin_inset Quotes gld
+\end_inset
+
+jordanifikacija
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Kako konstruiramo 
+\begin_inset Formula $J$
+\end_inset
+
+ in 
+\begin_inset Formula $P$
+\end_inset
+
+?
+ Izračunamo lastne vrednosti in pripadajoče korenske podprostore.
+ 
+\end_layout
+
+\begin_layout Itemize
+Naj bo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Za preglednost pišimo 
+\begin_inset Formula $N\coloneqq A-\lambda I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Izračunamo lastne vektorje in lastni podprostor 
+\begin_inset Formula $\Ker N^{r}$
+\end_inset
+
+ ter ga izrazimo z njegovo bazo,
+ recimo ji 
+\begin_inset Formula $B_{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Nato izberemo 
+\begin_inset Quotes gld
+\end_inset
+
+pomožne baze
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $\mathcal{B}_{1},\dots,\mathcal{B}_{r}$
+\end_inset
+
+,
+ ki pripadajo prostorom 
+\begin_inset Formula $\Ker N^{1},\dots,\Ker N^{r}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Pomožno bazo 
+\begin_inset Formula $\mathcal{B}_{r-1}$
+\end_inset
+
+ dopolnimo do baze 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+ z elementi 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+ 
+\begin_inset Formula $u_{1},\dots,u_{k_{1}}$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $\mathcal{B}_{r-1}\cup$
+\end_inset
+
+
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} $
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+popravek pomožne baze 
+\begin_inset Formula $\mathcal{B}_{r}$
+\end_inset
+
+
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Vektorje 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} \in\Ker N^{r}$
+\end_inset
+
+ pomnožimo z matriko 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ dobljeni 
+\begin_inset Formula $Nu_{1},\dots,Nu_{k_{1}}$
+\end_inset
+
+ ležijo v 
+\begin_inset Formula $\Ker N^{r-1}$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} $
+\end_inset
+
+ je linearno neodvisna.
+ Izberemo take 
+\begin_inset Formula $v_{1},\dots,v_{k_{2}}\in B_{r-1}$
+\end_inset
+
+,
+ ki dopolnijo LN 
+\begin_inset Formula $B_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{2}\right\} $
+\end_inset
+
+ do baze 
+\begin_inset Formula $\Ker N^{r-1}$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} \cup\left\{ v_{1},\dots,v_{k_{2}}\right\} $
+\end_inset
+
+ popravek pomožne baze 
+\begin_inset Formula $\mathcal{B}_{r-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Izberemo take 
+\begin_inset Formula $w_{1},\dots,w_{k_{3}}\in\mathcal{B}_{r-2}$
+\end_inset
+
+,
+ ki 
+\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\dots,N^{2}u_{k_{1}}Nv_{1},\dots,Nv_{k_{2}}\right\} $
+\end_inset
+
+ dopolnijo do baze 
+\begin_inset Formula $\Ker N^{r-2}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\cdots,N^{2}u_{k_{1}},Nv_{1},\dots,Nv_{k_{2}},w_{1},\dots,w_{k_{3}}\right\} $
+\end_inset
+
+ popravek pomožne baze 
+\begin_inset Formula $B_{r-2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Postopek ponavljamo,
+ dokler ne popravimo vseh možnih baz.
+\end_layout
+
+\begin_layout Standard
+Dobimo t.
+ i.
+ 
+\begin_inset Quotes gld
+\end_inset
+
+jordanske verige
+\begin_inset Quotes grd
+\end_inset
+
+.
+ Ena jordanska veriga je 
+\begin_inset Formula $\left(u,Nu,N^{2}u,\dots,N^{x}u\right)$
+\end_inset
+
+,
+ torej preslikanje elementa 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ ki začne kot dopolnitev baze korenskega podprostora 
+\begin_inset Formula $\Ker N^{x+1}$
+\end_inset
+
+ in je na koncu 
+\begin_inset Formula $x-$
+\end_inset
+
+krat preslikan z 
+\begin_inset Formula $N$
+\end_inset
+
+,
+ torej konča v korenskem podprostoru 
+\begin_inset Formula $\Ker N$
+\end_inset
+
+.
+ Nekatere verige se začno v največjem korenskem podprostoru 
+\begin_inset Formula $\Ker N^{r}$
+\end_inset
+
+,
+ nekatere šele kasneje,
+ v 
+\begin_inset Formula $\Ker N^{1}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\Ker N^{2}$
+\end_inset
+
+ ali pa 
+\begin_inset Formula $\Ker N^{3}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Imamo torej 
+\begin_inset Formula $k_{1}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k_{2}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r-1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $k_{3}$
+\end_inset
+
+ jordanskih verig dolžine 
+\begin_inset Formula $r-2$
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $k_{r}$
+\end_inset
+
+ jordanskih verig dolžine 1.
+ Skupaj je jordanskih verig 
+\begin_inset Formula $k_{1}+\cdots+k_{r}=\dim\Ker N$
+\end_inset
+
+.
+ Jordanskih verig za lastno vrednost 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ je torej toliko,
+ kot je njena geometrijska večkratnost.
+\end_layout
+
+\begin_layout Standard
+Vsaki jordanski verigi dolžine 
+\begin_inset Formula $k$
+\end_inset
+
+ pripada ena jordanska kletka velikosti 
+\begin_inset Formula $k\times k$
+\end_inset
+
+.
+ 
+\begin_inset Formula $k-$
+\end_inset
+
+vektorjev iz verige zložimo v 
+\begin_inset Formula $P$
+\end_inset
+
+ tako,
+ da je vektor z začetka verige (torej tisti iz popravljene baze večjega prostora) na levi strani v matriki.
+\end_layout
+
+\begin_layout Example*
+Poišči jordansko kanonično formo matrike
+\begin_inset Formula 
+\[
+A=\left[\begin{array}{cccc}
+0 & 1 & -1 & 2\\
+0 & 2 & 2 & 2\\
+0 & 0 & 2 & 0\\
+0 & 0 & 0 & 2
+\end{array}\right].
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Najprej izračunamo karakteristični polinom:
+ 
+\begin_inset Formula $\det\left(A-\lambda I\right)=x\left(x-2\right)^{3}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\lambda_{1}=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n_{1}=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{2}=2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n_{2}=3$
+\end_inset
+
+.
+ Lastni vektorji:
+ 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Lin\left\{ \left(1,0,0,0\right)\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)=\Lin\left\{ \left(3,0,-2,2\right),\left(1,2,0,0\right)\right\} $
+\end_inset
+
+.
+ Če bi dobili 4 lastne vektorje,
+ bi lahko matriko diagonalizirali.
+ Tako je ne moremo.
+ Ker 
+\begin_inset Formula $n_{1}=1$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $r_{1}$
+\end_inset
+
+ največ 
+\begin_inset Formula $1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A-0I\right)^{2}=\cdots$
+\end_inset
+
+.
+ Izračunamo korenske podprostore
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+za lastno vrednost 0:
+ 
+\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A\right)=\Ker\left(A^{2}\right)$
+\end_inset
+
+.
+ Dobimo eno verigo 
+\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
+\end_inset
+
+ dolžine 
+\begin_inset Formula $1$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+za lastno vrednost 2:
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)^{2}=\Lin\left\{ \left(1,0,0,2\right),\left(-1,0,1,0\right),\left(1,2,0,0\right)\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $\Ker\left(A-2I\right)^{3}=\Ker\left(A-2I\right)^{2}$
+\end_inset
+
+.
+ Opazimo,
+ da je 
+\begin_inset Formula $\left(1,0,0,2\right)$
+\end_inset
+
+ dopolnitev baze 
+\begin_inset Formula $\Ker\left(A-2I\right)$
+\end_inset
+
+ do baze 
+\begin_inset Formula $\Ker\left(A-2I\right)^{2}$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $\left\{ \left(1,0,0,2\right)\right\} $
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+popravljena baza
+\begin_inset Quotes grd
+\end_inset
+
+ 
+\begin_inset Formula $N^{2}$
+\end_inset
+
+.
+ Preslikamo 
+\begin_inset Formula $\left(A-2I\right)\left(1,0,0,2\right)=\left(2,4,0,0\right)$
+\end_inset
+
+,
+ kar tvori verigo dolžine 2 
+\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
+\end_inset
+
+.
+ Edini linearno neodvisen od 
+\begin_inset Formula $\left(2,4,0,0\right)$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathcal{B}_{1}$
+\end_inset
+
+ je 
+\begin_inset Formula $\left(3,0,-2,2\right)$
+\end_inset
+
+,
+ zato je slednji začetek zadnje tretje verige dolžine 1 
+\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Tri verige,
+ ki jih dobimo,
+ so 
+\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$
+\end_inset
+
+ za lastno vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$
+\end_inset
+
+ ter 
+\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$
+\end_inset
+
+ obe za lastno vrednost 2.
+ Zložimo jih v matriko 
+\begin_inset Formula $P$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+P=\left[\begin{array}{cccc}
+1 & 1 & 2 & 3\\
+0 & 0 & 4 & 0\\
+0 & 0 & 0 & -2\\
+0 & 2 & 0 & 2
+\end{array}\right]
+\]
+
+\end_inset
+
+V matriko 
+\begin_inset Formula $J$
+\end_inset
+
+ pa zložimo kletke pripadajočih velikosti:
+\begin_inset Formula 
+\[
+J=\left[\begin{array}{cccc}
+0 &  &  & 0\\
+ & 2 & 1\\
+ &  & 2\\
+0 &  &  & 2
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+In velja 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+ (
+\begin_inset Formula $P^{-1}$
+\end_inset
+
+ izračunamo z Gaussom).
+\end_layout
+
+\begin_layout Subsubsection
+Funkcije matrik
+\end_layout
+
+\begin_layout Standard
+Če poznamo razcep 
+\begin_inset Formula $A=PJP^{-1}$
+\end_inset
+
+,
+ prevedemo računanje potenc 
+\begin_inset Formula $A$
+\end_inset
+
+ na računanje potenc matrike 
+\begin_inset Formula $J$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula 
+\[
+A^{n}=\left(PJP^{-1}\right)\left(PJP^{-1}\right)\cdots\left(PJP^{-1}\right)=PJP^{-1}PJP^{-1}\cdots PJP^{-1}=PJ^{n}P^{-1}.
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $J$
+\end_inset
+
+ bločno diagonalna matrika,
+ sestavljena iz jordanskih kletk,
+ se potenciranje 
+\begin_inset Formula $J$
+\end_inset
+
+ prevede na potenciranje kletk,
+ kajti 
+\begin_inset Formula 
+\[
+J^{n}=\left[\begin{array}{ccc}
+J_{1}^{n} &  & 0\\
+ & \ddots\\
+0 &  & J_{m}^{n}
+\end{array}\right].
+\]
+
+\end_inset
+
+Potenciranje jordanske kletke:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]^{n}=\left(\lambda I+\left[\begin{array}{ccc}
+1 &  & 0\\
+ & \ddots\\
+0 &  & 1
+\end{array}\right]\right)^{n}=\left(\lambda I+N\right)^{n}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\binom{n}{0}\left(\lambda I\right)^{n}N^{0}+\binom{n}{1}\left(\lambda N\right)^{n-1}N^{1}+\cdots+\binom{n}{n}\left(\lambda N\right)^{0}N^{n}=\binom{n}{0}\lambda^{n}+\binom{n}{1}\lambda^{n-1}N^{1}+\cdots+\binom{n}{n}N^{n}
+\]
+
+\end_inset
+
+Poraja se vprašanje,
+ kako potencirati 
+\begin_inset Formula $N=\left[\begin{array}{ccccc}
+0 & 1 &  &  & 0\\
+ & \ddots & \ddots\\
+ &  &  & \ddots\\
+ &  &  & \ddots & 1\\
+0 &  &  &  & 0
+\end{array}\right]$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $N^{2}=\left[\begin{array}{ccccc}
+0 & 0 & 1 &  & 0\\
+ & \ddots & \ddots & \ddots\\
+ &  &  & \ddots & 1\\
+ &  &  & \ddots & 0\\
+0 &  &  &  & 0
+\end{array}\right]$
+\end_inset
+
+ in tako dalje (
+\begin_inset Quotes gld
+\end_inset
+
+diagonalo
+\begin_inset Quotes grd
+\end_inset
+
+ enic pomikamo gor in desno).
+ Za 
+\begin_inset Formula $r\times r$
+\end_inset
+
+ jordansko kletko,
+ kadar 
+\begin_inset Formula $n\geq r$
+\end_inset
+
+ (sicer dobimo le prvih nekaj naddiagonal),
+ sledi
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]^{n}=\left[\begin{array}{ccccc}
+\lambda^{n} & n\lambda^{n-1} & \cdots & \binom{n}{r-2}\lambda^{n-r+2} & \binom{n}{r-1}\lambda^{n-r+1}\\
+ & \lambda^{n} & n\lambda^{n-1} & \ddots & \binom{n}{r-2}\lambda^{n-r+2}\\
+ &  & \ddots & \ddots & \vdots\\
+ &  &  & \lambda^{n} & n\lambda^{n-1}\\
+0 &  &  &  & \lambda^{n}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za računanje poljubne funkcije jordanske kletke pa velja predpis
+\begin_inset Formula 
+\[
+f\left(\left[\begin{array}{ccccc}
+\lambda & 1 &  &  & 0\\
+ & \lambda & 1\\
+ &  & \ddots & \ddots\\
+ &  &  & \lambda & 1\\
+0 &  &  &  & \lambda
+\end{array}\right]\right)=\left[\begin{array}{ccccc}
+f\left(\lambda\right) & f'\left(\lambda\right) & \frac{f''\left(\lambda\right)}{2} & \cdots & \frac{f^{\left(k-1\right)\left(\lambda\right)}}{\left(k-1\right)!}\\
+ & f\left(\lambda\right) & f'\left(\lambda\right) & \ddots & \cdots\\
+ &  & \ddots & \ddots & \frac{f''\left(\lambda\right)}{2}\\
+ &  &  & f\left(\lambda\right) & f'\left(\lambda\right)\\
+0 &  &  &  & f\left(\lambda\right)
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+In torej za računanje poljubne funkcije poljubne matrike 
+\begin_inset Formula $f\left(A\right)=f\left(PJP^{-1}\right)=Pf\left(J\right)P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Vektorski prostori s skalarnim produktom
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad poljem 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ nenujno končno razsežen.
+ Preslikavi 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{R}$
+\end_inset
+
+ pravimo skalarni produtkt,
+ če zadošča naslednjim lastnostim:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+pozitivna definitnost:
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle >0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+simetričnost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\left\langle u,v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+linearnost v prvem faktorju:
+ 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+linearnost v drugem faktorju.
+ 
+\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle =\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle =\beta_{1}\left\langle v,v_{1}\right\rangle +\beta_{2}\left\langle v,v_{2}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Skalarni produkt z 0:
+ 
+\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Alternativna formulacija 1:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Dokazujemo ekvivalenco:
+ alternativna formulacija 1 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ originalna definicija 1.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Leftarrow\right)$
+\end_inset
+
+ Predpostavimo 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \geq0$
+\end_inset
+
+ in izjavo negirajmo:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \leq0\Rightarrow v=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\Rightarrow\right)$
+\end_inset
+
+ Predpostavimo 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+ 
+\end_layout
+
+\end_deeper
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Primeri vektorskih prostorov s skalarnim produktom:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ s standardnim skalarnim produktom:
+ 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ z nestandardnim skalarnim produktom:
+ Za pojubne 
+\begin_inset Formula $\gamma_{1}>0,\dots,\gamma_{n}>0$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\beta_{1}+\cdots+\gamma_{n}\alpha_{n}\beta_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen primer s standardnim skalarnim produktom:
+ 
+\begin_inset Formula $V=C\left[a,b\right]\sim$
+\end_inset
+
+ zvezne 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle =\int_{a}^{b}f\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+ Zveznost je potrebna za dokaz aksioma 1,
+ sicer za neznano neničelno funkcijo 
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+1 & ;x=0\\
+0 & ;\text{drugače}
+\end{cases}$
+\end_inset
+
+ velja 
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)g\left(x\right)dx=0$
+\end_inset
+
+.
+ Temu pravimo standardni skalarni produkt v 
+\begin_inset Formula $C\left[a,b\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen primer z nestandardnim skalarnim produktom:
+ Naj bo 
+\begin_inset Formula $w:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ ki zadošča 
+\begin_inset Formula $\forall x\in\left[a,b\right]:w\left(x\right)>0$
+\end_inset
+
+.
+ Ostalo kot prej.
+ 
+\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle _{w}=\int_{a}^{b}f\left(x\right)g\left(x\right)w\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Vektorski prostor s skalarnnim produktom je tak par 
+\begin_inset Formula $\left(V,\left\langle \cdot,\cdot\right\rangle \right)$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle $
+\end_inset
+
+ skalarni produkt na 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ To je torej vektorski prostor,
+ za katerega izberemo in fiksiramo skalarni produkt.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad poljem 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ nenujno končno razsežen.
+ Preslikavi 
+\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{C}$
+\end_inset
+
+ pravimo skalarni produtkt,
+ če zadošča naslednjim lastnostim:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+pozitivna definitnost:
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle >0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+konjugirana simetričnost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\overline{\left\langle u,v\right\rangle }$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+linearnost v prvem faktorju:
+ 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+konjugirana linearnost v drugem faktorju.
+ 
+\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\overline{\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle }=\overline{\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\overline{\left\langle v_{1},v\right\rangle }+\overline{\beta_{2}}\overline{\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\left\langle v,v_{1}\right\rangle +\overline{\beta_{2}}\left\langle v,v_{2}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Skalarni produkt z 0:
+ 
+\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Alternativna formulacija 1:
+ 
+\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri vektorskih prostorov s skalarnim produktom:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+standardni skalarni produkt na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+nestandardni skalarni produkt na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+:
+ Za neke 
+\begin_inset Formula $\gamma_{1}\in\mathbb{\mathbb{R}}^{+},\dots,\gamma_{n}\in\mathbb{R}^{+}$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\overline{\beta_{1}}+\cdots+\gamma_{n}\alpha_{n}\overline{\beta_{n}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen vektorski prostor na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ s standardnim skalarnim produktom:
+ Naj bo 
+\begin_inset Formula $V=C\left(\left[a,b\right],\mathbb{C}\right)$
+\end_inset
+
+ —
+ 
+\begin_inset Formula $f=g+ih$
+\end_inset
+
+ za 
+\begin_inset Formula $g,h\in C\left[a,b\right]$
+\end_inset
+
+ (zvezni funkciji iz 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ v 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+).
+ Definiramo 
+\begin_inset Formula $\left\langle f_{1},f_{2}\right\rangle =\int_{a}^{b}f_{1}\left(x\right)\overline{f_{2}\left(x\right)}dx=\int_{a}^{b}\left(g_{1}+ih_{1}\right)\left(x\right)\left(g_{2}-ih_{2}\right)\left(x\right)dx=\int_{a}^{b}\left(g_{1}g_{2}+g_{1}g_{2}\right)\left(x\right)dx+i\int_{a}^{b}\left(h_{1}g_{2}-g_{1}h_{2}\right)xdx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+neskončno razsežen vektorski prostor na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ z nestandardnim skalarnim produktom:
+ Isto kot zgoraj,
+ le da spet množimo z nekimi funkcijami,
+ kot pri realnem skalarnem produktu.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Norma
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor s skalarnim produktom.
+ 
+\begin_inset Formula $\forall v\in V:\left|\left|v\right|\right|=\sqrt{\left\langle v,v\right\rangle }$
+\end_inset
+
+ je norma 
+\begin_inset Formula $v$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Paragraph
+Osnovne lastnosti norme
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\left|\left|v\right|\right|>0\Leftrightarrow v\not=0\right)\wedge\left|\left|0\right|\right|=0$
+\end_inset
+
+ sledi iz prvega aksioma skalarnega produkta
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall\alpha\in F,v\in V:\left|\left|\alpha v\right|\right|=\left|\alpha\right|\left|\left|v\right|\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+trikotniška neenakost:
+ 
+\begin_inset Formula $\forall u,v\in V:\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
+\end_inset
+
+ sledi iz Cauchy-Schwarzove neenakosti na običajen način.
+\end_layout
+
+\begin_layout Claim*
+Cauchy-Schwarz.
+ Za 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor s skalarnim produktom velja 
+\begin_inset Formula $\forall v\in V:\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Za 
+\begin_inset Formula $v=0$
+\end_inset
+
+ očitno velja 
+\begin_inset Formula $0=0$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula 
+\[
+w=u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v
+\]
+
+\end_inset
+
+po prvi lastnosti velja
+\begin_inset Formula 
+\[
+0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle 
+\]
+
+\end_inset
+
+Oglejmo si
+\begin_inset Formula 
+\[
+\left\langle w,v\right\rangle =\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\left\langle u,v\right\rangle -\left\langle \frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\cancel{\left\langle u,v\right\rangle }-\frac{\cancel{\left\langle u,v\right\rangle }}{\cancel{\left\langle v,v\right\rangle }}\cancel{\left\langle v,v\right\rangle }=0
+\]
+
+\end_inset
+
+In se vrnimo k prejšnji enačbi:
+\begin_inset Formula 
+\[
+0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle =\left\langle w,u\right\rangle -0=\left\langle w,u\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,u\right\rangle =\left\langle u,u\right\rangle -\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }\left\langle v,u\right\rangle =\left|\left|u\right|\right|^{2}-\frac{\left\langle u,v\right\rangle \overline{\left\langle u,v\right\rangle }}{\left|\left|v\right|\right|^{2}}=\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0\leq\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}\leq\left|\left|u\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left\langle u,v\right\rangle \right|^{2}\leq\left|\left|u\right|\right|^{2}\left|\left|v\right|\right|^{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Z normo lahko izrazimo skalarni produkt:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+V 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\left(\left|\left|u+v\right|\right|^{2}-\left|\left|u-v\right|\right|^{2}\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+V 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle u,v\right\rangle =\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokaz v 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+.
+ Oglejmo si 
+\begin_inset Formula 
+\[
+\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u+i^{k}v,u+i^{k}v\right\rangle =\left\langle u,u+i^{k}v\right\rangle +i^{k}\left\langle v,u+i^{k}v\right\rangle =\overline{\left\langle u+i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u+i^{k}v,v\right\rangle }=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\overline{\left\langle u,u\right\rangle }+\overline{\left\langle i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u,v\right\rangle }+i^{k}\overline{\left\langle i^{k}v,v\right\rangle }=\left\langle u,u\right\rangle +\left\langle u,i^{k}v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left\langle v,i^{k}v\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left(-\left(i^{k}\right)\right)\left\langle v,v\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle 
+\]
+
+\end_inset
+
+Dodajmo vsoto:
+\begin_inset Formula 
+\[
+\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\sum_{k=0}^{3}i^{k}\left(\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle \right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\sum_{k=0}^{3}i^{k}\left\langle u,u\right\rangle +\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +\sum_{k=0}^{3}i^{k}i^{k}\left\langle v,u\right\rangle +\sum_{k=0}^{3}i^{k}\left\langle v,v\right\rangle =0+\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +0+0,
+\]
+
+\end_inset
+
+kajti 
+\begin_inset Formula $\sum_{k=0}^{3}i^{k}=1+i+\left(-1\right)+\left(-i\right)=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\sum_{k=0}^{3}i^{2k}=1+\left(-1\right)+1+\left(-1\right)=0$
+\end_inset
+
+.
+ Nadaljujmo:
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+=\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle =\sum_{k=0}^{3}1\left\langle u,v\right\rangle =4\left\langle u,v\right\rangle 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Ortogonalne množice in ortogonalne baze
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP 
+\begin_inset Formula $\forall u,v\in V:u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+trivialne opombe.
+ 
+\begin_inset Formula $\forall v\in V:v\perp\vec{0}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\forall v\in V:v\not=0\Leftrightarrow v\not\perp v$
+\end_inset
+
+ (prvi aksiom skalarnega produkta),
+ 
+\begin_inset Formula $\forall v\in V:u\perp v\Leftrightarrow v\perp u$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP in 
+\begin_inset Formula $v_{1},\dots,v_{k}\in V$
+\end_inset
+
+.
+ Množica 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ je:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+ortogonalna,
+ če 
+\begin_inset Formula $v_{1}\not=0\wedge\cdots\wedge v_{k}\not=0$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall i,j\in\left\{ 1..k\right\} :i\not=j\Rightarrow v_{i}\perp v_{j}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+normirana,
+ če 
+\begin_inset Formula $\forall v\in\left\{ v_{1},\dots,v_{k}\right\} :\left|\left|v\right|\right|=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+ortonormirana,
+ če je ortogonalna in ortonormirana hkrati.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+Iz (ortogonalne) množice 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ dobimo (orto)normirano tako,
+ da vsak element delimo z njegovo normo.
+ 
+\begin_inset Formula $\left\{ \frac{v_{1}}{\left|\left|v_{1}\right|\right|},\dots,\frac{v_{k}}{\left|\left|v_{k}\right|\right|}\right\} $
+\end_inset
+
+ je vedno normirana.
+\end_layout
+
+\begin_layout Claim*
+Vsaka ortogonalna množica je linearno neodvisna.
+\end_layout
+
+\begin_layout Proof
+Denimo,
+ da je 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ ortogonalna.
+ Vzemimo take 
+\begin_inset Formula $\alpha_{1},\dots,\alpha_{k}\ni:\alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k}=0$
+\end_inset
+
+.
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+alpha_1=
+\backslash
+cdots=
+\backslash
+alpha_k=0$}
+\end_layout
+
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\forall i\in\left\{ 1..k\right\} :0=\left\langle 0,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k},v\right\rangle =\alpha_{1}\left\langle v_{1},v_{i}\right\rangle +\cdots+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cdots+\alpha_{k}\left\langle v_{k},v_{i}\right\rangle =\cdots
+\]
+
+\end_inset
+
+Ker je množica ortogonalna,
+ je 
+\begin_inset Formula $\left\langle v_{l},v_{k}\right\rangle =0\Leftrightarrow l\not=k$
+\end_inset
+
+.
+ Nadaljujmo ...
+\begin_inset Formula 
+\[
+\cdots=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+Ker 
+\begin_inset Formula $\left\langle v_{i},v_{i}\right\rangle $
+\end_inset
+
+ ni 0,
+ ker je 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ neničeln (da,
+ tudi to je del definicije ortogonalnosti),
+ je 
+\begin_inset Formula $\alpha_{i}=0$
+\end_inset
+
+.
+ In to za vsak 
+\begin_inset Formula $i$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ni pa vsaka ortogonalna množica ogrodje.
+ Ortogonalni množici,
+ ki je ogrodje,
+ rečemo ortogonalna baza (LN sledi iz ortogonalnost).
+\end_layout
+
+\begin_layout Subsubsection
+Fourierov razvoj
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP,
+ 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} =\mathcal{B}$
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $V$
+\end_inset
+
+ in 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ poljuben element.
+ Kako razvijemo 
+\begin_inset Formula $v$
+\end_inset
+
+ po 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+,
+ vedoč,
+ da je ta baza ortogonalna?
+ Postopek imenujemo Fourierov razvoj.
+\end_layout
+
+\begin_layout Standard
+Ker je 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ ogrodje,
+ 
+\begin_inset Formula $\exists\alpha_{1},\dots,\alpha_{n}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Množimo skalarno z 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\quad\quad\quad\quad/\cdot v_{i}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle v,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle v,v_{i}\right\rangle =\cancel{\alpha_{1}\left\langle v_{1},v_{i}\right\rangle }+\cancel{\cdots}+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cancel{\cdots}+\cancel{\alpha_{n}\left\langle v_{n},v_{i}\right\rangle }=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }=\alpha_{i}
+\]
+
+\end_inset
+
+Torej 
+\begin_inset Formula $\forall v\in V$
+\end_inset
+
+ velja 
+\begin_inset Formula $v=\sum_{i=1}^{n}\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}$
+\end_inset
+
+.
+ Koeficientu 
+\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\left|\left|v_{i}\right|\right|^{2}}$
+\end_inset
+
+ pravimo Fourierov koeficient.
+ Če je baza ortonormirana,
+ je Fourierov koeficient 
+\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\cancel{\left|\left|v_{i}\right|\right|^{2}}}=\left\langle v,v_{i}\right\rangle $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Parsevalova identiteta
+\end_layout
+
+\begin_layout Theorem*
+Parsevalova identiteta.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP in 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $
+\end_inset
+
+ njegova ortogonalna baza.
+ Tedaj 
+\begin_inset Formula $\forall v\in V:$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\frac{\left|\left\langle v,v_{i}\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }.
+\]
+
+\end_inset
+
+Če je baza ortonormirana,
+ se enačba očitno poenostavi v
+\begin_inset Formula 
+\[
+\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\left|\left\langle v,v_{i}\right\rangle \right|^{2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left|\left|v\right|\right|^{2}=\left\langle v,v\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\right\rangle =$
+\end_inset
+
+ (uporabimo linearnost v 1.
+ in konjugirano linearnost v 2.
+ faktorju)
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+= & \alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle  & + & \cancel{\cdots} & + & \cancel{\alpha_{1}\overline{\alpha_{n}}\left\langle v_{1},v_{n}\right\rangle } & +\\
+ & \vdots &  &  &  & \vdots\\
++ & \cancel{\alpha_{n}\overline{\alpha_{1}}\left\langle v_{n},v_{1}\right\rangle } & + & \cancel{\cdots} & + & \alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle  & =
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle +\cdots+\alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle =\left|\alpha_{1}\right|^{2}\left|\left|v_{1}\right|\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}\left|\left|v_{n}\right|\right|^{2}=
+\]
+
+\end_inset
+
+Vstavimo formule za koeficiente po Fourierjevem razvoju:
+\begin_inset Formula 
+\[
+=\left|\frac{\left\langle v,v_{1}\right\rangle }{\left|\left|v_{1}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{1}\right|\right|^{2}}+\cdots+\left|\frac{\left\langle v,v_{n}\right\rangle }{\left|\left|v_{n}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{n}\right|\right|^{2}}=\frac{\left|\left\langle v,v_{1}\right\rangle \right|^{2}}{\left|\left|v_{1}\right|\right|}+\cdots+\frac{\left|\left\langle v,v_{n}\right\rangle \right|^{2}}{\left|\left|v_{n}\right|\right|}=\sum_{i=1}^{n}\frac{\left|\left\langle v_{i},v\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Projekcija na podprostor
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $W$
+\end_inset
+
+ podprostor 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Za vsak 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ želimo izračunati njegovo ortogonalno projekcijo na 
+\begin_inset Formula $W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Vektor 
+\begin_inset Formula $v'\in W$
+\end_inset
+
+ je ortogonalna projekcija vektorja 
+\begin_inset Formula $v\in V$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall w\in W:\left|\left|v-v'\right|\right|\leq\left|\left|v-w\right|\right|$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $v'$
+\end_inset
+
+ je najbližje 
+\begin_inset Formula $v$
+\end_inset
+
+ izmed vseh elementov 
+\begin_inset Formula $W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Zadošča preveriti,
+ da je 
+\begin_inset Formula $v-v'$
+\end_inset
+
+ ortogonalen na vse elemente 
+\begin_inset Formula $W$
+\end_inset
+
+ (pitagorov izrek),
+ kajti v tem primeru (če predpostavimo 
+\begin_inset Formula $\left(v'-w\right)\perp\left(v-v'\right)$
+\end_inset
+
+) velja
+\begin_inset Formula 
+\[
+\left|\left|v-w\right|\right|^{2}=\left|\left|v-v'+v'-w\right|\right|=\left|\left|v-v'\right|\right|^{2}+\left|\left|v'-w\right|\right|^{2}\geq\left|\left|v-v'\right|\right|^{2}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Dokaz pitagovorega izreka:
+ 
+\begin_inset Formula $\left|\left|a+b\right|\right|^{2}=\left\langle a+b,a+b\right\rangle =\left\langle a,a\right\rangle +\cancel{\left\langle a,b\right\rangle +\left\langle b,a\right\rangle }+\left\langle b,b\right\rangle =\left|\left|a\right|\right|^{2}+\left|\left|b\right|\right|^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $\left\{ w_{1},\dots,w_{k}\right\} $
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Formula za ortogonalno projekcijo se glasi:
+\begin_inset Formula 
+\[
+v'=\frac{\left\langle v,w_{1}\right\rangle }{\left\langle w_{1},w_{1}\right\rangle }w_{1}+\cdots+\frac{\left\langle v,w_{k}\right\rangle }{\left\langle w_{k},w_{k}\right\rangle }=\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Dokažimo,
+ da je 
+\begin_inset Formula $v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}$
+\end_inset
+
+ pravokoten na vse elemente 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Zaradi linearnosti skalarnega produkta zadošča preveriti,
+ da je pravokoten na bazo 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall j\in\left\{ 1..k\right\} $
+\end_inset
+
+ velja (spomnimo se,
+ da je 
+\begin_inset Formula $\left\langle w_{i},w_{j}\right\rangle =0\Leftrightarrow i\not=j$
+\end_inset
+
+,
+ zato po drugem enačaju ostane le še en člen vsote):
+\begin_inset Formula 
+\[
+\left\langle v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }\left\langle w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\frac{\left\langle v,w_{j}\right\rangle }{\cancel{\left\langle w_{j},w_{j}\right\rangle }}\cancel{\left\langle w_{j},w_{j}\right\rangle }=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle v,w_{j}\right\rangle -\left\langle v,w_{j}\right\rangle =0
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Obstoj ortogonalne baze —
+ Gram-Schmidtova ortogonalizacija
+\end_layout
+
+\begin_layout Standard
+Radi bi dokazali,
+ da ima vsak KRVPSSP ortogonalno bazo in da je moč vsako ortogonalno množico dopolniti do ortogonalne baze.
+ Konstruktiven dokaz 
+\begin_inset Formula $\ddot{\smile}!$
+\end_inset
+
+ —
+ postopek,
+ imenovan Gram-Schmidtova ortogonalizacija,
+ iz poljubne baze naredi ortogonalno.
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ njegova poljubna baza.
+ Naj bo 
+\begin_inset Formula $v_{1}\coloneqq u_{1}$
+\end_inset
+
+,
+\begin_inset Formula 
+\[
+v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=u_{2}-u_{2}'
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }v_{2}=u_{3}-u_{3}'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{n}\coloneqq u_{n}-\sum_{i=1}^{n-1}\frac{\left\langle u_{n},v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=u_{n}-u_{n}'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Trdimo,
+ da je 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ ortogonalna baza za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Opazimo,
+ da je 
+\begin_inset Formula $u_{2}'$
+\end_inset
+
+ ortogonalna projekcija 
+\begin_inset Formula $u_{2}$
+\end_inset
+
+ na 
+\begin_inset Formula $\Lin\left\{ v_{1}\right\} $
+\end_inset
+
+,
+ 
+\begin_inset Formula $u_{3}'$
+\end_inset
+
+ ortogonalna projekcija 
+\begin_inset Formula $u_{3}$
+\end_inset
+
+ na 
+\begin_inset Formula $\Lin\left\{ v_{1},v_{2}\right\} $
+\end_inset
+
+,
+ ...,
+ 
+\begin_inset Formula $u_{n}'$
+\end_inset
+
+ pa ortogonalna projekcija na 
+\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n-1}\right\} $
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+v_{2}=u_{2}-u_{2}'\perp\Lin\left\{ v_{1}\right\} \text{, torej }v_{2}\perp v_{1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}=u_{3}-u_{3}'\perp\Lin\left\{ v_{1},v_{2}\right\} \text{, torej }v_{3}\perp v_{1},v_{3}\perp v_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{n}=u_{n}-u_{n}'\perp\Lin\left\{ v_{1},\dots,v_{n-1}\right\} \text{, torej }v_{n}\perp v_{1},\dots,v_{n}\perp v_{n-1},
+\]
+
+\end_inset
+
+kar pomeni,
+ da so 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+ paroma ortogonalni.
+ Toda vprašanje je,
+ ali so neničelni,
+ kajti to je,
+ ne boste verjeli,
+ prav tako pogoj za ortogonalno množico.
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :$
+\end_inset
+
+ dokažimo neničelnost 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+:-)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $v_{1}$
+\end_inset
+
+ je neničeln,
+ ker je enak 
+\begin_inset Formula $u_{1}$
+\end_inset
+
+,
+ ki je element baze 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ 
+\begin_inset Formula $v_{2}$
+\end_inset
+
+ je neničeln,
+ ker je 
+\begin_inset Formula $v_{2}=u_{2}-\alpha v_{1}$
+\end_inset
+
+ in 
+\begin_inset Formula $u_{2}\not=\alpha v_{1}$
+\end_inset
+
+,
+ ker sta linearno neodvisna,
+ ker tvorita ortogonalno množico.
+ 
+\begin_inset Formula $v_{3}$
+\end_inset
+
+ je neničeln,
+ ker 
+\begin_inset Formula $v_{3}=u_{3}-\left(\beta v_{1}+\gamma v_{2}\right)$
+\end_inset
+
+ in ker so 
+\begin_inset Formula $v_{1},v_{2},u_{3}$
+\end_inset
+
+ LN,
+ 
+\begin_inset Formula $u_{3}\not=\left(\beta v_{1}+\gamma v_{2}\right)$
+\end_inset
+
+.
+ In tako dalje.
+\end_layout
+
+\begin_layout Paragraph*
+Dopolnitev ortogonalne množice do baze
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k}\right\} $
+\end_inset
+
+ ortogonalna množica,
+ torej je linearno neodvisna,
+ torej jo lahko dopolnimo do baze.
+ 
+\begin_inset Formula $\left\{ u_{k+1},\dots,u_{n}\right\} $
+\end_inset
+
+ je dopolnitev do baze.
+ Toda slednja še ni ortogonalna.
+ A nič ne de,
+ uporabimo lahko Gram-Schmidtovo ortogonalizacijo na 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{k},u_{k+1},\dots,u_{n}\right\} $
+\end_inset
+
+ in dobimo ortogonalno bazo 
+\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
+\end_inset
+
+.
+ Opazimo,
+ da ker so po predpostavki 
+\begin_inset Formula $u_{1},\dots,u_{k}$
+\end_inset
+
+ ortogonalni,
+ velja 
+\begin_inset Formula $v_{1}=u_{1},\dots,v_{k}=u_{k}$
+\end_inset
+
+ (po GS).
+\end_layout
+
+\begin_layout Example*
+primer GS ortogonalizacije iz analize.
+ Naj bo 
+\begin_inset Formula $V=\mathbb{R}\left[x\right]_{\leq3}$
+\end_inset
+
+.
+ Baza:
+ 
+\begin_inset Formula $u_{1}=1,u_{2}=x,u_{3}=x^{2},u_{4}=x^{3}$
+\end_inset
+
+,
+ skalarni produkt naj bo 
+\begin_inset Formula $\left\langle p,q\right\rangle =\int_{-1}^{1}p\left(x\right)q\left(x\right)dx$
+\end_inset
+
+.
+ Konstruirajmo pripadajočo ortogonalno bazo 
+\begin_inset Formula $v_{1},\dots,v_{4}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+v_{1}\coloneqq u_{1}=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=x-\frac{\int_{-1}^{1}xdx}{\int_{-1}^{1}dx}=x-0=x=u_{2}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }=x^{2}-\frac{\int_{-1}^{1}x^{2}dx}{\int_{-1}^{1}dx}-\frac{\int_{-1}^{1}x^{3}dx}{\int_{-1}^{1}x^{2}dx}x=\cdots=x²-\frac{1}{3}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+v_{4}\coloneqq\cdots=x^{2}-\frac{3}{5}x
+\]
+
+\end_inset
+
+Sklep:
+ 
+\begin_inset Formula $\left\{ 1,x,x^{2}-\frac{1}{3},x^{2}-\frac{3}{5}x\right\} $
+\end_inset
+
+ je ortogonalna baza za ta vektorski prostor s tem skalarnim produktom.
+ Normirajmo jo!
+ Norme teh baznih vektorjev po vrsti so 
+\begin_inset Formula $\sqrt{2},\sqrt{\frac{2}{3}},\sqrt{\frac{8}{45}},\sqrt{\frac{8}{175}}$
+\end_inset
+
+.
+ Pripadajoča ortonormirana baza je torej 
+\begin_inset Formula $\left\{ \frac{1}{\sqrt{2}},\frac{x}{\sqrt{\frac{2}{3}}},\frac{x^{2}-\frac{1}{3}}{\sqrt{\frac{8}{45}}},\frac{x^{2}-\frac{3}{5}x}{\sqrt{\frac{8}{175}}}\right\} .$
+\end_inset
+
+ Normiranje bi sicer prineslo lepše formule,
+ vendar bi v račune prineslo te objektivno grde konstante.
+\end_layout
+
+\begin_layout Subsubsection
+Ortogonalni komplement
+\end_layout
+
+\begin_layout Definition
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP nad 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $S\subseteq V$
+\end_inset
+
+.
+ Ortogonalni komplement 
+\begin_inset Formula $S$
+\end_inset
+
+ je množica 
+\begin_inset Formula $S^{\perp}$
+\end_inset
+
+.
+ Vsebuje vse tiste vektorje iz 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ ki so ortogonalni na 
+\begin_inset Formula $S$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $S^{\perp}\coloneqq\left\{ v\in V;\forall s\in S:v\perp s\right\} =\left\{ v\in V;v\perp S\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall S\subseteq V:S^{\perp}$
+\end_inset
+
+ je podprostor 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazati je treba 
+\begin_inset Formula $\forall u_{1},u_{2}\in S^{\perp},\alpha_{1},\alpha_{2}\in F:\alpha_{1}v_{1}+\alpha_{2}v_{2}\in S^{\perp}$
+\end_inset
+
+.
+ Po definiciji ortogonalnega komplementa velja 
+\begin_inset Formula 
+\[
+\forall s\in S:\left\langle u_{1},s\right\rangle =0\wedge\left\langle u_{2},s\right\rangle =0\Longrightarrow0=\alpha_{1}\left\langle u_{1},s\right\rangle +\alpha_{2}\left\langle u_{2},s\right\rangle =\left\langle \alpha_{1}u_{1}+\alpha_{2}u_{2},s\right\rangle \Longrightarrow\alpha_{1}u_{1}+\alpha_{2}u_{2}\in S^{\perp}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+ortogonalni razcep.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $W$
+\end_inset
+
+ vektorski podprostor 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Potem velja 
+\begin_inset Formula $V=W\oplus W^{\perp}$
+\end_inset
+
+ (ortogonalni razcep glede na 
+\begin_inset Formula $W$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ pojuben,
+ 
+\begin_inset Formula $V^{\perp}$
+\end_inset
+
+ pa ortogonalna projekcija 
+\begin_inset Formula $V$
+\end_inset
+
+ na 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Potem velja,
+ da je 
+\begin_inset Formula $v=v-v'+v'$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $v-v'$
+\end_inset
+
+ pravokoten na 
+\begin_inset Formula $W$
+\end_inset
+
+,
+ 
+\begin_inset Formula $v'$
+\end_inset
+
+ pa element 
+\begin_inset Formula $v'$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $v\in W\oplus W^{\perp}$
+\end_inset
+
+.
+ Vsota je direktna,
+ kajti 
+\begin_inset Formula $\forall v\in W\cap W^{\perp}:v\perp v\Leftrightarrow v\perp v\Leftrightarrow\left\langle v,v\right\rangle =0\Leftrightarrow v=0\Longrightarrow W\cap W^{\perp}=\left\{ 0\right\} $
+\end_inset
+
+ (po karakterizaciji direktnih vsot).
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $W$
+\end_inset
+
+ vektorski podprostor v 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\left(W^{\perp}\right)^{\perp}=W$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po definiciji ortogonalnega komplementa je 
+\begin_inset Formula $W\subseteq\left(W^{\perp}\right)^{\perp}$
+\end_inset
+
+,
+ ker 
+\begin_inset Formula $W\perp W^{\perp}$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $\dim W=\dim\left(W^{\perp}\right)^{\perp}$
+\end_inset
+
+.
+ Ortogonalni razcep glede na 
+\begin_inset Formula $W$
+\end_inset
+
+ je 
+\begin_inset Formula $V=W\oplus W^{\perp}\Rightarrow\dim W+\dim W^{\perp}=\dim V$
+\end_inset
+
+,
+ ortogonalni razcep glede na 
+\begin_inset Formula $W^{\perp}$
+\end_inset
+
+ pa je 
+\begin_inset Formula $V=W^{\perp}\oplus\left(W^{\perp}\right)^{\perp}\Rightarrow\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}=\dim V$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\dim V=\dim V
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\dim W+\dim W^{\perp}=\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\dim W^{\perp}=\dim\left(W^{\perp}\right)^{\perp}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Alternativni dokaz:
+ Naj bodo 
+\begin_inset Formula $w_{1},\dots,w_{k}$
+\end_inset
+
+ OB za 
+\begin_inset Formula $W$
+\end_inset
+
+.
+ Dopolnimo jo do OB za 
+\begin_inset Formula $V$
+\end_inset
+
+ z GS z 
+\begin_inset Formula $w_{k+1},\dots,w_{n}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $w_{k+1},\dots,w_{n}$
+\end_inset
+
+ OB za 
+\begin_inset Formula $W^{\perp}$
+\end_inset
+
+ in ker je 
+\begin_inset Formula $w_{1},\dots,w_{n}$
+\end_inset
+
+ njena dopolnitev do OB 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $w_{1},\dots,w_{k}$
+\end_inset
+
+ OB za 
+\begin_inset Formula $\left(W^{\perp}\right)^{\perp}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $W^{\perp}=\left(W^{\perp}\right)^{\perp}$
+\end_inset
+
+,
+ saj imata isti ortogonalni bazi.
+\end_layout
+
+\begin_layout Subsection
+Adjungirana linearna preslikava
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ vektorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Vemo,
+ da je 
+\begin_inset Formula $F$
+\end_inset
+
+ vekrorski prostor nad 
+\begin_inset Formula $F$
+\end_inset
+
+.
+ Linearnim preslikavam 
+\begin_inset Formula $V\to F$
+\end_inset
+
+ pravimo linearni funkcionali na 
+\begin_inset Formula $V$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ VPSSP in 
+\begin_inset Formula $F\in\left\{ \mathbb{R},\mathbb{C}\right\} $
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $w\in V$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\varphi:V\to F$
+\end_inset
+
+ (torej je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ linearni funkcional),
+ ki slika 
+\begin_inset Formula $v\mapsto\left\langle v,w\right\rangle $
+\end_inset
+
+.
+ Preslikava je po aksiomu 3 za skalarni produkt linearna.
+\end_layout
+
+\begin_layout Theorem*
+Rieszov izrek o reprezentaciji linearnih funkcionalov.
+ Naj bo 
+\begin_inset Formula $V$
+\end_inset
+
+ KRVPSSP.
+ Za vsak linearen funkcional 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ na 
+\begin_inset Formula $V$
+\end_inset
+
+ obstaja natanko en vektor 
+\begin_inset Formula $w\in V\ni:\forall v\in V:\varphi\left(v\right)=\left\langle v,w\right\rangle $
+\end_inset
+
+.
+ ZDB slednja konstrukcija nam da vse linearne funkcionale.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo enolično eksistenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Eksistenca 
+\begin_inset Formula $w$
+\end_inset
+
+:
+ Vzemimo poljubno OB 
+\begin_inset Formula $w_{1},\dots,w_{n}$
+\end_inset
+
+ za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\forall v\in V:v=\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}$
+\end_inset
+
+.
+ (fourierov razvoj po OB).
+ Ker je 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ linearna,
+ velja 
+\begin_inset Formula 
+\[
+\varphi\left(v\right)=\varphi\left(\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}\right)\overset{\text{linearna}}{=}\left\langle v,w_{1}\right\rangle \varphi w_{1}+\cdots+\left\langle v,w_{n}\right\rangle \varphi w_{n}\overset{\text{konj. hom. v 2. fakt.}}{=}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle v,\overline{\varphi w_{1}}w_{1}\right\rangle +\cdots+\left\langle v,\overline{\varphi w_{n}}w_{n}\right\rangle \overset{\text{konj. ad. v 2. fakt.}}{=}\left\langle v,\left(\varphi w_{1}\right)w_{1}+\cdots+\left(\varphi w_{n}\right)w_{n}\right\rangle 
+\]
+
+\end_inset
+
+Za dan 
+\begin_inset Formula $\varphi$
+\end_inset
+
+ smo konstruirali eksplicitno formulo za iskani 
+\begin_inset Formula $w$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Enoličnost 
+\begin_inset Formula $w$
+\end_inset
+
+:
+ PDDRAA 
+\begin_inset Formula $\forall v\in V:\varphi\left(v\right)=\left\langle v,w_{1}\right\rangle =\left\langle v,w_{2}\right\rangle $
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\forall v\in V:\left\langle v,w_{1}-w_{2}\right\rangle =0$
+\end_inset
+
+.
+ Vzemimo konkreten 
+\begin_inset Formula $v=w_{1}-w_{2}$
+\end_inset
+
+ in ga vstavimo v formulo 
+\begin_inset Formula $\left\langle v,w_{1}-w_{2}\right\rangle =0=\left\langle w_{1}-w_{2},w_{1}-w_{2}\right\rangle =0\Rightarrow w_{1}-w_{2}=0\Rightarrow w_{1}=w_{2}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Definition*
+Naj bosta 
+\begin_inset Formula $U,V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ linearna.
+ Adjungirana linearna preslikava,
+ pripadajoča 
+\begin_inset Formula $L$
+\end_inset
+
+,
+ je taka 
+\begin_inset Formula $L^{*}:V\to U$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $
+\end_inset
+
+.
+ Levi skalarni produkt je tisti iz 
+\begin_inset Formula $V$
+\end_inset
+
+,
+ desni pa tisti iz 
+\begin_inset Formula $U$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Da lahko pišemo 
+\begin_inset Formula $L^{*}$
+\end_inset
+
+,
+ trdimo,
+ da je 
+\begin_inset Formula $L^{*}$
+\end_inset
+
+ vedno obstaja in to vselej enolično.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo enolično eksistenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Enoličnost:
+ Naj bosta 
+\begin_inset Formula $L^{*}$
+\end_inset
+
+ in 
+\begin_inset Formula $L^{\circ}$
+\end_inset
+
+ dve adjungirani linearni preslikavi za 
+\begin_inset Formula $L$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle $
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula 
+\[
+\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\left\langle u,L^{*}v-L^{\circ}v\right\rangle 
+\]
+
+\end_inset
+
+Za vsaka 
+\begin_inset Formula $u$
+\end_inset
+
+ in 
+\begin_inset Formula $v$
+\end_inset
+
+.
+ Sedaj vstavimo 
+\begin_inset Formula $u=L^{*}v-L^{\circ}v$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+0=\left\langle L^{*}v-L^{\circ}v,L^{*}v-L^{\circ}v\right\rangle \Longrightarrow L^{*}v-L^{\circ}v=0\Longrightarrow\forall v\in V:L^{*}v=L^{\circ}v
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Eksistenca:
+ Naj bosta 
+\begin_inset Formula $U,V$
+\end_inset
+
+ KRVPSSP in 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ poljuben.
+ Vpeljimo linearni funkcional 
+\begin_inset Formula $\varphi:U\to F$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $u\mapsto\left\langle Lu,v\right\rangle $
+\end_inset
+
+.
+ Prepričajmo se,
+ da je ta funkcional linearna preslikava:
+ 
+\begin_inset Formula 
+\[
+\varphi\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\left\langle L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Lu_{1}+\alpha_{2}Lu_{2},v\right\rangle =\alpha_{1}\left\langle Lu_{1},v\right\rangle +\alpha_{2}\left\langle Lu_{2},v\right\rangle 
+\]
+
+\end_inset
+
+Uporabimo Rieszov izrek za funkcional 
+\begin_inset Formula $\varphi$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists!w\in U\ni:\forall u\in U:\varphi u=\left\langle u,w\right\rangle $
+\end_inset
+
+.
+ Vpeljimo 
+\begin_inset Formula $L^{*}v=w$
+\end_inset
+
+,
+ s čimer za poljuben 
+\begin_inset Formula $v$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $L^{*}v$
+\end_inset
+
+.
+ Dokažimo,
+ da je dobljena preslikava linearna:
+ 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$L^{*}
+\backslash
+left(
+\backslash
+beta_{1}v_{1}+
+\backslash
+beta_{2}v_{2}
+\backslash
+right)=
+\backslash
+beta_{1}L^{*}v_{1}+
+\backslash
+beta_{2}L^{*}v_{2}$}
+\end_layout
+
+\end_inset
+
+.
+ Vzemimo pojuben 
+\begin_inset Formula $u\in U$
+\end_inset
+
+ in računajmo (fino bi bilo dobiti nič):
+\begin_inset Formula 
+\[
+\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\beta_{1}L^{*}v_{1}-\beta_{2}L^{*}v_{2}\right\rangle \overset{\text{kl2f}}{=}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle \overset{\text{lin }L^{*}}{=}\left\langle u,\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle +\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =0
+\]
+
+\end_inset
+
+Ker to velja za vsak 
+\begin_inset Formula $u$
+\end_inset
+
+,
+ velja tudi za 
+\begin_inset Formula $u=L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)$
+\end_inset
+
+,
+ torej dobimo
+\begin_inset Formula 
+\[
+\left\langle L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right),L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =0
+\]
+
+\end_inset
+
+torej po prvem aksiomu za skalarni produtk velja linearnost:
+\begin_inset Formula 
+\[
+L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)=\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Naj bo 
+\begin_inset Formula $A\in M_{m\times n}\left(F\right)$
+\end_inset
+
+ s pripadajočo linearno preslikavo 
+\begin_inset Formula $L_{A}=F^{n}\to F^{m}$
+\end_inset
+
+,
+ ki slika 
+\begin_inset Formula $v\mapsto Av$
+\end_inset
+
+.
+ Kako izgleda matrika 
+\begin_inset Formula $L_{A^{*}}$
+\end_inset
+
+?
+ Odgovor je odvisen od izbire skalarnega produkta.
+ Izberimo standardni skalarni produkt v 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $F^{m}$
+\end_inset
+
+ in 
+\begin_inset Formula $L_{A^{*}}:F^{m}\to F^{n}$
+\end_inset
+
+ definiramo z 
+\begin_inset Formula $v\mapsto A^{*}v$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $A^{*}=\overline{A^{T}}$
+\end_inset
+
+,
+ torej transponiranka 
+\begin_inset Formula $A$
+\end_inset
+
+ z vsemi elementi konjugiranimi.
+ Izkaže se,
+ da je potemtakem 
+\begin_inset Formula $L_{A^{*}}$
+\end_inset
+
+ adjungirana linearna preslikava od 
+\begin_inset Formula $L_{A}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Matrika adjungirane linearne preslikave
+\end_layout
+
+\begin_layout Standard
+Naj bosta 
+\begin_inset Formula $U,V$
+\end_inset
+
+ KRVPSSP in naj bo 
+\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
+\end_inset
+
+ ONB za 
+\begin_inset Formula $U$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
+\end_inset
+
+ ONB za 
+\begin_inset Formula $V$
+\end_inset
+
+.
+ Vzemimo linearno preslikavo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+.
+ Izpeljimo zvezo med 
+\begin_inset Formula $L$
+\end_inset
+
+ in 
+\begin_inset Formula $L^{*}$
+\end_inset
+
+ glede na bazi 
+\begin_inset Formula $\mathcal{B}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathcal{C}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+ za 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$
+\end_inset
+
+ za 
+\begin_inset Formula $L^{*}:V\to U$
+\end_inset
+
+.
+ Izračunajmo 
+\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
+\end_inset
+
+ tako,
+ da uporabimo fourierov razvoj:
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+Lu_{1} & = & \left\langle Lu_{1},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{1},v_{m}\right\rangle v_{m}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+Lu_{n} & = & \left\langle Lu_{n},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{n},v_{m}\right\rangle v_{m}
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
+\left\langle Lu_{1},v_{1}\right\rangle  & \cdots & \left\langle Lu_{n},v_{1}\right\rangle \\
+\vdots &  & \vdots\\
+\left\langle Lu_{1},v_{m}\right\rangle  & \cdots & \left\langle Lu_{n},v_{m}\right\rangle 
+\end{array}\right]=\left[\begin{array}{ccc}
+\left\langle u_{1},L^{*}v_{1}\right\rangle  & \cdots & \left\langle u_{n},L^{*}v_{1}\right\rangle \\
+\vdots &  & \vdots\\
+\left\langle u_{1},L^{*}v_{m}\right\rangle  & \cdots & \left\langle u_{n},L^{*}v_{m}\right\rangle 
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Sedaj izračunajmo še 
+\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$
+\end_inset
+
+ spet s fourierovim razvojem in primerjajmo istoležne koeficiente:
+\begin_inset Formula 
+\[
+\begin{array}{ccccccc}
+L^{*}v_{1} & = & \left\langle L^{*}v_{1},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle L^{*}v_{1},u_{n}\right\rangle u_{n}\\
+\vdots &  & \vdots &  &  &  & \vdots\\
+L^{*}v_{m} & = & \left\langle Lv_{m},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle Lv_{m},u_{n}\right\rangle u_{n}
+\end{array}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[L\right]_{\mathcal{B}\leftarrow\mathcal{C}}=\left[\begin{array}{ccc}
+\left\langle L^{*}v_{1},u_{1}\right\rangle  & \cdots & \left\langle L^{*}v_{m},u_{1}\right\rangle \\
+\vdots &  & \vdots\\
+\left\langle L^{*}v_{1},u_{n}\right\rangle  & \cdots & \left\langle L^{*}v_{m},u_{n}\right\rangle 
+\end{array}\right]=\left[\begin{array}{ccc}
+\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{1},L^{*}v_{m}\right\rangle }\\
+\vdots &  & \vdots\\
+\overline{\left\langle u_{n},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle }
+\end{array}\right]=\left[\begin{array}{ccc}
+\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{1}\right\rangle }\\
+\vdots &  & \vdots\\
+\overline{\left\langle u_{1},L^{*}v_{m}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle }
+\end{array}\right]^{T}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\overline{\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}}^{T}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Kako izgleda lastnost 
+\begin_inset Formula $\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $
+\end_inset
+
+?
+ Naj bo 
+\begin_inset Formula $u\in F^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $v\in F^{m}$
+\end_inset
+
+ in 
+\begin_inset Formula $A=m\times n$
+\end_inset
+
+ matrika.
+ Ali za standardna skalarna produkta v 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $F^{m}$
+\end_inset
+
+ 
+\begin_inset Formula $\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle $
+\end_inset
+
+ velja tudi za matrike,
+ če vzamemo 
+\begin_inset Formula $A^{*}=\overline{A}^{T}=\overline{A^{T}}$
+\end_inset
+
+?
+ Pa preverimo (ja,
+ velja):
+\begin_inset Formula 
+\[
+\left\langle u,v\right\rangle =u_{1}\overline{v_{1}}+\cdots+u_{n}\overline{v_{n}}=\left[\begin{array}{ccc}
+\overline{v_{1}} & \cdots & \overline{v_{n}}\end{array}\right]\left[\begin{array}{c}
+u_{1}\\
+\vdots\\
+u_{n}
+\end{array}\right]=v^{*}u
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle Au,v\right\rangle =v^{*}Au
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle u,A^{*}v\right\rangle =\left(A^{*}v\right)^{*}u=v^{*}\left(A^{*}\right)^{*}u=v^{*}Au
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Lastnosti adjungiranja:
+ 
+\begin_inset Formula $\left(\alpha A+\beta B\right)^{*}=\overline{\alpha}A^{*}+\overline{\beta}B^{*}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(AB\right)^{*}=B^{*}A^{*}$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(A^{*}\right)^{*}=A$
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME DOKAŽI
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Jedro in slika adjungirane linearne preslikave
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ linearna.
+ Velja 
+\begin_inset Formula $\Ker\left(L^{*}\right)=\left(\Slika L\right)^{\perp}$
+\end_inset
+
+ in 
+\begin_inset Formula $\Slika\left(L^{*}\right)=\left(\Ker L\right)^{\perp}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $v\in\Ker L^{*}$
+\end_inset
+
+ za 
+\begin_inset Formula $L^{*}:V\to U$
+\end_inset
+
+.
+ Velja
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+v\in\Ker\left(L^{*}\right)\Leftrightarrow L^{*}v=0\Leftrightarrow\forall u\in U:\left\langle u,L^{*}v\right\rangle =0\Leftrightarrow\forall u\in U:\left\langle Lu,v\right\rangle =0\Leftrightarrow\forall w\in\Slika L:\left\langle w,v\right\rangle =0\Leftrightarrow v\in\left(\Slika L\right)^{\perp}
+\]
+
+\end_inset
+
+Velja torej 
+\begin_inset Formula $\Ker L^{*}=\left(\Slika L\right)^{\perp}\Rightarrow\Ker L=\left(\Slika L^{*}\right)^{\perp}\Rightarrow\left(\Ker L\right)^{\perp}=\Slika L^{*}\Rightarrow\left(\Ker L^{*}\right)^{\perp}=\Slika L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Za 
+\begin_inset Formula $L:U\to V$
+\end_inset
+
+ velja 
+\begin_inset Formula $\Ker\left(L^{*}L\right)=\Ker L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Vzemimo poljuben 
+\begin_inset Formula $u\in U$
+\end_inset
+
+ in dokazujemo enakost množic (obe vsebovanosti):
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\supseteq\right)$
+\end_inset
+
+ Če 
+\begin_inset Formula $u\in\Ker L\Rightarrow Lu=0\overset{\text{množimo z }L^{*}}{\Longrightarrow}L^{*}Lu=L^{*}u=0\Rightarrow u\in\Ker L^{*}L\Rightarrow\Ker L\subseteq\Ker L^{*}L$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(\subseteq\right)$
+\end_inset
+
+ Če 
+\begin_inset Formula $u\in\Ker L^{*}L\Rightarrow L^{*}Lu=0\Rightarrow\left\langle u,L^{*}Lu\right\rangle =0\Rightarrow\left\langle Lu,Lu\right\rangle =0\Rightarrow Lu=0\Rightarrow u\in\Ker L\Rightarrow\Ker L^{*}L\subseteq\Ker L$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Corollary*
+\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L^{*}\left(L^{*}\right)^{*}\right)=\Slika\left(\left(L^{*}L\right)^{*}\right)=\left(\Ker L^{*}L\right)^{\perp}=\left(\Ker L\right)^{\perp}=\Slika L^{*}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Lastne vrednosti adjungirane linearne preslikave.
+\end_layout
+
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $\overline{\lambda}$
+\end_inset
+
+ lastna vrednost za 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+.
+ ZDB 
+\begin_inset Formula $\det\left(A-\lambda I\right)=0\Rightarrow\det\left(A-\lambda\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $B=A-\lambda I$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $B^{*}=A^{*}-\overline{\lambda}I^{*}=A^{*}-\overline{\lambda}I$
+\end_inset
+
+.
+ Radi bi dokazali 
+\begin_inset Formula $\det B=0\Rightarrow\det B^{*}=0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $B^{*}=\overline{B}^{T}\Rightarrow\det B^{*}=\det\overline{B}^{T}=\det\overline{B}=\overline{\det B}\Rightarrow\det B=0\Rightarrow\det B^{*}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Iz te formule izvemo tudi karakteristični polimom 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $p_{A^{*}}\left(x\right)=\det\left(A^{*}-xI\right)\Rightarrow p_{A}\left(\overline{x}\right)=\det\left(A-\overline{x}I\right)=\det\left(A^{*}-xI\right)^{*}=\overline{\det\left(A^{*}-xI\right)}=\overline{p_{A^{*}}\left(x\right)}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{p_{A}\left(\overline{x}\right)}$
+\end_inset
+
+.
+ Torej,
+ če je 
+\begin_inset Formula $p_{A}\left(x\right)=c_{0}x^{0}+\cdots+x_{n}x^{n}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{c_{0}\overline{x^{0}}+\cdots+x_{n}\overline{x^{n}}}=\overline{c_{0}}x^{0}+\cdots+\overline{c_{n}}x^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Alternativen dokaz:
+ Najprej dokažimo 
+\begin_inset Formula $\dim\Ker B^{*}=\dim\Ker B$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\dim\Ker B^{*}=\dim\left(\Slika B\right)^{\perp}=n-\dim\Slika B=\dim\Ker B$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\Ker\left(B\right)\not=0\Leftrightarrow\Ker\left(B^{*}\right)\not=0$
+\end_inset
+
+,
+ torej so lastne vrednosti 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ konjugirane lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Med lastnimi vektorji 
+\begin_inset Formula $A$
+\end_inset
+
+ in lastnimi vektorji 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ (žal) ni posebne zveze.
+ Primer:
+ 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+1 & 2\\
+i & 1
+\end{array}\right]$
+\end_inset
+
+ ima lastne vektorje 
+\begin_inset Formula $\vec{v_{1}}=\left[\begin{array}{c}
+1-i\\
+-1
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{v_{2}}=\left[\begin{array}{c}
+1-i\\
+1
+\end{array}\right]$
+\end_inset
+
+,
+ 
+\begin_inset Formula $A^{*}=\left[\begin{array}{cc}
+1 & 1\\
+2 & -i
+\end{array}\right]$
+\end_inset
+
+ pa lastne vektorje 
+\begin_inset Formula $\vec{v_{1}'}=\left[\begin{array}{c}
+1-i\\
+-2
+\end{array}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{v_{2}'}=\left[\begin{array}{c}
+1-i\\
+2
+\end{array}\right]$
+\end_inset
+
+.
+ Med temi vektorji ni nobenih kolinearnosti.
+ Obstajajo pa zveze v nekaterih zanimivih primerih:
+\end_layout
+
+\begin_layout Claim*
+Če matrika 
+\begin_inset Formula $A$
+\end_inset
+
+ zadošča 
+\begin_inset Formula $A^{*}A=AA^{A}$
+\end_inset
+
+ (pravimo 
+\begin_inset Formula $A$
+\end_inset
+
+ je normalna),
+ iz 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+ sledi 
+\begin_inset Formula $A^{*}v=\overline{\lambda}v$
+\end_inset
+
+,
+ torej imata 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ iste lastne vrednosti.
+\end_layout
+
+\begin_layout Proof
+Če velja 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $Av-\lambda v=\left(A-\lambda I\right)v=Bv=0\Rightarrow v\in\Ker B$
+\end_inset
+
+.
+ Če velja 
+\begin_inset Formula $A^{*}v=\overline{\lambda}v$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $A^{*}v-\overline{\lambda}v=\left(A^{*}-\overline{\lambda}I\right)v=B^{*}v=0\Rightarrow v\in\Ker B^{*}$
+\end_inset
+
+.
+ Dokazati je treba še 
+\begin_inset Formula $\Ker B=\Ker B^{*}$
+\end_inset
+
+:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Ali velja 
+\begin_inset Formula $A^{*}A=AA^{*}\Rightarrow B^{*}B=BB^{*}$
+\end_inset
+
+?
+ Ja.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $B^{*}B=\left(A^{*}-\overline{\lambda}I\right)\left(A-\lambda I\right)=A^{*}A-\overline{\lambda}A-\lambda A^{*}+\overline{\lambda}\lambda I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $BB^{*}=\left(A-\lambda I\right)\left(A^{*}-\overline{\lambda}I\right)=AA^{*}-\overline{\lambda}A-\lambda A^{*}+\lambda\overline{\lambda}I$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Ali velja 
+\begin_inset Formula $B^{*}B=BB^{*}\Rightarrow\Ker B=\Ker B^{*}$
+\end_inset
+
+?
+ Iz 
+\begin_inset Formula $B^{*}B=BB^{*}$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker\left(BB^{*}\right)\Rightarrow\Ker\left(B\right)=\Ker\left(B^{*}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\Ker B=\Ker B^{*}\Rightarrow\forall v\in V:Av=\lambda v\Leftrightarrow A^{*}v=\overline{\lambda}v$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Subsubsection
+Normalne matrike
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A$
+\end_inset
+
+ je normalna 
+\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Dokazali smo že,
+ da za normalne matrike velja,
+ da imata 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ iste lastne vektorje,
+ kar v splošnem ne velja.
+\end_layout
+
+\begin_layout Claim*
+Lastni vektorji,
+ ki pripadajo različnim lastnim vrednostim normalne matrike,
+ so paroma ortogonalni.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A^{*}A=AA^{*}$
+\end_inset
+
+ za neko 
+\begin_inset Formula $A$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $Au=\lambda u$
+\end_inset
+
+ in 
+\begin_inset Formula $Av=\mu v$
+\end_inset
+
+ in 
+\begin_inset Formula $\mu\not=\lambda$
+\end_inset
+
+.
+ 
+\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+\mu\left\langle u,v\right\rangle =\left\langle u,\overline{\mu}v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle Au,v\right\rangle =\left\langle \lambda u,v\right\rangle =\lambda\left\langle u,v\right\rangle 
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(\mu-\lambda\right)\left\langle u,v\right\rangle =0\wedge u\not=\lambda\Rightarrow\left\langle u,v\right\rangle =0\Leftrightarrow u\perp v
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+Vsako normalno matriko se da diagonalizirati.
+\end_layout
+
+\begin_layout Proof
+Dokažimo,
+ da je jordanska forma normalne matrike diagonalna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ vsi korenski podprostori so lastni.
+ 
+\begin_inset Formula $\forall m,\lambda:\Ker\left(A-I\lambda\right)^{m}=\Ker\left(A-I\lambda\right)$
+\end_inset
+
+.
+ Zadošča dokazati za 
+\begin_inset Formula $m=2$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $m=2$
+\end_inset
+
+ in 
+\begin_inset Formula $B=A-I\lambda$
+\end_inset
+
+.
+ Dokažimo 
+\begin_inset Formula $\Ker B^{2}=\Ker B$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Če v 
+\begin_inset Formula $\Ker\left(A\right)=\Ker\left(A^{*}A\right)$
+\end_inset
+
+ vstavimo 
+\begin_inset Formula $A=B^{2}$
+\end_inset
+
+,
+ dobimo 
+\begin_inset Formula $\Ker B^{2}=\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ normalna,
+ je 
+\begin_inset Formula $B$
+\end_inset
+
+ normalna,
+ torej 
+\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=\left(B^{*}B\right)^{2}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)=\Ker\left(B^{*}B\right)^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Če v 
+\begin_inset Formula $\Ker A^{*}A=\Ker A$
+\end_inset
+
+ vstavimo 
+\begin_inset Formula $A=B^{*}B$
+\end_inset
+
+,
+ dobimo 
+\begin_inset Formula $\Ker B^{*}BB^{*}B=\Ker\left(B^{*}B\right)^{2}=\Ker B^{*}B$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Zopet upoštevamo 
+\begin_inset Formula $\Ker A^{*}A=\Ker A$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker B$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Ko dokažemo 
+\begin_inset Formula $B$
+\end_inset
+
+ normalna 
+\begin_inset Formula $\Rightarrow B^{*}$
+\end_inset
+
+ normalna,
+ bo iz 
+\begin_inset Formula $\Ker B^{2}=\Ker B$
+\end_inset
+
+ sledilo 
+\begin_inset Formula $\Ker B^{4}=\Ker B$
+\end_inset
+
+.
+ Preverimo,
+ a je 
+\begin_inset Formula $B^{2}$
+\end_inset
+
+ normalna,
+ če je 
+\begin_inset Formula $B$
+\end_inset
+
+ normalna:
+ 
+\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=BB^{*}BB^{*}=BBB^{*}B^{*}=B^{2}\left(B^{2}\right)^{*}$
+\end_inset
+
+.
+ Sedaj vemo 
+\begin_inset Formula $\Ker B=\Ker B^{2}=\Ker B^{4}=\Ker B^{8}=\cdots$
+\end_inset
+
+.
+ Vemo pa tudi,
+ da
+\begin_inset Formula 
+\[
+\Ker B\subseteq\Ker B^{2}\subseteq\Ker B^{3}\subseteq\Ker B^{4}\subseteq\Ker B^{5}\subseteq\Ker B^{6}\subseteq\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker B\subseteq\Ker B\subseteq\Ker B^{3}\subseteq\Ker B\subseteq\Ker B^{5}\subseteq\Ker B\subseteq\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\Ker B=\Ker B^{2}=\Ker B^{3}=\Ker B^{4}=\Ker B^{5}=\cdots
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\forall v:\Ker B^{m}=\Ker B
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Torej za vsako normalno matriko 
+\begin_inset Formula $A\exists$
+\end_inset
+
+ diagonalna 
+\begin_inset Formula $D$
+\end_inset
+
+ in obrnljiva 
+\begin_inset Formula $P$
+\end_inset
+
+ z ortonormiranimi stolpci,
+ da velja 
+\begin_inset Formula $AP=PD$
+\end_inset
+
+,
+ 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+.
+ Diagonala 
+\begin_inset Formula $D$
+\end_inset
+
+ so lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ stolpci 
+\begin_inset Formula $P$
+\end_inset
+
+ pa so njeni lastni vektorji.
+ Lastni podprostori 
+\begin_inset Formula $\left(A-\lambda_{1}I\right),\dots,\left(A-\lambda_{n}I\right)$
+\end_inset
+
+ so medsebojno pravokotni.
+ Izberimo ONB za vsak lasten podprostor.
+ Unija teh ONB je ONB za 
+\begin_inset Formula $F^{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $F^{n}=\Ker\left(A-\lambda_{1}I\right)\oplus\cdots\oplus\Ker\left(A-\lambda_{n}I\right)$
+\end_inset
+
+.
+ Ta ONB so stolpci matrike 
+\begin_inset Formula $P$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+Ortogonalne/unitarne matrike
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $A$
+\end_inset
+
+ kvadratna z ON stolpci glede na standardni skalarni produkt.
+ Pravimo,
+ da je 
+\begin_inset Formula $A$
+\end_inset
+
+ unitarna (v kompleksnem primer) oziroma ortogonalna (v realnem primeru).
+\end_layout
+
+\begin_layout Claim*
+Za unitarno 
+\begin_inset Formula $A$
+\end_inset
+
+ velja 
+\begin_inset Formula $A^{*}A=AA^{*}=I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokazujmo za unitarno.
+ Za ortogonalno je dokaz podoben.
+ Naj bo 
+\begin_inset Formula $A=\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{nn}
+\end{array}\right]$
+\end_inset
+
+ unitarna.
+ To pomeni,
+ da za vsaka stolpca 
+\begin_inset Formula $a_{i}=\left(a_{1i},\dots,a_{ni}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $a_{j}=\left(a_{1j},\dots,a_{nj}\right)$
+\end_inset
+
+ velja za vsak 
+\begin_inset Formula $i,j\in\left\{ 1..n\right\} $
+\end_inset
+
+ velja 
+\begin_inset Formula $\left\langle \text{\left[\begin{array}{c}
+a_{1i}\\
+\text{\ensuremath{\vdots}}\\
+a_{ni}
+\end{array}\right],\left[\begin{array}{c}
+a_{1j}\\
+\vdots\\
+a_{nj}
+\end{array}\right]}\right\rangle =a_{1i}\overline{a_{1j}}+\cdots+a_{ni}\overline{a_{nj}}=\begin{cases}
+0 & ;i\not=j\\
+1 & ;i=j
+\end{cases}$
+\end_inset
+
+.
+ Oglejmo si 
+\begin_inset Formula 
+\[
+A^{*}A=\left[\begin{array}{ccc}
+\overline{a_{11}} & \cdots & \overline{a_{n1}}\\
+\vdots &  & \vdots\\
+\overline{a_{1n}} & \cdots & \overline{a_{nn}}
+\end{array}\right]\left[\begin{array}{ccc}
+a_{11} & \cdots & a_{1n}\\
+\vdots &  & \vdots\\
+a_{n1} & \cdots & a_{nn}
+\end{array}\right]=\left[\begin{array}{ccc}
+1 &  & 0\\
+ & \ddots\\
+0 &  & 1
+\end{array}\right]
+\]
+
+\end_inset
+
+Očitno je res,
+ ker je vsak element 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ konstruiran s skalarnim množenjem vrstice leve matrike (konjugirani stolpci 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ ker smo poprej matriko transponiralo) in stolpca desne,
+ za kar predpis smo poprej že razbrali.
+\end_layout
+
+\begin_layout Remark*
+Za nekvadratne unitarne velja le 
+\begin_inset Formula $A^{*}A=I$
+\end_inset
+
+,
+ 
+\begin_inset Formula $AA^{*}=I$
+\end_inset
+
+ pa zaradi nezmožnosti množenja zaradi nepravilnih dimenzij seveda ne velja.
+\end_layout
+
+\begin_layout Claim*
+Naslednje trditve so za 
+\begin_inset Formula $P$
+\end_inset
+
+ z ortonormiranimi stolpci ekvivalentne:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $P^{*}P=I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall$
+\end_inset
+
+ ONB 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $
+\end_inset
+
+ je ON množica
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\exists$
+\end_inset
+
+ ONB 
+\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $
+\end_inset
+
+ je ON množica
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokazujemo ekvivalenco:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(1\Rightarrow2\right)$
+\end_inset
+
+ 
+\begin_inset Formula $\left\langle Pu,Pv\right\rangle =\left\langle u,P^{*}Pv\right\rangle =\left\langle u,v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(2\Rightarrow3\right)$
+\end_inset
+
+ 
+\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left\langle Pu,Pu\right\rangle =\left\langle u,u\right\rangle =\left|\left|u\right|\right|^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(2\Rightarrow1\right)$
+\end_inset
+
+ 
+\begin_inset Formula 
+\[
+\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle \Rightarrow\left\langle u,P^{*}Pv\right\rangle -\left\langle u,v\right\rangle =0\Rightarrow\left\langle u,\left(P^{*}P-I\right)v\right\rangle =0
+\]
+
+\end_inset
+
+ Sedaj izberimo 
+\begin_inset Formula $u=\left(P^{*}P-I\right)v$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left\langle \left(P^{*}P-I\right)v,\left(P^{*}P-I\right)v\right\rangle =0\Rightarrow P^{*}P-I=0\Rightarrow P^{*}P=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(3\Rightarrow2\right)$
+\end_inset
+
+ Po predpostavki 
+\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$
+\end_inset
+
+ Izrazimo skalarni produkt z normo:
+ 
+\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula 
+\[
+\left\langle Pu,Pv\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|Pu+i^{k}Pv\right|\right|^{2}=\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|P\left(u+i^{k}v\right)\right|\right|^{2}\overset{\text{predpostavka}}{=}\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u,v\right\rangle 
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(5\Rightarrow4\right)$
+\end_inset
+
+ Vzemimo poljuben 
+\begin_inset Formula $u$
+\end_inset
+
+ in ga razvijmo po ONB 
+\begin_inset Formula $u_{1},\dots,u_{n}$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $u=\alpha_{1}u_{1}+\cdots+\alpha_{n}u_{n}$
+\end_inset
+
+.
+ Ker so 
+\begin_inset Formula $u_{i}$
+\end_inset
+
+ ONB,
+ velja 
+\begin_inset Formula $\left|\left|u\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots\left|\alpha_{n}\right|^{2}$
+\end_inset
+
+.
+ Ker so 
+\begin_inset Formula $Pu_{1}$
+\end_inset
+
+ ONB po predpostavki,
+ 
+\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}$
+\end_inset
+
+,
+ torej velja 
+\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\left|u\right|\right|^{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(2\Rightarrow4\right)$
+\end_inset
+
+ Ker so 
+\begin_inset Formula $u_{1},\dots,u_{n}$
+\end_inset
+
+ ONM,
+ velja 
+\begin_inset Formula $\left\langle u_{i},u_{j}\right\rangle =\begin{cases}
+1 & ;i=j\\
+0 & ;i\not=j
+\end{cases}$
+\end_inset
+
+.
+ Tudi 
+\begin_inset Formula $Pu_{1},\dots,Pu_{n}$
+\end_inset
+
+ ortonormirana,
+ kajti po predpostavki 
+\begin_inset Formula $2$
+\end_inset
+
+ velja 
+\begin_inset Formula $\left\langle Pu_{i},Pu_{2}\right\rangle =\begin{cases}
+1 & ;i=j\\
+0 & ;i\not=j
+\end{cases}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $\left(4\Rightarrow5\right)$
+\end_inset
+
+ Očitno.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Lastne vrednosti unitarne matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ se nahajajo na enotski krožnici v 
+\begin_inset Formula $\Im$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A$
+\end_inset
+
+ unitarna in naj bo 
+\begin_inset Formula $v$
+\end_inset
+
+ tak,
+ da 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\left\langle v,v\right\rangle =\left\langle Av,Av\right\rangle =\left\langle \lambda v,\lambda v\right\rangle =\lambda\overline{\lambda}\left\langle v,v\right\rangle \Rightarrow\lambda\overline{\lambda}=1\Rightarrow\left|\lambda\right|=1\Rightarrow\lambda=e^{i\varphi}$
+\end_inset
+
+ za nek 
+\begin_inset Formula $\varphi$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Iz unitarnosti sledi normalnost,
+ zato so lastni vektorji unitarne matrike,
+ ki pripadajo paroma različnim lastnim vrednostim,
+ pravokotni (isto,
+ kot pri normalnih matrikah).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Prav tako kot pri normalnih matrikah lahko unitarne diagonalitziramo v tokrat ortogonalni bazi.
+ Pri unitarnih so stolpci 
+\begin_inset Formula $P$
+\end_inset
+
+ še celo normirani.
+ 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+,
+ kjer je 
+\begin_inset Formula $P$
+\end_inset
+
+ unitarna,
+ torej 
+\begin_inset Formula $P^{*}=P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Remark*
+Očitno je,
+ da če je 
+\begin_inset Formula $A$
+\end_inset
+
+ unitarna,
+ velja 
+\begin_inset Formula $A^{*}=A^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Simetrične/hermitske-matrike"
+
+\end_inset
+
+Simetrične/hermitske matrike
+\end_layout
+
+\begin_layout Definition*
+Matrika nad 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ je simetrična,
+ če zanjo velja 
+\begin_inset Formula $A^{*}=A$
+\end_inset
+
+.
+ Matrika nad 
+\begin_inset Formula $\mathbb{C}$
+\end_inset
+
+ je hermitska,
+ če zanjo velja 
+\begin_inset Formula $A^{*}=A$
+\end_inset
+
+.
+ Linearni preslikavi,
+ pripadajoči hermitski/simetrični matriki,
+ pravimo sebiadjungirana.
+\end_layout
+
+\begin_layout Fact*
+Vsaka hermitska/simetrična matrika je normalna,
+ kajti 
+\begin_inset Formula $A^{*}A=AA=AA^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Lastne vrednosti hermitskih/simetričnih matrik so realne.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=A^{*}$
+\end_inset
+
+ in naj bo 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+ za nek neničeln 
+\begin_inset Formula $v$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\lambda\left\langle v,v\right\rangle =\left\langle \lambda v,v\right\rangle =\left\langle Av,v\right\rangle =\left\langle v,A^{*}v\right\rangle =\left\langle v,Av\right\rangle =\left\langle v,\lambda v\right\rangle =\overline{\lambda}\left\langle v,v\right\rangle $
+\end_inset
+
+.
+ Potemtakem 
+\begin_inset Formula $\lambda=\overline{\lambda}\Rightarrow\lambda\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Diagonalizacija je zopet enaka kot pri normalnih matrikah z dodatkom —
+ vsaka hermitska matrika je podobna realni diagonalni,
+ kar za normalne ni res —
+ normalne so lahko podobne kompleksnim diagonalnim matrikam.
+\end_layout
+
+\begin_layout Subsubsection
+Pozitivno (semi)definitne matrike
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A$
+\end_inset
+
+ je pozitivno semidefinitna 
+\begin_inset Formula $\sim A\geq0\Leftrightarrow A=A^{*}\wedge\forall v:\left\langle Av,v\right\rangle \geq0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je pozitivno definitna 
+\begin_inset Formula $\sim A>0\Leftrightarrow A=A^{*}\wedge\forall v\not=0:\left\langle Av,v\right\rangle >0$
+\end_inset
+
+.
+ S tem ko skalarni produkt primerjamo (
+\begin_inset Formula $>,\geq$
+\end_inset
+
+),
+ implicitno zahtevamo njegovo realnost.
+ Primerjalni operatorji namreč na kompleksnih številih niso definirani.
+\end_layout
+
+\begin_layout Example*
+Vzemimo poljubno nenujno kvadratno 
+\begin_inset Formula $B$
+\end_inset
+
+ in definirajmo 
+\begin_inset Formula $A=B^{*}B$
+\end_inset
+
+.
+ Potem je 
+\begin_inset Formula $A$
+\end_inset
+
+ pozitivno semidefinitna,
+ kajti 
+\begin_inset Formula $A^{*}=\left(B^{*}B\right)^{*}=B^{*}B=A$
+\end_inset
+
+ in 
+\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle B^{*}Bv,v\right\rangle =\left\langle Bv,Bv\right\rangle \geq0$
+\end_inset
+
+.
+ Če pa bi bili stolpci 
+\begin_inset Formula $B$
+\end_inset
+
+ linearno neodvisni,
+ pa bi veljalo 
+\begin_inset Formula $\forall v:v\not=0\Rightarrow\left\langle Av,v\right\rangle =\left\langle B^{*}Bv\right\rangle =\left\langle Bv,Bv\right\rangle >0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $A\geq0\Rightarrow$
+\end_inset
+
+ lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ so 
+\begin_inset Formula $\geq0$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A>0\Rightarrow$
+\end_inset
+
+ lastne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ so 
+\begin_inset Formula $>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $A\geq0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $Av=\lambda v$
+\end_inset
+
+ za nek 
+\begin_inset Formula $v\not=0$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda\left\langle v,v\right\rangle $
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $\left\langle Av,v\right\rangle \geq0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lambda\left\langle v,v\right\rangle \geq0$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\left\langle v,v\right\rangle >0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lambda\geq0$
+\end_inset
+
+.
+ Analogno za 
+\begin_inset Formula $A>0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Diagonalizacija je ista kot za normalna,
+ s tem da za diagonalno 
+\begin_inset Formula $D$
+\end_inset
+
+ velja še,
+ da je pozitivno (semi)definitna,
+ ko je 
+\begin_inset Formula $A$
+\end_inset
+
+ pozitivno semidefinitna.
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\forall A\geq0\exists B=B^{*},B\geq0\ni:B^{2}=A$
+\end_inset
+
+.
+ ZDB Za vsako pozitivno semidefinitno matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ obstajaja taka unitarna pozitivno semidefinitna 
+\begin_inset Formula $B$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $B^{2}=A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+ in 
+\begin_inset Formula $P^{*}=P^{-1}$
+\end_inset
+
+ in 
+\begin_inset Formula $D=\left[\begin{array}{ccc}
+\lambda_{1} &  & 0\\
+ & \ddots\\
+0 &  & \lambda_{n}
+\end{array}\right]$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $E=\left[\begin{array}{ccc}
+\sqrt{\lambda_{1}} &  & 0\\
+ & \ddots\\
+0 &  & \sqrt{\lambda_{n}}
+\end{array}\right]\geq0$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $B=PEP^{-1}=PEP^{*}$
+\end_inset
+
+.
+ Opazimo 
+\begin_inset Formula $B^{*}=B$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(PEP^{-1}\right)^{*}=\left(PEP^{*}\right)^{*}=PE^{*}P^{*}=PEP^{-1}=PEP^{*}$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $E^{*}=E$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $\forall a\in\mathbb{R}:\sqrt{a}\in\mathbb{R}$
+\end_inset
+
+.
+ Oglejmo si 
+\begin_inset Formula $B^{2}=PEP^{-1}PEP^{-1}=PE^{2}P^{-1}=PDP^{-1}$
+\end_inset
+
+.
+ Tako definiramo 
+\begin_inset Formula $\sqrt{A}=B$
+\end_inset
+
+ (tu 
+\begin_inset Formula $\sqrt{}$
+\end_inset
+
+ ni funkcija,
+ kot pri JKF,
+ temveč nov operator).
+\end_layout
+
+\begin_layout Claim*
+Naslednje trditve so ekvivalentne (zamenjamo lahko 
+\begin_inset Formula $\geq$
+\end_inset
+
+ in 
+\begin_inset Formula $>$
+\end_inset
+
+):
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $A\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A=A^{*}$
+\end_inset
+
+ in vse lastne vrednosti so 
+\begin_inset Formula $\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+ za nek unitaren 
+\begin_inset Formula $P$
+\end_inset
+
+ in diagonalen 
+\begin_inset Formula $D\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A=A^{*}$
+\end_inset
+
+ in obstaja 
+\begin_inset Formula $\sqrt{A}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A=B^{*}B$
+\end_inset
+
+ za neko nenujno kvadratno matriko 
+\begin_inset Formula $B$
+\end_inset
+
+ (za pozitivno definitno zahtevamo,
+ da ima 
+\begin_inset Formula $B$
+\end_inset
+
+ LN stolpce).
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+klasifikacija skalarnih produktov na 
+\begin_inset Formula $\mathbb{R}^{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $\left\langle u,v\right\rangle $
+\end_inset
+
+ standardni skalarni produkt na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $v=\left(\beta_{1},\dots,\beta_{n}\right)$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\left\langle u,v\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}=\left[\begin{array}{ccc}
+\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{c}
+\alpha_{1}\\
+\vdots\\
+\alpha_{n}
+\end{array}\right]=v^{*}\cdot u$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $A>0$
+\end_inset
+
+ definirajmo 
+\begin_inset Formula $\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$
+\end_inset
+
+.
+ Trdimo,
+ da je 
+\begin_inset Formula $\left[\cdot,\cdot\right]$
+\end_inset
+
+ spet skalarni produkt na 
+\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$
+\end_inset
+
+ in da je vsak skalarni produkt v 
+\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$
+\end_inset
+
+ take oblike.
+\end_layout
+
+\begin_layout Proof
+Dokazujemo oba dela trditve:
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\left[\cdot,\cdot\right]$
+\end_inset
+
+ je skalarni produkt
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+pozitivna semidefinitnost:
+ 
+\begin_inset Formula $\forall u\not=0:\left[u,u\right]=\left\langle Au,u\right\rangle \geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+konjutirana simetričnost:
+ 
+\begin_inset Formula $\forall u,v:\left[u,v\right]=\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\overline{\left\langle Av,u\right\rangle }=\overline{\left[v,u\right]}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Linearnost in homogenost:
+ 
+\begin_inset Formula $\forall\alpha_{1},\alpha_{2},u_{1}u_{2},v:\left[\alpha_{1}u_{1}+\alpha_{2}u_{2},v\right]=\left\langle A\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Au_{1}+\alpha_{2}Au_{2},v\right\rangle =\alpha_{1}\left\langle Au_{1},v\right\rangle +\alpha_{2}\left\langle Au_{2},v\right\rangle =\alpha_{1}\left[u,v\right]+\alpha_{2}\left[u,v\right]$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Za vsak skalarni produkt 
+\begin_inset Formula $\left[\cdot,\cdot\right]$
+\end_inset
+
+ na 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ obstaja taka pozitivno definitna matrika 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $e_{1},\dots,e_{n}$
+\end_inset
+
+ standardna baza za 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+ Definirajmo 
+\begin_inset Formula $A=\left[\begin{array}{ccc}
+\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
+\vdots &  & \vdots\\
+\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
+\end{array}\right]$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $A=A^{*}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+A^{*}=\left[\begin{array}{ccc}
+\overline{\left[e_{1},e_{1}\right]} & \cdots & \overline{\left[e_{1},e_{n}\right]}\\
+\vdots &  & \vdots\\
+\overline{\left[e_{n},e_{1}\right]} & \cdots & \overline{\left[e_{n},e_{n}\right]}
+\end{array}\right]=\left[\begin{array}{ccc}
+\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
+\vdots &  & \vdots\\
+\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
+\end{array}\right]=A
+\]
+
+\end_inset
+
+ Preveriti je treba še 
+\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=v^{*}Au$
+\end_inset
+
+.
+ 
+\begin_inset Formula $u=\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $v=\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula 
+\[
+\left[u,v\right]=\left[\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n},\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left(\alpha_{1}\overline{\beta_{1}}\left[e_{1},e_{1}\right]+\cdots+\alpha_{1}\overline{\beta_{n}}\left[e_{1},e_{n}\right]\right)+\cdots+\left(\alpha_{n}\overline{\beta_{1}}\left[e_{n},e_{1}\right]+\cdots+\alpha_{n}\overline{\beta_{n}}\left[e_{n},e_{n}\right]\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{ccc}
+\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{ccc}
+\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\
+\vdots &  & \vdots\\
+\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right]
+\end{array}\right]\left[\begin{array}{c}
+\alpha_{1}\\
+\vdots\\
+\alpha_{n}
+\end{array}\right]=v^{*}Au=\left\langle Au,v\right\rangle 
+\]
+
+\end_inset
+
+Da je 
+\begin_inset Formula $A$
+\end_inset
+
+ pozitivno definitna sledi,
+ saj mora za vsak neničeln 
+\begin_inset Formula $u$
+\end_inset
+
+ po aksiomu za pozitivno definitnost skalarnega produkta veljati 
+\begin_inset Formula $\left\langle Au,u\right\rangle >0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Subsubsection
+Singularni razcep (angl.
+ singular value decomposition —
+ SVD)
+\end_layout
+
+\begin_layout Standard
+Naj bo 
+\begin_inset Formula $A_{n\times n}$
+\end_inset
+
+ neka kompleksna ali realna matrika.
+ Tedaj je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ hermitska (
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Simetrične/hermitske-matrike"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+) matrika dimenzij 
+\begin_inset Formula $n\times n$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $\forall u:\left\langle A^{*}Au,u\right\rangle =\left\langle Au,Au\right\rangle \geq0$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ pozitivno semidefinitna,
+ torej so vse njene lastne vrednosti 
+\begin_inset Formula $\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Singularne vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ so kvadratni koreni lastnih vrednosti 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ normalna in 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ obstaja tak 
+\begin_inset Formula $v\not=0\ni:Av=\lambda v\Rightarrow A^{*}v=\overline{\lambda}v$
+\end_inset
+
+.
+ Odtod sledi,
+ da je 
+\begin_inset Formula $A^{*}Av=A^{*}\lambda v=\lambda A^{*}v=\lambda\overline{\lambda}v$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $\lambda$
+\end_inset
+
+ lastna vrednost matrike 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+.
+ Po definiciji singularne vrednosti je 
+\begin_inset Formula $\sqrt{\lambda\overline{\lambda}}=\sqrt{\left|\lambda\right|^{2}}=\left|\lambda\right|$
+\end_inset
+
+ singularna vrednost matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Potemtakem so singularne vrednostni normalnih matrik enake absolutnim vrednosti lastnih vrednosti.
+\end_layout
+
+\begin_layout Standard
+Nekatere lastne vrednosti so ničelne,
+ nekatere pa od nič strogo večje.
+ Koliko je katerih?
+ Število ničelnih singularnih vrednosti matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ je število ničelnih lastnih vrednosti matrike 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ hermitska,
+ je diagonalizabilna,
+ zato je algebraična večkratnost lastne vrednosti 0 enaka geometrijski večkratnosti lastne vrednosti 0,
+ slednja pa je definirana kot 
+\begin_inset Formula $\dim\Ker A^{*}A$
+\end_inset
+
+.
+ Ko upoštevamo 
+\begin_inset Formula $\dim\Ker A^{*}A=\dim\Ker A$
+\end_inset
+
+,
+ izvemo,
+ da je število ničelnih singularnih vrednosti matrike 
+\begin_inset Formula $A$
+\end_inset
+
+ njena ničnost (
+\begin_inset Formula $\n A$
+\end_inset
+
+).
+ Ker je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ velikosti 
+\begin_inset Formula $n\times n$
+\end_inset
+
+,
+ ima 
+\begin_inset Formula $A$
+\end_inset
+
+ 
+\begin_inset Formula $n$
+\end_inset
+
+ singularnih vrednosti,
+ torej je število neničelnih singularnih vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+ enako 
+\begin_inset Formula $n-\Ker A$
+\end_inset
+
+.
+ Upoštevajoč osnovni dimenzijski izrek jedra in slike,
+ velja 
+\begin_inset Formula $\n A+\rang A=n$
+\end_inset
+
+,
+ torej je neničelnih singularnih vrednosti 
+\begin_inset Formula $=\rang A=\dim\Slika A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matriko velja 
+\begin_inset Formula $\rang A\le\min\left\{ m,n\right\} $
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+posplošitev pojma diagonalne matrike na nekvadratne matrike.
+ Matrika 
+\begin_inset Formula $D_{m\times n}$
+\end_inset
+
+ je diagonalna,
+ če velja 
+\begin_inset Formula $\forall i:\left\{ 1..m\right\} ,j\in\left\{ 1..n\right\} :i\not=j\Rightarrow D_{i.j}=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri pravokotnih diagonalnih matrik:
+\begin_inset Formula 
+\[
+\left[\begin{array}{cccc}
+1 & 0 & 0 & 0\\
+0 & 2 & 0 & 0\\
+0 & 0 & 3 & 0
+\end{array}\right],\left[\begin{array}{ccc}
+1 & 0 & 0\\
+0 & 2 & 0\\
+0 & 0 & 3\\
+0 & 0 & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+singularni razcep.
+ Naj bo 
+\begin_inset Formula $A$
+\end_inset
+
+ kompleksna 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ matrika.
+ Potem obstajata taki unitarni 
+\begin_inset Formula $Q_{1},Q_{2}$
+\end_inset
+
+ in taka diagonalna 
+\begin_inset Formula $D$
+\end_inset
+
+ z diagonalci 
+\begin_inset Formula $\geq0\ni:A=Q_{1}DQ_{2}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Diagonalci 
+\begin_inset Formula $D$
+\end_inset
+
+ so ravno singularne vrednosti matrike 
+\begin_inset Formula $A$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $Q_{2}$
+\end_inset
+
+ unitarna,
+ je 
+\begin_inset Formula $Q_{2}^{*}=Q_{2}^{-1}\Rightarrow A=Q_{1}DQ_{2}^{*}$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{*}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $A^{*}=Q_{2}^{**}D^{*}Q_{1}^{*}=Q_{2}D^{*}Q_{1}^{*}$
+\end_inset
+
+ in 
+\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ podobna 
+\begin_inset Formula $D^{*}D$
+\end_inset
+
+,
+ diagonalci 
+\begin_inset Formula $D^{*}D$
+\end_inset
+
+ so lastne vrednosti 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ in stolpci 
+\begin_inset Formula $Q_{2}$
+\end_inset
+
+ so lastni vektorji 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+.
+ Diagonalci 
+\begin_inset Formula $D$
+\end_inset
+
+ so bodisi 0 bodisi kvadratni koreni od diagonalcev 
+\begin_inset Formula $D^{*}D$
+\end_inset
+
+,
+ torej kvadratni koreni lastnih vrednosti 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+,
+ torej singularne vrednosti od 
+\begin_inset Formula $A$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+obstoj singularnega razcepa.
+ Konstruirajmo 
+\begin_inset Formula $Q_{1},D,Q_{2}$
+\end_inset
+
+ in dokažimo veljavnost.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Konstrukcija 
+\begin_inset Formula $Q_{2}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $A$
+\end_inset
+
+ je 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ kompleksna.
+ Tvorimo 
+\begin_inset Formula $n\times n$
+\end_inset
+
+ matriko 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+.
+ Izračunajmo lastne vrednosti 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ in jih uredimo padajoče —
+ 
+\begin_inset Formula $\lambda_{1}\geq\cdots\geq\lambda_{n}$
+\end_inset
+
+.
+ Naj bodo 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ pripadajoči lastni vektorji —
+ 
+\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}$
+\end_inset
+
+.
+ Te 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ izberimo tako,
+ da so ortonormirani.
+ Lastni podprostori 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ so namreč paroma pravokotni,
+ saj je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ normalna,
+ saj je hermitska.
+ V vsakem podprostoru vzamemo ONB in 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ je unija teh ON baz.
+ Definiramo 
+\begin_inset Formula $Q_{2}=\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]$
+\end_inset
+
+.
+ Ker so 
+\begin_inset Formula $v_{1},\dots,v_{n}$
+\end_inset
+
+ ON,
+ je 
+\begin_inset Formula $Q_{2}$
+\end_inset
+
+ unitarna.
+\end_layout
+
+\begin_layout Itemize
+Konstrukcija 
+\begin_inset Formula $D$
+\end_inset
+
+:
+ Naj bo 
+\begin_inset Formula $r\coloneqq\rang A$
+\end_inset
+
+ (število ničelnih singularnih vrednosti 
+\begin_inset Formula $A$
+\end_inset
+
+).
+ Oglejmo si zaporedje lastnih vrednosti 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ 
+\begin_inset Formula $\lambda_{1}\geq\cdots>\lambda_{r+1}=\cdots=\lambda_{n}$
+\end_inset
+
+.
+ Lastne vrednosti po 
+\begin_inset Formula $r$
+\end_inset
+
+ so ničelne,
+ ostale pa večje od 0.
+ Lastne vrednosti
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ v tem vrstnem redu so singularne vrednosti matrike 
+\begin_inset Formula $A$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\sigma_{1}^{2}=\lambda_{1}\geq\cdots\geq\sigma_{r}^{2}=\lambda_{r}>\sigma_{r+1}^{2}=\cdots=\sigma_{n}^{2}=0$
+\end_inset
+
+.
+ Definiramo 
+\begin_inset Formula $D$
+\end_inset
+
+ kot 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ diagonalno matriko takole:
+\begin_inset Formula 
+\[
+D=\left[\begin{array}{cccccc}
+\sigma_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & \sigma_{r}\\
+ &  &  & \sigma_{r+1}=0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & \sigma_{n}=0
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Konstrukcija 
+\begin_inset Formula $Q_{1}$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\forall i\in\left\{ 1..r\right\} :u_{i}\coloneqq\frac{1}{\sigma_{1}}Av_{i}$
+\end_inset
+
+ za 
+\begin_inset Formula $v_{i}$
+\end_inset
+
+ lastne vektorje 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}=\sigma_{i}^{2}v_{i}$
+\end_inset
+
+.
+ Pokažimo,
+ da je 
+\begin_inset Formula $u_{1},\dots,u_{r}$
+\end_inset
+
+ ON množica.
+ 
+\begin_inset Formula $\forall i,j\in\left\{ 1..r\right\} :$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left\langle u_{i},u_{j}\right\rangle =\left\langle \frac{1}{\sigma_{j}}Av_{i},\frac{1}{\sigma_{j}}Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle Av_{i},Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle A^{*}Av_{i},v_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle \lambda_{i}v_{i},v_{j}\right\rangle =\frac{\lambda_{i}=\sigma_{i}^{\cancel{2}}}{\cancel{\sigma}_{i}\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{\sigma_{i}}{\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =\begin{cases}
+0 & ;i\not=j\\
+1 & ;i=j
+\end{cases}
+\]
+
+\end_inset
+
+Sedaj ONM 
+\begin_inset Formula $u_{1},\dots,u_{r}$
+\end_inset
+
+ z 
+\begin_inset Formula $u_{r+1},\dots,u_{n}$
+\end_inset
+
+ dopolnimo do ONB za 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+ (GS).
+ Definiramo 
+\begin_inset Formula $Q_{1}=\left[\begin{array}{ccc}
+u_{1} & \cdots & u_{n}\end{array}\right]$
+\end_inset
+
+.
+ Ker so stolpci ONB,
+ je matrika unitarna.
+\end_layout
+
+\begin_layout Itemize
+Sedaj preverimo,
+ da velja 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}\Leftrightarrow AQ_{2}=Q_{1}D$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+AQ_{2}=A\left[\begin{array}{ccc}
+v_{1} & \cdots & v_{n}\end{array}\right]=\left[\begin{array}{ccc}
+Av_{1} & \cdots & Av_{n}\end{array}\right]=\cdots
+\]
+
+\end_inset
+
+Upoštevamo,
+ da 
+\begin_inset Formula $i>r\Rightarrow\lambda_{i}=0\Rightarrow A^{*}Av_{i}=\lambda_{i}v_{i}=0\Rightarrow v_{i}\in\Ker A^{*}A\Rightarrow v_{i}\in\Ker A\Leftrightarrow Av_{i}=0$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\cdots=\left[\begin{array}{ccc}
+Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{cccccc}
+Av_{1} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]
+\]
+
+\end_inset
+
+Sedaj izračunajmo še 
+\begin_inset Formula 
+\[
+Q_{1}D=\left[\begin{array}{ccc}
+u_{1} & \cdots & u_{n}\end{array}\right]\left[\begin{array}{cccccc}
+\sigma_{1}\\
+ & \ddots\\
+ &  & \sigma_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+ &  &  &  &  & 0
+\end{array}\right]=\left[\begin{array}{cccccc}
+\sigma_{1}u_{1} & \cdots & \sigma_{r}u_{r} & 0 & \cdots & 0\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{cccccc}
+\cancel{\sigma_{1}\frac{1}{\sigma_{1}}}Av_{i} & \cdots & \cancel{\sigma_{r}\frac{1}{\sigma_{r}}}Av_{r} & 0 & \cdots & 0\end{array}\right]=\left[\begin{array}{cccccc}
+Av_{i} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]=AQ_{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+Poišči singularni razcep 
+\begin_inset Formula $A=\left[\begin{array}{cccc}
+1 & 1 & -1 & -1\\
+-1 & 0 & 1 & 0\\
+0 & -1 & 0 & 1
+\end{array}\right]$
+\end_inset
+
+.
+ Izračunajmo 
+\begin_inset Formula 
+\[
+A^{*}A=\left[\begin{array}{cccc}
+1 & 1 & -1 & -1\\
+-1 & 0 & 1 & 0\\
+0 & -1 & 0 & 1
+\end{array}\right]\left[\begin{array}{ccc}
+1 & -1 & 0\\
+1 & 0 & -1\\
+-1 & 1 & 0\\
+-1 & 0 & 1
+\end{array}\right]=\cdots=\left[\begin{array}{cccc}
+2 & 1 & -2 & -1\\
+1 & 2 & -1 & -2\\
+-2 & -1 & 2 & 1\\
+-1 & -2 & 1 & 2
+\end{array}\right]
+\]
+
+\end_inset
+
+Izračunajmo 
+\begin_inset Formula $p_{A^{*}A}\left(x\right)=\det\left(A^{*}A-xI\right)=\cdots=x^{2}\left(x-2\right)\left(x-6\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lambda_{1}=6$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{2}=2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{3}=0$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\lambda_{4}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\sigma_{1}=\sqrt{6}$
+\end_inset
+
+ in 
+\begin_inset Formula $\sigma_{2}=\sqrt{2}$
+\end_inset
+
+ ter 
+\begin_inset Formula $\sigma_{3}=\sigma_{4}=0$
+\end_inset
+
+.
+ (ujema se z dejstvom,
+ da je 
+\begin_inset Formula $\rang A=2$
+\end_inset
+
+).
+ Izračunajmo lastne vektorje 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+\lambda_{1}=6:\quad v_{1}'=\left[\begin{array}{c}
+1\\
+1\\
+-1\\
+-1
+\end{array}\right],\quad\left|\left|v_{1}'\right|\right|=6
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lambda_{2}=2:\quad v_{2}'=\left[\begin{array}{c}
+1\\
+-1\\
+-1\\
+1
+\end{array}\right],\quad\left|\left|v_{2}'\right|\right|=2
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\lambda_{3}=\lambda_{4}:\quad v_{3}'=\left[\begin{array}{c}
+1\\
+0\\
+1\\
+0
+\end{array}\right],v_{4}'=\left[\begin{array}{c}
+0\\
+1\\
+0\\
+2
+\end{array}\right],\quad\left|\left|v_{3}'\right|\right|=\sqrt{2},\left|\left|v_{4}'\right|\right|=\sqrt{2}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Z Gram-Schmidtom naredimo ortogonalno množico (v tem primeru so že ortogonalni) in jih normirajmo:
+\begin_inset Formula 
+\[
+v_{1}=\frac{1}{2}\left[\begin{array}{c}
+1\\
+1\\
+-1\\
+-1
+\end{array}\right],\quad v_{2}=\frac{1}{2}\left[\begin{array}{c}
+1\\
+-1\\
+-1\\
+1
+\end{array}\right],\quad v_{3}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
+1\\
+0\\
+1\\
+0
+\end{array}\right],\quad v_{4}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
+0\\
+1\\
+0\\
+1
+\end{array}\right].
+\]
+
+\end_inset
+
+Sestavimo 
+\begin_inset Formula 
+\[
+Q_{2}=\left[\begin{array}{cccc}
+\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\
+\frac{1}{2} & -\frac{1}{2} & 0 & \frac{1}{\sqrt{2}}\\
+-\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\
+-\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}}
+\end{array}\right],\quad D=\left[\begin{array}{cccc}
+\sqrt{6} &  &  & 0\\
+ & \sqrt{2}\\
+ &  & 0\\
+0 &  &  & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+Izračunamo 
+\begin_inset Formula $u_{1},\dots,u_{r}$
+\end_inset
+
+ za 
+\begin_inset Formula $Q_{1}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+u_{1}=\frac{1}{\sigma_{1}}Av_{1}=\frac{1}{\sqrt{6}}\left[\begin{array}{c}
+2\\
+-1\\
+-1
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+u_{2}=\frac{1}{\sigma_{2}}Av_{2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}
+0\\
+-1\\
+1
+\end{array}\right]
+\]
+
+\end_inset
+
+Dopolnimo ju do ONB za 
+\begin_inset Formula $\mathbb{R}^{3}$
+\end_inset
+
+ z Gram-Schmidtom (oz.
+ uganemo 
+\begin_inset Formula $\left[\begin{array}{c}
+1\\
+1\\
+1
+\end{array}\right]$
+\end_inset
+
+).
+ Dopolnitev normiramo:
+ 
+\begin_inset Formula $u_{3}=\frac{1}{\sqrt{3}}\left[\begin{array}{c}
+1\\
+1\\
+1
+\end{array}\right]$
+\end_inset
+
+ in vektorje vstavimo v 
+\begin_inset Formula $Q_{1}$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+Q_{1}=\left[\begin{array}{ccc}
+\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\\
+-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\
+-\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Iskani razcep je 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$
+\end_inset
+
+ (Izračunati je potrebno še en inverz —
+ 
+\begin_inset Formula $Q_{2}^{-1}$
+\end_inset
+
+ namreč).
+\end_layout
+
+\begin_layout Subsubsection
+Psevdoinverz —
+ Moore-Penroseov inverz
+\end_layout
+
+\begin_layout Standard
+Psevdoinverz je posplošitev inverza na nenujno kvadratne nenujno obrnljive matrike.
+ Najprej diagonalne matrike:
+ Njihov navaden inverz je takšen:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+d_{11} &  & 0\\
+ & \ddots\\
+0 &  & d_{nn}
+\end{array}\right]^{-1}=\left[\begin{array}{ccc}
+d_{11}^{-1} &  & 0\\
+ & \ddots\\
+0 &  & d_{nn}^{-1}
+\end{array}\right]
+\]
+
+\end_inset
+
+Kadar je diagonalec ničeln,
+ kot element polja nima multiplikativnega inverza.
+ Ideja za posplošeni inverz diagonelne matrike:
+ take diagonalce pustimo na 0,
+ torej na primer:
+\begin_inset Formula 
+\[
+\left[\begin{array}{ccc}
+1 &  & 0\\
+ & 2\\
+0 &  & 0
+\end{array}\right]^{+}=\left[\begin{array}{ccc}
+1 &  & 0\\
+ & \frac{1}{2}\\
+0 &  & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+Za nekvadratne diagonalne matrike pa takole:
+\begin_inset Formula 
+\[
+\left[\begin{array}{cccc}
+1 &  &  & 0\\
+ & 2\\
+0 &  & 0
+\end{array}\right]^{+}=\left[\begin{array}{ccc}
+1 &  & 0\\
+ & \frac{1}{2}\\
+ &  & 0\\
+0
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+posplošeni inverz diagonalne matrike.
+ Naj bo 
+\begin_inset Formula $D$
+\end_inset
+
+ diagonalna 
+\begin_inset Formula $m\times n$
+\end_inset
+
+ z neničelnimi diagonalci 
+\begin_inset Formula $d_{1},\dots d_{r}$
+\end_inset
+
+,
+ je 
+\begin_inset Formula $D^{+}$
+\end_inset
+
+ diagonalna 
+\begin_inset Formula $n\times m$
+\end_inset
+
+ z neničelnimi diagonalci 
+\begin_inset Formula $\frac{1}{d_{1}^{-1}},\dots,\frac{1}{d_{r}^{-1}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+Za diagonalno 
+\begin_inset Formula $D$
+\end_inset
+
+ opazimo 
+\begin_inset Formula $D^{++}=D$
+\end_inset
+
+ in za obrnljivo diagonalno 
+\begin_inset Formula $D$
+\end_inset
+
+ opazimo 
+\begin_inset Formula $D^{+}=D^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Sedaj bi radi pojem posplošili na nediagonalne matrike —
+ to storimo s pomočjo SVD.
+ 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{*}\ni:D$
+\end_inset
+
+ diagonalna in 
+\begin_inset Formula $Q_{1},Q_{2}$
+\end_inset
+
+ unitarni.
+ Tedaj velja 
+\begin_inset Formula $A$
+\end_inset
+
+ obrnljiva 
+\begin_inset Formula $\Leftrightarrow D$
+\end_inset
+
+ obrnljiva,
+ kajti 
+\begin_inset Formula $A^{-1}=Q_{2}D^{-1}D_{1}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Za splošen nenujno obrnljiv 
+\begin_inset Formula $A$
+\end_inset
+
+ definiramo 
+\begin_inset Formula $A^{+}\coloneqq Q_{2}D^{+}Q_{1}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Opazimo:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $A^{++}=\left(Q_{2}DQ_{1}^{-1}\right)^{+}=Q_{1}D^{++}Q_{2}^{-1}=Q_{1}DQ_{2}^{-1}=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $A$
+\end_inset
+
+ obrnljiva:
+ 
+\begin_inset Formula $A^{+}=A^{-1}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+osnovne lastnosti psevdoinverza.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $AA^{+}A=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(A^{+}A\right)^{*}=A^{+}A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A^{+}AA^{+}=A^{+}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(AA^{+}\right)^{*}=AA^{+}$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Dokažimo te 4 lastnosti najprej za 
+\begin_inset Formula $D$
+\end_inset
+
+ in nato za SVD.
+ Pri 
+\begin_inset Formula $D$
+\end_inset
+
+ predpostavimo,
+ da so ničle spodaj desno,
+ sicer obstaja permutacijska matrika,
+ ki je ortogonalna,
+ s katero lahko množimo 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ da jo pretvorimo v željeno obliko (in potem dokaz take 
+\begin_inset Formula $D$
+\end_inset
+
+ pade v primer SVD):
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Diagonalen primer.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $DD^{+}D=$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]\left[\begin{array}{cccccc}
+d_{1}^{-1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}^{-1}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{cccccc}
+1 &  &  &  &  & 0\\
+ & \ddots\\
+ &  & 1\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]=D
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $D^{+}DD^{+}=\cdots=D^{+}$
+\end_inset
+
+ na podoben način
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(DD^{+}\right)^{*}=\left[\begin{array}{cccccc}
+1 &  &  &  &  & 0\\
+ & \ddots\\
+ &  & 1\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]^{*}=\left[\begin{array}{cccccc}
+1 &  &  &  &  & 0\\
+ & \ddots\\
+ &  & 1\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]=DD^{+}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(D^{+}D\right)^{*}=\cdots=D^{+}D$
+\end_inset
+
+ podobno
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+Splošen primer 
+\begin_inset Formula $A$
+\end_inset
+
+ —
+ vstavimo 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $AA^{+}A=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{1}DD^{+}DQ_{2}^{*}=Q_{1}DQ_{2}^{*}=A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $A^{+}AA^{+}=\cdots=A^{+}$
+\end_inset
+
+ na podoben način
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(AA^{+}\right)^{*}=\left(Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}\right)^{*}=\left(Q_{1}DD^{+}Q_{1}^{*}\right)^{*}=Q_{1}\left(DD^{+}\right)^{*}Q_{1}^{*}=Q_{1}DD^{+}Q_{1}^{*}=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}=AA^{+}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(A^{+}A\right)^{*}=\cdots=A^{+}A$
+\end_inset
+
+ podobno
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Remark*
+\begin_inset Formula $A$
+\end_inset
+
+ obrnljiva 
+\begin_inset Formula $\Leftrightarrow D$
+\end_inset
+
+ obrnljiva,
+ torej 
+\begin_inset Formula $A^{+}=Q_{2}D^{+}Q_{1}^{-1}=Q_{2}D^{-1}Q_{1}^{-1}=A^{-1}$
+\end_inset
+
+.
+ Potemtakem za obrnljivo 
+\begin_inset Formula $A$
+\end_inset
+
+ velja 
+\begin_inset Formula $A^{+}=A^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Da je definicija dobra,
+ je treba dokazati,
+ da je 
+\begin_inset Formula $A^{+}$
+\end_inset
+
+ enoličen ne glede na SVD,
+ kajti SVD za 
+\begin_inset Formula $A$
+\end_inset
+
+ ni enoličen.
+ Naj bo 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=Q_{3}EQ_{4}^{-1}$
+\end_inset
+
+,
+ njen prvi psevsoinverz 
+\begin_inset Formula $B=Q_{2}D^{+}Q_{1}^{-1}$
+\end_inset
+
+ in njen drugi psevdoinverz 
+\begin_inset Formula $C=Q_{4}D^{+}Q_{3}^{-1}$
+\end_inset
+
+.
+ Ali velja 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$B
+\backslash
+overset{?}{=}C$}
+\end_layout
+
+\end_inset
+
+?
+ Velja 
+\begin_inset Formula 
+\[
+AB=\left(ACA\right)B=ACAB=\left(AC\right)^{*}\left(AB\right)^{*}=C^{*}A^{*}B^{*}A^{*}=C^{*}\left(ABA\right)^{*}=C^{*}A^{*}=\left(AC\right)^{*}=AC
+\]
+
+\end_inset
+
+in
+\begin_inset Formula 
+\[
+BA=B\left(ACA\right)=BACA=\left(BA\right)^{*}\left(CA\right)^{*}=A^{*}B^{*}A^{*}C^{*}=\left(ABA\right)^{*}C^{*}=A^{*}C^{*}=\left(CA\right)^{*}=CA
+\]
+
+\end_inset
+
+ter nazadnje še
+\begin_inset Formula 
+\[
+B=BAB=CAB=CAC=C.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Kako izračunamo 
+\begin_inset Formula $A^{+}$
+\end_inset
+
+ brez SVD?
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ pozitivno semidefinitna,
+ jo lahko diagonaliziramo v ortonormirani bazi:
+ 
+\begin_inset Formula $A=PDP^{-1}$
+\end_inset
+
+,
+ da ima 
+\begin_inset Formula $D$
+\end_inset
+
+ pozitivne diagonalce in da je 
+\begin_inset Formula $P^{-1}=P^{*}$
+\end_inset
+
+.
+ Opazimo,
+ da je to SVD od 
+\begin_inset Formula $A$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $Q_{1}=P,Q_{2}=P,D=D$
+\end_inset
+
+ in tedaj 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=PDP^{-1}$
+\end_inset
+
+.
+ Potemtakem je 
+\begin_inset Formula $A^{+}=PD^{+}P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Za splošno matriko 
+\begin_inset Formula $A$
+\end_inset
+
+ (nenujno pozitivno semidefinitno) pa velja 
+\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{+}A^{*}=A^{*}\left(AA^{*}\right)^{+}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ in 
+\begin_inset Formula $AA^{*}$
+\end_inset
+
+ sta pozitivno semidefinitni.
+\end_layout
+
+\begin_layout Proof
+Najprej bomo preverili za diagonalno,
+ nato za SVD:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Diagonalna 
+\begin_inset Formula $D_{n\times m}$
+\end_inset
+
+:
+ 
+\begin_inset Formula 
+\[
+D=\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+ &  &  &  &  & 0
+\end{array}\right],\quad D^{*}=\left[\begin{array}{cccccc}
+\overline{d_{1}} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & \overline{d_{r}}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right],\quad D^{*}D=\left[\begin{array}{cccccc}
+\frac{1}{\overline{d_{1}}d_{1}} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & \frac{1}{\overline{d_{r}}d_{r}}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right],
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left(D^{*}D\right)^{+}=\left[\begin{array}{cccccc}
+\frac{1}{\overline{d_{1}}d_{1}} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & \frac{1}{\overline{d_{r}}d_{r}}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right],\quad\left(D^{*}D\right)^{+}D^{*}=\left[\begin{array}{cccccc}
+\frac{\cancel{\overline{d_{1}}}}{\cancel{\overline{d_{1}}}d_{1}} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & \frac{\cancel{\overline{d_{r}}}}{\cancel{\overline{d_{r}}}d_{r}}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+0 &  &  &  &  & 0
+\end{array}\right]=D^{+}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+Za splošen 
+\begin_inset Formula $A$
+\end_inset
+
+ uporabimo SVD,
+ da to dokažemo:
+ 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(A^{*}A\right)^{+}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}$
+\end_inset
+
+.
+ Torej 
+\begin_inset Formula $\left(A^{*}A\right)^{+}A^{*}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}Q_{2}D^{*}Q_{1}^{*}=Q_{2}\left(D^{*}D\right)^{+}D^{*}Q_{1}^{*}=Q_{2}DQ_{1}^{*}=A^{+}$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Remark*
+V posebnih primerih lahko poenostavljamo dalje.
+ Recimo,
+ da ima 
+\begin_inset Formula $A$
+\end_inset
+
+ LN stolpce in je kvadratna 
+\begin_inset Formula $\Rightarrow\Ker A=\left\{ 0\right\} =\Ker A^{*}A$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ je obrnljiva in velja 
+\begin_inset Formula $\left(A^{*}A\right)^{-1}=\left(A^{*}A\right)^{+}$
+\end_inset
+
+.
+ Takrat torej velja 
+\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{-1}A^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Remark*
+To smo uporabili pri iskanju posplošene rešitve predoločenega sistema:
+ Za sistem 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ iščemo 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|$
+\end_inset
+
+ minimalen,
+ tedaj bo tak 
+\begin_inset Formula $\vec{x}$
+\end_inset
+
+ posplošena rešitev sistema.
+ Vemo,
+ da je posplošena reštev 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+ enaka rešitvi od 
+\begin_inset Formula $A^{*}A\vec{x}=A^{*}\vec{b}$
+\end_inset
+
+,
+ kajti,
+ če ima 
+\begin_inset Formula $A$
+\end_inset
+
+ LN stolpce,
+ je 
+\begin_inset Formula $A^{*}A$
+\end_inset
+
+ obrnljiva (s tem dokažemo,
+ da ima ta sistem vedno rešitev):
+\begin_inset Formula 
+\[
+A^{*}A\vec{x}=A^{*}\vec{b}\quad\quad\quad\quad/\cdot\left(A^{*}A\right)^{-1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\vec{x}=\left(A^{*}A\right)^{-1}A^{*}\vec{b}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\vec{x}=A^{+}\vec{b}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Uporaba psevdoinverza
+\end_layout
+
+\begin_layout Standard
+Vemo,
+ kaj je posplošena rešitev sistema 
+\begin_inset Formula $A\vec{x}=\vec{b}$
+\end_inset
+
+.
+ Problem je,
+ da ima sistem lahko več posplošenih rešitev (to se lahko zgodi,
+ če 
+\begin_inset Formula $A$
+\end_inset
+
+ nima LN stolpcev).
+ Med vsemi rešitvami iščemo tisto,
+ ki je najkrajša po normi —
+ 
+\begin_inset Formula $\left|\left|\vec{x}\right|\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Najkrajša posplošena rešitev sistema 
+\begin_inset Formula $Ax=b$
+\end_inset
+
+ je ravno 
+\begin_inset Formula $x=A^{+}b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo najprej za diagonalno matriko koeficientov,
+ nato pa še za splošen primer:
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $Dx=b$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+D_{m\times n}=\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+ &  &  &  &  & 0
+\end{array}\right],\quad x=\left[\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right],\quad b=\left[\begin{array}{c}
+b_{1}\\
+\vdots\\
+b_{m}
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\left|\left|Dx-b\right|\right|^{2}=\left|\left|\left[\begin{array}{cccccc}
+d_{1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+ &  &  &  &  & 0
+\end{array}\right]\left[\begin{array}{c}
+x_{1}\\
+\vdots\\
+x_{n}
+\end{array}\right]-\left[\begin{array}{c}
+b_{1}\\
+\vdots\\
+b_{m}
+\end{array}\right]\right|\right|^{2}=\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}+b_{r+1}^{2}+\cdots+b_{m}^{2}
+\]
+
+\end_inset
+
+Ta izraz doseže minimum,
+ ko 
+\begin_inset Formula $\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}=0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $x_{1}=\frac{b_{1}}{d_{1}},\dots,x_{r}=\frac{b_{r}}{d_{r}},x_{r+1}=\times,\dots,x_{n}=\times$
+\end_inset
+
+,
+ kjer 
+\begin_inset Formula $\times$
+\end_inset
+
+ predstavlja poljubno vrednost.
+ Najkrajša rešitev bo torej tista,
+ kjer 
+\begin_inset Formula $x_{r+1}=\cdots=x_{n}=0$
+\end_inset
+
+.
+ Trdimo,
+ da je 
+\begin_inset Formula $\left(\frac{b_{1}}{d_{1}},\cdots,\frac{b_{r}}{d_{r}},0,\cdots,0\right)=D^{+}b$
+\end_inset
+
+.
+ Preverimo:
+\begin_inset Formula 
+\[
+D_{n\times m}^{+}=\left[\begin{array}{cccccc}
+d_{1}^{-1} &  &  &  &  & 0\\
+ & \ddots\\
+ &  & d_{r}^{-1}\\
+ &  &  & 0\\
+ &  &  &  & \ddots\\
+ &  &  &  &  & 0
+\end{array}\right],\quad b=\left[\begin{array}{c}
+b_{1}\\
+\vdots\\
+b_{m}
+\end{array}\right],\quad D^{+}b=\left[\begin{array}{c}
+\frac{b_{1}}{d_{1}}\\
+\vdots\\
+\frac{b_{m}}{d_{m}}\\
+0\\
+\vdots\\
+0
+\end{array}\right]
+\]
+
+\end_inset
+
+Res je!
+\end_layout
+
+\begin_layout Itemize
+Splošen primer s SVD:
+ 
+\begin_inset Formula $A_{m\times n}=Q_{1}DQ_{2}^{*}$
+\end_inset
+
+,
+ kjer sta 
+\begin_inset Formula $Q_{1},Q_{2}$
+\end_inset
+
+ ortogonalni in 
+\begin_inset Formula $D$
+\end_inset
+
+ diagonalna.
+ Za tretji enačaj uporabimo dejstvo,
+ da množenje z ortogonalno matriko ohranja normo.
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\left|\left|Q_{2}^{*}x\right|\right|^{2}=\left\langle Q_{2}^{*}x,Q_{2}^{*}x\right\rangle =\left\langle x,Q_{2}Q_{2}^{*}x\right\rangle =\left\langle x,x\right\rangle =\left|\left|x\right|\right|^{2}$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula $\left|\left|Ax-b\right|\right|=\left|\left|Q_{1}DQ_{2}^{*}x-b\right|\right|=\left|\left|Q_{1}\left(DQ_{2}^{*}x-Q_{1}^{-1}b\right)\right|\right|=\left|\left|DQ_{2}^{*}x-Q_{1}^{-1}b\right|\right|=\left|\left|Dx'-c\right|\right|$
+\end_inset
+
+ za 
+\begin_inset Formula $x'=Q_{2}^{*}x$
+\end_inset
+
+ in 
+\begin_inset Formula $c=Q_{1}^{-1}b$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $Q_{2}$
+\end_inset
+
+ obrnljiva,
+ velja,
+ da če 
+\begin_inset Formula $x$
+\end_inset
+
+ preteče vse vektorje v 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+,
+ tudi 
+\begin_inset Formula $x'$
+\end_inset
+
+ preteče vse vektorje v 
+\begin_inset Formula $\mathbb{C}^{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+Potemtakem je 
+\begin_inset Formula $\min\left|\left|Ax-b\right|\right|=\min\left|\left|Dx'-c\right|\right|$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\left|\left|Ax-b\right|\right|$
+\end_inset
+
+ zavzame minimum v 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $\left|\left|Dx'-c\right|\right|$
+\end_inset
+
+ zavzame minimum v 
+\begin_inset Formula $x_{0}'=Q_{2}^{-1}x_{0}$
+\end_inset
+
+ in obratno,
+ če 
+\begin_inset Formula $\left|\left|Dx'-c\right|\right|$
+\end_inset
+
+ zavzame minimum v 
+\begin_inset Formula $x_{0}'$
+\end_inset
+
+,
+ potem 
+\begin_inset Formula $\left|\left|Ax-b\right|\right|$
+\end_inset
+
+ zavzame minimum v 
+\begin_inset Formula $x_{0}=Q_{2}x_{0}'$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $x\mapsto Q_{2}^{-1}x$
+\end_inset
+
+ bijektivna korespondenca med posplošenimi rešitvami 
+\begin_inset Formula $Ax-b$
+\end_inset
+
+ in posplošenimi rešitvami 
+\begin_inset Formula $Dx'-c$
+\end_inset
+
+.
+ Opazimo,
+ da preslikava ohranja normo,
+ torej 
+\begin_inset Formula $\left|\left|x_{0}'\right|\right|=\left|\left|Q_{2}x_{0}'\right|\right|=\left|\left|x_{0}\right|\right|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Od prej vemmo,
+ da je najkrajša posplošena rešitev 
+\begin_inset Formula $Dx_{0}-c$
+\end_inset
+
+ prav 
+\begin_inset Formula $x_{0}=D^{+}c$
+\end_inset
+
+.
+ Po zgornjem odstavku sledi,
+ da je 
+\begin_inset Formula $x_{0}=Q_{2}x_{0}'$
+\end_inset
+
+ najkrajša posplošena rešitev od 
+\begin_inset Formula $Ax=b$
+\end_inset
+
+.
+ Dobimo namreč 
+\begin_inset Formula $x_{0}=Q_{2}x_{0}'=Q_{2}D^{+}c=Q_{2}D^{+}Q_{1}^{-1}b=A^{+}b$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\end_deeper
+\begin_layout Subsection
+Kvadratne forme
+\end_layout
+
+\begin_layout Definition*
+Forma je homogen polinom,
+ torej tak,
+ v katerem imajo vsi monomi isto stopnjo.
+ Stopnja monoma je 
+\begin_inset Formula $\deg\left(\beta x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}\right)\coloneqq\alpha_{1}+\cdots+\alpha_{n}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Polinom je vsota monomov.
+ Stopnja polinoma je najvišja stopnja monoma v njem.
+\end_layout
+
+\begin_layout Example*
+Linearna forma v treh spremenljivkah:
+ 
+\begin_inset Formula $ax+by+cz=\left[\begin{array}{ccc}
+a & b & c\end{array}\right]\left[\begin{array}{c}
+x\\
+y\\
+z
+\end{array}\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Kvadratna forma je homogen polinom stopnje 2.
+ Primer kvadratne forme:
+\begin_inset Formula 
+\[
+ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc}
+x & y\end{array}\right]\left[\begin{array}{cc}
+a & b/2\\
+b/2 & a
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Kubična forma v treh spremenljivkah:
+\begin_inset Formula 
+\[
+ax^{3}+by^{3}+cz^{3}+dx^{2}y+ex^{2}z+fy^{2}x+gy^{2}x+iz^{2}x+jz^{2}y+kxyz
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Pravimo,
+ da sta matriki 
+\begin_inset Formula $A$
+\end_inset
+
+ in 
+\begin_inset Formula $B$
+\end_inset
+
+ kongruentni,
+ če obstaja obrnljiva 
+\begin_inset Formula $P\ni:B=PAP^{T}$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+Radi bi naredili klasifikacijo kvadratnih form.
+ Če naredimo primerno linearno zamenjavo koordinat,
+ se kvadratna forma poenostavi v 
+\begin_inset Formula $ex^{2}+fy^{2}$
+\end_inset
+
+ (mešani členi izginejo).
+\begin_inset Formula 
+\[
+x=\alpha x'+\beta y'
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+y=\gamma x'+\delta y'
+\]
+
+\end_inset
+
+zapišemo kot
+\begin_inset Formula 
+\[
+\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=\left[\begin{array}{cc}
+\alpha & \beta\\
+\gamma & \delta
+\end{array}\right]\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right],\quad\quad\overset{\text{transponiranje}}{\Longrightarrow}\quad\quad\left[\begin{array}{cc}
+x & y\end{array}\right]=\left[\begin{array}{cc}
+x' & y'\end{array}\right]\left[\begin{array}{cc}
+\alpha & \gamma\\
+\beta & \delta
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc}
+x & y\end{array}\right]\left[\begin{array}{cc}
+a & b/2\\
+b/2 & c
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=\left[\begin{array}{cc}
+x' & y'\end{array}\right]\left[\begin{array}{cc}
+\alpha & \gamma\\
+\beta & \delta
+\end{array}\right]\left[\begin{array}{cc}
+a & b/2\\
+b/2 & a
+\end{array}\right]\left[\begin{array}{cc}
+\alpha & \beta\\
+\gamma & \delta
+\end{array}\right]\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{cc}
+x' & y'\end{array}\right]P^{T}AP\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right]
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $A$
+\end_inset
+
+ simetrična,
+ lahko izberemo tako ortogonalno 
+\begin_inset Formula $P$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $P^{T}AP$
+\end_inset
+
+ diagonalna,
+ recimo 
+\begin_inset Formula $\left[\begin{array}{cc}
+d_{1} & 0\\
+0 & d_{2}
+\end{array}\right]$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+\left[\begin{array}{cc}
+x' & y'\end{array}\right]\left[\begin{array}{cc}
+d_{1} & 0\\
+0 & d_{2}
+\end{array}\right]\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right]=d_{1}\left(x'\right)^{2}+d_{2}\left(y'\right)^{2}
+\]
+
+\end_inset
+
+Kaj vemo o 
+\begin_inset Formula $2\times2$
+\end_inset
+
+ ortogonalnih matrikah?
+ 
+\begin_inset Formula $P=\left[\begin{array}{cc}
+a & b\\
+c & d
+\end{array}\right]\Rightarrow P^{T}P=\left[\begin{array}{cc}
+a & c\\
+b & d
+\end{array}\right]\left[\begin{array}{cc}
+a & b\\
+c & d
+\end{array}\right]=\left[\begin{array}{cc}
+a^{2}+c^{2} & ab+cd\\
+ab+cd & b^{2}+d^{2}
+\end{array}\right]$
+\end_inset
+
+.
+ Da je 
+\begin_inset Formula $P^{T}P=I$
+\end_inset
+
+,
+ mora veljati 
+\begin_inset Formula $ab+cd=0$
+\end_inset
+
+ in 
+\begin_inset Formula $a^{2}+c^{2}=b^{2}+d^{2}=1$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a=\cos\varphi$
+\end_inset
+
+,
+ 
+\begin_inset Formula $c=\sin\varphi$
+\end_inset
+
+,
+ 
+\begin_inset Formula $b=\cos\tau$
+\end_inset
+
+,
+ 
+\begin_inset Formula $d=\sin\tau$
+\end_inset
+
+.
+ Iz 
+\begin_inset Formula $\cos\left(\varphi+\tau\right)=0$
+\end_inset
+
+ sledi 
+\begin_inset Formula $\tau=\varphi\pm\frac{\pi}{2}$
+\end_inset
+
+,
+ torej je 
+\begin_inset Formula $P_{1}=\left[\begin{array}{cc}
+\cos\varphi & -\sin\varphi\\
+\sin\varphi & \cos\varphi
+\end{array}\right]$
+\end_inset
+
+ (vrtež za 
+\begin_inset Formula $\varphi$
+\end_inset
+
+) ali 
+\begin_inset Formula $P_{2}=\left[\begin{array}{cc}
+\cos\varphi & \sin\varphi\\
+\sin\varphi & -\cos\varphi
+\end{array}\right]$
+\end_inset
+
+ (zrcaljenje žez 
+\begin_inset Formula $\varphi/2$
+\end_inset
+
+).
+ 
+\begin_inset Formula $\det P_{1}=1$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\det P_{2}=-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Če je 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+v_{1} & v_{2}\end{array}\right]\left[\begin{array}{cc}
+d_{1} & 0\\
+0 & d_{2}
+\end{array}\right]\left[\begin{array}{cc}
+v_{1} & v_{2}\end{array}\right]^{-1}$
+\end_inset
+
+,
+ je tudi 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+v_{1} & -v_{2}\end{array}\right]\left[\begin{array}{cc}
+d_{1} & 0\\
+0 & d_{2}
+\end{array}\right]\left[\begin{array}{cc}
+v_{1} & -v_{2}\end{array}\right]^{-1}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $\left[\begin{array}{cc}
+v_{1} & v_{2}\end{array}\right]$
+\end_inset
+
+ ortogonalna,
+ je tudi 
+\begin_inset Formula $\left[\begin{array}{cc}
+v_{1} & -v_{2}\end{array}\right]$
+\end_inset
+
+ ortogonalna.
+ Če je 
+\begin_inset Formula $A$
+\end_inset
+
+ 
+\begin_inset Formula $2\times2$
+\end_inset
+
+ simetrična matrika,
+ lahko poiščemo tak vrtež 
+\begin_inset Formula $P=\left[\begin{array}{cc}
+\cos\varphi & -\sin\varphi\\
+\sin\varphi & \cos\varphi
+\end{array}\right]$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $A=P\left[\begin{array}{cc}
+d_{1} & 0\\
+0 & d_{2}
+\end{array}\right]P^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Povzetek:
+ 
+\begin_inset Formula $ax^{2}+bxy+cy^{2}\overset{\text{vrtež}}{\longrightarrow}d_{1}x^{2}+d_{2}y^{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Nariši krivuljo 
+\begin_inset Formula $4x^{2}+4xy+7y^{2}=1$
+\end_inset
+
+.
+ Pripadajoča kvadratna forma:
+ 
+\begin_inset Formula 
+\[
+\left[\begin{array}{cc}
+x & y\end{array}\right]\left[\begin{array}{cc}
+4 & 2\\
+2 & 7
+\end{array}\right]\left[\begin{array}{c}
+x\\
+y
+\end{array}\right]=1=\cdots
+\]
+
+\end_inset
+
+Radi bi se znebili mešanega člena:
+\begin_inset Formula 
+\[
+\cdots=\left[\begin{array}{cc}
+x' & y'\end{array}\right]P^{T}\left[\begin{array}{cc}
+4 & 2\\
+2 & 7
+\end{array}\right]P\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right]=1=\cdots
+\]
+
+\end_inset
+
+Iščemo tak vrtež 
+\begin_inset Formula $P$
+\end_inset
+
+,
+ da bo 
+\begin_inset Formula $P^{T}AP$
+\end_inset
+
+ diagonalna.
+ Izračunamo lastne vrednosti 
+\begin_inset Formula $A=\left[\begin{array}{cc}
+4 & 2\\
+2 & 7
+\end{array}\right]$
+\end_inset
+
+.
+ Lastni vrednosti sta 
+\begin_inset Formula $\left\{ 3,8\right\} $
+\end_inset
+
+.
+ Izračunamo lastna vektorja:
+ 
+\begin_inset Formula $\left\{ \left[\begin{array}{c}
+-2\\
+1
+\end{array}\right],\left[\begin{array}{c}
+1\\
+2
+\end{array}\right]\right\} $
+\end_inset
+
+.
+ Sta že ortogonalna,
+ treba ju je še normirati:
+ 
+\begin_inset Formula $\left\{ \left[\begin{array}{c}
+-\frac{2}{\sqrt{5}}\\
+\frac{1}{\sqrt{5}}
+\end{array}\right],\left[\begin{array}{c}
+\frac{1}{\sqrt{5}}\\
+\frac{2}{\sqrt{5}}
+\end{array}\right]\right\} $
+\end_inset
+
+.
+ Izdelamo vrzež:
+ 
+\begin_inset Formula $P=\left[\begin{array}{cc}
+\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}}\\
+\frac{2}{\sqrt{5}} & \frac{2}{\sqrt{5}}
+\end{array}\right]$
+\end_inset
+
+.
+ Izračunamo kot vrteža:
+ 
+\begin_inset Formula $\frac{\sin\varphi}{\cos\varphi}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\arctan2\approx63,4^{\circ}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ogledamo si torej kvadratno formo 
+\begin_inset Formula $\left[\begin{array}{cc}
+x' & y'\end{array}\right]\left[\begin{array}{cc}
+8 & 0\\
+0 & 3
+\end{array}\right]\left[\begin{array}{c}
+x'\\
+y'
+\end{array}\right]=8x'^{2}+3y'^{2}=1$
+\end_inset
+
+,
+ kar je elipsa (
+\begin_inset Formula $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$
+\end_inset
+
+) s polosema 
+\begin_inset Formula $\frac{1}{\sqrt{8}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\frac{1}{\sqrt{3}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Elipso narišemo in jo v koordinatnem sistemu zavrtimo v negativno smer za 
+\begin_inset Formula $63,4^{\circ}$
+\end_inset
+
+.
+ Po zasuku je risba te krivulje risba naše prvotne kvadratne forme.
+\end_layout
+
+\begin_layout Part
+Vaja za ustni izpit
+\end_layout
+
+\begin_layout Standard
+Ustni izpit je sestavljen iz treh vprašanj.
+ Sekcije so zaporedna vprašanja na izpitu,
+ podsekcije so učiteljevi naslovi iz Primerov vprašanj,
+ podpodsekcije pa so dejanska vprašanja,
+ kot so se pojavila na dosedanjih izpitih.
+\end_layout
+
+\begin_layout Section
+Prvo vprašanje
+\end_layout
+
+\begin_layout Standard
+Prvo vprašanje je iz 1.
+ semestra.
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $\det AB=\det A\det B$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Baze vektorskega prostora
+\end_layout
+
+\begin_layout Subsubsection
+Linearno neodvisne množice
+\end_layout
+
+\begin_layout Subsubsection
+Ogrodje
+\end_layout
+
+\begin_layout Subsubsection
+Definicija baze
+\end_layout
+
+\begin_layout Subsubsection
+Dimenzija prostora
+\end_layout
+
+\begin_layout Subsection
+Cramerovo pravilo
+\end_layout
+
+\begin_layout Subsubsection
+Trditev in dokaz
+\end_layout
+
+\begin_layout Subsection
+Obrnljive matrike
+\end_layout
+
+\begin_layout Subsubsection
+Definicija obrnljivosti
+\end_layout
+
+\begin_layout Subsubsection
+Produkt obrnljivih matrik je obrnljiva matrika
+\end_layout
+
+\begin_layout Subsubsection
+Karakterizacija obrnljivih matrik z dokazom
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $\Ker A=\left\{ 0\right\} \Leftrightarrow A$
+\end_inset
+
+ obrnljiva
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $A$
+\end_inset
+
+ ima desni inverz 
+\begin_inset Formula $\Rightarrow A$
+\end_inset
+
+ obrnljiva
+\end_layout
+
+\begin_layout Subsubsection
+Formula za inverz matrike z dokazom
+\end_layout
+
+\begin_layout Subsection
+Vektorski podprostori
+\end_layout
+
+\begin_layout Subsection
+Elementarne matrike
+\end_layout
+
+\begin_layout Subsection
+Pod-/predoločeni sistem
+\end_layout
+
+\begin_layout Subsubsection
+Definicija,
+ iskanje posplošene rešitve z izpeljavo
+\end_layout
+
+\begin_layout Subsubsection
+Moč ogrodja 
+\begin_inset Formula $\geq$
+\end_inset
+
+ moč LN množice
+\end_layout
+
+\begin_layout Subsubsection
+Vsak poddoločen sistem ima netrivialno rešitev
+\end_layout
+
+\begin_layout Standard
+Posledica prejšnje trditve.
+\end_layout
+
+\begin_layout Subsection
+Regresijska premica
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsection
+Vektorski/mešani produkt
+\end_layout
+
+\begin_layout Subsection
+Grupe/polgrupe
+\end_layout
+
+\begin_layout Subsubsection
+Definicija in lastnosti grupe
+\end_layout
+
+\begin_layout Subsubsection
+Definicija homomorfizma
+\end_layout
+
+\begin_layout Subsubsection
+Primeri homomorfizmov z dokazi
+\end_layout
+
+\begin_layout Subsubsection
+Definicija permutacijske grupe in dokaz,
+ da je grupa
+\end_layout
+
+\begin_layout Subsubsection
+Primeri grup
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da so ortogonalne matrike podgrupa v grupi obrnljivih matrik
+\end_layout
+
+\begin_layout Subsubsection
+Matrika permutacije
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da je preslikava,
+ ki permutaciji priredi matriko,
+ homomorfizem
+\end_layout
+
+\begin_layout Subsection
+Projekcija točke na premico/ravnino
+\end_layout
+
+\begin_layout Subsection
+\begin_inset Formula $\det A=\det A^{T}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Formula za inverz
+\end_layout
+
+\begin_layout Subsection
+Homogeni sistemi enačb
+\end_layout
+
+\begin_layout Section
+Drugo vprašanje
+\end_layout
+
+\begin_layout Standard
+Drugo vprašanje zajema snov linearnih preslikav/lastnih vrednosti.
+\end_layout
+
+\begin_layout Subsection
+Diagonalizacija
+\end_layout
+
+\begin_layout Subsubsection
+Definicija,
+ trditve
+\end_layout
+
+\begin_layout Subsection
+Prehod na novo bazo
+\end_layout
+
+\begin_layout Subsubsection
+Prehodna matrika in njene lastnosti
+\end_layout
+
+\begin_layout Subsubsection
+Predstavitev vektorjev in linearnih preslikav z različnimi bazami
+\end_layout
+
+\begin_layout Subsubsection
+Razvoj vektorja po eni in drugi bazi (prehod vektorja na drugo bazo)
+\end_layout
+
+\begin_layout Subsection
+Matrika linearne preslikave
+\end_layout
+
+\begin_layout Subsection
+Rang matrike
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da je rang število LN stolpcev
+\end_layout
+
+\begin_layout Subsubsection
+Dimenzijska formula za podprostore
+\end_layout
+
+\begin_layout Subsection
+\begin_inset Formula $\rang A=\rang A^{T}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Ekvivalentnost matrik
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da je relacija ekvivalenčna
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da je vsaka matrika ekvivalentna matriki 
+\begin_inset Formula $I_{r}$
+\end_inset
+
+,
+ t.
+ j.
+ bločni matriki,
+ katere zgornji levi blok je 
+\begin_inset Formula $I$
+\end_inset
+
+ dimenzije 
+\begin_inset Formula $r$
+\end_inset
+
+,
+ drugi trije bloki pa so ničelne matrike.
+\end_layout
+
+\begin_layout Subsection
+Jedro/slika
+\end_layout
+
+\begin_layout Subsection
+Minimalni poinom
+\end_layout
+
+\begin_layout Subsubsection
+Definicija karakterističnega in minimalnega polinoma
+\end_layout
+
+\begin_layout Subsection
+Cayley-Hamiltonov izrek
+\end_layout
+
+\begin_layout Subsubsection
+Trditev in dokaz
+\end_layout
+
+\begin_layout Subsection
+Korenski razcep
+\end_layout
+
+\begin_layout Subsubsection
+Definicija korenskih podprostorov
+\end_layout
+
+\begin_layout Subsubsection
+Presek različnih korenskih podprostorov je trivialen
+\end_layout
+
+\begin_layout Subsubsection
+Vsota korenskih podprostorov je direktna (se sklicuje na zgornjo trditev)
+\end_layout
+
+\begin_layout Subsection
+Osnovna formula rang 
+\begin_inset Formula $+$
+\end_inset
+
+ ničnost
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsection
+Funkcije matrik
+\end_layout
+
+\begin_layout Section
+Tretje vprašanje
+\end_layout
+
+\begin_layout Standard
+Tretje vprašanje zajema naslednje snovi:
+\end_layout
+
+\begin_layout Itemize
+vektorski prostori s skalarnim produktom,
+\end_layout
+
+\begin_layout Itemize
+adjungirana preslikava,
+\end_layout
+
+\begin_layout Itemize
+singularni razcep,
+\end_layout
+
+\begin_layout Itemize
+kvadratne forme.
+\end_layout
+
+\begin_layout Subsubsection
+Singularni razcep:
+ Konstrukcija 
+\begin_inset Formula $Q_{1},Q_{2},D$
+\end_inset
+
+ in dokaz 
+\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Ortogonalne/unitarne matrike
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz 
+\begin_inset Formula $AA^{*}=I$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Lastne vrednosti
+\end_layout
+
+\begin_layout Subsubsection
+Prehodna matrika iz ONB v drugo ONB ima ortogonalne stolpce (dokaz)
+\end_layout
+
+\begin_layout Subsection
+Kvadratne krivulje
+\end_layout
+
+\begin_layout Subsection
+Psevdoinverz
+\end_layout
+
+\begin_layout Subsubsection
+Definicija
+\end_layout
+
+\begin_layout Subsection
+Najkrajša posplošena rešitev sistema
+\end_layout
+
+\begin_layout Subsubsection
+Definicija,
+ trditev in dokaz
+\end_layout
+
+\begin_layout Subsection
+Simetrične matrike
+\end_layout
+
+\begin_layout Subsubsection
+Vse o simetričnih matrikah
+\end_layout
+
+\begin_layout Subsection
+Adjungirana linearna preslikava
+\end_layout
+
+\begin_layout Subsubsection
+Definicija in celotna formulacija
+\end_layout
+
+\begin_layout Subsubsection
+Rieszov izrek
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz obstoja in enoličnosti kot posledica Rieszovega izreka
+\end_layout
+
+\begin_layout Subsubsection
+Formula za matriko linearne preslikave in 
+\begin_inset Formula $\left\langle Au,v\right\rangle =v^{*}Au=\left\langle u,A^{*}v\right\rangle $
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Lastne vrednosti adjungirane matrike
+\end_layout
+
+\begin_layout Subsection
+Klasifikacija skalarnih produktov
+\end_layout
+
+\begin_layout Subsection
+Normalne matrike
+\end_layout
+
+\begin_layout Subsubsection
+Definicija,
+ lastnosti,
+ izreki,
+ dokazi
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $A$
+\end_inset
+
+ normalna 
+\begin_inset Formula $\Rightarrow A$
+\end_inset
+
+ in 
+\begin_inset Formula $A^{*}$
+\end_inset
+
+ imata isto množico lastnih vrednosti
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $\Ker\left(A-xI\right)=\Ker\left(A-\overline{x}I\right)$
+\end_inset
+
+ za normalno 
+\begin_inset Formula $A$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Ortogonalni komplement
+\end_layout
+
+\begin_layout Subsubsection
+Formula za ortogonalno projekcijo
+\end_layout
+
+\begin_layout Subsection
+Izrek o reprezentaciji linearnih funkcionalov
+\end_layout
+
+\begin_layout Subsection
+Pozitivno semidefinitne matrike
+\end_layout
+
+\begin_layout Subsubsection
+Definicija,
+ lastnosti.
+\end_layout
+
+\begin_layout Subsubsection
+Dokaz,
+ da imajo nenegativne lastne vrednosti.
+\end_layout
+
+\begin_layout Subsubsection
+Kvadratni koren pozitivno semidefinitne matrike.
+\end_layout
+
+\begin_layout Subsubsection
+\begin_inset Formula $A\geq0\Rightarrow A$
+\end_inset
+
+ sebiadjungirana
+\end_layout
+
+\begin_layout Subsection
+Ortogonalne in ortonormirane baze/Gram-Schmidt
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/p2/dn/DN06a_63230317.c b/šola/p2/dn/DN06a_63230317.c
new file mode 100644
index 0000000..a57b548
--- /dev/null
+++ b/šola/p2/dn/DN06a_63230317.c
@@ -0,0 +1,18 @@
+#include <stdio.h>
+int globina (int * a, int * b) {
+	int globina[2];
+	if (*a)
+		globina[0] = globina(a+*a, b);
+	if (*b)
+		globina[1] = globina(a, b+*b);
+	
+}
+int main (void) {
+	int n;
+	scanf("%d\n", &n);
+	int a[n];
+	int b[n];
+	for (int i = 0; i < n; i++)
+		scanf("%d %d\n", &a[i], &b[i]);
+	printf("%d\n", globina(a, b));
+}
diff --git a/šola/članki/dht/.gitignore b/šola/članki/dht/.gitignore
new file mode 100644
index 0000000..abd26b3
--- /dev/null
+++ b/šola/članki/dht/.gitignore
@@ -0,0 +1,3 @@
+*.svg
+*.png
+*.tsv
diff --git a/šola/članki/dht/dokument.lyx b/šola/članki/dht/dokument.lyx
new file mode 100644
index 0000000..13a6f2b
--- /dev/null
+++ b/šola/članki/dht/dokument.lyx
@@ -0,0 +1,2563 @@
+#LyX 2.4 created this file. For more info see https://www.lyx.org/
+\lyxformat 620
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass paper
+\begin_preamble
+% for subfigures/subtables
+\usepackage[caption=false,font=footnotesize]{subfig}
+\usepackage{textcomp}
+\usepackage[
+    type={CC},
+    modifier={by-sa},
+    version={3.0},
+]{doclicense}
+\usepackage{bera}% optional: just to have a nice mono-spaced font
+\usepackage{listings}
+\usepackage{xcolor}
+\lstset{
+	extendedchars=true,
+	literate={č}{{\v{c}}}1 {ž}{{\v{z}}}1 {š}{{\v{s}}}1,
+}
+
+\colorlet{punct}{red!60!black}
+\definecolor{background}{HTML}{EEEEEE}
+\definecolor{delim}{RGB}{20,105,176}
+\colorlet{numb}{magenta!60!black}
+
+\lstdefinelanguage{json}{
+    basicstyle=\normalfont\ttfamily,
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+    frame=lines,
+    backgroundcolor=\color{background},
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+     *{0}{{{\color{numb}0}}}{1}
+      {1}{{{\color{numb}1}}}{1}
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+      {3}{{{\color{numb}3}}}{1}
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+      {7}{{{\color{numb}7}}}{1}
+      {8}{{{\color{numb}8}}}{1}
+      {9}{{{\color{numb}9}}}{1}
+      {:}{{{\color{punct}{:}}}}{1}
+      {,}{{{\color{punct}{,}}}}{1}
+      {\{}{{{\color{delim}{\{}}}}{1}
+      {\}}{{{\color{delim}{\}}}}}{1}
+      {[}{{{\color{delim}{[}}}}{1}
+      {]}{{{\color{delim}{]}}}}{1},
+}
+\end_preamble
+\options journal
+\use_default_options false
+\maintain_unincluded_children no
+\language slovene
+\language_package babel
+\inputencoding utf8
+\fontencoding auto
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_roman_osf false
+\font_sans_osf false
+\font_typewriter_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures false
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command bibtex
+\index_command default
+\float_placement H
+\float_alignment class
+\paperfontsize default
+\spacing single
+\use_hyperref true
+\pdf_title "Kaj prenašamo s protokolom BitTorrent?"
+\pdf_author "Anton Lula Šijanec"
+\pdf_subject "Računalniška omrežja"
+\pdf_keywords "podazdeljena razpršilna tabela, porazdeljeni sistemi, omrežje P2P, podatkovno rudarjenje, BitTorrent"
+\pdf_bookmarks true
+\pdf_bookmarksnumbered true
+\pdf_bookmarksopen true
+\pdf_bookmarksopenlevel 1
+\pdf_breaklinks false
+\pdf_pdfborder true
+\pdf_colorlinks false
+\pdf_backref false
+\pdf_pdfusetitle false
+\pdf_quoted_options "pdfpagelayout=OneColumn, pdfnewwindow=true, pdfstartview=XYZ, plainpages=false"
+\papersize default
+\use_geometry true
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 0
+\use_formatted_ref 0
+\use_minted 0
+\use_lineno 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 0.5cm
+\topmargin 0.5cm
+\rightmargin 0.5cm
+\bottommargin 1.25cm
+\headheight 0.5cm
+\headsep 0.5cm
+\footskip 0.5cm
+\columnsep 0.5cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 2
+\papersides 1
+\paperpagestyle default
+\tablestyle default
+\tracking_changes false
+\output_changes false
+\change_bars false
+\postpone_fragile_content false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\docbook_table_output 0
+\docbook_mathml_prefix 1
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Kaj prenašamo s protokolom BitTorrent?
+\end_layout
+
+\begin_layout Author
+Anton Luka Šijanec,
+ 
+\begin_inset CommandInset href
+LatexCommand href
+name "anton@sijanec.eu"
+target "anton@sijanec.eu"
+type "mailto:"
+literal "true"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Institution
+Fakulteta za računalništvo in informatiko Univerze v Ljubljani
+\end_layout
+
+\begin_layout Abstract
+V članku predstavimo metodo za učinkovito in za omrežje neinvazivno metodo prenašanja metapodatkov iz pomožnega omrežja Kademlia mainline DHT protokola BitTorrent za izmenjavo datotek.
+ Sledi pregled/analiza z opisano metodo pridobljenih metapodatkov o datotekah na voljo v omrežju BitTorrent.
+\end_layout
+
+\begin_layout Abstract
+Porazdeljene razpršilne tabele (angl.
+ distributed hash table) so razpršilne tabele,
+ ki podatke,
+ ponavadi so to dokumenti,
+ strukturirani kot vrednost in njej pripadajoči ključ,
+ hranijo distribuirano na več vozliščih,
+ na katerih se podatki shranjujejo.
+ V računalniških sistemih se DHT uporablja za hrambo podatkov v omrežjih P2P (angl.
+ peer to peer),
+ kjer se podatki vseh uporabnikov enakomerno porazdelijo med vozlišča in so tako decentralizirani in preprosto dostopni članom omrežja.
+ Ker se podatki izmenjujejo znotraj omrežja na vozliščih,
+ ki z izvorom in destinacijo podatkov niso povezani,
+ jih lahko vozlišča v velikih količinah shranjujejo za potrebe statistične analize omrežja.
+\end_layout
+
+\begin_layout Abstract
+V raziskavi preverimo praktično zmožnost pridobivanja velike količine podatkov v omrežju BitTorrent za P2P izmenjavo datotek,
+ nato še analiziramo pridobljene podatke.
+ Vsaka poizvedba po seznamu imetnikov datotek vsebuje ključ podatka v DHT in se prenese preko okoli 
+\begin_inset Formula $\log_{2}n$
+\end_inset
+
+ vozlišč,
+ kjer je 
+\begin_inset Formula $n$
+\end_inset
+
+ število vseh uporabnikov v omrežju.
+ Ker vsaka poizvedba obišče tako veliko število vozlišč,
+ lahko med poizvedbo eno drugače nepovezano vozlišče prejme veliko obstoječih ključev v omrežju,
+ ki jih lahko uporabi za prenos metapodatkov v omrežju BitTorrent.
+\end_layout
+
+\begin_layout Abstract
+Osredotočili smo se le na na pridobivanje metapodatkov v omrežju BitTorrent,
+ samih datotek,
+ na katere se le-ti metapodatki sklicujejo in so v omrežju na voljo,
+ ker jih ponujajo drugi računalniki,
+ pa tako vsled tehničnih (njihove ogromne skupne velikosti) kot tudi pravnih razlogov (avtorsko zaščitena in protizakonita vsebina) nismo prenašali.
+ Metapodatki konceptualno sicer niso shranjeni v DHT (namesto metapodatkov o datotekah so v omrežju shranjeni seznami računalnikov,
+ od katerih si metapodatke lahko prenesemo),
+ vendar odkrivanje njihovega obstoja omogoči DHT.
+\end_layout
+
+\begin_layout Abstract
+S pridobljenimi metapodatki ugotovimo,
+ kateri odjemalci so najpopularnejši ter kakšna je razporeditev vsebine glede na tip datotek,
+ ki je na voljo preko protokola BitTorrent.
+\end_layout
+
+\begin_layout Keywords
+porazdeljena razpršilna tabela,
+ porazdeljeni sistemi,
+ omrežje P2P,
+\end_layout
+
+\begin_layout Keywords
+podatkovno rudarjenje,
+ BitTorrent
+\end_layout
+
+\begin_layout Section
+Introduction
+\end_layout
+
+\begin_layout Subsection
+Distribucija datotek po principu P2P
+\end_layout
+
+\begin_layout Standard
+Koncept P2P (angl.
+ 
+\shape italic
+peer-to-peer
+\shape default
+) predstavlja alternativen način distribucije identičnih datotek večim odjemalcem.
+ Namesto enega strežnika,
+ ki iste podatke odjemalcem pošlje vsakič znova,
+ v omrežjih P2P za distribucijo datotek vsak odjemalec podatke tako prejema kot tudi pošilja.
+ Odjemalec prejeto vsebino tudi sam deli drugim te vsebine željanim odjemalcem,
+ s čimer razbremeni ostale odjemalce.
+\end_layout
+
+\begin_layout Standard
+Odjemalec za druge izve s pomočjo centralnega strežnika ali pa drugačnega signalizacijskega protokola.
+ Ker se povezujejo neposredno,
+ medsebojno poznajo svoje internetne naslove.
+\end_layout
+
+\begin_layout Subsection
+Protokol BitTorrent
+\end_layout
+
+\begin_layout Standard
+Od 2008
+\begin_inset CommandInset citation
+LatexCommand cite
+key "harrison07"
+literal "false"
+
+\end_inset
+
+ je eden izmed popularnejših protokolov za P2P distribucijo BitTorrent,
+ razvit že 2001
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cohen01"
+literal "false"
+
+\end_inset
+
+.
+ Zaradi razširljive zasnove ga je moč dopolnjevati —
+ dodajati nove funkcije.
+ Sprva je BitTorrent temeljil na centralnih strežnikih za koordinacijo rojev,
+ od leta 2005 pa z uvedbo protokola DHT lahko deluje povsem neodvisno.
+\begin_inset CommandInset citation
+LatexCommand cite
+key "jones15"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\noindent
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="8" columns="3">
+<features tabularvalignment="middle" tabularwidth="100col%">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt" varwidth="true">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Pojem
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Angleško
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Razlaga
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+soležnik
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+peer
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+odjemni program na računalniku,
+ za povezavo nanj potrebujemo IP naslov in vrata
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+roj
+\begin_inset CommandInset citation
+LatexCommand cite
+key "dis"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+swarm
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+več soležnikov,
+ ki prenašajo datoteke nekega torrenta
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+torrent/
+\begin_inset Newline newline
+\end_inset
+
+metainfo
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+torrent/
+\begin_inset Newline newline
+\end_inset
+
+metainfo
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+datoteka z metapodatki datotek;
+ imena,
+ velikosti,
+ zgoščene vrednosti in drugo
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+sledilnik
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+tracker
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+koordinacijski strežnik z naslovi soležnikov v rojih
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+košček
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+piece
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+delček datoteke konstantne dolžine
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+infohash
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+infohash
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+zgoščena vrednost serializiranih podatkov pod ključem 
+\family typewriter
+info
+\family default
+ v torrentu,
+ ki unikatno opišejo ključne metapodatke o torrentu
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+objavi
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+announce
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+pošiljanje obvestila v DHT ali na sledilnik,
+ da se odjemalec želi priključiti nekemu roju
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Slovar pojmov BitTorrenta
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Za prenos je treba poznati metapodatke o obstoječih datotekah,
+ ki so shranjeni v t.
+ i.
+ obliki .torrent,
+ strojno berljivi z bencoding serializirani datoteki.
+ Vsebujejo vsaj imena in poti datotek ter njihove zgoščene vrednosti,
+ ime torrenta,
+ lastnosti prenosa in velikost koščka.
+\end_layout
+
+\begin_layout Standard
+V raziskavi ne iščemo soležnikov s sledilniki in ne prenašamo datotek,
+ temveč samo prenašamo in analiziramo metapodatke.
+\end_layout
+
+\begin_layout Subsection
+Protokol Kademlia mainline DHT
+\end_layout
+
+\begin_layout Standard
+V BitTorrent je za iskanje soležnikov v roju uporabljen DHT (angl.
+ 
+\shape italic
+distributed hash table
+\shape default
+),
+ ki odpravi odvisnost od sledilnika.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\noindent
+\align center
+\begin_inset Tabular
+<lyxtabular version="3" rows="6" columns="3">
+<features tabularvalignment="middle" tabularwidth="100col%">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt">
+<column alignment="center" valignment="top" width="0pt" varwidth="true">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Pojem
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Angleško
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+Razlaga
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+vozlišče
+\begin_inset CommandInset citation
+LatexCommand cite
+key "dis"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+node
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+odjemni program na računalniku
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+usmerjevalna
+\begin_inset Newline newline
+\end_inset
+
+tabela
+\begin_inset CommandInset citation
+LatexCommand cite
+key "dis"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+routing
+\begin_inset Newline newline
+\end_inset
+
+table
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+seznam vozlišč (IP,
+ vrata,
+ ID),
+ ki ga hrani posamezno vozlišče
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+ID vozlišča
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+node ID
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+160-bitna ob zagonu generirana naključna vozlišču pripadajoča številka
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+merilo
+\begin_inset Newline newline
+\end_inset
+
+razdalje
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+distance
+\begin_inset Newline newline
+\end_inset
+
+metric
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+funkcija (XOR),
+ ki izrazi razdaljo med vozliščema kot 160-bitno številko
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+koš
+\begin_inset CommandInset citation
+LatexCommand cite
+key "dis"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+bucket
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+element usmerjevalne tabele,
+ ki glede na merilo razdalje vsebuje bližnja vozlišča
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Slovar pojmov Kademlie.
+ Slovenski prevodi niso ustaljeni.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Na visokem nivoju gre za abstraktno razpršilno tabelo,
+ shranjeno porazdeljeno na velikem omrežju vozlišč.
+ Podpira naslednji operaciji
+\begin_inset CommandInset citation
+LatexCommand cite
+key "norberg08"
+literal "false"
+
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Paragraph
+
+\family typewriter
+get_peers
+\end_layout
+
+\begin_layout Standard
+Vrne seznam soležnikov (IP naslov in vrata) za torrent,
+ opisan z njegovim infohashom.
+\end_layout
+
+\begin_layout Paragraph
+
+\family typewriter
+announce
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+qbittorrent pravi sporoči,
+ v deluge in transmission nisem našel prevoda
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+V seznam soležnikov za torrent,
+ opisan z njegovim infohashom,
+ vstavi IP naslov in vrata pošiljatelja zahteve.
+\end_layout
+
+\begin_layout Standard
+V raziskavi s sodelovanjem v DHT prestrezamo obstoječe ključe v razpršilni tabeli,
+ z njimi pridobimo soležnike,
+ od katerih prenesemo metapodatke o torrentih za kasnejšo analizo.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/sola-gimb-4/inf/rn/predst/dht.svg
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Shematski prikaz DHT
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Opis standardov
+\end_layout
+
+\begin_layout Paragraph
+Serializacija bkodiranje 
+\family typewriter
+(bencoding)
+\end_layout
+
+\begin_layout Standard
+V BEP-0003
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cohen08"
+literal "false"
+
+\end_inset
+
+ je opisan bencoding.
+ Z njim je serializirana večina struktur BitTorrenta in Kademlie.
+ Bkodiranje je podobno bolj znanemu JSONu 
+\begin_inset CommandInset citation
+LatexCommand cite
+key "pezoa2016foundations"
+literal "false"
+
+\end_inset
+
+ —
+ vsebuje štiri podatkovne tipe:
+ niz,
+ število,
+ seznam in slovar.
+\end_layout
+
+\begin_layout Paragraph
+Datoteka metainfo/.torrent
+\end_layout
+
+\begin_layout Standard
+Za distribucijo vsebine s protokolom BitTorrent ustvarimo .torrent datoteko,
+ bkodiran slovar z metapodatki,
+ nujnimi za prenos datotek.
+ Za raziskavo so pomembni metapodatki pod ključem 
+\family typewriter
+info
+\family default
+:
+\begin_inset CommandInset citation
+LatexCommand cite
+key "cohen08"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+private
+\family default
+:
+ prepoved objavljanja preko DHT,
+ le preko sledilnikov (ti torrenti niso zajeti v raziskavi)
+\begin_inset CommandInset citation
+LatexCommand cite
+key "harrison08"
+literal "false"
+
+\end_inset
+
+ 
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+name
+\family default
+:
+ ime torrenta oz.
+ datoteke za enodatotečne torrente
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+piece length
+\family default
+:
+ velikost koščka —
+ datoteke so spojene skupaj in razdeljene na enako velike koščke
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+pieces
+\family default
+:
+ niz dolžine 
+\begin_inset Formula $20n$
+\end_inset
+
+ (
+\begin_inset Formula $n$
+\end_inset
+
+ je število koščkov) s SHA-1 vrednostmi koščkov
+\begin_inset CommandInset citation
+LatexCommand cite
+key "Eastlake2001"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+length
+\family default
+:
+ dolžina datoteke za enodatotečne torrente
+\end_layout
+
+\begin_layout Itemize
+
+\family typewriter
+files
+\family default
+:
+ seznam datotek v večdatotečnem torrentu —
+ vsaka datoteka je predstavljena kot slovar z 
+\family typewriter
+length
+\family default
+ in 
+\family typewriter
+path
+\family default
+.
+\end_layout
+
+\begin_layout Standard
+Kdor pozna infohash,
+ lahko od soležnika prenese metainfo in posledično tudi pripadajoče datoteke.
+ Roj najde in se vanj vključi s poizvedbo v DHT,
+ saj je infohash ključ v tej razpršilni tabeli
+\begin_inset CommandInset citation
+LatexCommand cite
+key "hazel08"
+literal "false"
+
+\end_inset
+
+.
+ Infohash običajno oblikujemo v t.
+ i.
+ magnetno povezavo (magnet URI):
+
+\family typewriter
+ magnet:?dn=
+\series bold
+ime torrenta
+\series default
+&xt=urn:btih:
+\series bold
+infohash
+\end_layout
+
+\begin_layout Standard
+Druga različica BitTorrenta ima drugačno metainfo strukturo s podobnimi podatki.
+ Uporablja SHA-256 in namesto 
+\family typewriter
+pieces
+\family default
+ uporablja 
+\family typewriter
+merkle hash tree
+\family default
+
+\begin_inset CommandInset citation
+LatexCommand cite
+key "v2"
+literal "false"
+
+\end_inset
+
+ za zgoščene vrednosti datotek.
+\end_layout
+
+\begin_layout Paragraph
+Graf DHT
+\end_layout
+
+\begin_layout Standard
+DHT vzdržuje sezname soležnikov v roju vseh obstoječih torrentov.
+ Vozlišča komunicirajo preko UDP in so del velikega usmerjenega grafa,
+ vsako s 
+\begin_inset Formula $K\log_{2}n$
+\end_inset
+
+ (konstanta 
+\begin_inset Formula $K=8$
+\end_inset
+
+,
+ 
+\begin_inset Formula $n$
+\end_inset
+
+ je število vseh vozlišč na svetu) povezavami —
+ vpisi v usmerjevalno tabelo.
+\end_layout
+
+\begin_layout Standard
+Vozlišče skrbi za urejeno usmerjevalno tabelo dosegljivih
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+dvosmerna komunikacija zaradi NAT ni samoumevna
+\end_layout
+
+\end_inset
+
+ vozlišč v koših;
+ 
+\begin_inset Formula $i$
+\end_inset
+
+ti koš hrani do 
+\begin_inset Formula $K$
+\end_inset
+
+ med 
+\begin_inset Formula $2^{i}$
+\end_inset
+
+ in 
+\begin_inset Formula $2^{i-1}$
+\end_inset
+
+ po merilu XOR oddaljenih vozlišč,
+ torej je shranjenih veliko bližnjih in malo zelo oddaljenih vozlišč.
+\begin_inset CommandInset citation
+LatexCommand cite
+key "maymounkov2002kademlia"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Paragraph
+Poizvedbe po grafu
+\end_layout
+
+\begin_layout Standard
+Sprehod po grafu med poljubnima vozliščema je torej dolg v povprečju 
+\begin_inset Formula $\log n$
+\end_inset
+
+ (
+\begin_inset Formula $n$
+\end_inset
+
+ kot prej).
+ Roj torrenta z infohashom 
+\begin_inset Formula $x$
+\end_inset
+
+ je shranjen na vozliščih z ID blizu 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ tedaj ima poizvedba po soležnikih/objavljanje soležnika časovno kompleksnost 
+\begin_inset Formula $O\left(\log n\right)$
+\end_inset
+
+.
+ Za pridobitev seznama soležnikov torrenta pošljemo bkodiran UDP paket tipa 
+\family typewriter
+get_peers
+\family default
+ 
+\begin_inset Formula $t$
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+odvisno od implementacije
+\end_layout
+
+\end_inset
+
+ vozliščem iz usmerjevalne tabele,
+ ki so blizu infohasha.
+ Pozvana vozlišča odgovorijo s seznamom do 
+\begin_inset Formula $K$
+\end_inset
+
+ temu infohashu najbližjih vozlišč in seznamom soležnikov za ta torrent,
+ če ga imajo.
+ Novodobljenim vozliščem spet pošljemo poizvedbo 
+\family typewriter
+get_peers
+\family default
+ in postopek nadaljujemo,
+ dokler ne najdemo nekaj infohashu najbližjih vozlišč.
+ V tista pošiljamo objave in od njih še naprej prejemamo informacije o roju.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\align center
+\begin_inset Graphics
+	filename Dht_example_SVG.svg
+	width 50col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+\begin_inset CommandInset label
+LatexCommand label
+name "fig:Usmerjevalna-tabela-za"
+
+\end_inset
+
+Usmerjevalna tabela za vozlišče 110 (koši so osenčeni)
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Metode
+\end_layout
+
+\begin_layout Standard
+Vsaka poizvedba obišče 
+\begin_inset Formula $\log n$
+\end_inset
+
+ vozlišč,
+ torej vsako vozlišče v DHT prejema ogromno ključev —
+ infohashov.
+ V raziskavi v C spišemo program travnik,
+ nepopolno implementacijo odjemalca BitTorrent s poudarkom na DHT.
+ Osredotočimo se na zajem metapodatkov iz ključev,
+ ki jih prejmemo s sodelovanjem v omrežju.
+\end_layout
+
+\begin_layout Standard
+Ko vozlišče,
+ na katerem teče travnik,
+ prejme paket z infohashom,
+ ga doda v seznam željenih torrentov.
+ Neprestano posodablja roje torrentov znanih infohashov in se povezuje na soležnike iz njih,
+ dokler mu ne uspe prenesti metapodatkov torrenta oziroma dokler ne obupa/preteče 256 sekundni TTL torrenta.
+ Izdeluje .torrent datoteke z najdenimi metapodatki in internetnim naslovom ter ime programske opreme soležnika,
+ od katerega je metapodatke prejel.
+ Ne objavlja se v roj,
+ ker ne redistribuira niti metapodatkov niti datotek.
+\end_layout
+
+\begin_layout Standard
+Da program prvič začne sodelovati z omrežjem —
+ da ga sosednja vozlišča vpišejo v svoje usmerjevalne tabele —
+ prenese metapodatke vgrajenega torrenta 
+\family typewriter
+Big Buck Bunny
+\family default
+.
+\end_layout
+
+\begin_layout Standard
+Za implementacijo spišemo knjižnico za bkodiranje in bdekodiranje,
+ knjižnico za DHT in nekaj funkcij za prenos metapodatkov od soležnikov preko TCP.
+\end_layout
+
+\begin_layout Standard
+Za obdelavo dobljenih torrent datotek uporabimo Jupyter Notebook
+\begin_inset CommandInset citation
+LatexCommand cite
+key "Kluyver2016jupyter"
+literal "false"
+
+\end_inset
+
+ in spišemo preprosto knjižnico za razčlenjevanje .torrent metainfo datotek,
+ ki jih generira travnik.
+\end_layout
+
+\begin_layout Section
+Rezultati
+\end_layout
+
+\begin_layout Subsection
+Zajem
+\end_layout
+
+\begin_layout Standard
+Podatke smo zajemali iz različnih lokacij in v različnih časovnih obdobjih.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float table
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Tabular
+<lyxtabular version="3" rows="4" columns="5">
+<features tabularvalignment="middle">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<column alignment="center" valignment="top">
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+mesto
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+datum
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+dni
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+torrentov
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+sek./torrent
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+T-2,
+ FTTH,
+ SI
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1.-2.
+ '23
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+16
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+47863
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+29
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+GRNET,
+ VPS,
+ GR
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+1.-2.
+ '23
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+31
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+412846
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+6
+\end_layout
+
+\end_inset
+</cell>
+</row>
+<row>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+T-2,
+ FTTH,
+ SI
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+6.
+ '24
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+5
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+62110
+\end_layout
+
+\end_inset
+</cell>
+<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
+\begin_inset Text
+
+\begin_layout Plain Layout
+7
+\end_layout
+
+\end_inset
+</cell>
+</row>
+</lyxtabular>
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Omrežne lokacije in časovna obdobja zajemov.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Metapodatki prvega zajema opisujejo 3084321 datotek v skupni velikosti 259 TiB,
+ metapodatki drugega zajema 17101702 datotek v velikosti 1881 TiB in metapodatki tretjega zajema 3725125 datotek v velikosti 345 TiB.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+januarja in februarja 2023:
+\end_layout
+
+\begin_layout Itemize
+16 dni:
+ domači optični priključek v Sloveniji (T-2):
+ 47863 torrentov (3084321 datotek,
+ 259 TiB,
+ 29 sekund na torrent)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+travnik
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+31 dni VPS v Grčiji (grNet)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+1
+\end_layout
+
+\end_inset
+
+:
+ 
+\family roman
+\series medium
+\shape up
+\size normal
+\emph off
+\nospellcheck off
+\bar no
+\strikeout off
+\xout off
+\uuline off
+\uwave off
+\noun off
+\color none
+412846
+\family default
+\series default
+\shape default
+\size default
+\emph default
+\nospellcheck default
+\bar default
+\strikeout default
+\xout default
+\uuline default
+\uwave default
+\noun default
+\color inherit
+ torrentov (17101702 datotek,
+ 1881 TiB,
+ 6 sekund na torrent)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+okeanos
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+XX dni VPS v Grčiji (grNet) 2:
+ 342220 torrentov ()
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+oliwerix
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Itemize
+5 dni junija 2024 na domačem optičnem priključku v Sloveniji (T-2):
+ 62110 torrentov (3725125 datotek,
+ 345 TiB,
+ 7 sekund na torrent)
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+2024b
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Primer strukture torrent datoteke z metapodatki
+\end_layout
+
+\begin_layout Standard
+Spodaj je iz bencoding v JSON pretvorjena metainfo datoteka prevzetega torrenta z infohashom 
+\family typewriter
+696802a16728636cd72617e4cd7b64e3ca314e71
+\family default
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+lstinputlisting[language=json,firstnumber=1,
+ numbers=none,
+ breaklines=true,
+ basicstyle=
+\backslash
+tiny]{../../../../sola-gimb-4/inf/rn/predst/torrent.json}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Analiza
+\end_layout
+
+\begin_layout Subsubsection
+Odjemalci,
+ od katerih so bili prejeti torrenti
+\end_layout
+
+\begin_layout Standard
+Imenom programom odstranimo različico in jim ročno normaliziramo ime
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+textmu Torrent
+\end_layout
+
+\end_inset
+
+ se sicer pojavi dvakrat,
+ enkrat ima znak mikro,
+ enkrat pa grško črko mu.
+ Unicode namreč ta dva znaka,
+ ki sicer izgledata identično,
+ hrani pod dvema različnima kodama.
+\end_layout
+
+\end_inset
+
+ ter prikažemo njihovo gostoto v populaciji.
+\begin_inset CommandInset citation
+LatexCommand cite
+key "Hunter:2007"
+literal "false"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/sola-gimb-4/inf/rn/dok/odjemalci_1_ods.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Reprezentacija odjemalcev,
+ ki predstavljajo vsaj en odstotek populacije,
+ na logaritemski skali
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Različice odjemalca 
+\family typewriter
+qBittorrent
+\family default
+ skozi čas
+\end_layout
+
+\begin_layout Standard
+Primerjava porazdelitve različic po zgornji analizi najpopularnejšega odjemalca na 
+\begin_inset Formula $\log_{10}$
+\end_inset
+
+ skali pokaže višanje različic skozi čas.
+ 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Različice smo razvrstili s pythonsko funkcijo 
+\family typewriter
+packaging.version.Version
+\family default
+.
+\end_layout
+
+\end_inset
+
+ V obeh letih smo prejeli torrente od skupno 88 različnih inačic qBittorrenta.
+ V 2023 smo največ torrentov prejeli od odjemalcev različice 4.5.0,
+ v 2024 pa od odjemalcev različice 4.6.3.
+ 
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/r/šola/članki/dht/verzije2324.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Graphics
+	filename /root/projects/r/šola/članki/dht/verzije2324promil.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Primerjava distribucij različic odjemalca 
+\family typewriter
+qBittorrent
+\family default
+ med 2023 (plavo) in 2024 (roza),
+ ki predstavljajo vsaj promil populacije (delež).
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Geolokacija IP naslovov odjemalcev
+\end_layout
+
+\begin_layout Standard
+Z uporabo podatkovne zbirke MaxMind GeoLite2
+\begin_inset CommandInset citation
+LatexCommand cite
+key "maxmindgeoip2"
+literal "false"
+
+\end_inset
+
+ IP naslovom,
+ od katerih smo prejeli torrente,
+ določimo izvirno državo.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename countries_procent.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Reprezentativnost držav,
+ iz katerih smo prenesli metainfo,
+ na linearni skali.
+ Prikazane so le države,
+ iz katerih izvira vsaj odstotek populacije.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Predstavnost ključev v prejetih slovarjih 
+\family typewriter
+info
+\end_layout
+
+\begin_layout Standard
+Poleg standardnih obveznih nekateri torrenti vsebujejo tudi dodatne metapodatke v slovarju info.
+ Pogostost slednjih prikazuje spodnji grafikon.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/sola-gimb-4/inf/rn/dok/vsi_ključi.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Reprezentacija ključev v slovarju 
+\family typewriter
+info
+\family default
+ na logaritemski skali
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsubsection
+Tipi datotek,
+ ki se prenašajo v torrentih
+\end_layout
+
+\begin_layout Standard
+Iz končnice datoteke izvemo tip datoteke.
+ Vsakemu torrentu priredimo reprezentativen tip,
+ tisti,
+ ki po velikosti prevladuje v torrentu.
+ Glede na število torrentov z nekim reprezentativnim tipom kvantificiramo pogostost tega datotečnega tipa za tipe,
+ ki zavzemajo vsaj promil populacije.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/sola-gimb-4/inf/rn/dok/reprezentativni_.1_ods.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Reprezentativni tipi torrentov,
+ ki predstavljajo vsaj en promil populacije,
+ na logaritemski skali
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Razvidno je,
+ da je večina torrentov namenjena prenosu videovsebin,
+ zvočnih datotek in stisnjenih arhivov.
+\end_layout
+
+\begin_layout Standard
+Če bi za določilo pojavnosti tipa uporabili število datotek,
+ bi prevladovali tipi vsebin,
+ ki so ponavadi preneseni kot kopica datotek,
+ denimo slike (diagram v prilogi na sliki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "fig:Pojavnost-tipa-kot"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+),
+ če pa bi za določilo pojavnosti tipa uporabili velikost datotek tega tipa,
+ pa bi prevladovali tisti tipi,
+ ki zasedajo več prostora.
+ V tem primeru bi npr.
+ videovsebine zaradi svoje velikosti občutno presegale digitalne knjige (diagram v prilogi na sliki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "fig:Pojavnost-tipa-kot-velikost"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Subsubsection
+\begin_inset CommandInset label
+LatexCommand label
+name "subsec:Porazdeljenost-infohashov"
+
+\end_inset
+
+Porazdeljenost infohashov
+\end_layout
+
+\begin_layout Plain Layout
+Zaradi delovanja poizvedb v DHT pričakujemo,
+ da je porazdelitev infohashov po celotnem sprektru števil z intervala 
+\begin_inset Formula $\left[0,2^{160}-1\right]$
+\end_inset
+
+ gostejša okoli IDja vozlišča,
+ ki ga je med prenašanjem imelo naše iskalno vozlišče.
+ IDje smo izbrali naključno na vsaki merilni lokaciji in jih med meritvijo tudi nekajkrat zamenjali,
+ zato spodnjem grafikonu opazimo vrhove tam,
+ kjer so bili naši IDji med zajemom podatkov.
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Razporeditev pridobljenih infohashov na spektru vseh infohashov
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Diskusija
+\end_layout
+
+\begin_layout Paragraph
+Statistična kvaliteta vzorca
+\end_layout
+
+\begin_layout Standard
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+V razdelku 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "subsec:Porazdeljenost-infohashov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ opazimo neenakomerno porazdeljenost infohashov,
+ kar je posledica načina vzorčenja torrentov.
+ Vsa populacija ima namreč zaradi delovanja zgoščevalne funkcije SHA-1 homogeno porazdeljene infohashe.
+ Kljub temu trdimo,
+ da pridobljen vzorec torrentov dobro predstavlja celotno populacijo.
+ 
+\end_layout
+
+\end_inset
+
+Zaradi lastnosti uniformne porazdelitve zgoščevalne funkcije
+\begin_inset CommandInset citation
+LatexCommand cite
+after "5.2"
+key "rfc4086"
+literal "false"
+
+\end_inset
+
+ mesto infohasha na intervalu vseh možnih infohashov ni odvisno od metapodatkov.
+ Kot posledico načina vzorčenja z DHT pričakujemo,
+ da je porazdelitev infohashov prejetih torrentov po celotnem sprektru števil z intervala 
+\begin_inset Formula $\left[0,2^{160}-1\right]$
+\end_inset
+
+ gostejša okoli IDja vozlišča,
+ ki ga je med prenašanjem imelo naše iskalno vozlišče.
+ IDje smo tekom raziskave izbrali naključno na vsaki merilni lokaciji in jih med meritvijo tudi nekajkrat zamenjali.
+ Kljub temu je vsled nepovezanosti vsebine in infohasha vzorec še vedno statistično reprezentativen.
+ Zajem ne more biti pristranski glede na metapodatke,
+ ker nikjer v procesu zajema ne obravnavamo torrentov glede na metainfo,
+ temveč le glede na infohash.
+\end_layout
+
+\begin_layout Paragraph
+Težava z zajemom podatkov
+\end_layout
+
+\begin_layout Standard
+Vsled majhne velikosti UDP paketov DHT glavno ozko grlo pri zajemu predstavlja število paketov,
+ ki jih mrežna oprema lahko posreduje v sekundi.
+ Domača optična povezava dopušča do okoli 2000 paketov na sekundo na naključno porazdeljene IP naslove,
+ odjemno mesto na VPS pa je imelo to omejitev veliko višjo,
+ zato smo tam v istem časovnem intervalu shranili veliko več torrent datotek.
+\end_layout
+
+\begin_layout Paragraph
+Etičnost in legitimnost rudarjenja podatkov
+\end_layout
+
+\begin_layout Standard
+Čeprav gre za izrazito osebne podatke,
+ se morajo uporabniki BitTorrent omrežja zavedati,
+ da so njihovi prenosi 
+\shape italic
+a priori
+\shape default
+ javni,
+ tudi če jih nihče aktivno ne zajema.
+ Nekateri BitTorrent odjemalci uporabnike ob prvem zagonu o tem obvestijo.
+\end_layout
+
+\begin_layout Section
+Priloge
+\end_layout
+
+\begin_layout Standard
+Izvorna koda programa travnik in ipynb datotek za analizo podatkov je na voljo na 
+\begin_inset CommandInset href
+LatexCommand href
+name "http://ni.šijanec.eu./sijanec/travnik"
+target "http://ni.šijanec.eu./sijanec/travnik"
+literal "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Korpus zajetih metapodatkov je na voljo na 
+\begin_inset CommandInset href
+LatexCommand href
+name "rsync://b.sjanec.eu./travnik"
+target "rsync://b.sijanec.eu./travnik"
+type "other"
+literal "false"
+
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename po_številu_datotek.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+\begin_inset CommandInset label
+LatexCommand label
+name "fig:Pojavnost-tipa-kot"
+
+\end_inset
+
+Pojavnost tipa kot število datotek
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+\begin_inset Graphics
+	filename /root/projects/sola-gimb-4/inf/rn/dok/po_velikosti_datotek.png
+	width 100col%
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+\begin_inset CommandInset label
+LatexCommand label
+name "fig:Pojavnost-tipa-kot-velikost"
+
+\end_inset
+
+Pojavnost tipa kot velikost datotek (tipi,
+ ki zavzamejo vsaj odstotek populacije)
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Plain Layout
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset CommandInset bibtex
+LatexCommand bibtex
+btprint "btPrintCited"
+bibfiles "/root/projects/r/šola/citati"
+options "IEEEtran"
+encoding "default"
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section*
+Viri slik
+\end_layout
+
+\begin_layout Itemize
+Slika 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "fig:Usmerjevalna-tabela-za"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+:
+ Limaner:
+ nespremenjena,
+ izvorna pod CC BY-SA
+\end_layout
+
+\begin_layout Section*
+Dovoljenje
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+doclicenseImage[imagewidth=2cm]
+\end_layout
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+TODO:
+ preštej datoteke v oliwerix,
+ še enkrat nariši vse grafe upoštevajoč vse torrente,
+ primerjaj verzije med travnik in 2024b
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
diff --git a/šola/članki/dht/makefile b/šola/članki/dht/makefile
new file mode 100644
index 0000000..5be42f0
--- /dev/null
+++ b/šola/članki/dht/makefile
@@ -0,0 +1,13 @@
+all: Dht_example_SVG.svg bt.svg
+
+Dht_example_SVG.svg:
+	wget https://upload.wikimedia.org/wikipedia/commons/6/63/Dht_example_SVG.svg
+
+infohash.png: # tole je treba pognati na strežniku b
+	./infohash.sh
+
+bt.svg:
+	wget -O- https://upload.wikimedia.org/wikipedia/commons/0/09/BitTorrent_network.svg | sed -e s/Downloader/Soležnik/ -e s/Uploader/Soležnik/ > bt.svg
+
+clean:
+	rm *.svg *.png *.tsv