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-rw-r--r--šola/ana2/teor.lyx864
-rw-r--r--šola/la/teor.lyx19983
2 files changed, 20847 insertions, 0 deletions
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/la/teor.lyx b/šola/la/teor.lyx
new file mode 100644
index 0000000..e25c8bf
--- /dev/null
+++ b/šola/la/teor.lyx
@@ -0,0 +1,19983 @@
+#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 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:
+
+\begin_inset Formula $\left(\text{vse obrnljive matrike fiksne dimenzije},\cdot\right)$
+\end_inset
+
+,
+
+\begin_inset Formula $\left(\text{vse permutacije neprazne končne množice},\circ\right)$
+\end_inset
+
+.
+\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
+
+.ž
+\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
+\[
+P^{-1}AP=\cdots=\left[\begin{array}{cc}
+\lambda_{i}I_{m_{i}} & B\\
+0 & C
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\series bold
+Dokaza ne razumem.
+ Obupam.
+\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 matrik,
+ katerih elementi so polinomi stopnje
+\begin_inset Formula $\leq1$
+\end_inset
+
+,
+ torej takele oblike:
+\begin_inset Formula
+\[
+\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}
+\]
+
+\end_inset
+
+
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+Ne razumem,
+ zakaj so tu matrike
+\begin_inset Formula $B$
+\end_inset
+
+ in ne skalarji.
+\end_layout
+
+\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 Caylay-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 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
generated by cgit v1.2.3 (git 2.25.1) at 2024-07-01 02:41:50 +0200