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Diffstat (limited to 'šola/la')
-rw-r--r-- | šola/la/teor.lyx | 25345 |
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diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx new file mode 100644 index 0000000..fcb46f9 --- /dev/null +++ b/šola/la/teor.lyx @@ -0,0 +1,25345 @@ +#LyX 2.4 created this file. For more info see https://www.lyx.org/ +\lyxformat 620 +\begin_document +\begin_header +\save_transient_properties true +\origin unavailable +\textclass article +\begin_preamble +\usepackage{hyperref} +\usepackage{siunitx} +\usepackage{pgfplots} +\usepackage{listings} +\usepackage{multicol} +\sisetup{output-decimal-marker = {,}, quotient-mode=fraction, output-exponent-marker=\ensuremath{\mathrm{3}}} +\usepackage{amsmath} +\usepackage{tikz} +\newcommand{\udensdash}[1]{% + \tikz[baseline=(todotted.base)]{ + \node[inner sep=1pt,outer sep=0pt] (todotted) {#1}; + \draw[densely dashed] (todotted.south west) -- (todotted.south east); + }% +}% +\DeclareMathOperator{\Lin}{\mathcal Lin} +\DeclareMathOperator{\rang}{rang} +\DeclareMathOperator{\sled}{sled} +\DeclareMathOperator{\Aut}{Aut} +\DeclareMathOperator{\red}{red} +\DeclareMathOperator{\karakteristika}{char} +\DeclareMathOperator{\Ker}{Ker} +\DeclareMathOperator{\Slika}{Ker} +\DeclareMathOperator{\sgn}{sgn} +\DeclareMathOperator{\End}{End} +\DeclareMathOperator{\n}{n} +\DeclareMathOperator{\Col}{Col} +\usepackage{algorithm,algpseudocode} +\providecommand{\corollaryname}{Posledica} +\end_preamble +\use_default_options true +\begin_modules +enumitem +theorems-ams +\end_modules +\maintain_unincluded_children no +\language slovene +\language_package default +\inputencoding auto-legacy +\fontencoding auto +\font_roman "default" "default" +\font_sans "default" "default" +\font_typewriter "default" "default" +\font_math "auto" "auto" +\font_default_family default +\use_non_tex_fonts false +\font_sc false +\font_roman_osf false +\font_sans_osf false +\font_typewriter_osf false +\font_sf_scale 100 100 +\font_tt_scale 100 100 +\use_microtype false +\use_dash_ligatures true +\graphics default +\default_output_format default +\output_sync 0 +\bibtex_command default +\index_command default +\float_placement class +\float_alignment class +\paperfontsize default +\spacing single +\use_hyperref false +\papersize default +\use_geometry true +\use_package amsmath 1 +\use_package amssymb 1 +\use_package cancel 1 +\use_package esint 1 +\use_package mathdots 1 +\use_package mathtools 1 +\use_package mhchem 1 +\use_package stackrel 1 +\use_package stmaryrd 1 +\use_package undertilde 1 +\cite_engine basic +\cite_engine_type default +\biblio_style plain +\use_bibtopic false +\use_indices false +\paperorientation portrait +\suppress_date false +\justification false +\use_refstyle 1 +\use_formatted_ref 0 +\use_minted 0 +\use_lineno 0 +\index Index +\shortcut idx +\color #008000 +\end_index +\leftmargin 2cm +\topmargin 2cm +\rightmargin 2cm +\bottommargin 2cm +\headheight 2cm +\headsep 2cm +\footskip 1cm +\secnumdepth 3 +\tocdepth 3 +\paragraph_separation indent +\paragraph_indentation default +\is_math_indent 0 +\math_numbering_side default +\quotes_style german +\dynamic_quotes 0 +\papercolumns 1 +\papersides 1 +\paperpagestyle default +\tablestyle default +\tracking_changes false +\output_changes false +\change_bars false +\postpone_fragile_content false +\html_math_output 0 +\html_css_as_file 0 +\html_be_strict false +\docbook_table_output 0 +\docbook_mathml_prefix 1 +\end_header + +\begin_body + +\begin_layout Title +Teorija linearne algebre za ustni izpit — + IŠRM 2023/24 +\end_layout + +\begin_layout Author + +\noun on +Anton Luka Šijanec +\end_layout + +\begin_layout Date +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +today +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Abstract +Povzeto po zapiskih s predavanj prof. + Cimpriča. +\end_layout + +\begin_layout Standard +\begin_inset CommandInset toc +LatexCommand tableofcontents + +\end_inset + + +\end_layout + +\begin_layout Part +Teorija +\end_layout + +\begin_layout Section +Prvi semester +\end_layout + +\begin_layout Subsection +Vektorji v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Identificaramo +\begin_inset Formula $n-$ +\end_inset + +terice realnih števil, + točke v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + +, + množice paroma enakih geometrijskih vektorjev. +\end_layout + +\begin_layout Standard +Osnovne operacije z vektorji: + Vsota (po komponentah) in množenje s skalarjem (po komponentah), + kjer je skalar realno število. +\end_layout + +\begin_layout Standard +Lastnosti teh računskih operacij: + asociativnost in komutativnost vsote, + aditivna enota, + +\begin_inset Formula $-\vec{a}=\left(-1\right)\cdot\vec{a}$ +\end_inset + +, + leva in desna distributivnost, + homogenost, + multiplikativna enota. +\end_layout + +\begin_layout Subsubsection +Linearna kombinacija vektorjev +\end_layout + +\begin_layout Definition* +Linearna kombinacija vektorjev +\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$ +\end_inset + + je izraz oblike +\begin_inset Formula $\alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}}$ +\end_inset + +, + kjer so +\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$ +\end_inset + + skalarji. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Množico vseh linearnih kombinacij vektorjev +\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$ +\end_inset + + označimo z +\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} $ +\end_inset + + in ji pravimo linearna ogrinjača (angl. + span). + +\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} =\left\{ \alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}};\forall\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\right\} $ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Linearna neodvisnost vektorjev +\end_layout + +\begin_layout Paragraph* +Ideja +\end_layout + +\begin_layout Standard +En vektor je linearno neodvisen, + če ni enak +\begin_inset Formula $\vec{0}$ +\end_inset + +. + Dva, + če ne ležita na isti premici. + Trije, + če ne ležijo na isti ravnini. +\end_layout + +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:odvisni" + +\end_inset + +Vektorji +\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$ +\end_inset + + so linearno odvisni, + če se da enega izmed njih izraziti z linearno kombinacijo preostalih +\begin_inset Formula $n-1$ +\end_inset + + vektorjev. + Vektorji so linearno neodvisni, + če niso linearno odvisni (in obratno). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:vsi0" + +\end_inset + +Vektorji +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + so linearno neodvisni, + če za vsake skalarje, + ki zadoščajo +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$ +\end_inset + +, + velja +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$ +\end_inset + +. + ZDB poleg +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$ +\end_inset + + ne obstajajo nobeni drugi +\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$ +\end_inset + +, + kjer bi veljalo +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:kvečjemu1" + +\end_inset + + +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + so linearno neodvisni, + če se da vsak vektor na kvečjemu en način izraziti kot linearno kombinacijo +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Theorem* +Te tri definicije so ekvivalentne. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\ref{def:odvisni}\Rightarrow\ref{def:vsi0}\right)$ +\end_inset + + Recimo, + da so +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + linearno odvisni v smislu +\begin_inset CommandInset ref +LatexCommand ref +reference "def:odvisni" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. + Dokažimo, + da so tedaj linearno odvisni tudi v smislu +\begin_inset Formula $\ref{def:vsi0}$ +\end_inset + +. + Obstaja tak +\begin_inset Formula $i$ +\end_inset + +, + da lahko +\begin_inset Formula $v_{i}$ +\end_inset + + izrazimo z linearno kombinacijo preostalih, + torej +\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + + za neke +\begin_inset Formula $\alpha$ +\end_inset + +. + Sledi +\begin_inset Formula $0=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\left(-1\right)v_{i}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + +, + kar pomeni, + da obstaja linearna kombinacija, + ki je enaka 0, + toda niso vsi koeficienti 0 (že koeficient pred +\begin_inset Formula $v_{i}$ +\end_inset + + je +\begin_inset Formula $-1$ +\end_inset + +), + tedaj so vektorji po definiciji +\begin_inset CommandInset ref +LatexCommand ref +reference "def:vsi0" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + linearno odvisni. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\ref{def:vsi0}\Rightarrow\ref{def:odvisni}\right)$ +\end_inset + + Recimo, + da so +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + linearno odvisno v smislu +\begin_inset Formula $\ref{def:vsi0}$ +\end_inset + +. + Tedaj obstajajo +\begin_inset Formula $\alpha$ +\end_inset + +, + ki niso vse 0, + da velja +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$ +\end_inset + +. + Tedaj +\begin_inset Formula $\exists i\ni:\alpha_{i}\not=0$ +\end_inset + + in velja +\begin_inset Formula +\[ +\alpha_{i}v_{i}=-\alpha_{1}v_{1}-\cdots-\alpha_{i-1}v_{i-1}-\alpha_{i+1}v_{i+1}-\cdots-\alpha_{n}v_{n}\quad\quad\quad\quad/:\alpha_{i} +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{i}=-\frac{\alpha_{1}}{\alpha_{i}}v_{i}-\cdots-\frac{\alpha_{i-1}}{\alpha_{i}}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_{i}}v_{i+1}-\cdots-\frac{\alpha_{n}}{\alpha_{i}}v_{n}\text{,} +\] + +\end_inset + +s čimer smo +\begin_inset Formula $v_{i}$ +\end_inset + + izrazili kot linearno kombinacijo preostalih vektorjev. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\ref{def:vsi0}\Leftrightarrow\ref{def:kvečjemu1}\right)$ +\end_inset + + Naj bodo +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + LN. + Recimo, + da obstaja +\begin_inset Formula $v$ +\end_inset + +, + ki se ga da na dva načina izraziti kot linearno kombinacijo +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{n}v_{n}$ +\end_inset + +. + Sledi +\begin_inset Formula $0=\left(\alpha_{1}-\beta_{1}\right)v_{1}+\cdots+\left(\alpha_{n}-\beta_{n}\right)v_{n}$ +\end_inset + +. + Po definiciji +\begin_inset CommandInset ref +LatexCommand ref +reference "def:vsi0" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + velja +\begin_inset Formula $\forall i:\alpha_{i}-\beta_{i}=0\Leftrightarrow\alpha_{i}=\beta_{i}$ +\end_inset + +, + torej sta načina, + s katerima izrazimo +\begin_inset Formula $v$ +\end_inset + +, + enaka, + torej lahko +\begin_inset Formula $v$ +\end_inset + + izrazimo na kvečjemu en način z +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +, + kar ustreza definiciji +\begin_inset CommandInset ref +LatexCommand ref +reference "def:kvečjemu1" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Ogrodje in baza +\end_layout + +\begin_layout Definition* +Vektorji +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + so ogrodje (angl. + span), + če +\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} =\mathbb{R}^{n}\Leftrightarrow\forall v\in\mathbb{R}^{n}\exists\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Vektorji +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + so baza, + če so LN in ogrodje +\begin_inset Formula $\Leftrightarrow\forall v\in\mathbb{R}^{n}:\exists!\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + + ZDB vsak vektor +\begin_inset Formula $\in\mathbb{R}^{n}$ +\end_inset + + se da na natanko en način izraziti kot LK +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primer baze je standardna baza +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + +: + +\begin_inset Formula $\left\{ \left(1,0,0,\dots,0\right),\left(0,1,0,\dots,0\right),\left(0,0,1,\dots,0\right),\left(0,0,0,\dots,1\right)\right\} $ +\end_inset + +. + To pa ni edina baza. + Primer nestandardne baze v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + + je +\begin_inset Formula $\left\{ \left(1,1,1\right),\left(0,1,1\right),\left(0,0,1\right)\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Norma in skalarni produkt +\end_layout + +\begin_layout Definition* +Norma vektorja +\begin_inset Formula $v=\left(\alpha_{1},\dots,\alpha_{n}\right)$ +\end_inset + + je definirana z +\begin_inset Formula $\left|\left|v\right|\right|=\sqrt{\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}}$ +\end_inset + +. + Geometrijski pomen norme je dolžina krajevnega vektorja z glavo v +\begin_inset Formula $v$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Osnovne lastnosti norme: + +\begin_inset Formula $\left|\left|v\right|\right|\geq0$ +\end_inset + +, + +\begin_inset Formula $\left|\left|v\right|\right|=0\Rightarrow v=\vec{0}$ +\end_inset + +, + +\begin_inset Formula $\left|\left|\alpha v\right|\right|=\left|\alpha\right|\cdot\left|\left|v\right|\right|$ +\end_inset + +, + +\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$ +\end_inset + + (trikotniška neenakost) +\end_layout + +\begin_layout Definition* +Skalarni produkt +\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right),v=\left(\beta_{1},\dots,\beta_{n}\right)$ +\end_inset + + označimo z +\begin_inset Formula $\left\langle u,v\right\rangle \coloneqq\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$ +\end_inset + +. + Obstaja tudi druga oznaka in pripadajoča drugačna definicija +\begin_inset Formula $u\cdot v\coloneqq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$ +\end_inset + +, + kjer je +\begin_inset Formula $\varphi$ +\end_inset + + kot med +\begin_inset Formula $u,v$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Velja +\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Uporabimo kosinusni izrek, + ki pravi, + da v trikotniku s stranicami dolžin +\begin_inset Formula $a,b,c$ +\end_inset + + velja +\begin_inset Formula $c^{2}=a^{2}+b^{2}-2ab\cos\varphi$ +\end_inset + +, + kjer je +\begin_inset Formula $\varphi$ +\end_inset + + kot med +\begin_inset Formula $b$ +\end_inset + + in +\begin_inset Formula $c$ +\end_inset + +. + Za vektorja +\begin_inset Formula $v$ +\end_inset + + in +\begin_inset Formula $u$ +\end_inset + + z vmesnim kotom +\begin_inset Formula $\varphi$ +\end_inset + + torej velja +\begin_inset Formula +\[ +\left|\left|u-v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi. +\] + +\end_inset + +Obenem velja +\begin_inset Formula $\left|\left|u\right|\right|^{2}=\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}=\left\langle u,u\right\rangle $ +\end_inset + +, + torej lahko zgornjo enačbo prepišemo v +\begin_inset Formula +\[ +\left\langle u-v,u-v\right\rangle =\left\langle u,u\right\rangle +\left\langle v,v\right\rangle -2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi. +\] + +\end_inset + +Naj bo +\begin_inset Formula $w=u,v$ +\end_inset + +. + Iz prihodnosti si izposodimo obe linearnosti in simetričnost. + +\begin_inset Formula +\[ +\left\langle u-v,u-v\right\rangle =\left\langle u-v,w\right\rangle =\left\langle u,w\right\rangle -\left\langle v,w\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle +\] + +\end_inset + + Prišli smo do enačbe +\begin_inset Formula +\[ +\cancel{\left\langle u,u\right\rangle }-2\left\langle u,v\right\rangle +\cancel{\left\langle v,v\right\rangle }=\cancel{\left\langle u,u\right\rangle }+\cancel{\left\langle v,v\right\rangle }-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\quad\quad\quad\quad/:-2 +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle u,v\right\rangle =\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi. +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Paralelogramska identiteta. + +\begin_inset Formula $\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}$ +\end_inset + + ZDB vsota kvadratov dolžin obeh diagonal je enota vsoti kvadratov dolžin vseh štirih stranic. +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +\left|\left|u+v\right|\right|^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u+v\right\rangle +\left\langle v,u+v\right\rangle =\left\langle u,u\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle +\left\langle v,v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|u-v\right|\right|^{2}=\left\langle u-v,u-v\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left\langle u,u\right\rangle +2\left\langle v,v\right\rangle =2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Cauchy-Schwarzova neenakost. + +\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$ +\end_inset + + +\end_layout + +\begin_layout Proof +\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|=\left|\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\right|=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\left|\cos\varphi\right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$ +\end_inset + +, + kajti +\begin_inset Formula $\left|\cos\varphi\right|\in\left[0,1\right]$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Trikotniška neenakost. + +\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$ +\end_inset + + +\end_layout + +\begin_layout Proof +Sledi iz Cauchy-Schwarzove. + Velja +\begin_inset Formula +\[ +-\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/\cdot2 +\] + +\end_inset + + +\begin_inset Formula +\[ +-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq2\left|\left\langle u,v\right\rangle \right|\leq2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq\cancel{2\left|\left\langle u,v\right\rangle \right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq}2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2} +\] + +\end_inset + +uporabimo kosinusni izrek na levi strani enačbe, + desno pa zložimo v kvadrat: +\begin_inset Formula +\[ +\left|\left|u+v\right|\right|^{2}\leq\left(\left|\left|u\right|\right|+\left|\left|v\right|\right|\right)^{2}\quad\quad\quad\quad/\sqrt{} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right| +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Za neničelna vektorja velja +\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$ +\end_inset + +, + kar je 0 +\begin_inset Formula $\Leftrightarrow\varphi=\pi=90°$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Vektorski in mešani produkt +\end_layout + +\begin_layout Standard +Definirana sta le za vektorje v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $u=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right),v=\left(\beta_{1},\beta_{2},\beta_{3}\right)$ +\end_inset + +. + +\begin_inset Formula $u\times v=\left(\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2},\alpha_{3}\beta_{1}-\alpha_{1}\beta_{3},\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph +Geometrijski pomen +\end_layout + +\begin_layout Standard +Vektor +\begin_inset Formula $u\times v$ +\end_inset + + je pravokoten na +\begin_inset Formula $u$ +\end_inset + + in +\begin_inset Formula $v$ +\end_inset + +, + njegova dolžina je +\begin_inset Formula $\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\sin\varphi$ +\end_inset + +, + kar je ploščina paralelograma, + ki ga oklepata +\begin_inset Formula $u$ +\end_inset + + in +\begin_inset Formula $v$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Pravilo desnega vijaka nam je v pomoč pri doložanju usmeritve vektorskega produkta. + Če iztegnjen kazalec desne roke predstavlja +\begin_inset Formula $u$ +\end_inset + + in iztegnjen sredinec +\begin_inset Formula $v$ +\end_inset + +, + iztegnjen palec kaže v smeri +\begin_inset Formula $u\times v$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Lagrangeva identiteta. + +\begin_inset Formula $\left|\left|u\times v\right|\right|+\left\langle u,v\right\rangle ^{2}=\left|\left|u\right|\right|^{2}\cdot\left|\left|v\right|\right|^{2}$ +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +DOKAZ??????? +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Definition* +Mešani produkt vektorjev +\begin_inset Formula $u,v,w$ +\end_inset + + je skalar +\begin_inset Formula $\left\langle u\times v,w\right\rangle $ +\end_inset + +. + Oznaka: + +\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle $ +\end_inset + +. +\end_layout + +\begin_layout Paragraph* +Geometrijski pomen +\end_layout + +\begin_layout Standard +Volumen paralelpipeda, + ki ga določajo +\begin_inset Formula $u,v,w$ +\end_inset + +. + Razlaga: + +\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle =\left|\left|u\times v\right|\right|\cdot\left|\left|w\right|\right|\cdot\cos\varphi$ +\end_inset + +; + +\begin_inset Formula $\left|\left|u\times v\right|\right|$ +\end_inset + + je namreč ploščina osnovne ploskve, + +\begin_inset Formula $\left|\left|w\right|\right|\cdot\cos\varphi$ +\end_inset + + pa je višina paralelpipeda. +\end_layout + +\begin_layout Claim* +Osnovne lastnosti vektorskega produkta so +\begin_inset Formula $u\times u=0$ +\end_inset + +, + +\begin_inset Formula $u\times v=-\left(v\times u\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\alpha u+\beta v\right)\times w=\alpha\left(u\times w\right)+\beta\left(v\times w\right)$ +\end_inset + + (linearnost) +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Claim* +Osnovne lastnosti mešanega produkta so linearnost v vsakem faktorju, + menjava dveh faktorjev spremeni predznak ( +\begin_inset Formula $\left[u,v,w\right]=-\left[v,u,w\right]$ +\end_inset + +), + cikličen pomik ne spremeni vrednosti ( +\begin_inset Formula $\left[u,v,w\right]=\left[v,w,u\right]=\left[w,u,v\right]$ +\end_inset + +). +\end_layout + +\begin_layout Subsubsection +Premica v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Premico lahko podamo z +\end_layout + +\begin_layout Itemize +dvema različnima točkama +\end_layout + +\begin_layout Itemize +s točko +\begin_inset Formula $\vec{r_{0}}$ +\end_inset + + in neničelnim smernim vektorjem +\begin_inset Formula $\vec{p}$ +\end_inset + +. + Premica je tako množica točk +\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+t\vec{p};\forall t\in\mathbb{R}\right\} $ +\end_inset + +. + Taki enačbi premice rečemo parametrična. +\end_layout + +\begin_layout Itemize +s točko in normalo (v +\begin_inset Formula $\mathbb{R}^{2}$ +\end_inset + +; + v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + potrebujemo točko in +\begin_inset Formula $n-1$ +\end_inset + + normal) +\end_layout + +\begin_layout Standard +Nadaljujmo s parametričnim zapisom +\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$ +\end_inset + +. + Če točke zapišemo po komponentah, + dobimo parametrično enačbo premice po komponentah: + +\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+t\left(p_{1},p_{2},p_{3}\right)$ +\end_inset + +. +\begin_inset Formula +\[ +x=x_{0}+tp_{1} +\] + +\end_inset + + +\begin_inset Formula +\[ +y=y_{0}+tp_{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +z=z_{0}+tp_{3} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Sedaj lahko iz vsake enačbe izrazimo +\begin_inset Formula $t$ +\end_inset + + in dobimo normalno enačbo premice v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + +: +\begin_inset Formula +\[ +t=\frac{x-x_{0}}{p_{1}}=\frac{y-y_{0}}{p_{2}}=\frac{z-z_{0}}{p_{3}}\text{, oziroma v splošnem za premico v \ensuremath{\mathbb{R}^{n}}: }t=\frac{x_{1_{0}}-x_{1}}{p_{1}}=\cdots=\frac{x_{n_{0}}-x_{n}}{p_{n}} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Osnovne naloge s premicami so projekcija točke na premico, + zrcaljenje točke čez premico in razdalja med točko in premico. +\end_layout + +\begin_layout Paragraph* +Iskanje projekcije dane točke na dano premico +\end_layout + +\begin_layout Standard +(skica prepuščena bralcu) +\begin_inset Formula $\vec{r_{1}}$ +\end_inset + + projiciramo na +\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$ +\end_inset + + in dobimo +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + +. + Za +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + + vemo, + da leži na premici, + torej +\begin_inset Formula $\exists t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}$ +\end_inset + +. + Poleg tega vemo, + da je +\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$ +\end_inset + + pravokoten na premico oz. + njen smerni vektor +\begin_inset Formula $\vec{p}$ +\end_inset + +, + torej +\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0$ +\end_inset + +. + Ti dve enačbi združimo, + da dobimo +\begin_inset Formula $t$ +\end_inset + +, + ki ga nato vstavimo v prvo enačbo: +\begin_inset Formula +\[ +\left\langle \vec{r_{0}}+t\vec{p}-\vec{r_{1},}\vec{p}\right\rangle =0\Longrightarrow\left\langle \vec{r_{0}},\vec{p}\right\rangle +t\left\langle \vec{p},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0\Longrightarrow t=\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle } +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}=\vec{r_{0}}+\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }\vec{p} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Spotoma si lahko izpišemo obrazec za oddaljenost točke od premice: + +\begin_inset Formula $a=\left|\left|\vec{r_{1}'}-\vec{r_{1}}\right|\right|$ +\end_inset + + in obrazec za zrcalno sliko ( +\begin_inset Formula $\vec{r_{1}''}$ +\end_inset + +): + +\begin_inset Formula $\vec{r_{1}'}=\frac{\vec{r_{1}''}+\vec{r_{1}}}{2}\Longrightarrow\vec{r_{1}''}=2\vec{r_{1}'}-\vec{r_{1}}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Ravnine v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + +\end_layout + +\begin_layout Standard +Ravnino lahko podamo +\end_layout + +\begin_layout Itemize +s tremi nekolinearnimi točkami +\end_layout + +\begin_layout Itemize +s točko na ravnini in dvema neničelnima smernima vektorjema, + ki sta linarno neodvisna. + Ravnina je tako množica točk +\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q};\forall s,t\in\mathbb{R}\right\} $ +\end_inset + +. + Taki enačbi ravnine rečemo parametrična. +\end_layout + +\begin_layout Itemize +s točko in na ravnini in normalo (v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + +; + v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + poleg točke potrebujemo +\begin_inset Formula $n-2$ +\end_inset + + normal) +\end_layout + +\begin_layout Standard +Nadaljujmo s parametričnim zapisom +\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$ +\end_inset + +. + Če točke zapišemo po komponentah, + dobimo parametrično enačbo ravnine po komponentah: + +\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+s\left(p_{1},p_{2},p_{3}\right)+t\left(q_{1},q_{2},q_{3}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +x=x_{0}+sp_{1}+tq_{1} +\] + +\end_inset + + +\begin_inset Formula +\[ +y=y_{0}+sp_{2}+tq_{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +z=y_{0}+sp_{3}+tq_{3} +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Normalna enačba ravnine v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + + +\end_layout + +\begin_layout Standard +(skica prepuščena bralcu) Vemo, + da je +\begin_inset Formula $\vec{n}$ +\end_inset + + (normala) pravokotna na vse vektorje v ravnini, + tudi na +\begin_inset Formula $\vec{r}-\vec{r_{0}}$ +\end_inset + + za poljuben +\begin_inset Formula $\vec{r}$ +\end_inset + + na ravnini. + Velja torej normalna enačba ravnine: + +\begin_inset Formula $\left\langle \vec{r}-\vec{r_{0}},\vec{n}\right\rangle =0$ +\end_inset + +. + Razpišimo jo po komponentah, + da na koncu dobimo normalno enačbo ravnine po komponentah: +\begin_inset Formula +\[ +\left\langle \left(x,y,z\right)-\left(x_{0},y_{0},z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle =0=\left\langle \left(x-x_{0},y-y_{0},z-z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +n_{1}\left(x-x_{0}\right)+n_{2}\left(y-y_{0}\right)+n_{3}\left(z-z_{0}\right)=0=n_{1}x-n_{1}x_{0}+n_{2}y-n_{2}y_{0}+n_{3}z-n_{3}z_{0}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +n_{1}x+n_{2}y+n_{3}z=n_{1}x_{0}+n_{2}y_{0}+n_{3}z_{0}=d +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Iskanje pravokotne projekcije dane točke na dano ravnino +\end_layout + +\begin_layout Standard +(skica prepuščena bralcu) Projicirati želimo +\begin_inset Formula $\vec{r_{1}}$ +\end_inset + + v +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + + na ravnini +\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$ +\end_inset + +. + Vemo, + da +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + + leži na ravnini, + zato +\begin_inset Formula $\exists s,t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+s\vec{p}+t\vec{q}$ +\end_inset + +. + Poleg tega vemo, + da je +\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$ +\end_inset + + pravokoten na ravnino oz. + na +\begin_inset Formula $\vec{p}$ +\end_inset + + in na +\begin_inset Formula $\vec{q}$ +\end_inset + + hkrati, + torej +\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{q}\right\rangle $ +\end_inset + +. + Vstavimo +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + + iz prve enačbe v drugo in dobimo +\begin_inset Formula +\[ +\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{q}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle \vec{r_{0}},\vec{p}\right\rangle +s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}},\vec{q}\right\rangle +s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle -\left\langle \vec{r_{1}},\vec{q}\right\rangle +\] + +\end_inset + +dobimo sistem dveh enačb +\begin_inset Formula +\[ +s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{p}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{q}\right\rangle +\] + +\end_inset + +sistem rešimo in dobljena +\begin_inset Formula $s,t$ +\end_inset + + vstavimo v prvo enačbo zgoraj, + da dobimo +\begin_inset Formula $\vec{r_{1}'}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Regresijska premica +\end_layout + +\begin_layout Standard +Regresijska premica je primer uporabe zgornje naloge. + V ravnini je danih +\begin_inset Formula $n$ +\end_inset + + točk +\begin_inset Formula $\left(x_{1},y_{1}\right),\dots,\left(x_{n},y_{n}\right)$ +\end_inset + +. + Iščemo tako premico +\begin_inset Formula $y=ax+b$ +\end_inset + +, + ki se najbolj prilega tem točkam. + Prileganje premice točkam merimo z metodo najmanjših kvadratov: + naj bo +\begin_inset Formula $d_{i}$ +\end_inset + + navpična razdalja med +\begin_inset Formula $\left(x_{i},y_{i}\right)$ +\end_inset + + in premico +\begin_inset Formula $y=ax+b$ +\end_inset + +, + torej razdalja med točkama +\begin_inset Formula $\left(x_{i},y_{i}\right)$ +\end_inset + + in +\begin_inset Formula $\left(x_{i},ax_{i}+b\right)$ +\end_inset + +, + kar je +\begin_inset Formula $\left|y_{i}-ax_{i}-b\right|$ +\end_inset + +. + Minimizirati želimo vsoto kvadratov navpičnih razdalj, + torej izraz +\begin_inset Formula $d_{1}^{2}+\cdots+d_{n}^{2}=\left(y_{1}-ax_{1}-b\right)^{2}+\cdots+\left(y_{n}-ax_{n}-b\right)^{2}=\left|\left|\left(y_{1}-ax_{1}-b,\dots,y_{n}-ax_{n}-b\right)\right|\right|^{2}=\left|\left|\left(y_{1},\dots,y_{n}\right)-a\left(x_{1},\dots,x_{n}\right)-b\left(1,\dots,1\right)\right|\right|^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če je torej +\begin_inset Formula $\vec{r}=\vec{0}+a\left(x_{1},\dots,x_{n}\right)+b\left(1,\dots,1\right)$ +\end_inset + + hiperravnina v +\begin_inset Formula $n-$ +\end_inset + +dimenzionalnem prostoru, + bo norma, + ki jo želimo minimizirati, + najmanjša tedaj, + ko +\begin_inset Formula $a,b$ +\end_inset + + izberemo tako, + da najdemo projekcijo +\begin_inset Formula $\left(y_{1},\dots,y_{n}\right)$ +\end_inset + + na to hiperravnino (skica prepuščena bralcu). + Rešimo sedaj nalogo projekcije točke na ravnino: +\end_layout + +\begin_layout Standard +Označimo +\begin_inset Formula $\vec{y}\coloneqq\left(y_{1},\dots,y_{n}\right)$ +\end_inset + +, + +\begin_inset Formula $\vec{x}\coloneqq\left(x_{1},\dots,x_{n}\right)$ +\end_inset + +, + +\begin_inset Formula $\vec{1}=\left(1,\dots,1\right)$ +\end_inset + +. + Vemo, + da +\begin_inset Formula $\vec{y}-a\vec{x}-b\vec{1}\perp\vec{x},\vec{1}$ +\end_inset + +, + torej +\begin_inset Formula $\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{x}\right\rangle =0=\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{1}\right\rangle $ +\end_inset + + in dobimo sistem enačb +\begin_inset Formula +\[ +\left\langle \vec{y},\vec{x}\right\rangle =a\left\langle \vec{x},\vec{x}\right\rangle +b\left\langle \vec{1},\vec{x}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle \vec{y},\vec{1}\right\rangle =a\left\langle \vec{x},\vec{1}\right\rangle +b\left\langle \vec{1},\vec{1}\right\rangle . +\] + +\end_inset + +V sistem sedaj vstavimo definicije točk +\begin_inset Formula $\left(x_{i},y_{i}\right)$ +\end_inset + + in ga nato delimo s številom točk, + da dobimo sistem s povprečji, + ki ga nato rešimo (izluščimo +\begin_inset Formula $a,b$ +\end_inset + +): +\begin_inset Formula +\[ +\sum_{i=1}^{n}y_{i}x_{i}=a\sum_{i=i}^{n}x_{i}^{2}+b\sum_{i=1}^{n}x_{i}\quad\quad\quad\quad/:n +\] + +\end_inset + + +\begin_inset Formula +\[ +\sum_{i=1}^{n}y_{i}=a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}1=a\sum_{i=1}^{n}x_{i}+bn\quad\quad\quad\quad/:n +\] + +\end_inset + + +\begin_inset Formula +\[ +\overline{yx}=a\overline{x^{2}}+b\overline{x} +\] + +\end_inset + + +\begin_inset Formula +\[ +\overline{y}=a\overline{x}+b\Longrightarrow\overline{y}-a\overline{x}=b +\] + +\end_inset + + +\begin_inset Formula +\[ +\overline{yx}=a\overline{x^{2}}+\left(\overline{y}-a\overline{x}\right)\overline{x}=a\overline{x^{2}}+\overline{y}\cdot\overline{x}-a\overline{x}^{2}\Longrightarrow a\left(\overline{x^{2}}-\overline{x}^{2}\right)=\overline{yx}-\overline{y}\cdot\overline{x}\Longrightarrow a=\frac{\overline{yx}-\overline{y}\cdot\overline{x}}{\overline{x^{2}}-\overline{x}^{2}} +\] + +\end_inset + + +\end_layout + +\begin_layout Subsection +Sistemi linearnih enačb +\end_layout + +\begin_layout Standard +Ta sekcija, + z izjemo prve podsekcije, + je precej dobesedno povzeta po profesorjevi beamer skripti. +\end_layout + +\begin_layout Subsubsection +Linearna enačba +\end_layout + +\begin_layout Definition* +\begin_inset Formula $\sim$ +\end_inset + + je enačba oblike +\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=b$ +\end_inset + + in vsebuje koeficiente, + spremenljivke in desno stran. + Množica rešitev so vse +\begin_inset Formula $n-$ +\end_inset + +terice realnih števil, + ki zadoščajo enačbi +\begin_inset Formula $R=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n};a_{1}x_{1}+\cdots+a_{n}x_{n}=b\right\} $ +\end_inset + +. + Če so vsi koeficienti 0, + pravimo, + da je enačba trivialna, + sicer (torej čim je en koeficient neničeln) je netrivialna. +\end_layout + +\begin_layout Remark* +Za trivialno enačbo velja +\begin_inset Formula $R=\begin{cases} +\emptyset & ;b\not=0\\ +\mathbb{R}^{n} & ;b=0 +\end{cases}$ +\end_inset + +. + Za netrivialno enačbo pa velja +\begin_inset Formula $a_{i}\not=0$ +\end_inset + +, + torej: +\begin_inset Formula +\[ +a_{1}x_{1}+\cdots+a_{i}x_{i}+\cdots+a_{n}x_{n}=b +\] + +\end_inset + + +\begin_inset Formula +\[ +a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=b-a_{i}x_{i}=-a_{i}\left(x_{i}-\frac{b}{a_{i}}\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i-1},x_{i}-\frac{b}{a_{i}},x_{i+1},\dots,x_{n}\right)\right\rangle =0=\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i},\dots,x_{n}\right)-\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)\right\rangle +\] + +\end_inset + +Tu lahko označimo +\begin_inset Formula $\vec{n}\coloneqq\left(a_{i},\dots,a_{n}\right)$ +\end_inset + +, + +\begin_inset Formula $\vec{r}=\left(x_{1},\dots,x_{i},\dots,x_{n}\right)$ +\end_inset + +, + +\begin_inset Formula $\vec{r_{0}}=\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)$ +\end_inset + + in dobimo +\begin_inset Formula $\left\langle \vec{n},\vec{r}-\vec{r_{0}}\right\rangle $ +\end_inset + +, + kar je normalna enačba premice v +\begin_inset Formula $\mathbb{R}^{2}$ +\end_inset + +, + normalna enačba ravnine v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + + oziroma normalna enačba hiperravnine v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Sistem linearnih enačb +\end_layout + +\begin_layout Definition* +Sistem +\begin_inset Formula $m$ +\end_inset + + linearnih enačb z +\begin_inset Formula $n$ +\end_inset + + spremenljivkami je sistem enačb oblike +\begin_inset Formula +\[ +\begin{array}{ccccccc} +a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\ +\vdots & & & & \vdots & & \vdots\\ +a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m} +\end{array}. +\] + +\end_inset + + +\end_layout + +\begin_layout Fact* +Množica rešitev sistema je +\begin_inset Formula $\mathbb{R}^{n}\Leftrightarrow\forall i,j:a_{i,j}=b_{i}=0$ +\end_inset + +. + Sicer je množica rešitev presek hiperravnin v +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + — + rešitev posameznih enačb. + To vključuje tudi primer prazne množice rešitev, + saj je takšna na primer presek dveh vzporednih hiperravnin. +\end_layout + +\begin_layout Example* +Množica rešitev +\begin_inset Formula $2\times2$ +\end_inset + + sistema je lahko +\end_layout + +\begin_layout Itemize +cela ravnina +\end_layout + +\begin_layout Itemize +premica v ravnini +\end_layout + +\begin_layout Itemize +točka v ravnini +\end_layout + +\begin_layout Itemize +prazna množica +\end_layout + +\begin_layout Remark* +Enako velja za množico rešitev +\begin_inset Formula $3\times2$ +\end_inset + + sistema. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Množica rešitev sistema +\begin_inset Formula $2\times3$ +\end_inset + + pa ne more biti točka v prostoru, + lahko pa je cel prostor, + ravnina v prostoru, + premica v prostoru ali prazna množica. +\end_layout + +\begin_layout Paragraph* +Algebraičen pomen rešitev sistema +\end_layout + +\begin_layout Standard +Rešitve sistema +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\begin{array}{ccccccc} +a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\ +\vdots & & & & \vdots & & \vdots\\ +a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m} +\end{array} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +lahko zapišemo kot linearno kombinacijo stolpecv sistema in spremenljivk: +\begin_inset Formula +\[ +\left(b_{1},\dots,b_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)=x_{1}\left(a_{1,1},\dots,a_{m,1}\right)+\cdots+x_{n}\left(a_{1,n},\dots,a_{m,n}\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{b}=x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}} +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Klasifikacija sistemov linearnih enačb +\end_layout + +\begin_layout Standard +Sisteme linearnih enačb delimo glede na velikost na +\end_layout + +\begin_layout Itemize +kvadratne (toliko enačb kot spremenljivk), +\end_layout + +\begin_layout Itemize +poddoločene (več spremenljivk kot enačb), +\end_layout + +\begin_layout Itemize +predoločene (več enačb kot spremenljivk); +\end_layout + +\begin_layout Standard +glede na rešljivost na +\end_layout + +\begin_layout Itemize +nerešljive (prazna množica rešitev), +\end_layout + +\begin_layout Itemize +enolično rešljive (množica rešitev je singleton), +\end_layout + +\begin_layout Itemize +neenolično rešljive (moč množice rešitev je več kot 1); +\end_layout + +\begin_layout Standard +glede na obliko desnih strani na +\end_layout + +\begin_layout Itemize +homogene (vektor desnih stani je ničeln) +\end_layout + +\begin_layout Itemize +nehomogene (vektor desnih strani je neničen) +\end_layout + +\begin_layout Remark* +Če sta +\begin_inset Formula $\vec{x}$ +\end_inset + + in +\begin_inset Formula $\vec{y}$ +\end_inset + + dve različni rešitvi sistema, + je rešitev sistema tudi +\begin_inset Formula $\left(1-t\right)\vec{x}+t\vec{y}$ +\end_inset + + za vsak realen +\begin_inset Formula $t$ +\end_inset + +, + torej ima vsak neenolično rešljiv sistem neskončno rešitev. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Pogosto (a nikakor ne vedno) se zgodi, + da je kvadraten sistem enolično rešljiv, + predoločen sistem nerešljiv, + poddoločen sistem pa neenolično rešljiv. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Homogen sistem je vedno rešljiv, + saj obstaja trivialna rešitev +\begin_inset Formula $\vec{0}$ +\end_inset + +. + Vprašanje pri homogenih sistemih je torej, + kdaj je enolično in kdaj neenolično rešljiv. + Dokazali bomo, + da je vsak poddoločen homogen sistem linearnih enačb neenolično rešljiv. +\end_layout + +\begin_layout Subsubsection +Reševanje sistema +\end_layout + +\begin_layout Standard +Sisteme lahko rešujemo z izločanjem spremenljivk. + Iz ene enačbe izrazimo spremenljivko in jo vstavimo v druge enačbe, + da izrazimo zopet nove spremenljivke, + ki jih spet vstavimo v nove enačbe, + iz katerih spremenljivk še nismo izražali in tako naprej, + vse dokler ne pridemo do zadnjega možnega izražanja (dodatno branje prepuščeno bralcu). +\end_layout + +\begin_layout Standard +Sisteme pa lahko rešujemo tudi z Gaussovo metodo. + Trdimo, + da se rešitev sistema ne spremeni, + če na njem uporabimo naslednje elementarne vrstične transformacije: +\end_layout + +\begin_layout Itemize +menjava vrstnega reda enačb, +\end_layout + +\begin_layout Itemize +množenje enačbe z neničelno konstanto, +\end_layout + +\begin_layout Itemize +prištevanje večkratnika ene enačbe k drugi. +\end_layout + +\begin_layout Standard +Z Gaussovo metodo (dodatno branje prepuščeno bralcu) mrcvarimo razširjeno matriko sistema, + dokler ne dobimo reducirane kvadratne stopničaste forme (angl. + row echelon), + ki izgleda takole ( +\begin_inset Formula $\times$ +\end_inset + + reprezentira poljubno realno številko, + +\begin_inset Formula $0$ +\end_inset + + ničlo in +\begin_inset Formula $1$ +\end_inset + + enico): +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\left[\begin{array}{ccccccccccccccccc|c} +0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\ +\vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\ + & & & \vdots & \vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \cdots & \times\\ + & & & & & & & \vdots & \vdots & & \vdots & 0 & 0 & \cdots & 0 & 1 & \cdots & \times\\ + & & & & & & & & & & & \vdots & \vdots & & \vdots & 0 & \cdots & \vdots\\ +\vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & & \vdots & \vdots & & \vdots\\ +0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots & \times +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Homogeni sistemi +\end_layout + +\begin_layout Definition* +Sistem je homogen, + če je vektor desnih strani ničeln. +\end_layout + +\begin_layout Standard +Vedno ima rešitev +\begin_inset Formula $\vec{0}$ +\end_inset + +. + Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev. + Splošna rešitev nehomogenega sistema je vsota partikularne rešitve tega nehomogenega sistema in splošne rešitve njemu prirejenega homogenega sistema. +\end_layout + +\begin_layout Remark* +V tem razdelku nehomogen sistem pomeni nenujno homogen sistem (torej splošen sistem linearnih enačb), + torej je vsak homogen sistem nehomogen. +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "claim:Vpoddol-hom-sist-ima-ne0-reš" + +\end_inset + +Vsak poddoločen homogen sistem ima vsaj eno netrivialno rešitev. +\end_layout + +\begin_layout Proof +Dokaz z indukcijo po številu enačb. +\end_layout + +\begin_deeper +\begin_layout Paragraph* +Baza +\end_layout + +\begin_layout Standard +\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=0$ +\end_inset + + za +\begin_inset Formula $n\geq2$ +\end_inset + +. + Če je +\begin_inset Formula $a_{n}=0$ +\end_inset + +, + je netrivialna rešitev +\begin_inset Formula $\left(0,\dots,0,1\right)$ +\end_inset + +, + sicer pa +\begin_inset Formula $\left(0,\dots,0,-a_{n},a_{n-1}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph* +Korak +\end_layout + +\begin_layout Standard +Denimo, + da velja za vse poddoločene homogene sisteme z +\begin_inset Formula $m-1$ +\end_inset + + vrsticami. + Vzemimo poljuben homogen sistem z +\begin_inset Formula $n>m$ +\end_inset + + stolpci (da je poddoločen). + Če je +\begin_inset Formula $a_{n}=0$ +\end_inset + +, + je netriviačna rešitev +\begin_inset Formula $\left(0,\dots,0,1\right)$ +\end_inset + +, + sicer pa iz ene od enačb izrazimo +\begin_inset Formula $x_{n}$ +\end_inset + + s preostalimi spremenljivkami. + Dobljen izraz vstavimo v preostalih +\begin_inset Formula $m-1$ +\end_inset + + enačb z +\begin_inset Formula $n-1$ +\end_inset + + spremenljivkami in dobljen sistem uredimo. + Po I. + P. + ima slednji netrivialno rešitev +\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1}\right)$ +\end_inset + +. + To rešitev vstavimo v izraz za +\begin_inset Formula $x_{n}$ +\end_inset + + in dobimo +\begin_inset Formula $\alpha_{n}$ +\end_inset + + in s tem +\begin_inset Formula $\left(\alpha_{1},\dots,\alpha_{n-1},\alpha_{n}\right)$ +\end_inset + + kot netrivialno rešitev sistema z +\begin_inset Formula $m$ +\end_inset + + vrsticami. +\end_layout + +\end_deeper +\begin_layout Claim* +Linearna kombinacija dveh rešitev homogenega sistema je spet njegova rešitev. +\end_layout + +\begin_layout Proof +Če sta +\begin_inset Formula $\left(s_{1},\dots,s_{n}\right)$ +\end_inset + + in +\begin_inset Formula $\left(t_{1},\dots,t_{n}\right)$ +\end_inset + + dve rešitvi homogenega sistema, + velja za +\begin_inset Formula $\vec{s}$ +\end_inset + + +\begin_inset Formula $\forall i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\left(s_{1},\dots,s_{n}\right)\right\rangle =a_{i,1}s_{1}+\cdots+a_{i,n}s_{n}=0$ +\end_inset + + in enako za +\begin_inset Formula $\vec{t}$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\forall\alpha,\beta\in\mathbb{R},i:\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =0$ +\end_inset + +. +\begin_inset Formula +\[ +\left\langle \left(a_{i,1},\dots,a_{i,n}\right),\alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right)\right\rangle =\left\langle \alpha\left(s_{1},\dots,s_{n}\right)+\beta\left(t_{1},\dots,t_{n}\right),\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\alpha\left\langle \vec{s},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\beta\left\langle \vec{t},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\alpha0+\beta0 +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Splošna rešitev +\begin_inset Formula $\vec{x}$ +\end_inset + + rešljivega nehomogenega sistema s partikularno rešitvijo +\begin_inset Formula $\vec{p}$ +\end_inset + + je +\begin_inset Formula $\vec{x}=\vec{p}+\vec{h}$ +\end_inset + +, + kjer je +\begin_inset Formula $\vec{h}$ +\end_inset + + rešitev temu sistemu prirejenega homogenega sistema (desno stvar smo prepisali z ničlami). +\end_layout + +\begin_layout Remark* +Trdimo, + da je množica rešitev nehomogenega sistema samo množica rešitev prirejenega homogenega sistema, + premaknjena za partikularno rešitev nehomogenega sistema. +\end_layout + +\begin_layout Proof +Velja +\begin_inset Formula $\forall i:\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}\wedge\left\langle \vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =0$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\forall i:\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}$ +\end_inset + +. +\begin_inset Formula +\[ +\left\langle \vec{p}+\vec{h},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =\left\langle \vec{p},\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle +\left\langle \vec{h}\left(a_{i,1},\dots,a_{i,n}\right)\right\rangle =b_{i}+0=b_{i} +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Predoločeni sistemi +\end_layout + +\begin_layout Standard +Predoločen sistem, + torej tak z več enačbami kot spremenljivkami, + je običajno, + a ne nujno, + nerešljiv. +\end_layout + +\begin_layout Definition* +Posplošena rešitev sistema linearnih enačb je taka +\begin_inset Formula $n-$ +\end_inset + +terica števil +\begin_inset Formula $\left(x_{1},\dots x_{n}\right)$ +\end_inset + +, + za katero je vektor levih strani +\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$ +\end_inset + + najbližje vektorju desnih strani +\begin_inset Formula $\left(b_{1},\dots,b_{n}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Če je sistem rešljiv, + se njegova rešitev ujema s posplošeno rešitvijo. + Po metodi najmanjših kvadratov želimo minimizirati izraz +\begin_inset Formula $\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n}-b_{1}\right)^{2}+\cdots+\left(a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}-b_{n}\right)^{2}$ +\end_inset + + oziroma kvadrat norme razlike +\begin_inset Formula $\left|\left|x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}-\vec{b}\right|\right|^{2}$ +\end_inset + +. +\begin_inset Foot +status open + +\begin_layout Plain Layout +Z +\begin_inset Formula $\vec{a_{i}}$ +\end_inset + + označujemo stolpične vektorje sistema, + torej +\begin_inset Formula $\vec{a_{i}}=\left(a_{1,i},\dots,a_{m,i}\right)$ +\end_inset + +. +\end_layout + +\end_inset + + Podobno kot pri regresijski premici želimo pravokotno projicirati +\begin_inset Formula $\vec{b}$ +\end_inset + + na +\begin_inset Formula $\Lin\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $ +\end_inset + +. + Iščemo torej take skalarje +\begin_inset Formula $\left(x_{1},\dots,x_{n}\right)$ +\end_inset + +, + da je +\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b}\perp\vec{a_{1}},\dots,\vec{a_{n}}$ +\end_inset + + (hkrati pravokotna na vse vektorje, + ki določajo to linearno ogrinjačo). + Preuredimo skalarne produkte in zopet dobimo sistem enačb: +\begin_inset Formula +\[ +\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{1}}\right\rangle =\cdots=\left\langle \vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}-\vec{b},\vec{a_{n}}\right\rangle =0 +\] + +\end_inset + + +\begin_inset Formula +\[ +x_{1}\left\langle \vec{a_{1}},\vec{a_{1}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{1}}\right\rangle =\left\langle \vec{b},\vec{a_{1}}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +x_{1}\left\langle \vec{a_{1}},\vec{a_{n}}\right\rangle +\cdots+x_{n}\left\langle \vec{a_{n}},\vec{a_{n}}\right\rangle =\left\langle \vec{b},\vec{a_{n}}\right\rangle +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Izkaže se, + da je zgornji sistem vedno rešljiv. + Enolično takrat, + ko so +\begin_inset Formula $\left\{ \vec{a_{1}},\dots,\vec{a_{n}}\right\} $ +\end_inset + + linearno neodvisni. + Če je neenolično rešljiv, + pa poiščemo njegovo najkrajšo rešitev. +\end_layout + +\begin_layout Subsubsection +Poddoločeni sistemi +\end_layout + +\begin_layout Claim* +Poddoločen sistem, + torej tak, + ki ima več spremenljivk kot enačb, + ima neskončno rešitev, + čim je rešljiv. +\end_layout + +\begin_layout Proof +Sledi iz zgornjih dokazov, + da ima vsak poddoločen homogen sistem neskončno rešitev in da je +\begin_inset Formula $\vec{p}+\vec{h}$ +\end_inset + + splošna rešitev nehomogenega sistema, + če je +\begin_inset Formula $\vec{p}$ +\end_inset + + partikularna rešitev tega sistema in +\begin_inset Formula $\vec{h}$ +\end_inset + + splošna rešitev prirejenega homogenega sistema. +\end_layout + +\begin_layout Remark* +Seveda je lahko poddoločen sistem nerešljiv. + Trivialen primer: + +\begin_inset Formula $x+y+z=1$ +\end_inset + +, + +\begin_inset Formula $x+y+z=2$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kadar ima sistem neskončno rešitev, + nas često zanima najkrajša (recimo zadnja opomba v prejšnji sekciji). + Geometrijski gledano je najkrajša rešitev pravokotna projekcija izhodišča na presek hiperravnin, + ki so množica rešitve sistema. + Vsaka enačba določa eno hiperravnino v normalni obliki, + torej +\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}$ +\end_inset + +. + Projekcija izhodišča na hiperravnino v normalni obliki je presečišče premice, + ki gre skozi izhodišče in je pravokotna na ravnino, + torej +\begin_inset Formula $\vec{r}=t\vec{n_{i}}$ +\end_inset + +, + in ravnine same. + Vstavimo drugo enačbo v prvo in dobimo +\begin_inset Formula $\left\langle t\vec{n_{i}},\vec{n_{i}}\right\rangle =b_{i}$ +\end_inset + + in izrazimo +\begin_inset Formula $t=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }$ +\end_inset + +, + s čimer dobimo +\begin_inset Formula $\vec{r}=\frac{b}{\left\langle \vec{n_{i}},\vec{n_{i}}\right\rangle }\vec{n_{i}}$ +\end_inset + +. + Doslej je to le projekcija na eno hiperravnino. +\end_layout + +\begin_layout Standard +Za pravokotno projekcijo na presek hiperravnin pa najprej določimo ravnino, + ki je pravokotna na vse hiperravnine sistema, + torej +\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$ +\end_inset + +, + in najdimo presek te ravnine z vsemi hiperravninami. + To storimo tako, + da enačbo ravnine vstavimo v enačbe hiperravnin in jih uredimo: + +\begin_inset Formula $\left\langle \vec{r},\vec{n_{i}}\right\rangle =b_{i}\sim\left\langle t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}\sim t_{1}\left\langle \vec{n_{1}},\vec{n_{i}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{i}}\right\rangle =b_{i}$ +\end_inset + +. + To nam da sistem enačb +\begin_inset Formula +\[ +t_{1}\left\langle \vec{n_{1}},\vec{n_{1}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{1}}\right\rangle =b_{1} +\] + +\end_inset + + +\begin_inset Formula +\[ +\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +t_{1}\left\langle \vec{n_{1}},\vec{n_{m}}\right\rangle +\cdots+t_{m}\left\langle \vec{n_{m}},\vec{n_{m}}\right\rangle =b_{m} +\] + +\end_inset + +Rešimo sistem in dobimo +\begin_inset Formula $\left(t_{1},\dots,t_{m}\right)$ +\end_inset + +, + kar vstavimo v enačbo ravnine +\begin_inset Formula $\vec{r}=t_{1}\vec{n_{1}}+\cdots+t_{m}\vec{n_{m}}$ +\end_inset + +, + da dobimo najkrajšo rešitev. +\end_layout + +\begin_layout Subsection +Matrike +\end_layout + +\begin_layout Definition* +\begin_inset Formula $m\times n$ +\end_inset + + matrika je element +\begin_inset Formula $\left(\mathbb{R}^{n}\right)^{m}$ +\end_inset + +, + torej +\begin_inset Formula $A=\left(\left(a_{1,1},\dots,a_{1,n}\right),\dots,\left(a_{m,1},\dots,a_{m,n}\right)\right)$ +\end_inset + +. + Ima +\begin_inset Formula $m$ +\end_inset + + vrstic in +\begin_inset Formula $n$ +\end_inset + + stolpcev, + zato jo pišemo takole: +\begin_inset Formula +\[ +A=\left[\begin{array}{ccc} +a_{1,1} & \cdots & a_{1,n}\\ +\vdots & & \vdots\\ +a_{m,1} & \cdots & a_{m,n} +\end{array}\right] +\] + +\end_inset + +Matrikam velikosti +\begin_inset Formula $1\times n$ +\end_inset + + pravimo vrstični vektor, + matrikam velikosti +\begin_inset Formula $m\times1$ +\end_inset + + pa stolpični vektor. + Obe vrsti običajno identificiramo z vektorji. + +\begin_inset Formula $\left[1\right]$ +\end_inset + + identificiramo z 1. + Na preseku +\begin_inset Formula $i-$ +\end_inset + +te vrstice in +\begin_inset Formula $j-$ +\end_inset + +tega stolpca matrike se nahaja element +\begin_inset Formula $a_{i,j}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Seštevanje matrik je definirano le za matrike enakih dimenzij. + Vsota matrik +\begin_inset Formula $A+B$ +\end_inset + + je matrika +\begin_inset Formula +\[ +A+B=\left[\begin{array}{ccc} +a_{1,1}+b_{1,1} & \cdots & a_{1,n}+b_{1,n}\\ +\vdots & & \vdots\\ +a_{m,1}+b_{m.1} & \cdots & a_{m,n}+b_{m,n} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Ničelna matrika 0 je aditivna enota. +\begin_inset Formula +\[ +0=\left[\begin{array}{ccc} +0 & \cdots & 0\\ +\vdots & & \vdots\\ +0 & \cdots & 0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Produkt matrike s skalarjem. +\begin_inset Formula +\[ +A\cdot\alpha=\alpha\cdot A=\left[\begin{array}{ccc} +\alpha a_{1,1} & \cdots & \alpha a_{1,n}\\ +\vdots & & \vdots\\ +\alpha a_{m,1} & \cdots & \alpha a_{m,n} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Produkt dveh matrik +\begin_inset Formula $A_{m\times n}\cdot B_{n\times p}=C_{m\times p}$ +\end_inset + +. + Velja +\begin_inset Formula $c_{i,j}=\sum_{k=1}^{n}a_{i,k}b_{j,k}$ +\end_inset + +. + (razmislek prepuščen bralcu) +\end_layout + +\begin_layout Remark* +Kvadratna matrika identiteta +\begin_inset Formula $I$ +\end_inset + + je multiplikativna enota: + +\begin_inset Formula $i_{ij}=\begin{cases} +0 & ;i\not=j\\ +1 & ;i=j +\end{cases}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Transponiranje matrike +\begin_inset Formula $A_{m\times n}^{T}=B_{n\times m}$ +\end_inset + +. + +\begin_inset Formula $b_{ij}=a_{ji}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Lastnosti transponiranja: + +\begin_inset Formula $\left(A^{T}\right)^{T}=A$ +\end_inset + +, + +\begin_inset Formula $\left(A+B\right)^{T}=A^{T}+B^{T}$ +\end_inset + +, + +\begin_inset Formula $\left(\alpha A\right)^{T}=\alpha A^{T}$ +\end_inset + +, + +\begin_inset Formula $\left(AB\right)^{T}=B^{T}A^{T}$ +\end_inset + +, + +\begin_inset Formula $I^{T}=I$ +\end_inset + +, + +\begin_inset Formula $0^{T}=0$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Matrični zapis sistema linearnih enačb +\end_layout + +\begin_layout Standard +Matrika koeficientov vsebuje koeficiente, + imenujmo jo +\begin_inset Formula $A$ +\end_inset + + (ena vrstica matrike je ena enačba v sistemu). + Stolpični vektor spremenljivk vsebuje spremenljivke +\begin_inset Formula $\vec{x}=\left(x_{1},\dots,x_{n}\right)$ +\end_inset + +. + Vektor desne strani vsebuje desne strani +\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{m}\right)$ +\end_inset + +. + Sistem torej zapišemo kot +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Tudi Gaussovo metodo lahko zapišemo matrično. + Trem elementarnim preoblikovanjem, + ki ne spremenijo množice rešitev, + priredimo ustrezne t. + i. + elementarne matrike: +\end_layout + +\begin_layout Itemize +\begin_inset Formula $E_{i,j}\left(\alpha\right)$ +\end_inset + +: + identiteta, + ki ji na +\begin_inset Formula $i,j-$ +\end_inset + +to mesto prištejemo +\begin_inset Formula $\alpha$ +\end_inset + +. + Ustreza prištevanju +\begin_inset Formula $\alpha-$ +\end_inset + +kratnika +\begin_inset Formula $j-$ +\end_inset + +te vrstice k +\begin_inset Formula $i-$ +\end_inset + +ti vrstici. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $P_{ij}$ +\end_inset + +: + v +\begin_inset Formula $I$ +\end_inset + + zamenjamo +\begin_inset Formula $i-$ +\end_inset + +to in +\begin_inset Formula $j-$ +\end_inset + +to vrstico. + Ustreza zamenjavi +\begin_inset Formula $i-$ +\end_inset + +te in +\begin_inset Formula $j-$ +\end_inset + +te vrstice. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $E_{i}\left(\alpha\right)$ +\end_inset + +: + v +\begin_inset Formula $I$ +\end_inset + + pomnožiš +\begin_inset Formula $i-$ +\end_inset + +to vrstico z +\begin_inset Formula $\alpha$ +\end_inset + +. + Ustreza množenju +\begin_inset Formula $i-$ +\end_inset + +te vrstice s skalarjem +\begin_inset Formula $\alpha$ +\end_inset + +. +\end_layout + +\begin_layout Fact* +Vsako matriko je moč z levim množenjem z elementarnimi matrikami (Gaussova metoda) prevesti na reducirano vrstično stopničasto formo/obliko. + ZDB +\begin_inset Formula $\forall A\in M\left(\mathbb{R}\right)\exists E_{1},\dots,E_{k}\ni:R=E_{1}\cdot\cdots\cdot E_{k}\cdot A$ +\end_inset + + je r. + v. + s. + f. + Ko rešujemo sistem s temi matrikami množimo levo in desno stran sistema. +\end_layout + +\begin_layout Subsubsection +Postopek iskanja posplošene rešitve predoločenega sistema +\end_layout + +\begin_layout Enumerate +Sistem +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + z leve pomnožimo z +\begin_inset Formula $A^{T}$ +\end_inset + + in dobimo sistem +\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Poiščemo običajno rešitev dobljenega sistema, + za katero se izkaže, + da vselej obstaja (dokaz v 2. + semestru). +\end_layout + +\begin_layout Enumerate +Dokažemo, + da je običajna rešitev +\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$ +\end_inset + + enaka posplošeni rešitvi +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}$ +\end_inset + + bi radi minimizirali. + Naj bo +\begin_inset Formula $\vec{x_{0}}$ +\end_inset + + običajna rešitev sistema +\begin_inset Formula $A^{T}A\vec{x}=A^{T}\vec{b}$ +\end_inset + +. +\begin_inset Formula +\[ +\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $\vec{u}=A\vec{x}-A\vec{x_{0}}$ +\end_inset + + in +\begin_inset Formula $\vec{v}=A\vec{x_{0}}-\vec{b}$ +\end_inset + +. + Trdimo, + da +\begin_inset Formula $\vec{u}\perp\vec{v}$ +\end_inset + +, + torej +\begin_inset Formula $\left\langle \vec{u},\vec{v}\right\rangle =0$ +\end_inset + +. + Dokažimo: +\begin_inset Formula +\[ +\left\langle A\vec{x}-A\vec{x_{0}},A\vec{x_{0}}-\vec{b}\right\rangle =\left\langle A\left(\vec{x}-\vec{x_{0}}\right),A\vec{x_{0}}-\vec{b}\right\rangle =\left(A\vec{x_{0}}-\vec{b}\right)^{T}A\left(\vec{x}-\vec{x_{0}}\right)=\left(A\vec{x_{0}}-\vec{b}\right)^{T}\left(A^{T}\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left(A^{T}\left(A\vec{x_{0}}-\vec{b}\right)\right)^{T}\left(\vec{x}-\vec{x_{0}}\right)=\left(A^{T}A\vec{x_{0}}-A^{T}\vec{b}\right)\left(\vec{x}-\vec{x_{0}}\right)\overset{\text{predpostavka o }\vec{x_{0}}}{=}0\left(\vec{x}-\vec{x_{0}}\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Ker sedaj vemo, + da sta +\begin_inset Formula $\vec{u}$ +\end_inset + + in +\begin_inset Formula $\vec{v}$ +\end_inset + + pravokotna, + lahko uporabimo Pitagorov izrek, + ki za njiju pravi +\begin_inset Formula $\left|\left|\vec{u}+\vec{v}\right|\right|^{2}=\left|\left|\vec{u}\right|\right|^{2}+\left|\left|\vec{v}\right|\right|^{2}$ +\end_inset + +. + V naslednjih izpeljavah je +\begin_inset Formula $\vec{x}$ +\end_inset + + poljuben, + +\begin_inset Formula $\vec{x_{0}}$ +\end_inset + + pa kot prej. +\begin_inset Formula +\[ +\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}+A\vec{x_{0}}-\vec{b}\right|\right|^{2}=\left|\left|A\vec{x}-A\vec{x_{0}}\right|\right|^{2}+\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|A\vec{x}-\vec{b}\right|\right|^{2}\geq\left|\left|A\vec{x_{0}}-\vec{b}\right|\right|^{2}, +\] + +\end_inset + +kar pomeni , + da je +\begin_inset Formula $\vec{x_{0}}$ +\end_inset + + manjši ali enak kot vsi ostale +\begin_inset Formula $n-$ +\end_inset + +terice spremenljivk. +\end_layout + +\begin_layout Subsubsection +Najkrajša rešitev sistema +\end_layout + +\begin_layout Standard +Ta sekcija je precej dobesedno povzeta po profesorjevi beamer skripti. +\end_layout + +\begin_layout Standard +Sistem +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + je lahko neenolično rešljiv. + Tedaj nas često zanima po normi najkrajša rešitev sistema. +\end_layout + +\begin_layout Claim* +Najkrajša rešitev sistema +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + je +\begin_inset Formula $A^{T}\vec{y_{0}}$ +\end_inset + +, + kjer je +\begin_inset Formula $\vec{y_{0}}$ +\end_inset + + poljubna rešitev sistema +\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $\vec{x_{0}}$ +\end_inset + + poljubna rešitev sistema +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + in +\begin_inset Formula $\vec{y_{0}}$ +\end_inset + + poljubna rešitev sistema +\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$ +\end_inset + +. + Dokazali bi radi, + da velja +\begin_inset Formula $\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\leq\left|\left|\vec{x_{0}}\right|\right|^{2}$ +\end_inset + +. + Podobno, + kot v prejšnji sekciji: +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +\left|\left|\vec{x_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|u+v\right|\right|^{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Dokažimo, + da sta +\begin_inset Formula $\vec{u}=\vec{x_{0}}-A^{T}\vec{y_{0}}$ +\end_inset + + in +\begin_inset Formula $\vec{v}=A^{T}\vec{y_{0}}$ +\end_inset + + pravokotna, + da lahko uporabimo pitagorov izrek v drugi vrstici: +\begin_inset Formula +\[ +\left\langle \vec{x_{0}}-A^{T}\vec{y_{0}},A^{T}\vec{y_{0}}\right\rangle =\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)^{T}A^{T}\vec{y_{0}}=\left(A\left(\vec{x_{0}}-A^{T}\vec{y_{0}}\right)\right)^{T}\vec{y_{0}}=\left(A\vec{x_{0}}-AA^{T}\vec{y_{0}}\right)^{T}\vec{y_{0}}=\left(\vec{b}-\vec{b}\right)^{T}\vec{y_{0}}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|u+v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}+A^{T}\vec{y_{0}}\right|\right|^{2}=\left|\left|\vec{x_{0}}-A^{T}\vec{y_{0}}\right|\right|^{2}+\left|\left|A^{T}\vec{y_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right| +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|\vec{x_{0}}\right|\right|^{2}\geq\left|\left|A^{T}\vec{y_{0}}\right|\right| +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Iz rešljivosti +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + sledi rešljivost +\begin_inset Formula $AA^{T}\vec{y}=\vec{b}$ +\end_inset + +, + toda to znamo dokazati šele v drugem semestru. +\end_layout + +\begin_layout Subsubsection +Inverzi matrik +\end_layout + +\begin_layout Definition* +Matrika +\begin_inset Formula $B$ +\end_inset + + je inverz matrike +\begin_inset Formula $A$ +\end_inset + +, + če velja +\begin_inset Formula $AB=I$ +\end_inset + + in +\begin_inset Formula $BA=I$ +\end_inset + +. + Matrika +\begin_inset Formula $A$ +\end_inset + + je obrnljiva, + če ima inverz, + sicer je neobrnljiva. +\end_layout + +\begin_layout Claim* +Če inverz obstaja, + je enoličen. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $B_{1}$ +\end_inset + + in +\begin_inset Formula $B_{2}$ +\end_inset + + inverza +\begin_inset Formula $A$ +\end_inset + +. + Velja +\begin_inset Formula $AB_{1}=B_{1}A=AB_{2}=B_{2}A=I$ +\end_inset + +. + +\begin_inset Formula $B_{1}=B_{1}I=B_{1}\left(AB_{2}\right)=\left(B_{1}A\right)B_{2}=IB_{2}=B_{2}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Če inverz +\begin_inset Formula $A$ +\end_inset + + obstaja, + ga označimo z +\begin_inset Formula $A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri obrnljivih matrik: +\end_layout + +\begin_deeper +\begin_layout Itemize +Identična matrika +\begin_inset Formula $I$ +\end_inset + +: + +\begin_inset Formula $I\cdot I=I$ +\end_inset + +, + +\begin_inset Formula $I^{-1}=I$ +\end_inset + + +\end_layout + +\begin_layout Itemize +Elementarne matrike: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $E_{ij}\left(\alpha\right)\cdot E_{ij}\left(-\alpha\right)=I$ +\end_inset + +, + torej +\begin_inset Formula $E_{ij}\left(\alpha\right)^{-1}=E_{ij}\left(-\alpha\right)$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $P_{ij}\cdot P_{ij}=I$ +\end_inset + +, + torej +\begin_inset Formula $P_{ij}^{-1}=P_{ij}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $E_{i}\left(\alpha\right)\cdot E_{i}\left(\alpha^{-1}\right)=I$ +\end_inset + +, + torej +\begin_inset Formula $E_{i}\left(\alpha\right)^{-1}=E_{i}\left(\alpha^{-1}\right)$ +\end_inset + + +\end_layout + +\end_deeper +\end_deeper +\begin_layout Claim* +Produkt obrnljivih matrik je obrnljiva matrika. +\end_layout + +\begin_layout Proof +Naj bodo +\begin_inset Formula $A_{1},\dots,A_{n}$ +\end_inset + + obrnljive matrike, + torej po definiciji velja +\begin_inset Formula $A_{1}\cdot\cdots\cdot A_{n}\cdot A_{n}^{-1}\cdot\cdots\cdot A_{1}^{-1}=A_{n}\cdot\cdots\cdot A_{1}\cdot A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}=I$ +\end_inset + +. + Opazimo, + da velja +\begin_inset Formula $\left(A_{1}\cdot\cdots\cdot A_{n}\right)^{-1}=A_{1}^{-1}\cdot\cdots\cdot A_{n}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Vsaka obrnljiva matrika je produkt elementarnih matrik. + Dokaz sledi kasneje. +\end_layout + +\begin_layout Example* +Primeri neobrnljivih matrik: +\end_layout + +\begin_deeper +\begin_layout Itemize +Ničelna matrika, + saj pri množenju s katerokoli matriko pridela ničelno matriko in velja +\begin_inset Formula $I\not=0$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Matrike z ničelnim stolpcem/vrstico. +\end_layout + +\begin_deeper +\begin_layout Proof +Naj ima +\begin_inset Formula $A$ +\end_inset + + vrstico samih ničel. + Tedaj za vsako +\begin_inset Formula $B$ +\end_inset + + velja, + da ima +\begin_inset Formula $AB$ +\end_inset + + vrstico samih ničel (očitno po definiciji množenja). + +\begin_inset Formula $AB$ +\end_inset + + zato ne more biti +\begin_inset Formula $I$ +\end_inset + +, + saj +\begin_inset Formula $I$ +\end_inset + + ne vsebuje nobene vrstice samih ničel. + Podobno za ničelni stolpec. +\end_layout + +\end_deeper +\begin_layout Itemize +Nekvadratne matrike +\end_layout + +\begin_deeper +\begin_layout Proof +Naj ima +\begin_inset Formula $A_{m\times n}$ +\end_inset + + več vrstic kot stolpcev ( +\begin_inset Formula $m>n$ +\end_inset + +). + PDDRAA obstaja +\begin_inset Formula $B$ +\end_inset + +, + da +\begin_inset Formula $AB=I$ +\end_inset + +. + Uporabimo Gaussovo metodo na +\begin_inset Formula $A$ +\end_inset + +. + Z levim množenjem +\begin_inset Formula $A$ +\end_inset + + z nekimi elementarnimi matrikami lahko pridelamo RVSO. + +\begin_inset Formula $E_{1}\cdots E_{n}A=R$ +\end_inset + +. + +\begin_inset Formula $E_{1}\cdots E_{n}AB=E_{1}\cdots E_{n}I=E_{1}\cdots E_{n}=RB$ +\end_inset + +. + Toda +\begin_inset Formula $R$ +\end_inset + + ima po konstrukciji ničelno vrstico (je namreč +\begin_inset Formula $A$ +\end_inset + + podobna RVSO in a ima več vrstic kot stolpcev). + Potemtakem ima tudi +\begin_inset Formula $RB$ +\end_inset + + ničelno vrstico, + torej je neobrnljiva, + toda +\begin_inset Formula $RB$ +\end_inset + + je enak produktu elementarnih matrik, + torej bi morala biti obrnljiva. + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + +\end_layout + +\end_deeper +\end_deeper +\begin_layout Remark* +Iz +\begin_inset Formula $AB=I$ +\end_inset + + ne sledi nujno +\begin_inset Formula $BA=I$ +\end_inset + +. + Primer: + +\begin_inset Formula $A=\left[\begin{array}{ccc} +1 & 0 & 0\\ +0 & 1 & 0 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $B=\left[\begin{array}{cc} +1 & 0\\ +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $AB=\left[\begin{array}{cc} +1 & 0\\ +0 & 1 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $BA=\left[\begin{array}{ccc} +1 & 0 & 0\\ +0 & 1 & 0\\ +0 & 0 & 0 +\end{array}\right]$ +\end_inset + +. + Velja pa to za kvadratne matrike. + Dokaz kasneje. +\end_layout + +\begin_layout Subsubsection +Karakterizacija obrnljivih matrik +\end_layout + +\begin_layout Theorem* +Za vsako kvadratno matriko +\begin_inset Formula $A$ +\end_inset + + so naslednje trditve ekvivalentne: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + je obrnljiva +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + ima levi inverz ( +\begin_inset Formula $\exists B\ni:BA=I$ +\end_inset + +) +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + ima desni inverz ( +\begin_inset Formula $\exists B\ni:AB=I$ +\end_inset + +) +\end_layout + +\begin_layout Enumerate +stolpci +\begin_inset Formula $A$ +\end_inset + + so linearno neodvisni +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +stolpci +\begin_inset Formula $A$ +\end_inset + + so ogrodje +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:VbEx:Ax=b" + +\end_inset + + +\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +RVSO +\begin_inset Formula $A$ +\end_inset + + je +\begin_inset Formula $I$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + je produkt elementarnih matrik +\end_layout + +\end_deeper +\begin_layout Standard +Shema dokaza teh ekvivalenc je zanimiv graf. + Bralcu je prepuščena njegova skica. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco. +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(1\Rightarrow2\right)$ +\end_inset + + Sledi iz definicije. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(1\Rightarrow3\right)$ +\end_inset + + Sledi iz definicije. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow5\right)$ +\end_inset + + Naj +\begin_inset Formula $\exists B\ni:BA=I$ +\end_inset + +. + Dokažimo, + da +\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$ +\end_inset + +. + Pa dajmo: + +\begin_inset Formula $A\vec{x}=0\Rightarrow B\left(A\vec{x}\right)=B0=0=\left(BA\right)\vec{x}=I\vec{x}=\vec{x}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(3\Rightarrow7\right)$ +\end_inset + + Naj +\begin_inset Formula $\exists B\ni:AB=I$ +\end_inset + +. + Dokažimo, + da +\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$ +\end_inset + +. + Vzemimo +\begin_inset Formula $\vec{x}=B\vec{b}$ +\end_inset + +. + Tedaj +\begin_inset Formula $A\vec{x}=AB\vec{b}=I\vec{b}=\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(5\Rightarrow4\right)$ +\end_inset + + Naj +\begin_inset Formula $\forall\vec{x}:A\vec{x}=0\Rightarrow\vec{x}=0$ +\end_inset + +. + Dokažimo, + da so stolpci +\begin_inset Formula $A$ +\end_inset + + linearno neodvisni. + Naj bo +\begin_inset Formula $A=\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{mn} +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $\vec{x}=\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right]$ +\end_inset + +. + Tedaj +\begin_inset Formula $A\vec{x}=\left[\begin{array}{ccc} +a_{11}x_{1} & \cdots & a_{1n}x_{n}\\ +\vdots & & \vdots\\ +a_{n1}x_{1} & \cdots & a_{mn}x_{n} +\end{array}\right]=\left[\begin{array}{c} +a_{11}\\ +\vdots\\ +a_{m1} +\end{array}\right]x_{1}+\cdots+\left[\begin{array}{c} +a_{1n}\\ +\vdots\\ +a_{mn} +\end{array}\right]x_{n}=\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}$ +\end_inset + +. + Po definiciji +\begin_inset CommandInset ref +LatexCommand ref +reference "def:vsi0" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + za linearno neodvisnost mora veljati +\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=0\Rightarrow x_{1}=\cdots=x_{n}=0$ +\end_inset + +. + Ravno to pa smo predpostavili. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(7\Rightarrow6\right)$ +\end_inset + + Uporabimo iste oznake kot zgoraj. + Za poljuben +\begin_inset Formula $\vec{b}$ +\end_inset + + iščemo tak +\begin_inset Formula $\vec{x}$ +\end_inset + +, + da je +\begin_inset Formula $\vec{a_{1}}x_{1}+\cdots+\vec{a_{n}}x_{n}=A\vec{x}=\vec{b}$ +\end_inset + + (definicija ogrodja). + Po predpostavki +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:VbEx:Ax=b" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + velja, + da +\begin_inset Formula $\forall\vec{b}\exists\vec{x}\ni:A\vec{x}=\vec{b}$ +\end_inset + +. + Torej po predpostavki najdemo ustrezen +\begin_inset Formula $\vec{x}$ +\end_inset + + za poljuben +\begin_inset Formula $\vec{b}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(4\Rightarrow8\right)$ +\end_inset + + Za dokaz uvedimo nekaj lem, + ki dokažejo trditev. +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:kom1" + +\end_inset + +Če ima +\begin_inset Formula $A_{n\times n}$ +\end_inset + + LN stolpce in če je +\begin_inset Formula $C_{n\times n}$ +\end_inset + + obrnljiva, + ima tudi +\begin_inset Formula $CA$ +\end_inset + + LN stolpce. +\end_layout + +\begin_layout Proof +Naj bodo +\begin_inset Formula $a_{1},\dots,a_{n}$ +\end_inset + + stolpci +\begin_inset Formula $A$ +\end_inset + +. + Velja +\begin_inset Formula $Ax=0\Rightarrow x=0$ +\end_inset + +. + Dokazati želimo, + da +\begin_inset Formula $CAx=0\Rightarrow x=0$ +\end_inset + +. + Predpostavimo +\begin_inset Formula $CAx=0$ +\end_inset + +. + Množimo obe strani z +\begin_inset Formula $C^{-1}$ +\end_inset + +. + +\begin_inset Formula $C^{-1}CAx=C^{-1}0\sim IAx=0\sim Ax=0\Rightarrow x=0$ +\end_inset + +. +\end_layout + +\begin_layout Lemma +Če ima +\begin_inset Formula $A$ +\end_inset + + LN stolpce, + ima njena RVSO LN stolpce. +\end_layout + +\begin_layout Proof +Po Gaussu obstajajo take elementarne +\begin_inset Formula $E_{1},\dots,E_{n}$ +\end_inset + +, + da je +\begin_inset Formula $E_{n}\cdots E_{1}A=R$ +\end_inset + + RVSO. + Po lemi +\begin_inset CommandInset ref +LatexCommand ref +reference "lem:kom1" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + ima +\begin_inset Formula $E_{1}A$ +\end_inset + + LN stolpce, + prav tako +\begin_inset Formula $E_{2}E_{1}A$ +\end_inset + + in tako dalje, + vse do +\begin_inset Formula $E_{n}\cdots E_{1}A=R$ +\end_inset + +. +\end_layout + +\begin_layout Lemma +Če ima RVSO +\begin_inset Formula $R$ +\end_inset + + LN stolpce, + je enaka identiteti. +\end_layout + +\begin_layout Proof +PDDRAA +\begin_inset Formula $R\not=I$ +\end_inset + +. + Tedaj ima bodisi ničelni stolpec bodisi stopnico, + daljšo od 1. + Če ima ničelni stolpec, + ni LN. + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. + Če ima stopnico, + daljšo od 1, + kar pomeni, + da v vrstici takoj za prvo enico obstajajo neki neničelni +\begin_inset Formula $\times-$ +\end_inset + +i, + pa je stolpec z nekim neničelnim +\begin_inset Formula $\times-$ +\end_inset + +om linearna kombinacija ostalih stolpcev, + torej stolpci +\begin_inset Formula $R$ +\end_inset + + niso LN +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(6\Rightarrow8\right)$ +\end_inset + + Predpostavimo, + da so stolpci +\begin_inset Formula $A$ +\end_inset + + ogrodje in dokazujemo, + da RVSO +\begin_inset Formula $A$ +\end_inset + + je +\begin_inset Formula $I$ +\end_inset + +. +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:68kom1" + +\end_inset + +Če so stolpci +\begin_inset Formula $A_{n\times n}$ +\end_inset + + ogrodje in če je +\begin_inset Formula $C_{n\times n}$ +\end_inset + + obrnljica, + so tudi stolpci +\begin_inset Formula $CA$ +\end_inset + + ogrodje. +\end_layout + +\begin_layout Proof +Naj bodo stolpci +\begin_inset Formula $A$ +\end_inset + + ogrodje. + Torej +\begin_inset Formula $\forall b\exists x\ni:Ax=C^{-1}b$ +\end_inset + +. + Množimo obe strani z +\begin_inset Formula $C^{-1}$ +\end_inset + +. + +\begin_inset Formula $\forall b\exists x\ni:CAx=b$ +\end_inset + + — + stolpci +\begin_inset Formula $CA$ +\end_inset + + so ogrodje. +\end_layout + +\begin_layout Lemma +Če so stolpci +\begin_inset Formula $A$ +\end_inset + + ogrodje, + so stolpci njene RVSO ogrodje. +\end_layout + +\begin_layout Proof +Po Gaussu obstajajo take elementarne +\begin_inset Formula $E_{1},\dots,E_{n}$ +\end_inset + +, + da je +\begin_inset Formula $E_{n}\cdots E_{1}A=R$ +\end_inset + + RVSO. + Po lemi +\begin_inset CommandInset ref +LatexCommand ref +reference "lem:68kom1" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + so stolpci +\begin_inset Formula $E_{1}A$ +\end_inset + + ogrodje in tudi stolpci +\begin_inset Formula $E_{2}E_{1}A$ +\end_inset + + so ogrodje in tako dalje vse do +\begin_inset Formula $R$ +\end_inset + +. +\end_layout + +\begin_layout Lemma +Če so stolpci RVSO +\begin_inset Formula $R$ +\end_inset + + ogrodje, + je +\begin_inset Formula $R=I$ +\end_inset + +. +\end_layout + +\begin_layout Proof +PDDRAA +\begin_inset Formula $R\not=I$ +\end_inset + +. + Tedaj ima bodisi ničelni stolpec bodisi stopnico, + daljšo od 1. + Če ima ničelni stolpec, + stolpci niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora). + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + Če ima stopnico, + daljšo od 1, + pa je stolpec z nekim neničelnim +\begin_inset Formula $\times-$ +\end_inset + +om linearna kombinacija ostalih stolpcev, + torej stolpci +\begin_inset Formula $R$ +\end_inset + + niso ogrodje zaradi enoličnosti moči baze (dimenzije prostora) +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. +\begin_inset Note Note +status open + +\begin_layout Plain Layout +(tegale ne razumem zares dobro, + niti med predavanji nismo dokazali) mogoče čim ima stopnico, + daljšo od 1, + ima ničelno vrstico? +\end_layout + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(8\Rightarrow9\right)$ +\end_inset + + Predpostavimo, + da je +\begin_inset Formula $R\coloneqq\text{RVSO}\left(A\right)=I$ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $A$ +\end_inset + + produkt elementarnih matrik. + Po Gaussu obstajajo take elementarne matrike +\begin_inset Formula $E_{1},\dots,E_{n}$ +\end_inset + +, + da +\begin_inset Formula $E_{n}\cdots E_{1}A=R$ +\end_inset + +. + Elementarne matrike so obrnljive, + zato množimo z leve najprej z +\begin_inset Formula $E_{n}^{-1}$ +\end_inset + +, + nato z +\begin_inset Formula $E_{n-1}^{-1}$ +\end_inset + +, + vse do +\begin_inset Formula $E_{1}^{-1}$ +\end_inset + + in dobimo +\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}R$ +\end_inset + +. + Upoštevamo, + da je inverz elementarne matrike elementarna matrika in da je +\begin_inset Formula $R=I$ +\end_inset + +. + Tedaj +\begin_inset Formula $A=E_{1}^{-1}\cdots E_{n}^{-1}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Claim* +\begin_inset Formula $A$ +\end_inset + + je obrnljiva +\begin_inset Formula $\Leftrightarrow A^{T}$ +\end_inset + + obrnljiva. +\end_layout + +\begin_layout Proof +Velja +\begin_inset Formula $AB=I\Leftrightarrow\left(AB\right)^{T}=I^{T}\Leftrightarrow B^{T}A^{T}=I$ +\end_inset + + in +\begin_inset Formula $BA=I\Leftrightarrow\left(BA\right)^{T}=I^{T}\Leftrightarrow A^{T}B^{T}=I$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +\begin_inset Formula $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$ +\end_inset + + in vrstice so LN in ogrodje. +\end_layout + +\begin_layout Remark* +Inverz +\begin_inset Formula $A$ +\end_inset + + lahko izračunamo po Gaussu. + Zapišemo razširjeno matriko +\begin_inset Formula $\left[A,I\right]$ +\end_inset + + in na obeh applyamo iste elementarne transformacije, + da +\begin_inset Formula $A$ +\end_inset + + pretvorimo v RVSO. + Če je +\begin_inset Formula $A$ +\end_inset + + obrnljiva, + dobimo na levi identiteto, + na desni pa +\begin_inset Formula $A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Determinante +\end_layout + +\begin_layout Definition* +Vsaki kvadratni matriki +\begin_inset Formula $A$ +\end_inset + + priredimo število +\begin_inset Formula $\det A$ +\end_inset + +. + Definicija za +\begin_inset Formula $1\times1$ +\end_inset + + matrike: + +\begin_inset Formula $\det\left[a\right]\coloneqq a$ +\end_inset + +. + Rekurzivna definicija za +\begin_inset Formula $n\times n$ +\end_inset + + matrike: +\begin_inset Formula +\[ +\det\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]=\sum_{k=1}^{n}\left(-1\right)^{k+1}a_{1k}\det A_{1k}, +\] + +\end_inset + +kjer +\begin_inset Formula $A_{ij}$ +\end_inset + + predstavja +\begin_inset Formula $A$ +\end_inset + + brez +\begin_inset Formula $i-$ +\end_inset + +te vrstice in +\begin_inset Formula $j-$ +\end_inset + +tega stolpca. + Tej formuli razvoja se reče +\begin_inset Quotes gld +\end_inset + +razvoj determinante po prvi vrstici +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $2\times2$ +\end_inset + + determinanta. + +\begin_inset Formula $\det\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]=ad-bc$ +\end_inset + +. + Geometrijski pomen je ploščina paralelograma, + ki ga razpenjata +\begin_inset Formula $\left(c,d\right)$ +\end_inset + + in +\begin_inset Formula $\left(a,b\right)$ +\end_inset + +, + kajti ploščina bi bila +\begin_inset Formula $\left(a+c\right)\left(b+d\right)-2bc-2\frac{cd}{2}-2\frac{ab}{2}=\cancel{ab}+\cancel{cb}+ad+\cancel{cd}-\cancel{2}bc-\cancel{cd}-\cancel{ab}=ad-bc$ +\end_inset + +. + Če zamenjamo vrstni red vektorjev, + pa dobimo za predznak napačen rezultat, + torej je ploščina enaka +\begin_inset Formula $\left|\det\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]\right|$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $3\times3$ +\end_inset + + determinanta. +\begin_inset Formula +\[ +\det\left[\begin{array}{ccc} +a_{11} & a_{12} & a_{13}\\ +a_{21} & a_{22} & a_{23}\\ +a_{31} & a_{32} & a_{33} +\end{array}\right]=a_{11}\det\left[\begin{array}{cc} +a_{22} & a_{23}\\ +a_{32} & a_{33} +\end{array}\right]-a_{12}\det\left[\begin{array}{cc} +a_{21} & a_{23}\\ +a_{31} & a_{33} +\end{array}\right]+a_{13}\left[\begin{array}{cc} +a_{21} & a_{22}\\ +a_{31} & a_{32} +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +a_{11}\left(a_{22}a_{33}-a_{23}a_{32}\right)-a_{12}\left(a_{21}a_{33}-a_{23}a_{31}\right)+a_{13}\left(a_{21}a_{32}-a_{22}a_{31}\right) +\] + +\end_inset + +To si lahko zapomnimo s Saurusovim pravilom. + Pripišemo na desno stran prva dva stolpca in seštejemo produkte po šestih diagonalah. + Naraščajoče diagonale (tiste s pozitivnim koeficientom, + če bi jih risali kot premice v ravnini) prej negiramo. + Geometrijski +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Vektorski produkt. + Velja: +\begin_inset Formula +\[ +\left\langle \left(x,y,z\right),\left(a_{21},a_{22},a_{23}\right)\times\left(a_{31},a_{32},a_{33}\right)\right\rangle =\det\left[\begin{array}{ccc} +x & y & z\\ +a_{21} & a_{22} & a_{23}\\ +a_{31} & a_{32} & a_{33} +\end{array}\right], +\] + +\end_inset + +torej je +\begin_inset Formula $\left[\left(a_{11},a_{12},a_{13}\right),\left(a_{21},a_{22},a_{23}\right),\left(a_{31},a_{32},a_{33}\right)\right]$ +\end_inset + + (mešani produkt) determinanta matrike +\begin_inset Formula $A$ +\end_inset + +, + torej je +\begin_inset Formula $\left|\det A\right|$ +\end_inset + + ploščina paralelpipeda, + ki ga razpenjajo trije vrstični vektorji +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Računanje determinant +\end_layout + +\begin_layout Standard +Determinante računati po definiciji je precej zahtevno (bojda +\begin_inset Formula $O\left(n!\right)$ +\end_inset + +) za +\begin_inset Formula $n\times n$ +\end_inset + + determinanto. + Boljšo računsko zahtevnost dobimo z Gaussovo metodo. + Oglejmo si najprej posplošeno definicijo determinante: + +\begin_inset Quotes gld +\end_inset + +razvoj po poljubni +\begin_inset Formula $i-$ +\end_inset + +ti vrstici +\begin_inset Quotes grd +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\det A=\sum_{j=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij} +\] + +\end_inset + + +\begin_inset Quotes grd +\end_inset + +razvoj po poljubnem +\begin_inset Formula $j-$ +\end_inset + +tem stolpcu +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula +\[ +\det A=\sum_{i=1}^{n}\left(-1\right)^{i+j}a_{ij}\det A_{ij} +\] + +\end_inset + +Ti dve formuli sta še vedno nepolinomske zahtevnosti, + uporabni pa sta v primerih, + ko imamo veliko ničel na kaki vrstici/stolpcu. + Determinanta zgornjetrikotne matrike je po tej formuli produkt diagonalcev. +\end_layout + +\begin_layout Standard +Kako pa se determinanta obnaša pri elementarnih vrstičnih transformacijah iz Gaussove metode? +\end_layout + +\begin_layout Itemize +menjava vrstic +\begin_inset Formula $\Longrightarrow$ +\end_inset + + determinanti se spremeni predznak +\end_layout + +\begin_layout Itemize +množenje vrstice z +\begin_inset Formula $\alpha\Longrightarrow$ +\end_inset + + determinanta se pomnoži z +\begin_inset Formula $\alpha$ +\end_inset + + +\end_layout + +\begin_layout Itemize +prištevanje večkratnika ene vrstice k drugi +\begin_inset Formula $\Longrightarrow$ +\end_inset + + determinanta se ne spremeni +\end_layout + +\begin_layout Standard +Časovna zahtevnost Gaussove metode je bojda polinomska +\begin_inset Formula $O\left(n^{3}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Ideja dokaza veljavnosti Gaussove metode: + Indukcija po velikosti matrike. +\end_layout + +\begin_layout Standard +Baza: + +\begin_inset Formula $2\times2$ +\end_inset + + matrike +\end_layout + +\begin_layout Standard +Korak: + Razvoj po vrstici, + ki je elementarna transformacija ne spremeni, + dobiš +\begin_inset Formula $n$ +\end_inset + + +\begin_inset Formula $\left(n-1\right)\times\left(n-1\right)$ +\end_inset + + determimant, + ki so veljavne po I. + P. +\end_layout + +\begin_layout Subsubsection +Lastnosti determinante +\end_layout + +\begin_layout Claim* +Velja +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\det\left(AB\right)=\det A\det B$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\det A^{T}=\det A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\det\left[\begin{array}{cc} +A & B\\ +0 & C +\end{array}\right]=\det A\det C$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Dokazujemo tri trditve +\end_layout + +\begin_deeper +\begin_layout Enumerate +Dokazujemo +\begin_inset Formula $\det\left(AB\right)=\det A\det B$ +\end_inset + +. + Obravnavajmo dva posebna primera: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + je elementarna: + obrat pomeni množenje determinante z +\begin_inset Formula $-1$ +\end_inset + +, + množenje vrstice z +\begin_inset Formula $\alpha$ +\end_inset + + množi determinanto z +\begin_inset Formula $\alpha$ +\end_inset + +, + prištevanje večkratnika vrstice k drugi vrstici množi determinanto z +\begin_inset Formula $1$ +\end_inset + +. + Očitno torej trditev velja, + če je +\begin_inset Formula $A$ +\end_inset + + elementarna. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A$ +\end_inset + + ima ničelno vrstico: + tedaj ima tudi +\begin_inset Formula $AB$ +\end_inset + + ničelno vrstico in je +\begin_inset Formula $\det A=0$ +\end_inset + + in +\begin_inset Formula $\det AB=0$ +\end_inset + +, + torej očitno trditev velja, + če ima +\begin_inset Formula $A$ +\end_inset + + ničelno vrstico. +\end_layout + +\begin_layout Standard +Obravnavajmo še splošen primer: + Po Gaussovi metodi obstajajo take elementarne +\begin_inset Formula $E_{1},\dots,E_{n}$ +\end_inset + +, + da je +\begin_inset Formula $E_{n}\cdots E_{1}A=R$ +\end_inset + + RVSO. + Ker je +\begin_inset Formula $A$ +\end_inset + + kvadratna, + je tudi +\begin_inset Formula $R$ +\end_inset + + kvadratna. + Ločimo dva primera: +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $R=I$ +\end_inset + +. + Tedaj +\begin_inset Formula $\det\left(E_{n}\cdots E_{1}AB\right)=\det E_{n}\cdots\det E_{1}\det AB=\det\left(RB\right)=\det\left(IB\right)=\det B$ +\end_inset + + +\begin_inset Formula +\[ +\det I=\det R=\det E_{n}\cdots\det E_{1}\det A\quad\quad\quad\quad/\cdot\det B +\] + +\end_inset + + +\begin_inset Formula +\[ +\det I\det B=\det B=\det E_{n}\cdots\det E_{1}\det A\det B +\] + +\end_inset + + +\begin_inset Formula $\det B$ +\end_inset + + zapišimo na dva načina ne levo in desno stran enačbe. + +\begin_inset Formula +\[ +\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B +\] + +\end_inset + + +\begin_inset Formula +\[ +\det AB=\det A\det B +\] + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Remark* +\begin_inset Formula $\exists A^{-1}\Leftrightarrow\det A\not=0$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Proof +Predpostavimo +\begin_inset Formula $A$ +\end_inset + + je obrnljiva. + Tedaj +\begin_inset Formula $\exists A^{-1}=B\ni:AB=I\overset{\circ\det}{\Longrightarrow}\det\left(AB\right)=\det I$ +\end_inset + +. + PDDRAA +\begin_inset Formula $\det A\not=0$ +\end_inset + +, + tedaj +\begin_inset Formula $\det AB=\det B\det A=0\not=\det I=1$ +\end_inset + +. + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + +\end_layout + +\begin_layout Proof +Predpostavimo sedaj +\begin_inset Formula $A$ +\end_inset + + ni obrnljiva. + Tedaj +\begin_inset Formula $\nexists A^{-1}\Rightarrow\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=R$ +\end_inset + + ima ničelno vrstico. + Uporabimo isti razmislek kot spodaj, + torej +\begin_inset Formula $\det R=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$ +\end_inset + +. + Ker so determinante elementarnih matrik vse neničelne, + mora biti +\begin_inset Formula $\det A$ +\end_inset + + ničeln, + da je produkt ničeln. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Enumerate +\begin_inset Formula $R$ +\end_inset + + ima ničelno vrstico. + Tedaj +\begin_inset Formula $\det\left(R\right)=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$ +\end_inset + +. + Ker so determinante elementarnih matrik vse neničelne, + mora biti +\begin_inset Formula $\det A$ +\end_inset + + ničeln, + da je produkt ničeln. +\end_layout + +\end_deeper +\begin_layout Enumerate +Dokazujemo +\begin_inset Formula $\det A^{T}=\det A$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Enumerate +Če je +\begin_inset Formula $A$ +\end_inset + + elementarna matrika, + to drži: + +\begin_inset Formula $\det P_{ij}=-1=\det P_{ij}^{T}=\det P_{ij}$ +\end_inset + +, + +\begin_inset Formula $\det E_{i}\left(\alpha\right)=\alpha=\det E_{i}\left(\alpha\right)^{T}=\det E_{i}\left(\alpha\right)$ +\end_inset + +, + +\begin_inset Formula $\det E_{ij}\left(\alpha\right)=1=\det E_{ji}\left(\alpha\right)^{T}=\det E_{ij}\left(\alpha\right)^{T}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Če ima +\begin_inset Formula $A$ +\end_inset + + ničelno vrstico, + to drži, + saj ima tedaj +\begin_inset Formula $A^{T}$ +\end_inset + + ničeln stolpec in +\begin_inset Formula $\det A=0=\det A^{T}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Splošen primer: + Po Gaussovi metodi +\begin_inset Formula $\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=\text{RVSO}\left(A\right)=R$ +\end_inset + +. + Zopet ločimo dva primera: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $R=I$ +\end_inset + +. + +\begin_inset Formula $\det R=\det R^{T}=1$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $R$ +\end_inset + + ima ničelno vrstico. + +\begin_inset Formula $\det R=\det R^{T}=0$ +\end_inset + + +\end_layout + +\begin_layout Standard +Sedaj vemo, + da +\begin_inset Formula $\det R=\det R^{T}$ +\end_inset + +. + Računajmo: +\begin_inset Formula +\[ +\det R=\det R^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left(E_{n}\cdots E_{1}A\right)=\det\left(E_{n}\cdots E_{1}A\right)^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left(E_{n}\cdots E_{1}A\right)=\det\left(A^{T}E_{1}^{T}\cdots E_{n}^{T}\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A=\det A^{T}\cancel{\det E_{1}^{T}}\cdots\cancel{\det E_{n}^{T}} +\] + +\end_inset + + +\begin_inset Formula +\[ +\det A=\det A^{T} +\] + +\end_inset + + +\end_layout + +\end_deeper +\end_deeper +\begin_layout Enumerate +Dokazujemo +\begin_inset Formula $\det\left[\begin{array}{cc} +A & B\\ +0 & C +\end{array}\right]=\det A\det C$ +\end_inset + +. + Levi izraz v enačbi vsebuje t. + i. + bločno matriko. + Upoštevamo poprej dokazano multiplikativnost determinante in opazimo, + da pri bločnem množenju matrik velja +\begin_inset Formula +\[ +\left[\begin{array}{cc} +A & B\\ +0 & C +\end{array}\right]=\left[\begin{array}{cc} +I & 0\\ +0 & C +\end{array}\right]\left[\begin{array}{cc} +A & B\\ +0 & I +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left[\begin{array}{cc} +A & B\\ +0 & C +\end{array}\right]=\det\left[\begin{array}{cc} +I & 0\\ +0 & C +\end{array}\right]\det\left[\begin{array}{cc} +A & B\\ +0 & I +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left[\begin{array}{cc} +A & B\\ +0 & C +\end{array}\right]=\det C\det A +\] + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Standard +Pojasnilo: + Za +\begin_inset Formula $\det C$ +\end_inset + + si razpišemo bločno matriko, + za +\begin_inset Formula $\det A$ +\end_inset + + si zopet razpišemo bločno matriko in nato z Gaussovimi transformacijami z enicami iz spodnjega desnega bloka izničimo zgornji desni blok ( +\begin_inset Formula $B$ +\end_inset + +). +\end_layout + +\end_deeper +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Cramerjevo pravilo — + eksplicitna formula za rešitve kvadratnega sistema linearnih enačb +\end_layout + +\begin_layout Standard +Radi bi dobili eksplicitne formule za komponente rešitve +\begin_inset Formula $x_{i}$ +\end_inset + + kvadratnega sistema linearnih enačb +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + +. + Izpeljimo torej eksplicitno formulo . + Druga/srednja matrika je identična, + v kateri smo +\begin_inset Formula $i-$ +\end_inset + +ti stolpec zamenjali z vektorjem spremenljivk +\begin_inset Formula $\vec{x}$ +\end_inset + + (to označimo z +\begin_inset Formula $I_{i}\left(\vec{x}\right)$ +\end_inset + +), + tretja/desna matrika pa je matrika koeficientov v kateri smo +\begin_inset Formula $i-$ +\end_inset + +ti stolpec zamenjali z vektorjem desnih strani +\begin_inset Formula $b$ +\end_inset + + (to označimo z +\begin_inset Formula $A_{i}\left(\vec{x}\right)$ +\end_inset + +). +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]\left[\begin{array}{ccccccc} +1 & & 0 & x_{1} & & & 0\\ + & \ddots & & \vdots\\ + & & 1 & x_{i-1}\\ + & & & x_{i}\\ + & & & x_{i+1} & 1\\ + & & & \vdots & & \ddots\\ +0 & & & x_{n} & 0 & & 1 +\end{array}\right]=\left[\begin{array}{ccccc} +a_{11} & \cdots & b_{1} & \cdots & a_{1n}\\ +\vdots & \ddots & \vdots & \iddots & \vdots\\ +a_{i1} & & b_{i} & & a_{in}\\ +\vdots & \iddots & \vdots & \ddots & \vdots\\ +a_{n1} & \cdots & b_{n} & \cdots & a_{nn} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]\left[\begin{array}{ccccccc} +1 & & 0 & x_{1} & & & 0\\ + & \ddots & & \vdots\\ + & & 1 & x_{i-1}\\ + & & & x_{i}\\ + & & & x_{i+1} & 1\\ + & & & \vdots & & \ddots\\ +0 & & & x_{n} & 0 & & 1 +\end{array}\right]=\left[\begin{array}{ccccc} +a_{11} & \cdots & a_{11}x_{1}+\cdots+a_{1n}x_{n} & \cdots & a_{1n}\\ +\vdots & \ddots & \vdots & \iddots & \vdots\\ +a_{i1} & & a_{i1}x_{1}+\cdots+a_{in}x_{n} & & a_{in}\\ +\vdots & \iddots & \vdots & \ddots & \vdots\\ +a_{n1} & \cdots & a_{n1}x_{1}+\cdots+a_{nn}x_{n} & \cdots & a_{nn} +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +AI_{i}\left(\vec{x}\right)=A_{i}\left(\vec{b}\right)\quad\quad\quad\quad/\det +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left(AI_{i}\left(\vec{x}\right)\right)=\det A_{i}\left(\vec{b}\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +\det A\det I_{i}\left(\vec{x}\right)=\det A_{i}\left(\vec{b}\right) +\] + +\end_inset + +Izračunamo +\begin_inset Formula $\det I_{i}\left(\vec{x}\right)$ +\end_inset + + z razvojem po +\begin_inset Formula $i-$ +\end_inset + +ti vrstici. +\begin_inset Formula +\[ +\det A\cdot x_{i}=\det A_{i}\left(\vec{b}\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A} +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Formula-za-inverz-matrike" + +\end_inset + +Formula za inverz matrike +\end_layout + +\begin_layout Standard +Za dano obrnljivo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + iščemo eksplicitno formulo za celice +\begin_inset Formula $X$ +\end_inset + +, + da velja +\begin_inset Formula $AX=I$ +\end_inset + +. + Ideja: + najprej bomo problem prevedli na reševanje sistemov linearnih enačb in uporabili Cramerjevo pravilo ter končno poenostavili formule. + Naj bodo +\begin_inset Formula $\vec{x_{1}},\dots,\vec{x_{n}}$ +\end_inset + + stolpci +\begin_inset Formula $X$ +\end_inset + + in +\begin_inset Formula $\vec{i_{1}},\dots,\vec{i_{n}}$ +\end_inset + +. + Potemtakem je +\begin_inset Formula $\left[\begin{array}{ccc} +A\vec{x_{1}} & \cdots & A\vec{x_{n}}\end{array}\right]=A\left[\begin{array}{ccc} +\vec{x_{1}} & \cdots & \vec{x_{n}}\end{array}\right]=AX=I=\left[\begin{array}{ccc} +\vec{i_{1}} & \cdots & \vec{i_{n}}\end{array}\right]$ +\end_inset + +. + Primerjajmo sedaj stolpce na obeh straneh: + +\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :A\vec{x_{i}}=\vec{i_{i}}$ +\end_inset + +. + ZDB za vsak stolpec +\begin_inset Formula $X$ +\end_inset + + smo dobili sistem +\begin_inset Formula $n\times n$ +\end_inset + + linearnih enačb. + Te sisteme +\begin_inset Formula $A\vec{x_{j}}=\vec{i_{j}}$ +\end_inset + + +\begin_inset Foot +status open + +\begin_layout Plain Layout +Tokrat uporabimo indeks +\begin_inset Formula $j$ +\end_inset + +, + ker z njim reprezentiramo stolpec in ponavadi, + ko govorimo o elementu +\begin_inset Formula $x_{ij}$ +\end_inset + + matrike +\begin_inset Formula $X$ +\end_inset + +, + z +\begin_inset Formula $i$ +\end_inset + + označimo vrstico. +\end_layout + +\end_inset + + bomo rešili s Cramerjevim pravilom. +\begin_inset Formula +\[ +x_{ij}=\left(\vec{x}_{j}\right)_{i}=\frac{\det A_{i}\left(\vec{i_{j}}\right)}{\det A}\overset{\text{razvoj po \ensuremath{j-}ti vrstici}}{=}\frac{\det A_{ji}\cdot\left(-1\right)^{j+i}}{\det A} +\] + +\end_inset + + +\begin_inset Formula +\[ +X=A^{-1}=\left[\begin{array}{ccc} +\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\ +\vdots & & \vdots\\ +\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A} +\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc} +\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\ +\vdots & & \vdots\\ +\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n} +\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T}, +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +kjer +\begin_inset Formula $\tilde{A}$ +\end_inset + + pravimo kofaktorska matrika. +\end_layout + +\begin_layout Subsection +Algebrske strukture +\end_layout + +\begin_layout Subsubsection +Uvod +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $M$ +\end_inset + + neprazna množica. + Operacija na +\begin_inset Formula $M$ +\end_inset + + pove, + kako iz dveh elementov +\begin_inset Formula $M$ +\end_inset + + dobimo nov element +\begin_inset Formula $M$ +\end_inset + +. + Na primer, + če +\begin_inset Formula $a,b\in M$ +\end_inset + +, + je +\begin_inset Formula $a\circ b$ +\end_inset + + nov element +\begin_inset Formula $M$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Operacija na +\begin_inset Formula $M$ +\end_inset + + je funkcija +\begin_inset Formula $\circ:M\times M\to M$ +\end_inset + +, + kjer je +\begin_inset Formula $M\times M$ +\end_inset + + kartezični produkt (urejeni pari). + +\begin_inset Formula $\left(a,b\right)\mapsto\circ\left(a,b\right)$ +\end_inset + +, + slednje pa označimo z +\begin_inset Formula $\circ\left(a,b\right)=a\circ b$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Na isti množici imamo lahko več različno definiranih operacij. + Ločimo jih tako, + da uvedemo pojem grupoida. +\end_layout + +\begin_layout Definition* +Grupoid je +\begin_inset Formula $\left(\text{neprazna množica},\text{izbrana operacija }\circ:M\times M\to M\right)$ +\end_inset + +. + Na primer +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Še posebej nas zanimajo operacije z lepimi lastnostmi, + denimo asociativnost, + komutativnost, + obstoj enot, + inverzov. +\end_layout + +\begin_layout Definition* +Grupoid, + katerega +\begin_inset Formula $\circ$ +\end_inset + + je asociativna +\begin_inset Formula $\Leftrightarrow\forall a,b,c\in M:\left(a\circ b\right)\circ c=a\circ\left(b\circ c\right)$ +\end_inset + +, + je polgrupa. + Tedaj skladnja dopušča pisanje brez oklepajev: + +\begin_inset Formula $a\circ b\circ c\circ d$ +\end_inset + + je nedvoumen/veljaven izraz, + ko je +\begin_inset Formula $\circ$ +\end_inset + + asociativna. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Komutativnost: + +\begin_inset Formula $\circ$ +\end_inset + + je komutativna +\begin_inset Formula $\Leftrightarrow\forall a,b\in M:a\circ b=b\circ a$ +\end_inset + +. + Grupoidom s komutativno operacijo pravimo, + da so komutativni. +\end_layout + +\begin_layout Example* +Asociativni in komutativni grupoidi (komutativne polgrupe): + +\begin_inset Formula $\left(\mathbb{N},\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{Q},\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{N},+\right)$ +\end_inset + + — + številske operacije. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Asociativni, + a ne komutativni grupoidi (nekomutativne polgrupe): + +\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$ +\end_inset + + — + množenje matrik. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Komutativni, + a ne asociativni grupoidi: + Jordanski produkt matrik: + +\begin_inset Formula $A\circ B=\frac{1}{2}\left(AB+BA\right)$ +\end_inset + +\SpecialChar endofsentence + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Niti komutativni niti asociativni grupoidi: + Vektorski produkt v +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + +: + +\begin_inset Formula $\left(\mathbb{R}^{3},\times\right)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $M\not=\emptyset$ +\end_inset + +. + +\begin_inset Formula $F$ +\end_inset + + naj bodo vse funkcije +\begin_inset Formula $M\to M$ +\end_inset + +, + +\begin_inset Formula $\circ$ +\end_inset + + pa kompozitum dveh funkcij. + Izkaže se, + da: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\left(F,\circ\right)$ +\end_inset + + je vedno polgrupa. +\end_layout + +\begin_deeper +\begin_layout Proof +Definicija kompozituma: + +\begin_inset Formula $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ +\end_inset + +. +\begin_inset Formula +\[ +\left(f\circ g\right)\circ h\overset{?}{=}f\circ\left(g\circ h\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall x:\left(\left(f\circ g\right)\circ h\right)\left(x\right)=\left(f\circ g\right)\left(h\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right) +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall x:\left(f\circ\left(g\circ h\right)\right)\left(x\right)=f\left(\left(g\circ h\right)\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right) +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Itemize +Čim ima +\begin_inset Formula $M$ +\end_inset + + vsaj tri elemente, + +\begin_inset Formula $\left(F,\circ\right)$ +\end_inset + + ni komutativna. +\end_layout + +\end_deeper +\begin_layout Definition* +Naj bo +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + grupoid. + Element +\begin_inset Formula $e\in M$ +\end_inset + + je enota, + če +\begin_inset Formula $\forall a\in M:e\circ a=a\wedge a\circ e=a$ +\end_inset + +. + Če velja le eno v konjunkciji, + je +\begin_inset Formula $e$ +\end_inset + + bodisi leva bodisi desna enota (respectively) in v takem primeru +\begin_inset Formula $e$ +\end_inset + + ni enota. +\end_layout + +\begin_layout Example* +Ali spodnji grupoidi imajo enoto in kakšna je? +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{R},+\right)$ +\end_inset + +: + enota je 0. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{N},\cdot\right)$ +\end_inset + +: + enota je 1. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{N},+\right)$ +\end_inset + +: + ni enote, + kajti +\begin_inset Formula $0\not\in\mathbb{N}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$ +\end_inset + +: + enota je +\begin_inset Formula $I_{n\times n}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Claim* +Vsak grupoid ima kvečjemu eno enoto. + Dve enoti v istem grupoidu sta enaki. + Še več: + vsaka leva enota je enaka vsaki desni enoti. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $e$ +\end_inset + + leva enota in +\begin_inset Formula $f$ +\end_inset + + desna enota, + torej +\begin_inset Formula $\forall a:e\circ a=a\wedge a\circ f=a$ +\end_inset + +. + Tedaj +\begin_inset Formula $e\circ f=f$ +\end_inset + + in +\begin_inset Formula $e\circ f=e$ +\end_inset + +. + Ker je vsaka leva enota vsaki desni, + sta poljubni enoti enaki. + Enota je, + če obstaja, + ena sama in je obenem edina leva in edina desna enota. +\end_layout + +\begin_layout Example* +Lahko se zgodi, + da obstaja poljubno različnih levih, + a nobene desne enote. + Primer so vse matrike oblike +\begin_inset Formula $\left[\begin{array}{cc} +a & b\\ +0 & 0 +\end{array}\right]$ +\end_inset + +. + Račun +\begin_inset Formula $\left[\begin{array}{cc} +a & b\\ +0 & 0 +\end{array}\right]\cdot\left[\begin{array}{cc} +c & d\\ +0 & 0 +\end{array}\right]=\left[\begin{array}{cc} +ac & ad\\ +0 & 0 +\end{array}\right]$ +\end_inset + + pokaže, + da so vsi elementi +\begin_inset Formula $\left[\begin{array}{cc} +1 & \times\\ +0 & 0 +\end{array}\right]$ +\end_inset + + leve enote. + Iz dejstva, + da je več (tu celo neskončno) levih enot, + sledi dejstvo, + da ni desnih. +\end_layout + +\begin_layout Definition* +Polgrupi z enoto pravimo monoid. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + monoid z enoto +\begin_inset Formula $e$ +\end_inset + +. + Inverz elementa +\begin_inset Formula $a\in M$ +\end_inset + + je tak +\begin_inset Formula $b\in M\ni:b\circ a=e\wedge a\circ b=e$ +\end_inset + +. + Elementu, + ki zadošča levi strani konjunkcije, + pravimo levi inverz +\begin_inset Formula $a$ +\end_inset + +, + elemetu, + ki zadošča desni strani konjunkcije, + pa desni inverz +\begin_inset Formula $a$ +\end_inset + +. + Inverz +\begin_inset Formula $a$ +\end_inset + + je torej tak element, + ki je hkrati levi in desni inverz +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Ni nujno, + da ima vsak element monoida inverz. + Primer je +\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$ +\end_inset + +; + niso vse matrike obrnljive. +\end_layout + +\begin_layout Claim* +Vsak element monoida ima kvečjemu en inverz. + Vsak levi inverz je enak vsakemu desnemu. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $b$ +\end_inset + + levi in +\begin_inset Formula $c$ +\end_inset + + desni inverz +\begin_inset Formula $a$ +\end_inset + +, + torej +\begin_inset Formula $b\circ a=e=a\circ c$ +\end_inset + +. + Računajmo: + +\begin_inset Formula $b=b\circ e=b\circ\left(a\circ c\right)=\left(b\circ a\right)\circ c=e\circ c=c$ +\end_inset + +. + Če obstaja, + je torej inverz en sam, + in ta je edini levi in edini desni inverz. +\end_layout + +\begin_layout Definition* +Ker vemo, + da je inverz enoličen, + lahko vpeljemo oznako +\begin_inset Formula $a^{-1}$ +\end_inset + + za inverz elementa +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Ali v spodnjih monoidih obstajajo inverzi in kakšni so? +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{Z},+\right)$ +\end_inset + +: + inverz +\begin_inset Formula $a$ +\end_inset + + je +\begin_inset Formula $-a$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{Z},\cdot\right)$ +\end_inset + +: + inverz +\begin_inset Formula $1$ +\end_inset + + je +\begin_inset Formula $1$ +\end_inset + +, + inverz +\begin_inset Formula $-1$ +\end_inset + + je +\begin_inset Formula $-1$ +\end_inset + +, + ostali elementi pa inverza nimajo. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$ +\end_inset + +: + inverz +\begin_inset Formula $a$ +\end_inset + + je +\begin_inset Formula $\frac{1}{a}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Remark* +Če desnega inverza ni, + je lahko levih inverzov več. + Primer: + Naj bodo +\begin_inset Formula $M$ +\end_inset + + vse funkcije +\begin_inset Formula $\mathbb{N}\to\mathbb{N}$ +\end_inset + + in naj bo +\begin_inset Formula $\circ$ +\end_inset + + kompozitum funkcij. + Tedaj velja: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $f\in M$ +\end_inset + + ima levi inverz +\begin_inset Formula $\Leftrightarrow f$ +\end_inset + + injektivna. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f\in M$ +\end_inset + + ima desni inverz +\begin_inset Formula $\Leftrightarrow f$ +\end_inset + + surjektivna. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $f\in M$ +\end_inset + + ima inverz +\begin_inset Formula $\Leftrightarrow f$ +\end_inset + + bijektivna. +\end_layout + +\end_deeper +\begin_layout Example* +\begin_inset Formula $f\left(n\right)=n+1$ +\end_inset + + je injektivna, + a ne surjektivna. + Vsi za komponiranje levi inverzi +\begin_inset Formula $f$ +\end_inset + + so funkcije oblike +\begin_inset Formula $g\left(x\right)=\begin{cases} +x-1 & ;x>1\\ +\times & ;x=1 +\end{cases}$ +\end_inset + + ZDB +\begin_inset Formula $x$ +\end_inset + + lahko slikajo v karkoli, + pa bo +\begin_inset Formula $\left(g\circ f\right)$ +\end_inset + + še vedno funkcija identiteta. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +V +\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$ +\end_inset + + je vsak levi inverz tudi desni inverz. + To je res tudi za funkcije na končni množici, + toda ni res v splošnem. +\end_layout + +\begin_layout Definition* +Grupa je tak monoid, + v katerem ima vsak element inverz. + Daljše: + grupa je taka neprazna množica +\begin_inset Formula $G$ +\end_inset + + z operacijo +\begin_inset Formula $\circ$ +\end_inset + +, + ki zadošča asociativnosti, + obstaja enota in za vsak element obstaja njegov inverz. + Grupi s komutativno operacijo pravimo Abelova grupa. +\end_layout + +\begin_layout Example* +Nekaj abelovih grup: + +\begin_inset Formula $\left(\mathbb{Z},+\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),+\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{R}^{n},+\right)$ +\end_inset + +. + Nekaj neabelovih grup: + +\family roman +\series medium +\shape up +\size normal +\emph off +\nospellcheck off +\bar no +\strikeout off +\xout off +\uuline off +\uwave off +\noun off +\color none +vse obrnljive matrike fiksne dimenzije +\family default +\series default +\shape default +\size default +\emph default +\nospellcheck default +\bar default +\strikeout default +\xout default +\uuline default +\uwave default +\noun default +\color inherit +, + +\family roman +\series medium +\shape up +\size normal +\emph off +\nospellcheck off +\bar no +\strikeout off +\xout off +\uuline off +\uwave off +\noun off +\color none +vse permutacije neprazne končne množice +\family default +\series default +\shape default +\size default +\emph default +\nospellcheck default +\bar default +\strikeout default +\xout default +\uuline default +\uwave default +\noun default +\color inherit +. +\end_layout + +\begin_layout Subsubsection +Podstrukture +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + grupoid. + Reciumi, + da je +\begin_inset Formula $N$ +\end_inset + + neprazna podmnožica +\begin_inset Formula $M$ +\end_inset + +. + Pod temi pogoji se lahko zgodi, + da +\begin_inset Formula $\exists a,b\in N\ni:a\circ b\not\in N$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Oglejmo si grupoid +\begin_inset Formula $\left(\mathbb{Z},+\right)$ +\end_inset + +. + +\begin_inset Formula $N\subseteq\mathbb{Z}$ +\end_inset + + naj bodo liha cela števila. + +\begin_inset Formula $\forall a,b\in N:a+b\not\in N\Rightarrow\exists a,b\in N\ni:a+b\not\in N$ +\end_inset + +, + kajti vsota lihih števil je soda. +\end_layout + +\begin_layout Definition* +Pravimo, + da je podmnožica +\begin_inset Formula $N\subseteq M$ +\end_inset + + zaprta za +\begin_inset Formula $\circ$ +\end_inset + +, + če +\begin_inset Formula $\forall a,b\in N:a\circ b\in N$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Oglejmo si spet grupoid +\begin_inset Formula $\left(\mathbb{Z},+\right)$ +\end_inset + +. + +\begin_inset Formula $N\subseteq\mathbb{Z}$ +\end_inset + + naj bodo soda cela števila. + +\begin_inset Formula $N$ +\end_inset + + je zaprta za +\begin_inset Formula $+$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Takemu +\begin_inset Formula $N$ +\end_inset + +, + kjer je +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + z implicitno podedovano operacijo ( +\begin_inset Formula $a\circ_{N}b=a\circ b$ +\end_inset + +) pravimo podgrupoid +\begin_inset Formula $\left(N,\circ_{N}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Exercise* +Pokaži, + da je +\begin_inset Quotes gld +\end_inset + +general linear +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A\not=0\right\} $ +\end_inset + + grupa za matrično množenje. +\end_layout + +\begin_deeper +\begin_layout Standard +Asociativnost je dokazana zgoraj. + Enota je +\begin_inset Formula $I_{n}$ +\end_inset + +. + Inverzi obstajajo, + ker so determinante neničelne in tudi inverzi imajo neničelne determinante. + Preveriti je treba še vsebovanost, + torej +\begin_inset Formula $\forall A,B\in GL_{n}\left(\mathbb{R}\right):A\cdot B\in GL_{n}\left(\mathbb{R}\right)$ +\end_inset + +. + Vzemimo poljubni +\begin_inset Formula $A,B\in GL_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + torej +\begin_inset Formula $\det A\not=0\wedge\det B\not=0$ +\end_inset + +. + +\begin_inset Formula $\det\left(AB\right)=\det A\det B=0\Leftrightarrow\det A=0\vee\det B=0$ +\end_inset + +, + toda ker noben izmed izrazov disjunkcije ne drži, + determinanta +\begin_inset Formula $AB$ +\end_inset + + nikdar ni 0. + Enota +\begin_inset Formula $I$ +\end_inset + + je vsebovana v +\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + saj +\begin_inset Formula $\det I=1\not=0$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Exercise* +Ali je +\begin_inset Quotes gld +\end_inset + +special linear +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A=1\right\} $ +\end_inset + + grupa za matrično množenje? +\end_layout + +\begin_deeper +\begin_layout Standard +Vse lastnosti (razen vsebovanosti) smo preverili zgoraj. + Preveriti je treba vsebovanost, + torej ali +\begin_inset Formula $\forall A,B\in SL_{n}\left(\mathbb{R}\right):A\cdot B\in SL_{n}\left(\mathbb{R}\right)$ +\end_inset + +. + Vzemimo poljubni +\begin_inset Formula $A,B\in SL_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + torej +\begin_inset Formula $\det A=1\wedge\det B=1$ +\end_inset + +. + +\begin_inset Formula $\det\left(AB\right)=\det A\det B=1\cdot1=1$ +\end_inset + +. + Preveriti je treba še, + da so inverzi vsebovani. + Za poljubno +\begin_inset Formula $A\in SL_{n}\left(\mathbb{R}\right)$ +\end_inset + + je +\begin_inset Formula $\det A^{-1}=\frac{1}{\det A}=1$ +\end_inset + +, + ker je +\begin_inset Formula $\det A=1$ +\end_inset + +. + Enota +\begin_inset Formula $I$ +\end_inset + + je vsebovana v +\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + saj +\begin_inset Formula $\det I=1$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Fact* +Za podedovano operacijo +\begin_inset Formula $\circ_{N}$ +\end_inset + + v podstrukturi se asociativnost in komutativnost podedujeta, + ni pa nujno, + da če obstaja enota v +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + +, + obstaja enota tudi v +\begin_inset Formula $\left(N,\circ_{N}\right)$ +\end_inset + +. + Prav tako ni rečeno, + da se podeduje obstoj inverzov. +\end_layout + +\begin_layout Definition* +Če je +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + polgrupa (asociativen grupoid) in +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + pravimo, + da je +\begin_inset Formula $N$ +\end_inset + + podpolgrupa, + če je zaprta za +\begin_inset Formula $\circ$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:podmonoid" + +\end_inset + +Če je +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + monoid (polgrupa z enoto) in +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + je +\begin_inset Formula $N$ +\end_inset + + podmonoid, + če je zaprt za +\begin_inset Formula $\circ$ +\end_inset + + in vsebuje enoto iz +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + (prav tisto enoto, + glej primer +\begin_inset CommandInset ref +LatexCommand ref +reference "exa:nxnmonoid" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + spodaj). +\end_layout + +\begin_layout Example* +\begin_inset Formula $\left(\mathbb{N},\cdot\right)$ +\end_inset + + je monoid. + Soda števila so podpolgrupa (zaprta so za množenje), + niso pa podmonoid, + saj ne vsebujejo enice (enote). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example +\begin_inset CommandInset label +LatexCommand label +name "exa:nxnmonoid" + +\end_inset + + +\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$ +\end_inset + + je monoid za operacijo +\begin_inset Formula $\left(a,b\right)\circ\left(c,d\right)=\left(ac,bd\right)$ +\end_inset + +, + saj je enota +\begin_inset Formula $\left(1,1\right)$ +\end_inset + +. + +\begin_inset Formula $\left(\mathbb{N}\times\left\{ 0\right\} ,\circ\right)$ +\end_inset + + pa za +\begin_inset Formula $\circ$ +\end_inset + + kot prej je sicer podpolgrupa v +\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$ +\end_inset + + in ima enoto +\begin_inset Formula $\left(1,0\right)$ +\end_inset + +, + vendar, + ker +\begin_inset Formula $\left(1,0\right)\not=\left(1,1\right)$ +\end_inset + +, + to ni podmonoid. + Enota mora torej biti, + kot pravi definicija +\begin_inset CommandInset ref +LatexCommand ref +reference "def:podmonoid" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +, + ista kot enota v +\begin_inset Quotes gld +\end_inset + +starševski +\begin_inset Quotes grd +\end_inset + + strukturi. +\end_layout + +\begin_layout Definition* +Če je +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + grupa in +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + pravimo, + da je +\begin_inset Formula $N$ +\end_inset + + podgrupa +\begin_inset Formula $\Longleftrightarrow$ +\end_inset + + hkrati velja +\end_layout + +\begin_deeper +\begin_layout Itemize +je zaprta za +\begin_inset Formula $\circ$ +\end_inset + +, +\end_layout + +\begin_layout Itemize +vsebuje isto enoto kot +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + in +\end_layout + +\begin_layout Itemize +vsebuje inverz vsakega svojega elementa; + ti inverzi pa so itak po enoličnosti enaki inverzom iz +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Example* +special linear, + +\begin_inset Formula $SL_{n}$ +\end_inset + +, + grupa vseh matrik z determinanto enako 1, + je podgrupa +\begin_inset Quotes gld +\end_inset + +general linear +\begin_inset Quotes grd +\end_inset + +, + +\begin_inset Formula $GL_{n}$ +\end_inset + +, + grupe vseh obrnljivih +\begin_inset Formula $n\times n$ +\end_inset + + matrik, + kajti +\begin_inset Formula $\det I=1$ +\end_inset + +, + +\begin_inset Formula $\det$ +\end_inset + + je multiplikativna (glej vajo zgoraj) in +\begin_inset Formula $\det A=1\Leftrightarrow\det A^{-1}=1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +ortogonalne matrike, + +\begin_inset Formula $O_{n}$ +\end_inset + +, + vse +\begin_inset Formula $n\times n$ +\end_inset + + matrike +\begin_inset Formula $A$ +\end_inset + +, + ki zadoščajo +\begin_inset Formula $A^{T}A=I$ +\end_inset + +, + je podgrupa +\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + kajti: +\end_layout + +\begin_deeper +\begin_layout Itemize +Je zaprta: +\begin_inset Formula +\[ +A,B\in O_{n}\overset{?}{\Longrightarrow}AB\in O_{n} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(AB\right)^{T}\left(AB\right)\overset{?}{=}I +\] + +\end_inset + + +\begin_inset Formula +\[ +B^{T}\left(A^{T}A\right)B\overset{?}{=}I +\] + +\end_inset + + +\begin_inset Formula +\[ +I=I +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Vsebuje enoto +\begin_inset Formula $I$ +\end_inset + +: +\begin_inset Formula +\[ +I^{T}I=I +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Vsebuje inverze vseh svojih elementov: + Uporabimo +\begin_inset Formula $A^{T}A=I\Rightarrow A^{T}=A^{-1}$ +\end_inset + + +\begin_inset Formula +\[ +A\in O_{n}\overset{?}{\Longrightarrow}A^{-1}\in O_{n} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(A^{-1}\right)^{T}A^{-1}\overset{?}{=}I +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(A^{T}\right)^{T}A^{T}=AA^{T}=I +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Fact* +specialna ortogonalna grupa, + +\begin_inset Formula $SO_{n}\coloneqq O_{n}\cap SL_{n}$ +\end_inset + + je podgrupa +\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$ +\end_inset + +. + Dokazati je moč še bolj splošno, + namreč, + da je presek dveh podgrup spet podgrupa. + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +DOKAŽI????? +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + + grupa in +\begin_inset Formula $N\subseteq M$ +\end_inset + + neprazna. + Tedaj velja +\begin_inset Formula $N$ +\end_inset + + podgrupa +\begin_inset Formula $\Leftrightarrow\forall a,b\in N:a\circ b^{-1}\in N$ +\end_inset + + (zaprtost za odštevanje — + v abelovih grupah namreč običajno operacijo označimo s +\begin_inset Formula $+$ +\end_inset + + in označimo +\begin_inset Formula $a+b^{-1}=a-b$ +\end_inset + +). +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Naj bo +\begin_inset Formula $N$ +\end_inset + + podgrupa v +\begin_inset Formula $\left(M,\circ\right)$ +\end_inset + +. + Vzemimo +\begin_inset Formula $a,b\in N$ +\end_inset + +. + Upoštevamo +\begin_inset Formula $b\in N\Rightarrow b^{-1}\in N$ +\end_inset + + iz definicije podgrupe. + Torej velja +\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ b^{-1}\in N$ +\end_inset + +, + zopet iz definicije podgrupe. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Naj +\begin_inset Formula $\forall a,b\in N:a\circ b^{-1}\in N$ +\end_inset + +. + Preverimo lastnosti iz definicije podgrupe: +\end_layout + +\begin_deeper +\begin_layout Itemize +Vsebovanost enote: + Ker je +\begin_inset Formula $N$ +\end_inset + + neprazna, + vsebuje nek +\begin_inset Formula $a$ +\end_inset + +. + Po predpostavki je +\begin_inset Formula $a\circ a^{-1}\in N$ +\end_inset + +, + +\begin_inset Formula $a\circ a^{-1}$ +\end_inset + + pa je po definiciji inverza enota. +\end_layout + +\begin_layout Itemize +Vsebovanost inverzov: + Naj bo +\begin_inset Formula $a\in N$ +\end_inset + + poljuben. + Od prej vemo, + da +\begin_inset Formula $e\in N$ +\end_inset + +. + Po predpostavki, + ker +\begin_inset Formula $e,a\in N\Rightarrow e\circ a^{-1}\in N$ +\end_inset + +, + +\begin_inset Formula $e\circ a^{-1}$ +\end_inset + + pa je po definiciji enote +\begin_inset Formula $a^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Zaprtost: + Naj bosta +\begin_inset Formula $a,b\in N$ +\end_inset + + poljubna. + Od prej vemo, + da +\begin_inset Formula $b^{-1}\in N$ +\end_inset + +. + Po predpostavki, + ker +\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ\left(b^{-1}\right)^{-1}\in N$ +\end_inset + +, + +\begin_inset Formula $a\circ\left(b^{-1}\right)^{-1}$ +\end_inset + + pa je po definiciji inverza +\begin_inset Formula $a\circ b$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsubsection +Homomorfizmi +\end_layout + +\begin_layout Standard +\begin_inset Formula $\sim$ +\end_inset + + so operacije, + ki +\begin_inset Quotes gld +\end_inset + +ohranjajo strukturo +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Definition* +Naj bosta +\begin_inset Formula $\left(M_{1},\circ_{1}\right)$ +\end_inset + + in +\begin_inset Formula $\left(M_{2},\circ_{2}\right)$ +\end_inset + + dva grupoida. + Preslikava +\begin_inset Formula $f:M_{1}\to M_{2}$ +\end_inset + + je homomorfizem grupoidov, + če +\begin_inset Formula $\forall a,b\in M_{1}:f\left(a\circ_{1}b\right)=f\left(a\right)\circ_{2}f\left(b\right)$ +\end_inset + +. + Enaka definicija v polgrupah. + Za homomorfizem monoidov zahtevamo še, + da +\begin_inset Formula $f\left(e_{1}\right)=e_{2}$ +\end_inset + +, + kjer je +\begin_inset Formula $e_{1}$ +\end_inset + + enota +\begin_inset Formula $M_{1}$ +\end_inset + + in +\begin_inset Formula $e_{2}$ +\end_inset + + enota +\begin_inset Formula $M_{2}$ +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ +\end_inset + +, + ki slika +\begin_inset Formula $a\mapsto\left(a,0\right)$ +\end_inset + +. + +\begin_inset Formula $\circ_{1}$ +\end_inset + + naj bo množenje, + +\begin_inset Formula $\circ_{2}$ +\end_inset + + pa +\begin_inset Formula $\left(a,b\right)\circ_{2}\left(c,d\right)=\left(ac,bd\right)$ +\end_inset + + (množenje po komponentah). + +\begin_inset Formula $\left(1,1\right)$ +\end_inset + + je enota v +\begin_inset Formula $\mathbb{N\times\mathbb{N}}$ +\end_inset + +, + +\begin_inset Formula $1$ +\end_inset + + pa je enota v +\begin_inset Formula $\mathbb{N}$ +\end_inset + +. + +\begin_inset Formula $f$ +\end_inset + + je homomorfizem, + ker +\begin_inset Formula $f\left(a\circ_{1}b\right)=\left(a\cdot b,0\right)=\left(a,0\right)\circ_{2}\left(b,0\right)=f\left(a\right)\circ_{2}f\left(b\right)$ +\end_inset + +, + ni pa homomorfizem monoidov, + saj +\begin_inset Formula $f\left(1\right)=\left(1,0\right)\not=\left(1,1\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Za homomorfizem grup zahtevamo še, + da +\begin_inset Formula $f\left(a^{-1}\right)=f\left(a\right)^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Izkaže se, + da ohranjanje enote in inverzov pri homomorfizmih grup sledi že iz definicije homomorfizmov grupoidov. +\end_layout + +\begin_layout Claim* +Naj bosta +\begin_inset Formula $\left(M_{1},\circ_{1}\right)$ +\end_inset + + in +\begin_inset Formula $\left(M_{2},\circ_{2}\right)$ +\end_inset + + grupi. + Naj bo +\begin_inset Formula $f:M_{1}\to M_{2}$ +\end_inset + + preslikava, + ki je homomorfizem grupoidov. + Trdimo, + da slika enoto v enoto in inverze v inverze. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $e_{1}$ +\end_inset + + enota za +\begin_inset Formula $\left(M_{1},\circ_{1}\right)$ +\end_inset + + in +\begin_inset Formula $e_{2}$ +\end_inset + + enota za +\begin_inset Formula $\left(M_{2},\circ_{2}\right)$ +\end_inset + +. + Dokažimo, + da +\begin_inset Formula $f\left(e_{1}\right)\overset{?}{=}e_{2}$ +\end_inset + +. +\begin_inset Formula +\[ +f\left(e_{1}\right)=f\left(e_{1}\circ_{1}e_{1}\right)=f\left(e_{1}\right)\circ_{2}f\left(e_{1}\right)=f\left(e_{1}\right)^{-1}\circ f\left(e_{1}\right)\circ e_{2}=e_{2}\circ e_{2}=e_{2} +\] + +\end_inset + +Dokažimo še ohranjanje inverzov, + se pravi +\begin_inset Formula $b$ +\end_inset + + je inverz +\begin_inset Formula $a\overset{?}{\Longrightarrow}f\left(b\right)$ +\end_inset + + je inverz +\begin_inset Formula $f\left(a\right)$ +\end_inset + +. +\begin_inset Formula +\[ +a\circ_{1}b=e_{1}\overset{?}{\Longrightarrow}f\left(a\right)\circ_{2}f\left(b\right)=f\left(a\circ_{1}b\right)=f\left(e_{1}\right)=e_{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +b\circ_{1}a=e_{1}\overset{?}{\Longrightarrow}f\left(b\right)\circ_{2}f\left(a\right)=f\left(b\circ_{1}a\right)=f\left(e_{1}\right)=e_{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Example +\begin_inset CommandInset label +LatexCommand label +name "exa:primeri-homomorfizmov" + +\end_inset + +Primeri homomorfizmov. +\end_layout + +\begin_deeper +\begin_layout Enumerate +Determinanta: + +\begin_inset Formula $M_{n}\left(\mathbb{R}\right)\to\mathbb{R}$ +\end_inset + + je homomorfizem, + ker ima multiplikativno lastnost: + +\begin_inset Formula $\det\left(AB\right)=\det A\det B$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:permutacijska-matrika" + +\end_inset + + +\begin_inset Formula $S_{n}$ +\end_inset + + so vse permutacije množice +\begin_inset Formula $\left\{ 1..n\right\} $ +\end_inset + +. + Vsaki permutaciji +\begin_inset Formula $\sigma\in S_{n}$ +\end_inset + + priredimo permutacijsko matriko +\begin_inset Formula $P_{\sigma}\in M_{n}\left(\mathbb{R}\right)$ +\end_inset + + tako, + da vsebuje vektorje standardne baze +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + kot stolpce: +\begin_inset Formula +\[ +P_{\sigma}\coloneqq\left[\begin{array}{ccc} +\vec{e_{\sigma\left(1\right)}} & \cdots & \vec{e_{\sigma\left(n\right)}}\end{array}\right] +\] + +\end_inset + +Imamo preslikavo +\begin_inset Formula $S\to M_{n}\left(\mathbb{R}\right)$ +\end_inset + +, + ki slika +\begin_inset Formula $\sigma\mapsto P_{\sigma}$ +\end_inset + + in trdimo, + da je homomorfizem. + Dokažimo, + da je +\begin_inset Formula $\forall\sigma,\tau\in S_{n}:P_{\sigma\circ\tau}=P_{\sigma}\cdot P_{\tau}$ +\end_inset + +. + Opazimo, + da je +\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{i}}=\vec{e_{\sigma\left(i\right)}}$ +\end_inset + + (tu množimo matriko z vektorjem). + Če namesto +\begin_inset Formula $i$ +\end_inset + + pišemo +\begin_inset Formula $\tau\left(i\right)$ +\end_inset + +, + dobimo +\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{\tau\left(i\right)}}=\vec{e_{\left(\sigma\circ\tau\right)\left(i\right)}}$ +\end_inset + +. + Preverimo sedaj množenje +\begin_inset Formula $P_{\sigma}P_{\tau}=P_{\sigma}\left[\begin{array}{ccc} +\vec{e_{\tau\left(1\right)}} & \cdots & \vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc} +P_{\sigma}\vec{e_{\tau\left(1\right)}} & \cdots & P_{\sigma}\vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc} +\vec{e_{\left(\sigma\circ\tau\right)\left(1\right)}} & \cdots & \vec{e_{\left(\sigma\circ\tau\right)\left(n\right)}}\end{array}\right]=P_{\sigma\circ\tau}$ +\end_inset + +. + Preslikava je res homomorfizem. +\end_layout + +\end_deeper +\begin_layout Claim* +Kompozitum dveh homomorfizmov je tudi sam zopet homomorfizem. +\end_layout + +\begin_layout Proof +Imejmo tri grupoide in homomorfizma, + ki slikata med njimi takole: + +\begin_inset Formula $\left(M_{1},\circ_{1}\right)\overset{f}{\longrightarrow}\left(M_{2},\circ_{2}\right)\overset{g}{\longrightarrow}\left(M_{3},\circ_{3}\right)$ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $g\circ f$ +\end_inset + + spet homorfizem. +\begin_inset Formula +\[ +\left(g\circ f\right)\left(a\circ_{1}b\right)=g\left(f\left(a\circ_{1}b\right)\right)=g\left(f\left(a\right)\circ_{2}f\left(b\right)\right)=g\left(f\left(a\right)\right)\circ_{3}g\left(f\left(b\right)\right)=\left(g\circ f\right)\left(a\right)\circ_{3}\left(g\circ f\right)\left(b\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $S_{n}\overset{\sigma}{\longrightarrow}M_{n}\left(\mathbb{R}\right)\overset{\det}{\rightarrow}\mathbb{R}$ +\end_inset + +, + kjer je +\begin_inset Formula $\sigma$ +\end_inset + + preslikava iz točke +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:permutacijska-matrika" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + zgleda +\begin_inset CommandInset ref +LatexCommand ref +reference "exa:primeri-homomorfizmov" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + zgoraj. + +\begin_inset Formula $\sgn=\det\circ\sigma$ +\end_inset + +, + kjer je +\begin_inset Formula $\sgn$ +\end_inset + + parnost permutacije. + Preslikava +\begin_inset Formula $\sgn$ +\end_inset + + je homomorfizem, + ker je kompozitum dveh homomorfizmov. +\end_layout + +\begin_layout Definition* +Izomorfizem je preslikava, + ki je bijektivna in je homomorfizem. + Dve grupi sta izomorfni, + kadar med njima obstaja izomorfizem. +\end_layout + +\begin_layout Remark* +S stališča algebre sta dve izomorfni grupi v abstraktnem smislu enaki, + saj je izomorfizem zgolj reverzibilno preimenovanje elementov. +\end_layout + +\begin_layout Subsubsection +Bigrupoidi, + polkolobarji, + kolobarji +\end_layout + +\begin_layout Definition* +Neprazni množici +\begin_inset Formula $M$ +\end_inset + + z dvema operacijama +\begin_inset Formula $\circ_{1}$ +\end_inset + + in +\begin_inset Formula $\circ_{2}$ +\end_inset + + pravimo bigrupoid in ga označimo z +\begin_inset Formula $\left(M,\circ_{1},\circ_{2}\right)$ +\end_inset + +. + Običajno operaciji označimo z +\begin_inset Formula $+,\cdot$ +\end_inset + +, + tedaj bigrupoid pišemo kot +\begin_inset Formula $\left(M,+,\cdot\right)$ +\end_inset + +. +\end_layout + +\begin_layout Quotation +\begin_inset Quotes gld +\end_inset + +Če +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + + ena z drugo nimata nobene zveze, + je vseeno, + če ju študiramo skupaj ali posebej. +\begin_inset Quotes grd +\end_inset + + +\end_layout + +\begin_layout Definition* +Distributivnost je značilnost bigrupoida +\begin_inset Formula $\left(M,+,\cdot\right)$ +\end_inset + +. + Ločimo levo distributivnost: + +\begin_inset Formula $\forall a,b,c\in M:a\cdot\left(b+c\right)=a\cdot b+a\cdot c$ +\end_inset + + in desno distributivnost: + +\begin_inset Formula $\forall a,b,c\in M:\left(a+b\right)\cdot c=a\cdot c+b\cdot c$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Bigrupoid, + ki zadošča levi in desni distributivnosti, + je distributiven. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Distributiven bigrupoid, + je polkolobar, + če je +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + komutativna polgrupa. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Distributiven grupoid je kolobar, + če je +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + komutativna grupa. +\end_layout + +\begin_layout Example* +Primer polkolobarja, + ki ni kolobar, + je +\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$ +\end_inset + +. + Ni enote niti inverza za +\begin_inset Formula $+$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{N},+\right)$ +\end_inset + + pa je polgrupa. +\end_layout + +\begin_layout Standard +Kolobarje delimo glede na lastnosti operacije +\begin_inset Formula $\cdot$ +\end_inset + +: +\end_layout + +\begin_layout Definition* +Asociativen kolobar je tak, + kjer je +\begin_inset Formula $\cdot$ +\end_inset + + asociativna operacija +\begin_inset Formula $\sim\left(M,\cdot\right)$ +\end_inset + + je polgrupa. +\end_layout + +\begin_layout Example* +Primer kolobarja, + ki ni asociativen, + je +\begin_inset Formula $\left(\mathbb{R}^{3},+,\times\right)$ +\end_inset + +, + kjer je +\begin_inset Formula $\times$ +\end_inset + + vektorski produkt. + Primer kolobarja, + ki je asociativen, + je +\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$ +\end_inset + +, + kjer je +\begin_inset Formula $\cdot$ +\end_inset + + matrično množenje. +\end_layout + +\begin_layout Definition* +Asociativen kolobar z enoto je tak, + ki ima multiplikativno enoto, + torej enoto za drugo operacijo +\begin_inset Formula $\sim\left(M,\cdot\right)$ +\end_inset + + je monoid. + Tipično se enoto za +\begin_inset Formula $\cdot$ +\end_inset + + označi z 1, + enoto za +\begin_inset Formula $+$ +\end_inset + + pa z 0. +\end_layout + +\begin_layout Example* +Primer asociativnega kolobarja brez enote je +\begin_inset Formula $\left(\text{soda }\mathbb{N},+,\cdot\right)$ +\end_inset + +. + Primer asociativnega kolobarja z enoto je +\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +\begin_inset Formula $b$ +\end_inset + + je inverz +\begin_inset Formula $a$ +\end_inset + +, + če +\begin_inset Formula $b\cdot a=e$ +\end_inset + + in +\begin_inset Formula $a\cdot b=e$ +\end_inset + +, + kjer je +\begin_inset Formula $e$ +\end_inset + + multiplikativna enota kolobarja. +\end_layout + +\begin_layout Remark* +Element 0 nima nikoli inverza, + ker +\begin_inset Formula $\forall a\in M:0\cdot a=0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\cancel{0\cdot a}=\left(0+0\right)\cdot a=0\cdot a+\cancel{0\cdot a}$ +\end_inset + + (dokaz velja za kolobarje, + ne pa polkolobarje, + ker imamo pravilo krajšanja +\begin_inset Foot +status open + +\begin_layout Plain Layout +Dokaz v mojih Odgovorih na vprašanja za ustni izpit Diskretnih struktur 2 IŠRM +\end_layout + +\end_inset + + le, + kadar je +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + grupa). +\end_layout + +\begin_layout Definition* +Asociativen kolobar z enoto, + v katerem ima vsak neničen element inverz, + je obseg. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Kolobar je komutativen, + če je +\begin_inset Formula $\cdot$ +\end_inset + + komutativna operacija ( +\begin_inset Formula $+$ +\end_inset + + je itak po definiciji že komutativna). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Komutativen obseg je polje. +\end_layout + +\begin_layout Example* +Primeri polj: + +\begin_inset Formula $\left(\mathbb{Q},+,\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$ +\end_inset + +, + +\begin_inset Formula $\left(F\left[\mathbb{R}\right],+,\cdot\right)$ +\end_inset + +, + kjer je +\begin_inset Formula $F\left[\mathbb{R}\right]$ +\end_inset + + polje racionalnih funkcij. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Primer obsega, + ki ni polje: + +\begin_inset Formula $\left(\mathbb{H},+,\cdot\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Kvaternioni so +\begin_inset Formula $M_{2\times2}\left(\mathbb{R}\right)$ +\end_inset + + take oblike: + za +\begin_inset Formula $\alpha,\beta\in\mathbb{C}$ +\end_inset + + je +\begin_inset Formula $\mathbb{H}\coloneqq\left[\begin{array}{cc} +\alpha & \beta\\ +-\overline{\beta} & \overline{\alpha} +\end{array}\right]=\left[\begin{array}{cc} +a+bi & c+di\\ +-c+di & a-bi +\end{array}\right]=\left[\begin{array}{cc} +1 & 0\\ +0 & 1 +\end{array}\right]a+\left[\begin{array}{cc} +i & 0\\ +0 & -i +\end{array}\right]b+\left[\begin{array}{cc} +0 & 1\\ +-1 & 0 +\end{array}\right]c+\left[\begin{array}{cc} +0 & i\\ +i & 0 +\end{array}\right]d=1a+bi+cj+dk$ +\end_inset + + za +\begin_inset Formula $a,b,c,d\in\mathbb{R}$ +\end_inset + + in dimenzije +\begin_inset Formula $1,i,j,k$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primer kolobarja: + Naj bo +\begin_inset Formula $X$ +\end_inset + + neprazna množica in +\begin_inset Formula $R$ +\end_inset + + kolobar. + +\begin_inset Formula $R^{X}$ +\end_inset + + so vse funkcije +\begin_inset Formula $X\to R$ +\end_inset + +. + Naj bosta +\begin_inset Formula $f,g\in R^{X}$ +\end_inset + +. + Definirajmo operaciji: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $+$ +\end_inset + + +\begin_inset Formula $f+g\coloneqq\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\cdot$ +\end_inset + + +\begin_inset Formula $f\cdot g\coloneqq\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Subsubsection +Podkolobarji +\end_layout + +\begin_layout Definition* +Podbigrupoid od +\begin_inset Formula $\left(M,+,\cdot\right)$ +\end_inset + + je taka podmnožica +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + ki je zaprta za +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + + ZDB +\begin_inset Formula $N\subseteq M$ +\end_inset + + je podgrupoid v +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + in +\begin_inset Formula $\left(M,\cdot\right)$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Podkolobar kolobarja +\begin_inset Formula $\left(M,+,\cdot\right)$ +\end_inset + + je taka podmnožica +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + da je +\begin_inset Formula $N$ +\end_inset + + podgrupa v +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + in +\begin_inset Formula $N$ +\end_inset + + podgrupoid v +\begin_inset Formula $\left(N,\cdot\right)\Leftrightarrow N$ +\end_inset + + zaprta za +\begin_inset Formula $\cdot$ +\end_inset + +. + Skrajšana definicija je torej, + da je +\begin_inset Formula $\forall a,b\in N:a+b^{-1}\in N\wedge a\cdot b\in N$ +\end_inset + +, + torej zaprtost za odštevanje in množenje. +\end_layout + +\begin_layout Example* +Primeri podkolobarjev +\end_layout + +\begin_deeper +\begin_layout Itemize +v +\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Itemize +zgornjetrikotne matrike +\end_layout + +\begin_layout Itemize +diagonalne matrike +\end_layout + +\begin_layout Itemize +matrike s spodnjo vrstico ničelno +\end_layout + +\begin_layout Itemize +matrike z ničelnim +\begin_inset Formula $i-$ +\end_inset + +tim stolpcem +\end_layout + +\begin_layout Itemize +\begin_inset Formula $M_{n}\left(\mathbb{Z}\right)$ +\end_inset + +, + +\begin_inset Formula $M_{n}\left(\mathbb{Q}\right)$ +\end_inset + + +\end_layout + +\begin_layout Itemize +matrike oblike +\begin_inset Formula $\left[\begin{array}{cc} +a & b\\ +b & a +\end{array}\right]$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Itemize +v +\begin_inset Formula $\left(\mathbb{R}^{[a,b]},+,\cdot\right)$ +\end_inset + + (vse funkcije +\begin_inset Formula $\left[a,b\right]\to\mathbb{R}$ +\end_inset + + za seštevanje in množenje) +\end_layout + +\begin_deeper +\begin_layout Itemize +vse omejene funkcije +\end_layout + +\begin_layout Itemize +vse zvezne funkcije +\end_layout + +\begin_layout Itemize +vse odvedljive funkcije +\end_layout + +\end_deeper +\end_deeper +\begin_layout Definition* +Podobseg obsega +\begin_inset Formula $\left(M,+,\cdot\right)$ +\end_inset + + je taka +\begin_inset Formula $N\subseteq M$ +\end_inset + +, + da velja: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $N$ +\end_inset + + podgrupa v +\begin_inset Formula $\left(M,+\right)$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $N\setminus\left\{ 0\right\} $ +\end_inset + + podgrupa v +\begin_inset Formula $\left(M\setminus\left\{ 0\right\} ,\cdot\right)$ +\end_inset + + +\end_layout + +\begin_layout Standard +ZDB: + +\begin_inset Formula $N$ +\end_inset + + je zaprta za odštevanje (seštevanje z aditivnim inverzom) in za deljenje (množenje z multiplikativnim inverzom) z neničelnimi elementi. +\end_layout + +\end_deeper +\begin_layout Example* +Primeri podobsegov: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\mathbb{R}$ +\end_inset + + je podobseg v +\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\mathbb{Q}$ +\end_inset + + je podobseg v +\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Izkaže se, + da je najmanjše podpolje v +\begin_inset Formula $\mathbb{R}$ +\end_inset + +, + ki vsebuje +\begin_inset Formula $\mathbb{Q}$ +\end_inset + + in +\begin_inset Formula $\sqrt{3}$ +\end_inset + + množica +\begin_inset Formula $\left\{ a+b\sqrt{3};\forall a,b\in\mathbb{Q}\right\} $ +\end_inset + +. + Očitno je zaprt za odštevanje. + Za deljenje? +\end_layout + +\begin_deeper +\begin_layout Standard +\begin_inset Formula +\[ +\frac{a+b\sqrt{3}}{c+d\sqrt{3}}=\frac{\left(a+b\sqrt{3}\right)\left(c-d\sqrt{3}\right)}{\left(c+d\sqrt{3}\right)\left(c-d\sqrt{3}\right)}=\frac{ac-ad\sqrt{3}+bc\sqrt{3}-3bd}{c^{2}-3d^{2}}=\frac{ac-3bd+\left(bc-ad\right)\sqrt{3}}{c^{2}-3d^{2}}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\frac{ac-3bd}{c^{2}-3d^{2}}+\frac{bc-ad}{c^{2}-3d^{2}}\sqrt{3} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Subsubsection +Homomorfizmi kolobarjev +\end_layout + +\begin_layout Definition* +Naj bosta +\begin_inset Formula $\left(M_{1},+_{1},\cdot_{1}\right)$ +\end_inset + + in +\begin_inset Formula $\left(M_{2},+_{2},\cdot_{2}\right)$ +\end_inset + + kolobarja. + +\begin_inset Formula $f:M_{1}\to M_{2}$ +\end_inset + + je homomorfizem kolobarjev +\begin_inset Formula $\Leftrightarrow\forall a,b\in M_{1}:f\left(a+_{1}b\right)=f\left(a\right)+_{2}f\left(b\right)\wedge f\left(a\cdot_{1}b\right)=f\left(a\right)\cdot_{2}f\left(b\right)$ +\end_inset + +. + ZDB +\begin_inset Formula $f$ +\end_inset + + mora biti homomorfizem grupoidov +\begin_inset Formula $\left(M_{1},+_{1}\right)\to\left(M_{2},+_{2}\right)$ +\end_inset + + in +\begin_inset Formula $\left(M_{1},\cdot_{1}\right)\to\left(M_{2},\cdot_{2}\right)$ +\end_inset + +. + Za homomorfizem kolobarjev z enoto zahtevamo še +\begin_inset Formula $f\left(1_{1}\right)=1_{2}$ +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $f:M_{2}\left(\mathbb{R}\right)\to M_{3}\left(\mathbb{R}\right)$ +\end_inset + + s predpisom +\begin_inset Formula $\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]\mapsto\left[\begin{array}{ccc} +a & b & 0\\ +c & d & 0\\ +0 & 0 & 0 +\end{array}\right]$ +\end_inset + + je homomorfizem kolobarjev, + ni pa homomorfizem kolobarjev z enoto, + kajti +\begin_inset Formula $f\left(\left[\begin{array}{cc} +1 & 0\\ +0 & 1 +\end{array}\right]\right)=\left[\begin{array}{ccc} +1 & 0 & 0\\ +0 & 1 & 0\\ +0 & 0 & 0 +\end{array}\right]$ +\end_inset + +, + kar ni enota v +\begin_inset Formula $M_{3}\left(\mathbb{R}\right)$ +\end_inset + + ( +\begin_inset Formula $I_{3}$ +\end_inset + +) za implicitni operaciji +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $g:M_{n}\left(\mathbb{R}\right)\to M_{n}\left(\mathbb{R}\right)$ +\end_inset + + ki slika +\begin_inset Formula $A\mapsto S^{-1}AS$ +\end_inset + +, + kjer je +\begin_inset Formula $S$ +\end_inset + + neka fiksna obrnljiva matrika v +\begin_inset Formula $M_{n}\left(\mathbb{R}\right)$ +\end_inset + +. + Uporabimo implicitni operaciji +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + + za matrike. + Računa +\begin_inset Formula $g\left(A+B\right)=S^{-1}\left(A+B\right)S=S^{-1}AS+S^{-1}BS=g\left(A\right)+g\left(B\right)$ +\end_inset + + in +\begin_inset Formula $g\left(AB\right)=S^{-1}ABS=S^{-1}AIBS=S^{-1}ASS^{-1}BS=g\left(A\right)g\left(B\right)$ +\end_inset + + pokažeta, + da je +\begin_inset Formula $g$ +\end_inset + + homomorfizem kolobarjev, + celo z enoto, + kajti +\begin_inset Formula $g\left(I\right)=S^{-1}IS=S^{-1}S=I$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $h:\mathbb{C}\to M_{n}\left(\mathbb{R}\right)$ +\end_inset + + s predpisom +\begin_inset Formula $\alpha+\beta i\to\left[\begin{array}{cc} +\alpha & \beta\\ +-\beta & \alpha +\end{array}\right]$ +\end_inset + + je homomorfizem kolobarjev z enoto. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Kolobar ostankov +\begin_inset Formula $\mathbb{Z}_{n}\coloneqq\left\{ 0..\left(n-1\right)\right\} $ +\end_inset + + je asociativni kolobar z enoto. + Če je +\begin_inset Formula $p$ +\end_inset + + praštevilo, + pa je celo +\begin_inset Formula $\mathbb{Z}_{p}$ +\end_inset + + polje za implicitni operaciji seštevanje in množenja po modulu. +\end_layout + +\begin_layout Subsection +Vektorski prostori +\end_layout + +\begin_layout Standard +Ideja: + Vektorski prostor je Abelova grupa z dodatno strukturo — + množenje s skalarjem. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $\left(F,+,\cdot\right)$ +\end_inset + + polje. + Vektorski prostor z operacijama +\begin_inset Formula $V+V\to V$ +\end_inset + + in +\begin_inset Formula $F\cdot V\to V$ +\end_inset + + nad +\begin_inset Formula $F$ +\end_inset + + je taka +\begin_inset Formula $\left(V,+,\cdot\right)$ +\end_inset + +, + da velja: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\left(V,+\right)$ +\end_inset + + je abelova grupa: + komutativnost, + asociativnost, + enota, + aditivni inverzi +\end_layout + +\begin_layout Enumerate +Lastnosti množenja s skalarjem. + +\begin_inset Formula $\forall\alpha,\beta\in F,a,b\in V:$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\alpha\left(a+b\right)=\alpha a+\alpha b$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(\alpha+\beta\right)a=\alpha a+\beta a$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(\alpha\cdot\beta\right)\cdot a=\alpha\cdot\left(\beta\cdot a\right)$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $1\cdot a=a$ +\end_inset + + +\end_layout + +\begin_layout Standard +Alternativna abstraktna formulacija aksiomov množenja s skalarjem se glasi: +\end_layout + +\begin_layout Standard +\begin_inset Formula $\forall\alpha\in F$ +\end_inset + + priredimo preslikavo +\begin_inset Formula $\varphi_{\alpha}:V\to V$ +\end_inset + +, + ki pošlje +\begin_inset Formula $v\mapsto\alpha v$ +\end_inset + +. + Štiri zgornje aksiome množenja s skalarjem sedaj označimo z abstraktnimi formulacijami: +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\varphi_{\alpha}\left(a+b\right)\overset{\text{def.}}{=}\alpha\left(a+b\right)=\alpha b+\alpha b\overset{\text{def.}}{=}\varphi_{\alpha}\left(a\right)+\varphi_{\alpha}\left(b\right)$ +\end_inset + + — + vidimo, + da je +\begin_inset Formula $\varphi_{\alpha}$ +\end_inset + + homomorfizem iz +\begin_inset Formula $\left(V,+\right)$ +\end_inset + + v +\begin_inset Formula $\left(V,+\right)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\varphi_{\alpha+\beta}\left(a\right)\overset{\text{def.}}{=}\left(\alpha+\beta\right)a=\alpha a+\beta a\overset{\text{def.}}{=}\varphi_{\alpha}a+\varphi_{\beta}a$ +\end_inset + + — + torej +\begin_inset Formula $\varphi_{\alpha+\beta}=\varphi_{\alpha}+\varphi_{\beta}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\varphi_{\alpha\beta}a\overset{\text{def.}}{=}\left(\alpha\beta\right)a=\alpha\left(\beta a\right)\overset{\text{def.}}{=}\varphi_{\alpha}\left(\varphi_{\beta}\left(a\right)\right)=\left(\varphi_{\alpha}\circ\varphi_{\beta}\right)\left(a\right)$ +\end_inset + + — + torej +\begin_inset Formula $\varphi_{\alpha\beta}=\varphi_{\alpha}\circ\varphi_{\beta}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\varphi_{1}a\overset{\text{def.}}{=}1a=a$ +\end_inset + + — + torej +\begin_inset Formula $\varphi_{1}=id$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO ALTERNATIVNA DEFINICIJA VEKTORSKEGA PROSTORA Z GRUPO ENDOMORFIZMOV +\end_layout + +\end_inset + + +\end_layout + +\end_deeper +\end_deeper +\begin_layout Remark* +Če v definiciji vektorskega prostora zamenjamo polje +\begin_inset Formula $F$ +\end_inset + + s kolobarjem +\begin_inset Formula $F$ +\end_inset + +, + dobimo definicijo +\series bold +modula +\series default +nad +\begin_inset Formula $F$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri vektorskih prostorov: +\end_layout + +\begin_deeper +\begin_layout Itemize +standarden primer: + naj bo +\begin_inset Formula $F$ +\end_inset + + pojle in +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + +. + Naj bo +\begin_inset Formula $V=F^{n}$ +\end_inset + +, + +\begin_inset Formula $+$ +\end_inset + + seštevanje po komponentah in +\begin_inset Formula $\cdot$ +\end_inset + + množenje s skalarjem po komponentah. + Pod temi pogoji je +\begin_inset Formula $\left(V,+,\cdot\right)$ +\end_inset + + vektorski prostor — + ustreza vsem osmim aksiomom. +\end_layout + +\begin_layout Itemize +Naj bo +\begin_inset Formula $F$ +\end_inset + + polje in +\begin_inset Formula $n,m\in\mathbb{N}$ +\end_inset + +. + Naj bo +\begin_inset Formula $V\coloneqq M_{m,n}\left(\mathbb{F}\right)=m\times n$ +\end_inset + + matrike nad +\begin_inset Formula $F$ +\end_inset + +. + +\begin_inset Formula $+$ +\end_inset + + in +\begin_inset Formula $\cdot$ +\end_inset + + definiramo kot pri matrikah. +\end_layout + +\begin_layout Itemize +Naj bo +\begin_inset Formula $F$ +\end_inset + + polje, + +\begin_inset Formula $S\not=\emptyset$ +\end_inset + + množica. + Naj bo +\begin_inset Formula $V\coloneqq F^{S}$ +\end_inset + + (vse funkcije +\begin_inset Formula $S\to F$ +\end_inset + +). + Naj bosta +\begin_inset Formula $\varphi,\tau:S\to F$ +\end_inset + +. + Definirajmo +\begin_inset Formula $\forall s\in S$ +\end_inset + + operaciji +\begin_inset Formula $\left(\varphi+\tau\right)\left(s\right)=\varphi\left(s\right)+\tau\left(s\right)$ +\end_inset + + in +\begin_inset Formula $\left(\varphi\cdot\tau\right)\left(s\right)=\varphi\left(s\right)\cdot\tau\left(s\right)$ +\end_inset + +. + Tedaj je +\begin_inset Formula $V$ +\end_inset + + vektorski prostor. + Ta definicija je podobna kot definiciji z +\begin_inset Formula $n-$ +\end_inset + +terico elementov polja, + saj lahko +\begin_inset Formula $n-$ +\end_inset + +terico identificiramo s funkcijo +\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} \to F$ +\end_inset + +, + toda ta primer dovoli neskončno razsežne vektorske prostore, + saj +\begin_inset Formula $S$ +\end_inset + + ni nujno končna, + +\begin_inset Formula $n-$ +\end_inset + +terica pa nekako implicitno je, + saj +\begin_inset Formula $n\in\mathbb{N}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Polinomi. + Naj bo +\begin_inset Formula $V\coloneqq F\left[x\right]$ +\end_inset + + (polinomi v spremenljivki +\begin_inset Formula $x$ +\end_inset + + s koeficienti v +\begin_inset Formula $F$ +\end_inset + +). + Seštevanje definirajmo po komponentah: + +\begin_inset Formula $\left(\alpha+\beta x+\gamma x^{2}\right)+\left(\pi+\tau x\right)=\left(\alpha+\pi+\left(\beta+\tau\right)x+\gamma x^{2}\right)$ +\end_inset + +, + množenje s skalarjem pa takole: + +\begin_inset Formula $\alpha\left(a+bx+cx^{2}\right)=\alpha a+\alpha bx+\alpha cx^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Naj bosta +\begin_inset Formula $V_{1}$ +\end_inset + + in +\begin_inset Formula $V_{2}$ +\end_inset + + dva vektorska prostora nad istim poljem +\begin_inset Formula $F$ +\end_inset + +. + Tvorimo nov vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + +, + ki mu pravimo +\begin_inset Quotes gld +\end_inset + +direktna vsota +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula $V_{1}$ +\end_inset + + in +\begin_inset Formula $V_{2}$ +\end_inset + + in ga označimo z +\begin_inset Formula $V_{1}\oplus V_{2}\coloneqq$ +\end_inset + + +\begin_inset Formula $\left\{ \left(v_{1},v_{2}\right);\forall v_{1}\in V_{1},v_{2}\in V:2\right\} $ +\end_inset + +. + Seštevamo po komponentah: + +\begin_inset Formula $\left(v_{1},v_{2}\right)+\left(v_{1}',v_{2}'\right)=\left(v_{1}+v_{1}',v_{2}+v_{2}'\right)$ +\end_inset + +, + s skalarjem pa množimo prvi komponento: + +\begin_inset Formula $\forall\alpha\in F:\alpha\left(v_{1},v_{2}\right)=\left(\alpha v_{1},v_{2}\right)$ +\end_inset + +. + Definicijo lahko posplošimo na +\begin_inset Formula $n$ +\end_inset + + vektorskih prostorov. + Tedaj so elementi prostora urejene +\begin_inset Formula $n-$ +\end_inset + +terice. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Podprostori vekrorskih prostorov — + vektorski podprostori +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $\left(V,+,\cdot\right)$ +\end_inset + + vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + +. + Vektorski podprostor je taka neprazna podmnožica +\begin_inset Formula $V$ +\end_inset + +, + ki je zaprta za seštevanje in množenje s skalarjem. + Natančneje: + +\begin_inset Formula $\left(W,+,\cdot\right)$ +\end_inset + + je vektorski podprostor +\begin_inset Formula $\left(V,+,\cdot\right)\Longleftrightarrow$ +\end_inset + + velja hkrati: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $W\subseteq V$ +\end_inset + + in +\begin_inset Formula $W\not=\emptyset$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:zaprtost+" + +\end_inset + + +\begin_inset Formula $\forall a,b\in W:a+b\in W$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:zaprtostskalar" + +\end_inset + + +\begin_inset Formula $\forall a\in W,\alpha\in F:\alpha a\in W$ +\end_inset + + +\end_layout + +\begin_layout Standard +Lastnosti +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:zaprtost+" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + in +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:zaprtostskalar" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + je moč združiti v eno: + +\begin_inset Formula $\forall a_{i},a_{2}\in W,\alpha_{1},\alpha_{2}\in F:\alpha_{1}a_{1}+\alpha_{2}a_{2}\in W$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Z drugimi besedami je vektorski podprostor taka podmnožica, + ki vsebuje vse linearne kombinacije svojih elementov. + Odštevanje +\begin_inset Formula $a-b$ +\end_inset + + je poseben primer linearne kombinacije, + kajti +\begin_inset Formula $a_{1}-a_{2}=1a_{1}+\left(-1\right)a_{2}$ +\end_inset + +. + Sledi, + da mora biti +\begin_inset Formula $\left(W,+\right)$ +\end_inset + + podgrupa +\begin_inset Formula $\left(V,+\right)$ +\end_inset + +, + torej taka podmnožica +\begin_inset Formula $V$ +\end_inset + +, + ki je zaprta za odštevanje. +\end_layout + +\end_deeper +\begin_layout Example* +Primeri vektorskih podprostorov: +\end_layout + +\begin_deeper +\begin_layout Itemize +Naj bo +\begin_inset Formula $V=\mathbb{R}^{2}$ +\end_inset + + (ravnina). + Vsi vektorski podprostori +\begin_inset Formula $V$ +\end_inset + + so premice, + ki gredo skozi izhodišče, + izhodišče samo in cela ravnina. + Slednja sta t. + i. + trivialna podprostora. +\end_layout + +\end_deeper +\begin_layout Remark* +\begin_inset Formula $\forall\left(V,+,\cdot\right)$ +\end_inset + + vektorski prostor +\begin_inset Formula $:\left\{ 0\right\} ,V$ +\end_inset + + sta vektorska podprostora. + Imenujemo ju trivialna vektorska podprostora. +\end_layout + +\begin_layout Claim* +Vsak podprostor vsebuje aditivno enoto 0. +\end_layout + +\begin_layout Proof +Po definiciji je vsak vektorski podprostor neprazen, + torej +\begin_inset Formula $\exists w\in W$ +\end_inset + +. + Polje gotovo vsebuje aditivno enoto 0, + torej po aksiomu +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:zaprtostskalar" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + za podprostore sledi +\begin_inset Formula $0\cdot w\in W$ +\end_inset + +. + Dokažimo +\begin_inset Formula $0\cdot w\overset{?}{=}0$ +\end_inset + +: + +\begin_inset Formula $\cancel{0\cdot w}=\left(0+0\right)\cdot w=0\cdot w+\cancel{0\cdot w}$ +\end_inset + + (pravilo krajšanja v grupi), + torej +\begin_inset Formula $0=0\cdot w$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Množica rešitev homogene (desna stran je 0) linearne enačbe je vselej vektorski podprostor. +\end_layout + +\begin_layout Proof +Imamo +\begin_inset Formula $\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n}=0$ +\end_inset + +. + Če sta +\begin_inset Formula $\vec{a}=\left(a_{1},\dots,a_{n}\right)$ +\end_inset + + in +\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{n}\right)$ +\end_inset + + rešitvi, + velja +\begin_inset Formula $\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}=0$ +\end_inset + + in +\begin_inset Formula $\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}=0$ +\end_inset + +. + Vzemimo poljubna +\begin_inset Formula $\alpha,\beta\in F$ +\end_inset + + in si oglejmo +\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}$ +\end_inset + +: +\begin_inset Formula +\[ +\alpha\left(\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}\right)+\beta\left(\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}\right)=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +\alpha_{1}\left(\alpha a_{1}+\beta b_{1}\right)+\cdots+\alpha_{n}\left(\alpha a_{n}+\beta b_{n}\right)=0 +\] + +\end_inset + +Vzemimo koeficiente v oklepajih pred +\begin_inset Formula $\alpha_{i}$ +\end_inset + + v enačbi pred to vrstico in jih zložimo v vektor. + Tedaj je +\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}=\left(\alpha a_{1}+\beta b_{1},\dots,\alpha a_{n}+\beta b_{n}\right)$ +\end_inset + + spet rešitev homogene linearne enačbe. + Ker je linearna kombinacija elementov vektorskega podprostora spet element vektorskega podprostora, + je po definiciji množica rešitev homogene linearne enačbe res vselej vektorski podprostor. +\end_layout + +\begin_layout Remark* +Podoben računa velja tudi za množico rešitev sistema linearnih enačb, + kar sicer sledi tudi iz naslednje trditve. +\end_layout + +\begin_layout Claim* +Presek dveh podprostorov je tudi sam spet podprostor. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $W_{1},W_{2}$ +\end_inset + + podprostora v +\begin_inset Formula $V$ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $W_{1}\cap V_{2}$ +\end_inset + + spet podprostor. + Vzemimo poljubna +\begin_inset Formula $a,b\in W_{1}\cap W_{2}$ +\end_inset + + in poljubna +\begin_inset Formula $\alpha,\beta\in F$ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$ +\end_inset + +. + Vemo, + da +\begin_inset Formula $a,b\in W_{1}$ +\end_inset + + in +\begin_inset Formula $a,b\in W_{2}$ +\end_inset + +. + Ker je podprostor po definiciji zaprt za linearne kombinacije svojih elementov, + je +\begin_inset Formula $\alpha a+\beta b\in W_{1}$ +\end_inset + + in +\begin_inset Formula $\alpha a+\beta b\in W_{2}$ +\end_inset + +, + torej +\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$ +\end_inset + +, + torej je presek podprostorov res zaprt za LK svojih elementov in je s tem tudi sam podprostor. +\end_layout + +\begin_deeper +\begin_layout Remark* +Slednji dokaz lahko očitno posplošimo na več podprostorov. + Presek nikdar ni prazen, + saj vsi podprostori vsebujejo aditivno enoto 0 (dokaz za to je malce višje). +\end_layout + +\end_deeper +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Vsota-podprostorov" + +\end_inset + +Vsota podprostorov +\end_layout + +\begin_layout Definition* +Naj bosta +\begin_inset Formula $W_{1}$ +\end_inset + + in +\begin_inset Formula $W_{2}$ +\end_inset + + podprostora v +\begin_inset Formula $V$ +\end_inset + +. + Vsoto podprostorov +\begin_inset Formula $W_{1}$ +\end_inset + + in +\begin_inset Formula $W_{2}$ +\end_inset + + označimo z +\begin_inset Formula $W_{1}+W_{2}=\left\{ w_{1}+w_{w};\forall w_{1}\in W_{1},w_{2}\in W_{2}\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Vsota podprostorov je tudi sama spet podprostor. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $a,b\in W_{1}+W_{2}$ +\end_inset + + poljubna. + Tedaj po definiciji +\begin_inset Formula $a=a_{1}+a_{2}$ +\end_inset + +, + kjer +\begin_inset Formula $a_{1}\in W_{1}$ +\end_inset + + in +\begin_inset Formula $a_{2}\in W_{2}$ +\end_inset + +, + in +\begin_inset Formula $b=b_{1}+b_{2}$ +\end_inset + +, + kjer +\begin_inset Formula $b_{1}\in W_{1}$ +\end_inset + + in +\begin_inset Formula $b_{2}\in W_{2}$ +\end_inset + +. + +\begin_inset Formula $\forall\alpha,\beta\in F$ +\end_inset + +: +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +\alpha a+\beta b=\alpha\left(a_{1}+a_{2}\right)+\beta\left(b_{1}+b_{2}\right)=\alpha a_{1}+\alpha a_{2}+\beta b_{1}+\beta b_{2}=\left(\alpha a_{1}+\beta b_{1}\right)+\left(\alpha a_{2}+\beta b_{2}\right)\in W_{1}+W_{2}, +\] + +\end_inset + +kajti +\begin_inset Formula $\left(\alpha a_{1}+\beta b_{1}\right)\in W_{1}$ +\end_inset + + in +\begin_inset Formula $\left(\alpha a_{2}+\beta b_{2}\right)\in W_{2}$ +\end_inset + +, + saj sta to linearni kombinaciji elementov prostorov. + Njuna vsota pa je element +\begin_inset Formula $W_{1}+W_{2}$ +\end_inset + + po definiciji vsote podprostorov. +\end_layout + +\begin_layout Subsubsection +Baze +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad poljem +\begin_inset Formula $F$ +\end_inset + +. + Množica +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + je baza, + če je LN in če je ogrodje. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + je LN, + če za vsake +\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$ +\end_inset + +, + ki zadoščajo +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$ +\end_inset + + velja +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$ +\end_inset + +. + Ekvivalentni definiciji LN: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + je LN +\begin_inset Formula $\Leftrightarrow\forall v\in V$ +\end_inset + + se da kvečjemu na en način izraziti kot linearno kombinacijo +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + je LN +\begin_inset Formula $\Leftrightarrow\nexists v\in\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + +, + da bi se ga dalo izraziti kot LK preostalih elementov. +\end_layout + +\begin_layout Standard +Dokaz ekvivalentnosti teh definicij je enak tistemu za +\begin_inset Formula $V=\mathbb{R}^{n}$ +\end_inset + + višje. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + je ogrodje +\begin_inset Formula $\Leftrightarrow\forall v\in V$ +\end_inset + + se da na vsaj en način izraziti kot LK te množice +\begin_inset Formula $\Leftrightarrow\Lin\left\{ v_{1},\dots,v_{n}\right\} =V$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri baz: +\end_layout + +\begin_deeper +\begin_layout Itemize +standardna baza: + Naj bo +\begin_inset Formula $V=F^{n}$ +\end_inset + +. + +\begin_inset Formula $v_{1}=\left(1,0,0,\dots,0,0\right)$ +\end_inset + +, + +\begin_inset Formula $v_{2}=\left(0,1,0,\dots,0,0\right)$ +\end_inset + +, + ..., + +\begin_inset Formula $v_{n}=\left(0,0,0,\dots,0,1\right)$ +\end_inset + +. + Da je +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} \subseteq F^{n}$ +\end_inset + + res baza, + preverimo z determinanto ( +\begin_inset Formula $\det A\not=0\Leftrightarrow\exists A^{-1}\Leftrightarrow$ +\end_inset + + stolpci so baza prostora): +\begin_inset Formula +\[ +\det\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]=0\Leftrightarrow\left\{ v_{1},\dots,v_{n}\right\} \text{ \textbf{ni} baza} +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +baze v +\begin_inset Formula $F\left[x\right]_{<n}$ +\end_inset + + (polinomi stopnje, + manjše od +\begin_inset Formula $n$ +\end_inset + +) +\end_layout + +\begin_deeper +\begin_layout Itemize +standardna baza: + +\begin_inset Formula $\left\{ 1,x,x^{2},x^{3},\dots,x^{n-1}\right\} $ +\end_inset + + +\end_layout + +\begin_layout Itemize +vzemimo paroma različne +\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$ +\end_inset + + in definirajmo +\begin_inset Formula $p_{i}\left(x\right)=\left(x-\alpha_{1}\right)\cdots\left(x-\alpha_{i-1}\right)\left(x-\alpha_{i+1}\right)\cdots\left(x-\alpha_{n}\right)$ +\end_inset + + za vsak +\begin_inset Formula $i\in\left\{ 1..n\right\} $ +\end_inset + +, + kar je polimom stopnje +\begin_inset Formula $n-1$ +\end_inset + +. +\begin_inset Formula $\left\{ \alpha_{1}p_{1}\left(x\right),\dots,\alpha_{n}p_{n}\left(x\right)\right\} $ +\end_inset + + je baza za +\begin_inset Formula $F\left[x\right]_{<n}$ +\end_inset + +. + +\end_layout + +\begin_deeper +\begin_layout Proof +Dokazujemo, + da so LN in ogrodje: +\end_layout + +\begin_layout Itemize +LN: + +\begin_inset Formula $\beta_{1}p_{i}\left(x\right)+\cdots+\beta_{n}p_{n}\left(x\right)=0\overset{?}{\Longrightarrow}\beta_{1}=\cdots=\beta_{n}=0$ +\end_inset + +. + Opazimo, + da +\begin_inset Formula $p_{i}\left(\alpha_{j}\right)=0\Leftrightarrow i=j$ +\end_inset + +. + Torej če za +\begin_inset Formula $x$ +\end_inset + + vstavimo katerikoli +\begin_inset Formula $\alpha_{i}$ +\end_inset + +, + bodo vsi členi 0, + razen +\begin_inset Formula $\beta_{i}p_{i}\left(x\right)$ +\end_inset + +. + Ker pa +\begin_inset Formula $\alpha_{i}$ +\end_inset + + ni ničla +\begin_inset Formula $p_{i}\left(x\right)$ +\end_inset + +, + je +\begin_inset Formula $\beta_{i}=0$ +\end_inset + +, + čim je +\begin_inset Formula $\beta_{i}p_{i}\left(x\right)=0$ +\end_inset + +. + Preverjati je treba le +\begin_inset Formula $\alpha_{i}$ +\end_inset + +, + ker je dovolj najti eno vrednost spremenljivke, + v kateri se vrednosti polinomov ne ujemajo, + da lahko rečemo, + da polinomi niso isti. +\end_layout + +\begin_layout Itemize +ogrodje: + Trdimo, + da za vsak polimom velja formula +\begin_inset Formula $f\left(x\right)=\frac{f\left(\alpha_{1}\right)}{p_{1}\left(\alpha_{1}\right)}p_{1}\left(x\right)+\cdots+\frac{f\left(\alpha_{n}\right)}{p_{n}\left(\alpha_{n}\right)}p_{n}\left(x\right)$ +\end_inset + +. + Obe strani enačbe imata stopnjo največ +\begin_inset Formula $n-1$ +\end_inset + + in se ujemata v +\begin_inset Formula $n$ +\end_inset + + različnih točkah. + Če za +\begin_inset Formula $x$ +\end_inset + + vstavimo +\begin_inset Formula $\alpha_{i}$ +\end_inset + + za vsak +\begin_inset Formula $i$ +\end_inset + +, + dobimo 0 v vseh členih, + razen v +\begin_inset Formula $i-$ +\end_inset + +tem, + kjer se vrednost +\begin_inset Formula $f\left(x\right)$ +\end_inset + + ujema z vrednostjo +\begin_inset Formula $f\left(\alpha_{1}\right)$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\end_deeper +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Obstoj-baze" + +\end_inset + +Obstoj baze +\end_layout + +\begin_layout Standard +Omejimo se na končno razsežne vektorske prostore. +\end_layout + +\begin_layout Definition* +Vektorski prostor je končno razsežen, + če ima končno ogrodje: + +\begin_inset Formula $\exists n\in\mathbb{N}\exists v_{1},\dots,v_{n}\ni:V=\Lin\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Theorem* +obstoj baze. + Vsak končno razsežen vektorski prostor ima vsaj eno bazo. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVP in naj bo +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + njegovo ogrodje. + Ker ogrodje ni nujno LN, + naj bo +\begin_inset Formula $S$ +\end_inset + + minimalna/najmanjša podmnožica +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + +, + ki je še ogrodje za +\begin_inset Formula $V$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $S$ +\end_inset + + baza za +\begin_inset Formula $V$ +\end_inset + +. + Po konstrukciji je ogrodje, + dokažimo še, + da je LN: + PDDRAA +\begin_inset Formula $S$ +\end_inset + + je linearno odvisna. + Tedaj +\begin_inset Formula $\exists v_{i}\in S\ni:v_{i}$ +\end_inset + + je LK +\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $ +\end_inset + + ogrodje manjše moči, + kar bi bilo v protislovju s predpostavko. + Tedaj obstajajo koeficienti, + da velja +\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + +. + Vzemimo poljuben +\begin_inset Formula $v\in V$ +\end_inset + +. + Ker je +\begin_inset Formula $S$ +\end_inset + + ogrodje +\begin_inset Formula $V$ +\end_inset + +, + obstajajo neki koeficienti +\begin_inset Formula $\beta_{1},\dots,\beta_{n}$ +\end_inset + +, + da velja +\begin_inset Formula +\[ +v=\beta_{1}v_{1}+\cdots+\beta_{i}v_{i}+\cdots+\beta_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{i}\left(\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}\right)+\cdots+\beta_{n}v_{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left(\beta_{1}+\beta_{i}\alpha_{1}\right)v_{1}+\cdots+\left(\beta_{i-1}+\beta_{i}\alpha_{i-1}\right)v_{i-1}+\left(\beta_{i+1}+\beta_{i}\alpha_{i+1}\right)v_{i+1}+\cdots+\left(\beta_{n}+\beta_{i}\alpha_{n}\right)v_{n} +\] + +\end_inset + +To pa je +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +, + saj je bilo rečeno, + da je +\begin_inset Formula $S$ +\end_inset + + najmanjše ogrodje, + mi pa smo razvili poljuben +\begin_inset Formula $v$ +\end_inset + + po manjšem ogrodju. + Torej ima vsak KRVP bazo in vsako ogrodje ima podmnožico, + ki je baza. +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "enoličnost-moči-baze." + +\end_inset + +enoličnost moči baze. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVP z +\begin_inset Formula $n-$ +\end_inset + +elementno bazo. + Tedaj velja vse to: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + LN množica +\begin_inset Formula $A$ +\end_inset + + v +\begin_inset Formula $V$ +\end_inset + + ima +\begin_inset Formula $\leq n$ +\end_inset + + elementov +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + ogrodje v +\begin_inset Formula $V$ +\end_inset + + ima +\begin_inset Formula $\geq n$ +\end_inset + + elementov +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + baza v +\begin_inset Formula $V$ +\end_inset + + ima +\begin_inset Formula $n$ +\end_inset + + elementov +\end_layout + +\end_deeper +\begin_layout Proof +Dokaz je dolg. +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:Vsak-poddoločen-homogen" + +\end_inset + +Vsak poddoločen homogen sistem linearnih enačb ima netrivialno rešitev. +\end_layout + +\begin_deeper +\begin_layout Proof +Dokaz se nahaja pod identično trditvijo +\begin_inset CommandInset ref +LatexCommand vref +reference "claim:Vpoddol-hom-sist-ima-ne0-reš" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:ln<=ogr" + +\end_inset + +Če je +\begin_inset Formula $u_{1},\dots,u_{m}$ +\end_inset + + LN množica v +\begin_inset Formula $V$ +\end_inset + + in +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + ogrodje za +\begin_inset Formula $V$ +\end_inset + +, + je +\begin_inset Formula $m\leq n$ +\end_inset + +. + ZDB moč katerekoli LN množice je manjša ali enaka od kateregakoli ogrodja v +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Proof +RAAPDD +\begin_inset Formula $u_{1},\dots,u_{m}$ +\end_inset + + je LN, + +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + je ogrodje in +\begin_inset Formula $m>n$ +\end_inset + +. + Iščemo protislovje. + Vsakega od +\begin_inset Formula $u_{i}$ +\end_inset + + lahko razvijemo po +\begin_inset Formula $v$ +\end_inset + +. +\begin_inset Formula +\[ +\begin{array}{ccccccc} +u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\ +\vdots & & \vdots & & & & \vdots\\ +u_{m} & = & \alpha_{m1}v_{1} & + & \cdots & + & \alpha_{mn}v_{n} +\end{array} +\] + +\end_inset + + +\begin_inset Formula $\forall i\in\left\{ 1..m\right\} $ +\end_inset + + pomnožimo +\begin_inset Formula $i-$ +\end_inset + +to enačbo s skalarjem +\begin_inset Formula $x_{i}$ +\end_inset + + in jih seštejmo. + +\begin_inset Formula $\vec{x}$ +\end_inset + + so abstraktne spremenljivke. + Tedaj: +\begin_inset Formula +\[ +x_{1}u_{1}+\cdots+x_{m}u_{m}=x_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+x_{m}\left(\alpha_{m1}v_{1}+\cdots+\alpha_{mn}v_{n}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=v_{1}\left(\alpha_{11}x_{1}+\cdots+\alpha_{m1}x_{m}\right)+\cdots+v_{n}\left(\alpha_{1n}x_{1}+\cdots+\alpha_{mn}x_{m}\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Izenačimo koeficiente za +\begin_inset Formula $v_{i}$ +\end_inset + + z 0 in dobimo poddoločen homogen sistem enačb (ima +\begin_inset Formula $n$ +\end_inset + + enačb in +\begin_inset Formula $m$ +\end_inset + + spremenljivk, + po predpostavki pa velja +\begin_inset Formula $m>n$ +\end_inset + +): +\begin_inset Formula +\[ +\begin{array}{ccccccc} +\alpha_{11}x_{1} & + & \cdots & + & \alpha_{m1}x_{m} & = & 0\\ +\vdots & & & & \vdots & & \vdots\\ +\alpha_{1n}x_{1} & + & \cdots & + & \alpha_{mn}x_{m} & = & 0 +\end{array} +\] + +\end_inset + +Po lemi +\begin_inset CommandInset ref +LatexCommand vref +reference "lem:Vsak-poddoločen-homogen" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + ima ta sistem netrivialno rešitev, + recimo +\begin_inset Formula $\left(\mu_{1},\dots,\mu_{m}\right)$ +\end_inset + +. + Če to rešitev vstavimo v +\begin_inset Formula $u_{1}x_{1}+\cdots+u_{m}x_{m}$ +\end_inset + +, + dobimo +\begin_inset Formula $u_{1}\mu_{1}+\cdots+u_{m}\mu_{m}=0$ +\end_inset + +. + Ker so +\begin_inset Formula $u_{1},\dots,u_{m}$ +\end_inset + + LN, + so +\begin_inset Formula $\mu_{1}=\cdots=\mu_{m}=0$ +\end_inset + +, + kar je v +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + s predpostavko. +\end_layout + +\end_deeper +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + baza je ogrodje +\begin_inset Formula $\Rightarrow$ +\end_inset + + po lemi +\begin_inset CommandInset ref +LatexCommand vref +reference "lem:ln<=ogr" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + velja, + da ima vsaka LN množica manj ali enako elementov kot vsako ogrodje, + torej tudi manj ali enako kot +\begin_inset Formula $n$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + baza je LN +\begin_inset Formula $\Rightarrow$ +\end_inset + + po lemi +\begin_inset CommandInset ref +LatexCommand vref +reference "lem:ln<=ogr" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + velja, + da ima vsako ogrodje več ali enako elementov kot vsaka LN, + torej tudi več ali enako kot +\begin_inset Formula $n$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Sledi iz zgornjih dveh točk, + saj je baza tako ogrodje kot LN hkrati. +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVP. + Njegova dimenzija, + +\begin_inset Formula $\dim V$ +\end_inset + +, + je moč baze v +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $\dim F^{n}=n$ +\end_inset + +, + +\begin_inset Formula $\dim M_{m\times n}\left(\mathbb{F}\right)=m\cdot n$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Dopolnitev LN množice do baze +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor z dimenzijo +\begin_inset Formula $n$ +\end_inset + +. + Trdimo, + da +\end_layout + +\begin_deeper +\begin_layout Enumerate +ima vsaka LN množica +\begin_inset Formula $\leq n$ +\end_inset + + elementov, +\end_layout + +\begin_layout Enumerate +je vsaka LN množica v +\begin_inset Formula $V$ +\end_inset + + z +\begin_inset Formula $n$ +\end_inset + + elementi baza, +\end_layout + +\begin_layout Enumerate +lahko vsako LN množico v +\begin_inset Formula $V$ +\end_inset + + dopolnimo do baze. +\end_layout + +\end_deeper +\begin_layout Proof +Dokaz je dolg +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:večja-ln" + +\end_inset + +Če so +\begin_inset Formula $v_{1},\dots,v_{m}\in V$ +\end_inset + + LN in če +\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + +, + potem so tudi +\begin_inset Formula $v_{1},\dots,v_{m},v_{m+1}$ +\end_inset + + LN. +\end_layout + +\begin_deeper +\begin_layout Proof +Naj velja +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0$ +\end_inset + + za nek +\begin_inset Formula $\vec{\alpha}\in F^{m+1}$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\vec{a}=\vec{0}$ +\end_inset + +. + Če +\begin_inset Formula $\alpha_{m+1}=0$ +\end_inset + +, + sledi +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m}v_{m}=0$ +\end_inset + +, + ker pa so po predpostavki +\begin_inset Formula $v_{1},\dots,v_{m}$ +\end_inset + + LN, + je +\begin_inset Formula $\vec{\alpha}=\vec{0}$ +\end_inset + +. + Sicer pa, + če PDDRAA +\begin_inset Formula $\alpha_{m+1}\not=0$ +\end_inset + +, + lahko z +\begin_inset Formula $a_{m+1}$ +\end_inset + + delimo: +\begin_inset Formula +\[ +\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +\alpha_{m+1}v_{m+1}=-\alpha_{1}v_{1}-\cdots-\alpha_{m}v_{m} +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{m+1}=\frac{-\alpha_{1}}{\alpha_{m+1}}v_{1}+\cdots+\frac{-\alpha_{m}}{\alpha_{m+1}}v_{m} +\] + +\end_inset + +Tedaj pridemo do +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +, + saj smo +\begin_inset Formula $v_{m+1}$ +\end_inset + + izrazili kot LK +\begin_inset Formula $\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + +, + po predpostavki pa je vendar +\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +že dokazano z dokazom trditve +\begin_inset CommandInset ref +LatexCommand vref +reference "enoličnost-moči-baze." +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + v razdelku +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Obstoj-baze" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{Vsaka LN množica v +\backslash +ensuremath{V} z +\backslash +ensuremath{n} elementi je baza.} +\end_layout + +\end_inset + + PDDRAA +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + je LN, + ki ni baza. + Tedaj +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + ni ogrodje. + Tedaj +\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} \not=V$ +\end_inset + +. + Zatorej +\begin_inset Formula $\exists v_{n+1}\in V\ni:\left\{ v_{1},\dots,v_{n},v_{n+1}\right\} $ +\end_inset + + je LN, + kar je v +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + s trditvijo, + da ima vsaka +\begin_inset Formula $LN$ +\end_inset + + množica v +\begin_inset Formula $V$ +\end_inset + + kvečjemu +\begin_inset Formula $n$ +\end_inset + + elementov. +\end_layout + +\begin_layout Enumerate +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{Vsako LN množico v $V$ z $n$ elementi lahko dopolnimo do baze.} +\end_layout + +\end_inset + + Naj bo +\begin_inset Formula $v_{1},\dots,v_{m}$ +\end_inset + + LN množica v +\begin_inset Formula $V$ +\end_inset + +. + Vemo, + da je +\begin_inset Formula $m\leq n$ +\end_inset + +. + Če +\begin_inset Formula $m=n$ +\end_inset + +, + je +\begin_inset Formula $v_{1},\dots,v_{m}$ +\end_inset + + baza po zgornji trditvi. + Sicer pa je +\begin_inset Formula $m<n$ +\end_inset + +: + Tedaj +\begin_inset Formula $v_{1},\dots,v_{m}$ +\end_inset + + ni ogrodje, + sicer bi imeli neko LN množico z več elementi kot neko ogrodje, + saj ima po popraj dokazanem vsako ogrodje vsaj toliko elementov kot vsaka LN množica. + Ker +\begin_inset Formula $v_{1},\dots,v_{m}$ +\end_inset + + ni ogrodje, + +\begin_inset Formula $\exists v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + +. + Po lemi +\begin_inset CommandInset ref +LatexCommand ref +reference "lem:večja-ln" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + je torej +\begin_inset Formula $v_{1},\dots,v_{m+1}$ +\end_inset + + LN množica. + Če je +\begin_inset Formula $m+1=n$ +\end_inset + +, + je to že baza, + sicer ponavljamo dodajanje elementov, + dokler ne dodamo +\begin_inset Formula $k$ +\end_inset + + elementov in dosežemo +\begin_inset Formula $m+k=n$ +\end_inset + +. + Tedaj je to baza. + Naredili smo +\begin_inset Formula $k=m-n$ +\end_inset + + korakov. +\end_layout + +\end_deeper +\begin_layout Proof +Uporabna vrednost tega izreka sta dva nova izreka o dimenzijah podprostorov: +\end_layout + +\begin_layout Claim* +Če je +\begin_inset Formula $V$ +\end_inset + + je KRVP in +\begin_inset Formula $W$ +\end_inset + + njegov podprostor, + je +\begin_inset Formula $\dim W\leq\dim V$ +\end_inset + +. +\end_layout + +\begin_layout Proof +PDDRAA +\begin_inset Formula $\dim W>\dim V$ +\end_inset + +. + Čim ima baza +\begin_inset Formula $W$ +\end_inset + + večjo moč kot baza +\begin_inset Formula $V$ +\end_inset + +, + obstaja v +\begin_inset Formula $W$ +\end_inset + + LN množica z večjo močjo kot baza +\begin_inset Formula $V$ +\end_inset + +. + Toda ker je ta LN množica LN tudi v +\begin_inset Formula $V$ +\end_inset + +, + obstaja v +\begin_inset Formula $V$ +\end_inset + + LN množica z več elementi kot baza +\begin_inset Formula $V$ +\end_inset + +, + kar je v +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + s trditvijo +\begin_inset CommandInset ref +LatexCommand vref +reference "enoličnost-moči-baze." +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + v razdelku +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Obstoj-baze" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\begin_layout Claim* +dimenzijska formula za podprostore. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVP in +\begin_inset Formula $W_{1},W_{2}$ +\end_inset + + podprostora v +\begin_inset Formula $V$ +\end_inset + +. + Velja +\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}-\dim\left(W_{1}\cap W_{2}\right)$ +\end_inset + +. + Vsota vektorskih podprostorov je definirana v razdelku +\begin_inset CommandInset ref +LatexCommand vref +reference "subsec:Vsota-podprostorov" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\begin_layout Proof +Izberimo bazo +\begin_inset Formula $w_{1},\dots,w_{m}$ +\end_inset + + za +\begin_inset Formula $W_{1}\cap W_{2}$ +\end_inset + +. + Naj bo +\begin_inset Formula $u_{1},\dots,u_{k}$ +\end_inset + + njena dopolnitev do baze +\begin_inset Formula $W_{1}$ +\end_inset + + in +\begin_inset Formula $v_{1},\dots,v_{l}$ +\end_inset + + njena dopolnitev do baze +\begin_inset Formula $W_{2}$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$ +\end_inset + + baza za +\begin_inset Formula $W_{1}+W_{2}$ +\end_inset + +. + Tedaj bi namreč veljalo +\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=m+k+l$ +\end_inset + +, + +\begin_inset Formula $\dim\left(W_{1}\cap W_{2}\right)=m$ +\end_inset + +, + +\begin_inset Formula $\dim\left(W_{1}\right)=m+k$ +\end_inset + + in +\begin_inset Formula $\dim\left(W_{2}\right)=m+l$ +\end_inset + +. + Treba je dokazati še, + da je +\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$ +\end_inset + + baza za +\begin_inset Formula $W_{1}+W_{2}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +Je ogrodje? + Vzemimo poljuben +\begin_inset Formula $v\in W_{1}+W_{2}$ +\end_inset + +. + Po definiciji +\begin_inset Formula $W_{1}+W_{2}\exists z_{1}\in W_{1},z_{2}\in W_{2}\ni:v=z_{1}+z_{2}$ +\end_inset + +. + Razvijmo +\begin_inset Formula $z_{1}$ +\end_inset + + po bazi +\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$ +\end_inset + + za +\begin_inset Formula $W_{1}$ +\end_inset + + in +\begin_inset Formula $z_{2}$ +\end_inset + + po bazi +\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{k}$ +\end_inset + + za +\begin_inset Formula $W_{2}$ +\end_inset + +. + Takole: + +\begin_inset Formula $z_{1}=\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}$ +\end_inset + + in +\begin_inset Formula $z_{2}=\gamma_{1}w_{1}+\cdots\gamma_{m}w_{m}+\delta_{1}v_{1}+\cdots\delta_{l}v_{l}$ +\end_inset + +. + Torej +\begin_inset Formula $v=z_{1}+z_{2}=\left(\alpha_{1}+\gamma_{1}\right)w_{1}+\cdots+\left(\alpha_{m}+\gamma_{m}\right)w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\delta_{1}v_{1}+\cdots+\delta_{l}v_{l}\in\Lin\left\{ w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}\right\} $ +\end_inset + +. + Je ogrodje. +\end_layout + +\begin_layout Itemize +Je LN? + Naj bo +\begin_inset Formula +\[ +\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l} +\] + +\end_inset + +Leva stran enačbe je +\begin_inset Formula $\in W_{1}$ +\end_inset + +, + desna pa +\begin_inset Formula $\in W_{2}$ +\end_inset + +, + zatorej je element, + ki ga izraza na obeh straneh enačbe opisujeta, + +\begin_inset Formula $\in W_{1}\cap W_{2}$ +\end_inset + +. + Torej je +\begin_inset Formula $v_{1},\dots,v_{l}$ +\end_inset + + baza za +\begin_inset Formula $W_{1}\cap W_{1}$ +\end_inset + +. + Toda baza od +\begin_inset Formula $W_{1}\cap W_{2}$ +\end_inset + + je tudi +\begin_inset Formula $w_{1},\dots,w_{m}$ +\end_inset + +, + zatorej lahko ta element razpišemo po njej: +\begin_inset Formula +\[ +\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}=\delta_{1}w_{1}+\cdots+\delta_{m}v_{m} +\] + +\end_inset + + +\begin_inset Formula +\[ +\delta_{1}w_{1}+\cdots+\delta_{m}w_{m}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0 +\] + +\end_inset + +Toda +\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{l}$ +\end_inset + + je baza za +\begin_inset Formula $W_{2}$ +\end_inset + + po naši prejšnji definiciji, + torej je LN množica, + zato +\begin_inset Formula $\delta_{1}=\cdots=\delta_{m}=\gamma_{1}=\cdots=\gamma_{l}=0$ +\end_inset + +. + Ker +\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$ +\end_inset + +, + se lahko vrnemo k drugi enačbi te točke in to ugotovitev upoštevamo: +\begin_inset Formula +\[ +\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l} +\] + +\end_inset + + +\begin_inset Formula +\[ +\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=0 +\] + +\end_inset + +Toda +\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$ +\end_inset + + je baza za +\begin_inset Formula $W_{1}$ +\end_inset + + po naši prejšnji definiciji, + torej je LN množica, + zato +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=0$ +\end_inset + +. + Torej velja +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=\gamma_{1}=\cdots=\gamma_{l}=0$ +\end_inset + +, + torej je ta množica res LN. +\end_layout + +\end_deeper +\begin_layout Corollary* +Velja torej +\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim\left(W_{1}\right)+\dim\left(W_{2}\right)$ +\end_inset + +. + Enačaj velja +\begin_inset Formula $\Leftrightarrow W_{1}\cap W_{2}=\left\{ 0\right\} $ +\end_inset + +, + kajti +\begin_inset Formula $\dim\left(\left\{ 0\right\} \right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Definition +\begin_inset CommandInset label +LatexCommand label +name "def:vsota-je-direktna" + +\end_inset + +Pravimo, + da je vsota +\begin_inset Formula $W_{1}+W_{2}$ +\end_inset + + direktna, + če velja +\begin_inset Formula $W_{1}\cap W_{2}=\left\{ 0\right\} $ +\end_inset + + oziroma ekvivalentno če je +\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$ +\end_inset + + oziroma ekvivalentno +\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Prehod na novo bazo +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor dimenzije +\begin_inset Formula $n$ +\end_inset + +. + Recimo, + da imamo dve bazi v +\begin_inset Formula $V$ +\end_inset + +. + +\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + naj bo +\begin_inset Quotes gld +\end_inset + +stara baza +\begin_inset Quotes grd +\end_inset + +, + +\begin_inset Formula $C=\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + pa naj bo +\begin_inset Quotes gld +\end_inset + +nova baza +\begin_inset Quotes grd +\end_inset + +. + +\begin_inset Formula $\forall v\in V$ +\end_inset + + lahko razvijemo po +\begin_inset Formula $B$ +\end_inset + + in po +\begin_inset Formula $C$ +\end_inset + +. + Razvoj po +\begin_inset Formula $B$ +\end_inset + +: + +\begin_inset Formula $v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$ +\end_inset + +, + razvoj po +\begin_inset Formula $C$ +\end_inset + +: + +\begin_inset Formula $v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}$ +\end_inset + +. + Kakšna je zveza med +\begin_inset Formula $\vec{\beta}$ +\end_inset + + in +\begin_inset Formula $\vec{\gamma}$ +\end_inset + + v obeh razvojih? +\end_layout + +\begin_layout Standard +Uvedimo oznako +\begin_inset Formula $\left[v\right]_{B}$ +\end_inset + +, + to naj bodo koeficienti vektorja +\begin_inset Formula $v$ +\end_inset + + pri razvoju po +\begin_inset Formula $B$ +\end_inset + +. + +\begin_inset Formula $\left[v\right]_{B}=\left[\begin{array}{c} +\beta_{1}\\ +\vdots\\ +\beta_{n} +\end{array}\right]$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Vsak vektor stare baze razvijmo po novi bazi, + kjer +\begin_inset Formula $\left[u_{i}\right]_{C}=\left[\begin{array}{c} +\alpha_{i1}\\ +\vdots\\ +\alpha_{in} +\end{array}\right]$ +\end_inset + +: +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\begin{array}{ccccccc} +u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\ +\vdots & & \vdots & & & & \vdots\\ +u_{n} & = & a_{n1}v_{1} & + & \cdots & + & a_{nn}v_{n} +\end{array} +\] + +\end_inset + +Koeficiente +\begin_inset Formula $\alpha$ +\end_inset + + zložimo v tako imenovanj prehodno matriko +\begin_inset Formula $P_{C\leftarrow B}$ +\end_inset + +: +\begin_inset Formula +\[ +P_{C\leftarrow B}=\left[\begin{array}{ccc} +\left[u_{1}\right]_{C} & \cdots & \left[u_{n}\right]_{C}\end{array}\right]=\left[\begin{array}{ccc} +a_{11} & \cdots & a_{n1}\\ +\vdots & & \vdots\\ +a_{1n} & \cdots & a_{nn} +\end{array}\right] +\] + +\end_inset + +Sledi +\begin_inset Formula +\[ +v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}=\beta_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+\beta_{n}\left(\alpha_{n1}v_{1}+\cdots+\alpha_{nn}v_{n}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=v_{1}\left(\beta_{1}\alpha_{11}+\beta_{2}\alpha_{21}+\cdots+\beta_{n}\alpha_{n1}\right)+\cdots+v_{n}\left(\beta_{1}\alpha_{1n}+\beta_{2}\alpha_{2n}+\cdots+\beta_{n}\alpha_{nn}\right)= +\] + +\end_inset + +po drugi strani je +\begin_inset Formula $v$ +\end_inset + + tudi lahko razvit po novi bazi: +\begin_inset Formula +\[ +=v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n} +\] + +\end_inset + +Iz česar, + ker je razvoj po bazi enoličen, + sledi +\begin_inset Formula +\[ +\begin{array}{ccccccc} +\gamma_{1} & = & \beta_{1}\alpha_{11} & + & \cdots & + & \beta_{n}\alpha_{n1}\\ +\vdots & & \vdots & & & & \vdots\\ +\gamma_{n} & = & \beta_{1}a_{1n} & + & \cdots & + & \beta_{n}\alpha_{nn} +\end{array}, +\] + +\end_inset + +kar v matrični obliki zapišemo +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +\alpha_{11} & \cdots & \alpha_{n1}\\ +\vdots & & \vdots\\ +\alpha_{1n} & \cdots & \alpha_{nn} +\end{array}\right]\left[\begin{array}{c} +\beta_{1}\\ +\vdots\\ +\beta_{n} +\end{array}\right]=\left[\begin{array}{c} +\gamma_{1}\\ +\vdots\\ +\gamma_{n} +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +P_{C\leftarrow B}\left[v\right]_{B}=\left[v\right]_{C}. +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Velja: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $P_{B\leftarrow B}=I$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Naj bodo prehodi med bazami takšnile: + +\begin_inset Formula $B\overset{P_{C\leftarrow B}}{\longrightarrow}C\overset{P_{D\leftarrow C}}{\longrightarrow}D$ +\end_inset + +. + Potem je +\begin_inset Formula $P_{D\leftarrow B}=P_{C\leftarrow B}\cdot P_{C\leftarrow D}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $P_{B\leftarrow C}\cdot P_{C\leftarrow B}=I$ +\end_inset + +, + +\begin_inset Formula $\left(P_{B\leftarrow C}\right)^{-1}=P_{C\leftarrow B}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Naj bo +\begin_inset Formula $v\in F^{n}$ +\end_inset + + in +\begin_inset Formula $S$ +\end_inset + + standardna baza za +\begin_inset Formula $F^{n}$ +\end_inset + +. + Potem +\begin_inset Formula $\left[v\right]_{S}=\left[\begin{array}{c} +\alpha_{1}\\ +\vdots\\ +\alpha_{n} +\end{array}\right]=v$ +\end_inset + +. + Sledi +\begin_inset Formula $P_{S\leftarrow B}=\left[\begin{array}{ccc} +\left[u_{1}\right]_{S} & \cdots & \left[u_{n}\right]_{S}\end{array}\right]=\left[\begin{array}{ccc} +u_{1} & \cdots & u_{n}\end{array}\right]$ +\end_inset + + za +\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + +. + Sledi tudi +\begin_inset Formula $P_{S\leftarrow C}=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + +, + kjer so +\begin_inset Formula $v,u,B,C$ +\end_inset + + kot prej (kot definirano na začetku tega razdelka). +\end_layout + +\begin_layout Itemize +\begin_inset Formula $P_{C\leftarrow B}=P_{C\leftarrow S}\cdot P_{S\leftarrow B}$ +\end_inset + + (slednji dve točki veljata samo v +\begin_inset Formula $F^{n}$ +\end_inset + +, + kjer je standardna baza lepa in zapisljiva kot elementi v matriki) +\end_layout + +\end_deeper +\begin_layout Section +Drugi semester +\end_layout + +\begin_layout Subsection +Linearne preslikave +\end_layout + +\begin_layout Standard +Radi bi definirali homomorfizem vektorskih prostorov. + Homomorfizem za abelove grupe smo že definirali, + vektorski prostor pa je le abelova grupa z dodatno strukturo (množenje s skalarjem). +\end_layout + +\begin_layout Definition* +Preslikava +\begin_inset Formula $f:V_{1}\to V_{2}$ +\end_inset + + je homomorfizem vektorskih prostorov nad istim poljem oziroma linearna preslikava, + če je aditivna (homomorfizem) ( +\begin_inset Formula $\forall u,v\in V_{1}:f\left(u+_{1}v\right)=fu+_{2}fv$ +\end_inset + +) in če je homogena: + +\begin_inset Formula $\forall u\in V_{1},\alpha\in F:f\left(\alpha u\right)=\alpha f\left(u\right)$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Ekvivalentno je preverjati oba pogoja hkrati. + Če za +\begin_inset Formula $L:U\to V$ +\end_inset + + velja +\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in F,u_{1},u_{2}\in U:L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}$ +\end_inset + +, + je +\begin_inset Formula $L$ +\end_inset + + linearna preslikava. +\end_layout + +\begin_layout Example* +Vrtež za kot +\begin_inset Formula $\tau$ +\end_inset + + v ravnini: +\begin_inset Formula +\[ +\left[\begin{array}{c} +x\\ +y +\end{array}\right]\to\left[\begin{array}{cc} +\cos\tau & -\sin\tau\\ +\sin\tau & \cos\tau +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Linearna funkcija iz analize ni linearna preslikava. + Premik za vektor +\begin_inset Formula $w$ +\end_inset + + ni linearna preslikava. + Odvajanje in integriranje sta linearni preslikavi. +\end_layout + +\begin_layout Fact* +Vsaka linearna preslikava slika 0 v 0. +\end_layout + +\begin_layout Definition* +Bijektivni linearni preslikavi pravimo linearni izomorfizem. +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "claim:invLinIzJeLinIz" + +\end_inset + +Inverz linearnega izomorfizma je zopet linearni izomorfizem. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $L:U\to V$ +\end_inset + + bijektivna linearna preslikava med vektorskima prostoroma nad istim poljem +\begin_inset Formula $F$ +\end_inset + +. + Dokazati je treba, + da je +\begin_inset Formula $L^{-1}:V\to U$ +\end_inset + + spet linearna preslikava. + Ker je +\begin_inset Formula $L$ +\end_inset + + linearna, + velja +\begin_inset Formula +\[ +\forall\alpha_{1},\alpha_{2}\in F,v_{1},v_{2}\in V:L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=\alpha_{1}LL^{-1}v_{1}+\alpha_{2}LL^{-1}v_{2}=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right) +\] + +\end_inset + +Ker je +\begin_inset Formula $L$ +\end_inset + + injektivna, + iz +\begin_inset Formula $L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$ +\end_inset + + sledi +\begin_inset Formula $\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}=L^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $F^{n}$ +\end_inset + + je linearno izomorfen +\begin_inset Formula $n-$ +\end_inset + +razsežnem +\begin_inset Formula $V$ +\end_inset + + nad +\begin_inset Formula $F$ +\end_inset + + +\end_layout + +\begin_layout Claim* +Vsak +\begin_inset Formula $n-$ +\end_inset + +razsežen vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + + je linearno izomorfen +\begin_inset Formula $F^{n}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $V$ +\end_inset + + +\begin_inset Formula $n-$ +\end_inset + +razsežen vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + + in +\begin_inset Formula $B=\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + baza za +\begin_inset Formula $V$ +\end_inset + +. + Definirajmo preslikavo +\begin_inset Formula $\phi_{B}:F^{n}\to V$ +\end_inset + + s predpisom +\begin_inset Formula $\left(x_{1},\cdots,x_{1}\right)\mapsto x_{1}v_{1}+\cdots+x_{n}v_{n}$ +\end_inset + +. + Ker je +\begin_inset Formula $B$ +\end_inset + + ogrodje, + je +\begin_inset Formula $\phi_{B}$ +\end_inset + + surjektivna. + Ker je +\begin_inset Formula $B$ +\end_inset + + linearno neodvisna, + je +\begin_inset Formula $\phi_{B}$ +\end_inset + + injektivna. + Pokažimo še, + da je linearna presikava: +\begin_inset Formula +\[ +\phi_{B}\left(\alpha\left(x_{1},\dots,x_{n}\right)+\beta\left(y_{1},\dots,y_{n}\right)\right)=\phi_{B}\left(\alpha x_{1}+\beta y_{1},\dots,\alpha x_{n}+\beta x_{n}\right)=v_{1}\left(\alpha x_{1}+\beta y_{1}\right)+\cdots+v_{n}\left(\alpha x_{n}+\beta x_{n}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\alpha\left(v_{1}x_{1}+\cdots+v_{n}x_{n}\right)+\cdots+\beta\left(v_{1}y_{1}+\cdots+v_{n}y_{n}\right)=\alpha\phi_{B}\left(x_{1},\dots,x_{n}\right)+\beta\phi_{B}\left(y_{1},\dots y_{n}\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Matrika-linearne-preslikave" + +\end_inset + +Matrika linearne preslikave — + linearni izomorfizem +\begin_inset Formula $M_{m,n}\left(F\right)\to\mathcal{L}\left(F^{n},F^{m}\right)$ +\end_inset + + +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $F$ +\end_inset + + polje in +\begin_inset Formula $m,n\in F$ +\end_inset + +. + +\begin_inset Formula $\mathcal{L}\left(F^{n},F^{m}\right)$ +\end_inset + + je vektorski prostor linearnih preslikav iz +\begin_inset Formula $F^{n}\to F^{m}$ +\end_inset + +. + Seštevanje definiramo z +\begin_inset Formula $\left(L_{1}+L_{2}\right)u\coloneqq L_{1}u+L_{2}u$ +\end_inset + +, + množenje s skalarjem pa +\begin_inset Formula $\left(\alpha L\right)u=\alpha\left(Lu\right)$ +\end_inset + +. + Naj bo +\begin_inset Formula $M_{m,n}\left(F\right)$ +\end_inset + + vektorski prostor vseh +\begin_inset Formula $m\times n$ +\end_inset + + matrik nad +\begin_inset Formula $F$ +\end_inset + + z znanim seštevanjem in množenjem. + Obstaja linearni izomorfizem med tema dvema prostoroma. +\end_layout + +\begin_layout Proof +Oznake kot v trditvi. + Za vsako +\begin_inset Formula $m\times n$ +\end_inset + + matriko +\begin_inset Formula $A=\left[a_{i,j}\right]$ +\end_inset + + definirajmo preslikavo +\begin_inset Formula $L_{A}$ +\end_inset + + iz +\begin_inset Formula $F^{n}$ +\end_inset + + v +\begin_inset Formula $F^{m}$ +\end_inset + + takole: + +\begin_inset Formula $L_{A}\left(x_{1},\dots,x_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$ +\end_inset + +. + Po definiciji matričnega množenja ta preslikava ustreza +\begin_inset Formula $L_{A}\vec{x}=A\vec{x}$ +\end_inset + +. + Dokažimo, + da je linearni izomorfizem. +\end_layout + +\begin_deeper +\begin_layout Itemize +Linearnost: + +\begin_inset Formula $L_{\alpha A+\beta B}\vec{x}=\left(\alpha A+\beta B\right)\vec{x}=\alpha A\vec{x}+\beta B\vec{x}=\alpha L_{A}\vec{x}+\beta L_{B}\vec{x}=\left(\alpha L_{A}+\beta L_{B}\right)\vec{x}$ +\end_inset + + +\end_layout + +\begin_layout Itemize +Bijektivnost: + Konstruirajmo inverzno preslikavo (iz trditve +\begin_inset CommandInset ref +LatexCommand ref +reference "claim:invLinIzJeLinIz" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + vemo, + da bo linearna). + Vsaki linearni presikavi +\begin_inset Formula $L:F^{n}\to F^{m}$ +\end_inset + + priredimo +\begin_inset Formula $m\times n$ +\end_inset + + matriko +\begin_inset Formula $\left[\begin{array}{ccc} +Le_{1} & \cdots & Le_{n}\end{array}\right]$ +\end_inset + +, + kjer je +\begin_inset Formula $e_{1},\dots,e_{n}$ +\end_inset + + standardna baza za +\begin_inset Formula $F^{n}$ +\end_inset + +. + Pokažimo, + da je ta preslikava res inverz, + torej preverimo, + da je kompozitum +\begin_inset Formula $A\mapsto L_{A}\mapsto\left[\begin{array}{ccc} +L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]$ +\end_inset + + identiteta in da je +\begin_inset Formula $\left[\begin{array}{ccc} +L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]\mapsto L_{A}\mapsto A$ +\end_inset + + tudi identiteta. +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$A +\backslash +mapsto L_{A} +\backslash +mapsto +\backslash +left[ +\backslash +begin{array}{ccc}Le_{1} & +\backslash +cdots & Le_{n} +\backslash +end{array} +\backslash +right] +\backslash +overset{?}{=}id$} +\end_layout + +\end_inset + +: + +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]=\left[\begin{array}{ccc} +Ae_{1} & \cdots & Ae_{n}\end{array}\right]=A\left[\begin{array}{ccc} +e_{1} & \cdots & e_{n}\end{array}\right]=AI=A +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$ +\backslash +left[ +\backslash +begin{array}{ccc}L_{A}e_{1} & +\backslash +cdots & L_{A}e_{n} +\backslash +end{array} +\backslash +right] +\backslash +mapsto L_{A} +\backslash +mapsto A +\backslash +overset{?}{=}id$} +\end_layout + +\end_inset + +: +\begin_inset Formula +\[ +\forall x:L_{\left[\begin{array}{ccc} +Le_{1} & \cdots & Le_{n}\end{array}\right]}x=\left[\begin{array}{ccc} +Le_{1} & \cdots & Le_{n}\end{array}\right]\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right]=x_{1}Le_{1}+\cdots+x_{n}Le_{n}=L\left(x_{1}e_{1}+\cdots+x_{n}e_{n}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=L\left(x_{1}\left[\begin{array}{c} +1\\ +0\\ +\vdots\\ +0 +\end{array}\right]+\cdots+x_{n}\left[\begin{array}{c} +0\\ +\vdots\\ +0\\ +1 +\end{array}\right]\right)=Lx +\] + +\end_inset + + +\end_layout + +\end_deeper +\end_deeper +\begin_layout Proof +Vsaki linearni preslikavi med dvema vektorskima prostoroma sedaj lahko priredimo matriko. + Prirejanje je odvisno od izbire baz v obeh vektorskih prostorih. + Matrika namreč preslika koeficiente iz polja +\begin_inset Formula $F$ +\end_inset + +, + s katerimi je dan vektor, + ki ga z leve množimo z matriko, + razvit po +\begin_inset Quotes gld +\end_inset + +vhodni +\begin_inset Quotes grd +\end_inset + + bazi, + v koeficiente iz istega polja, + s katerimi je rezultantni vektor razvit po +\begin_inset Quotes gld +\end_inset + +izhodni +\begin_inset Quotes grd +\end_inset + + bazi. +\end_layout + +\begin_layout Proof +Naj bosta +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $V$ +\end_inset + + vektorska prostora nad istim poljem +\begin_inset Formula $F$ +\end_inset + + in naj bo +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna preslikava. + Izberimo bazo +\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + za +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + + za +\begin_inset Formula $V$ +\end_inset + +. + Razvijmo vektorje +\begin_inset Formula $Lu_{1},\dots,Lu_{n}$ +\end_inset + + po bazi +\begin_inset Formula $\mathcal{C}$ +\end_inset + +: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\ +\vdots & & \vdots & & & & \vdots\\ +Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m} +\end{array} +\] + +\end_inset + +Skalarje +\begin_inset Formula $\alpha_{i,j}$ +\end_inset + + sedaj zložimo v spodnjo matriko, + ki ji pravimo +\series bold +matrika linearne preslikave +\begin_inset Formula $L$ +\end_inset + + glede na bazi +\begin_inset Formula $\mathcal{B}$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}$ +\end_inset + + +\series default +. + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc} +\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[\begin{array}{ccc} +a_{1,1} & \cdots & \alpha_{n,1}\\ +\vdots & & \vdots\\ +\alpha_{1,m} & \cdots & \alpha_{n,m} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +\begin_inset Formula $L:\mathbb{R}\left[x\right]_{\leq3}\to\mathbb{R}\left[x\right]_{\leq2}$ +\end_inset + + — + linearna preslikava iz realnih polinomov stopnje kvečjemu 3 v realne polinome stopnje kvečjemu 2, + ki predstavlja odvajanje polinomov. + Bazi sta +\begin_inset Formula $\mathcal{B}=\left\{ 1,x,x^{2},x^{2}\right\} $ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ 1,x,x^{2}\right\} $ +\end_inset + +. +\begin_inset Formula +\[ +\begin{array}{ccccccc} +L\left(1\right) & = & 0 & + & 0x & + & 0x^{2}\\ +L\left(x\right) & = & 1 & + & 0x & + & 0x^{2}\\ +L\left(x^{2}\right) & = & 0 & + & 2x & + & 0x^{2}\\ +L\left(x^{3}\right) & = & 0 & + & 0x & + & 3x^{2} +\end{array} +\] + +\end_inset + +Zapišimo matriko te linearne preslikave: +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cccc} +0 & 1 & 0 & 0\\ +0 & 0 & 2 & 0\\ +0 & 0 & 0 & 3 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Prehodna matrika je poseben primer matrike linearne preslikave. + +\begin_inset Formula $L:V\to V$ +\end_inset + + z bazama +\begin_inset Formula $\mathcal{B}=\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + +, + kjer +\begin_inset Formula $L=id$ +\end_inset + +. +\begin_inset Formula +\[ +\begin{array}{ccccccc} +id\left(u_{1}\right) & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,n}v_{n}\\ +\vdots & & \vdots & & & & \vdots\\ +id\left(u_{n}\right) & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,n}v_{n} +\end{array} +\] + +\end_inset + +Zapišimo matriko te linearne preslikave: + +\begin_inset Formula $\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}=P_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Lastnosti matrik linearnih preslikav +\end_layout + +\begin_layout Standard +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_layout Theorem +\begin_inset CommandInset label +LatexCommand label +name "thm:osnovna-formula" + +\end_inset + +osnovna formula. + Posplošitev formule +\begin_inset Formula $\left[u\right]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}\cdot\left[u\right]_{\mathcal{B}}$ +\end_inset + + se glasi +\begin_inset Formula $\left[Lu\right]_{\mathcal{C}}=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}$ +\end_inset + + za linearno preslikavo +\begin_inset Formula $L:U\to V$ +\end_inset + +, + +\begin_inset Formula $u\in U$ +\end_inset + +, + kjer je +\begin_inset Formula $\mathcal{\mathcal{B}}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + baza za +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + + baza za +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Po korakih: +\end_layout + +\begin_deeper +\begin_layout Enumerate +Razvijmo +\begin_inset Formula $u$ +\end_inset + + po bazi +\begin_inset Formula $\mathcal{B}$ +\end_inset + +: + +\begin_inset Formula $u=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:Uporabimo-L-na" + +\end_inset + +Uporabimo +\begin_inset Formula $L$ +\end_inset + + na obeh straneh: + +\begin_inset Formula $Lu=L\left(\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}\right)=\beta_{1}Lu_{1}+\cdots+\beta_{n}Lu_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Razvijmo bazo +\begin_inset Formula $\mathcal{B}$ +\end_inset + +, + preslikano z +\begin_inset Formula $L$ +\end_inset + +, + po bazi +\begin_inset Formula $\mathcal{C}$ +\end_inset + +: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\ +\vdots & & \vdots & & & & \vdots\\ +Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m} +\end{array} +\] + +\end_inset + + +\end_layout + +\begin_layout Enumerate +Razvoj vstavimo v enačbo iz koraka +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:Uporabimo-L-na" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + in uredimo: +\begin_inset Formula +\[ +Lu=\beta_{1}\left(\alpha_{1,1}v_{1}+\cdots+\alpha_{1,m}v_{m}\right)+\cdots+\beta_{n}\left(\alpha_{n,1}v_{1}+\cdots+\alpha_{n,m}v_{m}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=v_{1}\left(\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\right)+\cdots+v_{m}\left(\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}\right) +\] + +\end_inset + + +\end_layout + +\begin_layout Enumerate +Odtod sledi: +\begin_inset Formula +\[ +\left[Lu\right]_{\mathcal{C}}=\left[\begin{array}{c} +\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\\ +\vdots\\ +\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m} +\end{array}\right]=\left[\begin{array}{ccc} +\alpha_{1,1} & \cdots & \alpha_{n,1}\\ +\vdots & & \vdots\\ +\alpha_{1,m} & \cdots & \alpha_{n,m} +\end{array}\right]\left[\begin{array}{c} +\beta_{1}\\ +\vdots\\ +\beta_{n} +\end{array}\right]=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Theorem +\begin_inset CommandInset label +LatexCommand label +name "thm:matrika-kompozituma-linearnih" + +\end_inset + +matrika kompozituma linearnih preslikav. + Posplošitev formule +\begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + + se glasi +\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + +. + Trdimo, + da je kompozitum linearnih preslikav spet linearna preslikava in da enačba velja. +\end_layout + +\begin_layout Proof +Najprej dokažimo, + da je kompozitum linearnih preslikav spet linearna preslikava. +\begin_inset Formula +\[ +\left(K\circ L\right)\left(\alpha u+\beta v\right)=K\left(L\left(\alpha u+\beta v\right)\right)=K\left(\alpha Lu+\beta Lv\right)=\alpha KLu+\beta KLv=\alpha\left(K\circ L\right)u+\beta\left(K\circ L\right)v +\] + +\end_inset + +Sedaj pa dokažimo še enačbo +\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + +. + Naj bosta +\begin_inset Formula $L:U\to V$ +\end_inset + + in +\begin_inset Formula $K:V\to W$ +\end_inset + + linearni preslikavi, + +\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + baza +\begin_inset Formula $U$ +\end_inset + +, + +\begin_inset Formula $\mathcal{C}$ +\end_inset + + baza +\begin_inset Formula $V$ +\end_inset + + in +\begin_inset Formula $\mathcal{D}$ +\end_inset + + baza +\begin_inset Formula $W$ +\end_inset + +. + Od prej vemo, + da: +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc} +\left[Lu_{1}\right]_{\mathcal{D}} & \cdots & \left[Lu_{n}\right]_{\mathcal{D}}\end{array}\right], +\] + +\end_inset + +zato pišimo +\begin_inset Formula +\[ +\left[K\circ L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc} +\left[\left(K\circ L\right)u_{1}\right]_{\mathcal{D}} & \cdots & \left[\left(K\circ L\right)u_{n}\right]_{\mathcal{D}}\end{array}\right]=\left[\begin{array}{ccc} +\left[KLu_{1}\right]_{\mathcal{D}} & \cdots & \left[KLu_{n}\right]_{\mathcal{D}}\end{array}\right]\overset{\text{izrek \ref{thm:osnovna-formula}}}{=} +\] + +\end_inset + + +\begin_inset Formula +\[ +\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}\left[\begin{array}{ccc} +\left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[\begin{array}{ccc} +\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}} +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Jedro in slika linearne preslikave +\end_layout + +\begin_layout Definition* +Naj bosta +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $V$ +\end_inset + + vektorska prostora nad istim poljem +\begin_inset Formula $F$ +\end_inset + + in +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna preslikava. + Jedro +\begin_inset Formula $L$ +\end_inset + + naj bo +\begin_inset Formula $\Ker L\coloneqq\left\{ u\in U;Lu=0\right\} $ +\end_inset + + (angl. + kernel/null space) in Slika/zaloga vrednosti +\begin_inset Formula $L$ +\end_inset + + naj bo +\begin_inset Formula $\Slika L\coloneqq\left\{ Lu;\forall u\in U\right\} $ +\end_inset + + (angl. + image/range). +\end_layout + +\begin_layout Claim* +Trdimo naslednje: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\Ker L$ +\end_inset + + je vektorski podprostor v +\begin_inset Formula $U$ +\end_inset + + (če vsebuje +\begin_inset Formula $\vec{a}$ +\end_inset + + in +\begin_inset Formula $\vec{b}$ +\end_inset + +, + vsebuje tudi vse LK +\begin_inset Formula $\vec{a}$ +\end_inset + + in +\begin_inset Formula $\vec{b}$ +\end_inset + +) +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\Slika L$ +\end_inset + + je vektorski podprostor v +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Proof +Dokazujemo dve trditvi: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\forall u_{1},u_{2}\in\Ker L,\alpha_{1},\alpha_{2}\in F\overset{?}{\Longrightarrow}\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$ +\end_inset + +. + Po predpostavki velja +\begin_inset Formula $Lu_{1}=0$ +\end_inset + + in +\begin_inset Formula $Lu_{2}=0$ +\end_inset + +, + torej +\begin_inset Formula $\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}=0$ +\end_inset + +. + Iz linearnosti +\begin_inset Formula $L$ +\end_inset + + sledi +\begin_inset Formula $L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=0$ +\end_inset + +, + torej +\begin_inset Formula $\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall v_{1},v_{2}\in\Slika L,\beta_{1},\beta_{2}\in F\overset{?}{\Longrightarrow}\beta_{1}v_{1}+\beta_{2}v_{2}\in\Slika L$ +\end_inset + +. + Po predpostavki velja +\begin_inset Formula $\exists u_{1},u_{2}\in U\ni:v_{1}=Lu_{1}\wedge v_{2}=Lu_{2}$ +\end_inset + +. + Velja torej +\begin_inset Formula $\beta_{1}v_{1}+\beta_{2}v_{2}=\beta_{1}Lu_{1}+\beta_{2}Lu_{2}\overset{\text{linearnost}}{=}L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)$ +\end_inset + + in +\begin_inset Formula $\beta_{1}u_{1}+\beta_{2}u_{2}\in U$ +\end_inset + +, + torej je +\begin_inset Formula $L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)\in\Slika L$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Definition* +Ničnost +\begin_inset Formula $L$ +\end_inset + + je +\begin_inset Formula $\n\left(L\right)\coloneqq\dim\Ker L$ +\end_inset + + (angl. + nullity) in rang +\begin_inset Formula $L$ +\end_inset + + je +\begin_inset Formula $\rang\left(L\right)=\dim\Slika L$ +\end_inset + + (angl. + rank). +\end_layout + +\begin_layout Remark* +Jedro in sliko smo definirali za linearne preslikave, + vendar ju lahko definiramo tudi za poljubno matriko +\begin_inset Formula $A$ +\end_inset + + nad poljem +\begin_inset Formula $F$ +\end_inset + +, + saj smo v +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Matrika-linearne-preslikave" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + dokazali linearni izomorfizem med +\begin_inset Formula $m\times n$ +\end_inset + + matrikami nad +\begin_inset Formula $F$ +\end_inset + + in linearnimi preslikavami +\begin_inset Formula $F^{n}\to F^{m}$ +\end_inset + +. +\begin_inset Formula +\[ +Au=\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{m1} & \cdots & a_{mn} +\end{array}\right]\left[\begin{array}{c} +u_{1}\\ +\vdots\\ +u_{n} +\end{array}\right]=\left[\begin{array}{c} +a_{11}u_{1}+\cdots+a_{1n}u_{n}\\ +\vdots\\ +a_{m1}u_{1}+\cdots+a_{mn}u_{n} +\end{array}\right]=\left[\begin{array}{c} +a_{11}\\ +\vdots\\ +a_{m1} +\end{array}\right]u_{1}+\cdots+\left[\begin{array}{c} +a_{1n}\\ +\vdots\\ +a_{mn} +\end{array}\right]u_{n} +\] + +\end_inset + +Iz tega je razvidno, + da je +\begin_inset Formula $\Slika A$ +\end_inset + + torej linearna ogrinjača stolpcev matrike +\begin_inset Formula $A$ +\end_inset + +. + Pravimo tudi, + da je +\begin_inset Formula $\Slika A$ +\end_inset + + stolpični prostor +\begin_inset Formula $A$ +\end_inset + + oziroma +\begin_inset Formula $\Col A$ +\end_inset + + (angl. + column space). + +\begin_inset Formula $\rang A=\dim\Slika A$ +\end_inset + + je torej največje število linearno neodvisnih stolpcev +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Linearna preslikava +\begin_inset Formula $L$ +\end_inset + + je injektivna ( +\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$ +\end_inset + +) +\begin_inset Formula $\Leftrightarrow\Ker L=\left\{ 0\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Predpostavimo, + da je +\begin_inset Formula $L$ +\end_inset + + injektivna, + torej +\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$ +\end_inset + +. + Vzemimo poljuben +\begin_inset Formula $u\in\Ker L$ +\end_inset + +. + Zanj velja +\begin_inset Formula $Lu=0=L0\Rightarrow u=0$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Predpostavimo, + +\begin_inset Formula $\Ker L=\left\{ 0\right\} $ +\end_inset + +. + Računajmo: + +\begin_inset Formula $Lu_{1}=Lu_{2}\Longrightarrow Lu_{1}-Lu_{2}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(u_{1}-u_{2}\right)=0\Longrightarrow u_{1}-u_{2}=0\Longrightarrow u_{1}=u_{2}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Theorem* +osnovna formula. + Naj bo +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna preslikava. + Tedaj je +\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim U$ +\end_inset + +, + torej +\begin_inset Formula $\n L+\rang L=\dim U$ +\end_inset + +. + Za matrike torej trdimo +\begin_inset Formula $\n A+\rang A=\dim F^{n}=n$ +\end_inset + + za +\begin_inset Formula $m\times n$ +\end_inset + + matriko +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Vemo, + da sta jedro in slika podprostora. + Naj bo +\begin_inset Formula $w_{1},\dots,w_{k}$ +\end_inset + + baza jedra in +\begin_inset Formula $u_{1},\dots,u_{l}$ +\end_inset + + njena dopolnitev do baze +\begin_inset Formula $U$ +\end_inset + +. + Torej +\begin_inset Formula $\dim U=k+l=\n L+l$ +\end_inset + +. + Treba je še dokazati, + da je +\begin_inset Formula $l=\rang A$ +\end_inset + +. + Konstruirajmo bazo za +\begin_inset Formula $\Slika L$ +\end_inset + +, + ki ima +\begin_inset Formula $l$ +\end_inset + + elementov in dokažimo, + da so +\begin_inset Formula $Lu_{1},\dots,Lu_{l}$ +\end_inset + + baza za +\begin_inset Formula $\Slika L$ +\end_inset + +: +\end_layout + +\begin_deeper +\begin_layout Itemize +Je ogrodje? + Vzemimo poljuben +\begin_inset Formula $v\in\Slika L$ +\end_inset + +. + Zanj obstaja nek +\begin_inset Formula $u\in U\ni:Lu=v$ +\end_inset + +, + ki ga lahko razvijemo po bazi +\begin_inset Formula $U$ +\end_inset + + takole +\begin_inset Formula $u=\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}$ +\end_inset + +. + Sedaj na obeh straneh uporabimo +\begin_inset Formula $L$ +\end_inset + + in upoštevamo linearnost: +\begin_inset Formula +\[ +v=Lu=L\left(\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}\right)=\alpha_{1}Lw_{1}+\cdots+\alpha_{k}Lw_{k}+\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l} +\] + +\end_inset + +Ker so +\begin_inset Formula $w_{i}$ +\end_inset + + baza +\begin_inset Formula $\Ker L$ +\end_inset + +, + so elementi +\begin_inset Formula $\Ker L$ +\end_inset + +, + torej je +\begin_inset Formula $Lw_{i}=0$ +\end_inset + + za vsak +\begin_inset Formula $i$ +\end_inset + +. + Tako poljuben +\begin_inset Formula $v\in\Slika L$ +\end_inset + + razpišemo z bazo velikosti +\begin_inset Formula $l$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Je LN? + Računajmo: + +\begin_inset Formula $\gamma_{1}Lu_{1}+\cdots+\gamma_{l}Lu_{l}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\right)=0\Longrightarrow\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\in\Ker L$ +\end_inset + +, + kar pomeni, + da ga je moč razviti po bazi +\begin_inset Formula $\Ker L$ +\end_inset + +: +\begin_inset Formula +\[ +\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}=\delta_{1}w_{1}+\cdots+\delta_{k}w_{k} +\] + +\end_inset + + +\begin_inset Formula +\[ +\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}-\delta_{1}w_{1}-\cdots-\delta_{k}w_{k}=0 +\] + +\end_inset + +Ker je +\begin_inset Formula $w_{1},\dots,w_{k},u_{1},\dots,u_{l}$ +\end_inset + + baza +\begin_inset Formula $U$ +\end_inset + +, + je LN, + zato velja +\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=w_{1}=\cdots=w_{k}=0$ +\end_inset + +, + kar pomeni, + da očitno velja +\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$ +\end_inset + +, + torej je res LN. +\end_layout + +\end_deeper +\begin_layout Remark* +Bralcu prav pride skica s 3. + strani zapiskov predavanja +\begin_inset Quotes gld +\end_inset + +LA1P FMF 2024-02-28 +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Paragraph* +Do preproste matrike preslikave z ustreznimi bazami +\end_layout + +\begin_layout Standard +Imenujmo sedaj +\begin_inset Formula $\mathcal{B}=\left\{ w_{1},\dots,w_{k},u_{1},\dots,u_{l}\right\} $ +\end_inset + + bazo za +\begin_inset Formula $U$ +\end_inset + +, + in +\begin_inset Formula $\mathcal{C}=\left\{ Lu_{1},\dots,Lu_{l},z_{1},\dots,z_{m}\right\} $ +\end_inset + + baza za +\begin_inset Formula $V$ +\end_inset + +, + kjer je +\begin_inset Formula $z_{1},\dots,z_{m}$ +\end_inset + + dopolnitev +\begin_inset Formula $Lu_{1},\dots,Lu_{l}$ +\end_inset + + do baze +\begin_inset Formula $V$ +\end_inset + +, + kajti +\begin_inset Formula $V$ +\end_inset + + je lahko večji kot samo +\begin_inset Formula $\Slika L$ +\end_inset + +, + in si oglejmo matriko naše preslikave +\begin_inset Formula $L:U\to V$ +\end_inset + +, + ki slika iz baze +\begin_inset Formula $\mathcal{B}$ +\end_inset + + v bazo +\begin_inset Formula $\mathcal{C}$ +\end_inset + +. + Najprej razpišimo preslikane elemente baze +\begin_inset Formula $\mathcal{B}$ +\end_inset + + po bazi +\begin_inset Formula $\mathcal{C}$ +\end_inset + +: +\begin_inset Formula +\[ +\begin{array}{ccccccccccccc} +Lu_{1} & = & 1\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\ +\vdots & & \vdots & & & & \vdots & & \vdots & & & & \vdots\\ +Lu_{l} & = & 0\cdot Lu_{1} & + & \cdots & + & 1\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\ +Lw_{1} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\ +\vdots & & \vdots & & & & \vdots & & \vdots & & & & \vdots\\ +Lw_{k} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m} +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cc} +I_{l} & 0\\ +0 & 0 +\end{array}\right] +\] + +\end_inset + +S primerno izbiro baz +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $V$ +\end_inset + + je torej matrika preslikave precej preprosta, + zgolj bločna matrika z identiteto, + veliko +\begin_inset Formula $\rang L$ +\end_inset + + in ničlami, + ki ustrezajo dimenzijam +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kaj pa, + če je +\begin_inset Formula $L$ +\end_inset + + matrika? + Recimo ji +\begin_inset Formula $A$ +\end_inset + +, + da je +\begin_inset Formula $L_{A}=L$ +\end_inset + + od prej. + Tedaj +\begin_inset Formula $A\in M_{p,n}\left(F\right)$ +\end_inset + +. + Naj bo +\begin_inset Formula $P=\left[\begin{array}{cccccc} +u_{1} & \cdots & u_{l} & w_{1} & \cdots & w_{k}\end{array}\right]$ +\end_inset + + matrika, + katere stolpci so baza +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $Q=\left[\begin{array}{cccccc} +Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]$ +\end_inset + + matrika, + katere stolpci so baza +\begin_inset Formula $V$ +\end_inset + +. + Po karakterizaciji obrnljivih matrik sta obrnljivi. + Tedaj +\begin_inset Formula +\[ +AP=\left[\begin{array}{cccccc} +Au_{1} & \cdots & Au_{l} & Aw_{1} & \cdots & Aw_{k}\end{array}\right]\overset{\text{jedro}}{=}\left[\begin{array}{cccccc} +Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +Q\left[\begin{array}{cc} +I_{l} & 0\\ +0 & 0 +\end{array}\right]=\left[\begin{array}{cccccc} +Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]\left[\begin{array}{cc} +I_{l} & 0\\ +0 & 0 +\end{array}\right]=\left[\begin{array}{cccccc} +Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +AP=Q\left[\begin{array}{cc} +I_{l} & 0\\ +0 & 0 +\end{array}\right]\Longrightarrow Q^{-1}AP=\left[\begin{array}{cc} +I_{l} & 0\\ +0 & 0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Ekvivalentnost matrik +\end_layout + +\begin_layout Definition* +Matriki +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + sta ekvivalentni (oznaka +\begin_inset Formula $A\sim B$ +\end_inset + + +\begin_inset Foot +status open + +\begin_layout Plain Layout +Isto oznako uporabljamo tudi za podobne matrike, + vendar podobnost ni enako kot ekvivalentnost. +\end_layout + +\end_inset + +) +\begin_inset Formula $\Leftrightarrow\exists$ +\end_inset + + obrnljivi +\begin_inset Formula $P,Q\ni:B=PAQ$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Dokazali smo, + da je vsaka matrika +\begin_inset Formula $A$ +\end_inset + + ekvivalentna matriki +\begin_inset Formula $\left[\begin{array}{cc} +I_{r} & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + +, + kjer je +\begin_inset Formula $r=\rang A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokažimo, + da je relacija +\begin_inset Formula $\sim$ +\end_inset + + ekvivalenčna: +\end_layout + +\begin_deeper +\begin_layout Itemize +refleksivnost: + +\begin_inset Formula $A\sim A$ +\end_inset + + velja. + Naj bo +\begin_inset Formula $A\in M_{m,n}\left(F\right)$ +\end_inset + +. + Tedaj +\begin_inset Formula $A=I_{m}AI_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +simetričnost: + +\begin_inset Formula $A\sim B\Rightarrow B\sim A$ +\end_inset + +, + kajti če velja +\begin_inset Formula $B=PAQ$ +\end_inset + + in sta +\begin_inset Formula $P$ +\end_inset + + in +\begin_inset Formula $Q$ +\end_inset + + obrnljivi, + velja +\begin_inset Formula $P^{-1}BQ^{-1}=A$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +tranzitivnost: + +\begin_inset Formula $A\sim B\wedge B\sim C\Rightarrow A\sim C$ +\end_inset + +, + kajti, + če velja +\begin_inset Formula $B=PAQ$ +\end_inset + + in +\begin_inset Formula $C=SBT$ +\end_inset + + in so +\begin_inset Formula $P,Q,S,T$ +\end_inset + + obrnljive, + velja +\begin_inset Formula $C=\left(SP\right)A\left(QT\right)$ +\end_inset + + in produkt obrnljivih matrik je obrnljiva matrika. +\end_layout + +\end_deeper +\begin_layout Theorem* +Dve matriki sta ekvivalentni natanko tedaj, + ko imata enako velikost in enak rang. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Po predpostavki imata +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + enako velikost in enak rang +\begin_inset Formula $r$ +\end_inset + +. + Od prej vemo, + da sta obe ekvivalentni +\begin_inset Formula $\left[\begin{array}{cc} +I_{r} & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + +, + ker pa je relacija ekvivalentnosti ekvivalenčna, + sta +\begin_inset Formula $A\sim B$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Po predpostavki +\begin_inset Formula $A\sim B$ +\end_inset + +, + torej +\begin_inset Formula $\exists P,Q\ni:B=PAQ$ +\end_inset + +. + Če je +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Formula $m\times n$ +\end_inset + +, + je +\begin_inset Formula $P$ +\end_inset + + +\begin_inset Formula $m\times m$ +\end_inset + + in +\begin_inset Formula $Q$ +\end_inset + + +\begin_inset Formula $n\times n$ +\end_inset + +, + zatorej je po definiciji matričnega množenja +\begin_inset Formula $B$ +\end_inset + + +\begin_inset Formula $m\times n$ +\end_inset + +. + Dokazati je treba še +\begin_inset Formula $\rang A=\rang B$ +\end_inset + +. +\begin_inset Formula +\[ +\rang B=\rang PAQ\overset{?}{=}\rang PA=\overset{?}{=}\rang A +\] + +\end_inset + +Dokažimo najprej +\begin_inset Formula $\rang PAQ=\rang PA$ +\end_inset + + oziroma +\begin_inset Formula $\rang CQ=\rang C$ +\end_inset + + za obrnljivo +\begin_inset Formula $Q$ +\end_inset + + in poljubno C. + Dokažemo lahko celo +\begin_inset Formula $\Slika CQ=\Slika C$ +\end_inset + +: + +\begin_inset Formula +\[ +\forall u:u\in\Slika CQ\Leftrightarrow\exists v\ni:u=\left(CQ\right)v\Leftrightarrow\exists v'\ni:u=Cv'\Leftrightarrow u\in\Slika C. +\] + +\end_inset + +Sedaj dokažimo še +\begin_inset Formula $\rang\left(PA\right)=\rang\left(A\right)$ +\end_inset + +. + Zadošča dokazati, + da je +\begin_inset Formula $\Ker\left(PA\right)=\Ker A$ +\end_inset + +, + kajti tedaj bi iz enakosti izrazov +\begin_inset Formula +\[ +\dim\Slika A+\dim\Ker A=\dim F^{n}=n +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim\Slika PA+\dim\Ker PA=\dim F^{n}=n +\] + +\end_inset + +dobili +\begin_inset Formula $\dim\Slika PA=\dim\Slika A$ +\end_inset + +. + Dokažimo torej +\begin_inset Formula $\Ker PA=\Ker A$ +\end_inset + +: +\begin_inset Formula +\[ +\forall u:u\in\Ker PA\Leftrightarrow PAu=0\overset{P\text{ obrnljiva}}{\Longleftrightarrow}Au=0\Leftrightarrow u\in\Ker A. +\] + +\end_inset + +Torej je res +\begin_inset Formula $\Ker PA=\Ker A$ +\end_inset + +, + torej je res +\begin_inset Formula $\rang PA=\rang A$ +\end_inset + +, + torej je res +\begin_inset Formula $\rang A=\rang B$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Podobnost matrik +\end_layout + +\begin_layout Definition* +Kvadratni matriki +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + sta podobni, + če +\begin_inset Formula $\exists$ +\end_inset + + taka obrnljiva matrika +\begin_inset Formula $P\ni:B=PAP^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Podobnost je ekvivalenčna relacija. +\end_layout + +\begin_layout Proof +Dokazujemo, + da je relacija ekvivalenčna, + torej: +\end_layout + +\begin_deeper +\begin_layout Itemize +refleksivna: + +\begin_inset Formula $A=IAI^{-1}=IAI=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +simetrična: + +\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +tranzitivna: + +\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Remark +\begin_inset CommandInset label +LatexCommand label +name "rem:nista-podobni" + +\end_inset + +Očitno velja podobnost +\begin_inset Formula $\Rightarrow$ +\end_inset + + ekvivalentnost, + toda obrat ne velja vedno. + Na primer +\begin_inset Formula $\left[\begin{array}{cc} +1 & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + sta ekvivalentni (sta enake velikosti in ranga), + toda nista podobni (dokaz kasneje). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Standard +Od prej vemo, + da je vsaka matrika ekvivalentna matriki +\begin_inset Formula $\left[\begin{array}{cc} +I_{r} & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + +, + kjer je +\begin_inset Formula $r$ +\end_inset + + njen rang. + A je vsaka kvadratna matrika podobna kakšni lepi matriki? + Ja. + Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje). + Toda a je vsaka kvadratna matrika podobna diagonalni matriki? + Ne. +\end_layout + +\begin_layout Definition* +Matrika +\begin_inset Formula $D$ +\end_inset + + je diagonalna +\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kdaj je matrika +\begin_inset Formula $A$ +\end_inset + + podobna neki diagonalni matriki? + Kdaj +\begin_inset Formula $\exists$ +\end_inset + + diagonalna +\begin_inset Formula $D$ +\end_inset + + in obrnljiva +\begin_inset Formula $P\ni:A=PDP^{-1}$ +\end_inset + +? + Izpeljimo iz nastavka. + +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda n +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $P=\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$ +\end_inset + +, + kjer sta +\begin_inset Formula $D$ +\end_inset + + in +\begin_inset Formula $P$ +\end_inset + + neznani. + Ker mora biti +\begin_inset Formula $P$ +\end_inset + + obrnljiva, + so njeni stolpični vektorji LN. +\begin_inset Formula +\[ +A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc} +\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda_{n} +\end{array}\right]\text{ in }P\text{ obrnljiva} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc} +\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN} +\] + +\end_inset + + +\begin_inset Formula +\[ +A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Porodi se naloga, + imenovana +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + +. + Iščemo pare +\begin_inset Formula $\left(\lambda,\vec{v}\right)$ +\end_inset + +, + ki zadoščajo enačbi +\begin_inset Formula $A\vec{v}=\lambda\vec{v}$ +\end_inset + +. + +\end_layout + +\begin_layout Definition* +Pravimo, + da je +\begin_inset Formula $\lambda$ +\end_inset + + je lastna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +, + če obstaja tak +\begin_inset Formula $\vec{v}\not=0$ +\end_inset + +, + da je +\begin_inset Formula $A\vec{v}=\lambda\vec{v}$ +\end_inset + +. + V tem primeru pravimo, + da je +\begin_inset Formula $\vec{v}$ +\end_inset + + lastni vektor, + ki pripada lastni vrednosti +\begin_inset Formula $\lambda$ +\end_inset + +. + Paru +\begin_inset Formula $\left(\lambda,\vec{v}\right)$ +\end_inset + +, + ki zadošča enačbi, + pravimo lastni par matrike +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Nalogo +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + + rešujemo v dveh korakih. + Najprej najdemo vse +\begin_inset Formula $\lambda$ +\end_inset + +, + nato za vsako poiščemo pripadajoče +\begin_inset Formula $\vec{v}$ +\end_inset + +, + ki za lastno vrednost obstajajo po definiciji. +\end_layout + +\begin_layout Standard +Za nek +\begin_inset Formula $v\not=0$ +\end_inset + + pišimo +\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$ +\end_inset + + za nek +\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$ +\end_inset + + ni obrnljiva +\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Polinom +\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$ +\end_inset + + je karakteristični polinom matrike +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Premislek zgoraj nam pove, + da so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + ničle +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Karakteristični polinom lahko nima nobene ničle: + +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +-1 & 0 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc} +-\lambda & 1\\ +-1 & -\lambda +\end{array}\right]=x^{2}+1$ +\end_inset + +, + katerega ničli sta +\begin_inset Formula $\lambda_{1}=i$ +\end_inset + + in +\begin_inset Formula $\lambda_{2}=-i$ +\end_inset + +, + ki nista realni števili. + V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti, + saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle. +\end_layout + +\begin_layout Standard +Kako pa iščemo lastne vektorje za lastno vrednost +\begin_inset Formula $\lambda$ +\end_inset + +? + Spomnimo se na +\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$ +\end_inset + +. + Rešiti moramo homogen sistem linearnih enačb. + Po definiciji so lastni vektorji neničelni, + zato nas trivialna rešitev ne zanima. +\end_layout + +\begin_layout Definition* +Množici +\begin_inset Formula $\Ker\left(A-\lambda I\right)$ +\end_inset + + pravimo lastni podprostor matrike +\begin_inset Formula $A$ +\end_inset + +, + ki pripada +\begin_inset Formula $\lambda$ +\end_inset + +. + Slednji vsebuje +\begin_inset Formula $\vec{0}$ +\end_inset + + in množico vektorjev, + ki so vsi lastni vektorji +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Exercise* +Izračunaj lastne vrednosti od +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +-1 & 0 +\end{array}\right]$ +\end_inset + +. + Od prej vemo, + da +\begin_inset Formula $\lambda_{1}=i$ +\end_inset + +, + +\begin_inset Formula $\lambda_{2}=-i$ +\end_inset + +. + Izračunajmo +\begin_inset Formula $\Ker\left(A-iI\right)$ +\end_inset + + in +\begin_inset Formula $\Ker\left(A+iI\right)$ +\end_inset + +: +\begin_inset Formula +\[ +\Ker\left(A-iI\right):\quad\left[\begin{array}{cc} +-i & 1\\ +-1 & -i +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c} +1\\ +i +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker\left(A+iI\right):\quad\left[\begin{array}{cc} +i & 1\\ +-1 & i +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c} +1\\ +-i +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c} +1\\ +i +\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c} +1\\ +-i +\end{array}\right]\right\} +\] + +\end_inset + + +\end_layout + +\begin_layout Exercise* +Vstavimo lastna vektorja v +\begin_inset Formula $P$ +\end_inset + + in lastne vrednosti v +\begin_inset Formula $D$ +\end_inset + + na pripadajoči mesti. + Dobimo obrnljivo +\begin_inset Formula $P$ +\end_inset + + in velja +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + +\begin_inset Formula +\[ +P=\left[\begin{array}{cc} +1 & 1\\ +i & -i +\end{array}\right],\quad D=\left[\begin{array}{cc} +i & 0\\ +0 & -i +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Exercise* +Temu početju pravimo +\begin_inset Quotes gld +\end_inset + +diagonalizacija matrike +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Primer matrike, + ki ni diagonalizabilna: + +\begin_inset Formula $A=\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + +. + +\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc} +-\lambda & 1\\ +0 & -\lambda +\end{array}\right]=\lambda^{2}$ +\end_inset + +. + Ničli/lastni vrednosti sta +\begin_inset Formula $\lambda_{1}=0$ +\end_inset + + in +\begin_inset Formula $\lambda_{2}=0$ +\end_inset + +. + Toda +\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c} +1\\ +0 +\end{array}\right]\right\} $ +\end_inset + + in +\begin_inset Formula $P=\left[\begin{array}{cc} +1 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + ni obrnljiva. + S tem dokažemo trditev v primeru +\begin_inset CommandInset ref +LatexCommand ref +reference "rem:nista-podobni" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. + +\begin_inset Formula $\left[\begin{array}{cc} +1 & 0\\ +0 & 0 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\left[\begin{array}{cc} +0 & 1\\ +0 & 0 +\end{array}\right]$ +\end_inset + + nista podobni, + ker je prva diagonalna, + druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati). +\end_layout + +\begin_layout Standard +Lastne vrednosti lahko definiramo tudi za linearne preslikave, + saj so linearne preslikave linearno izomorfne matrikam. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad +\begin_inset Formula $F=\mathbb{C}$ +\end_inset + + in +\begin_inset Formula $L:V\to V$ +\end_inset + + linearna preslikava. + Število +\begin_inset Formula $\lambda\in F$ +\end_inset + + je lastna vrednost +\begin_inset Formula $L$ +\end_inset + +, + le obstaja tak neničelni +\begin_inset Formula $v\in V$ +\end_inset + +, + da velja +\begin_inset Formula $Lv=\lambda v$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Kako pa rešujemo +\begin_inset Quotes gld +\end_inset + +Lastni problem +\begin_inset Quotes grd +\end_inset + + za linearne preslikave? + +\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$ +\end_inset + +. + Toda determinante linearne preslikave nismo definirali. + Lahko pa determinanto izračunamo na matriki, + ki pripada tej linearni preslikavi. + Toda dvem različnim bazam pripadata različni matriki linearne preslikave. + Dokazati je treba, + da sta determinanti dveh matrik, + pripadajočih eni linearni preslikavi, + enaki, + četudi sta matriki v različnih bazah. +\end_layout + +\begin_layout Lemma +\begin_inset CommandInset label +LatexCommand label +name "lem:Podobni-matriki-imata" + +\end_inset + +Podobni matriki imata isto determinanto. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $B=PAP^{-1}$ +\end_inset + + za neko obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + Tedaj +\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Proof +\begin_inset Formula $L:V\to V$ +\end_inset + + naj bo linearna preslikava, + +\begin_inset Formula $V$ +\end_inset + + prostor nad +\begin_inset Formula $F=\mathbb{C}$ +\end_inset + +, + +\begin_inset Formula $\mathcal{B}$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}$ +\end_inset + + pa bazi +\begin_inset Formula $V$ +\end_inset + +. + Priredimo matriki +\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$ +\end_inset + + in +\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$ +\end_inset + +. + Spomnimo se izreka +\begin_inset CommandInset ref +LatexCommand vref +reference "thm:matrika-kompozituma-linearnih" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +: + +\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + +. + +\begin_inset Formula $L=\left[id\circ L\circ id\right]$ +\end_inset + +, + zato +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$ +\end_inset + + za neko obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + Torej sta matriki +\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$ +\end_inset + + in +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$ +\end_inset + + podobni, + torej imata po lemi +\begin_inset CommandInset ref +LatexCommand vref +reference "lem:Podobni-matriki-imata" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + isto determinanto. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Proof +Alternativen dokaz, + da imata podobni matriki iste lastne vrednosti: + +\begin_inset Formula $A$ +\end_inset + + podobna +\begin_inset Formula +\[ +B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B}, +\] + +\end_inset + +torej so lastne vrednosti enake. + Kaj pa lastni vektorji? + Naj bo +\begin_inset Formula $v$ +\end_inset + + lastni vektor +\begin_inset Formula $A$ +\end_inset + +, + torej +\begin_inset Formula +\[ +Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv, +\] + +\end_inset + +torej za +\begin_inset Formula $v$ +\end_inset + + lastni vektor +\begin_inset Formula $A$ +\end_inset + + sledi, + da je +\begin_inset Formula $Pv$ +\end_inset + + lastni vektor +\begin_inset Formula $B$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Linearni transformaciji torej priredimo tako matriko, + ki ima v začetnem in končnem prostoru isto bazo. + Tedaj lahko izračunamo lastne pare na tej matriki. +\end_layout + +\begin_layout Theorem* +Schurov izrek. + Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki. +\end_layout + +\begin_layout Proof +Indukcija po velikosti matrike. +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza: + +\begin_inset Formula $A_{1\times1}$ +\end_inset + + je zgornjetrikotna. +\end_layout + +\begin_layout Itemize +Korak: + Po I. + P. + trdimo, + da je vsaka +\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$ +\end_inset + + podobna kaki zgornjetrikotni matriki. + Dokažimo še za poljubno +\begin_inset Formula $A_{n\times n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $v_{1}$ +\end_inset + + pripadajoči lastni vektor ter +\begin_inset Formula $v_{2},\dots,v_{n}$ +\end_inset + + dopolnitev +\begin_inset Formula $v_{1}$ +\end_inset + + do baze +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Potem je matrika +\begin_inset Formula $P=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + + obrnljiva. +\begin_inset Formula +\[ +AP=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc} +\lambda & a_{1,2} & \cdots & a_{1,n}\\ +0 & \vdots & & \vdots\\ +\vdots & \vdots & & \vdots\\ +0 & a_{m,n} & \cdots & a_{m.n} +\end{array}\right]=P\left[\begin{array}{cc} +\lambda & B\\ +0 & C +\end{array}\right] +\] + +\end_inset + +Po I. + P. + obstaja taka zgornjetrikotna +\begin_inset Formula $T$ +\end_inset + + in obrnljiva +\begin_inset Formula $Q$ +\end_inset + +, + da +\begin_inset Formula $C=QTQ^{-1}$ +\end_inset + +. +\begin_inset Formula +\[ +\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]=\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]^{-1}\left[\begin{array}{cc} +\lambda & B\\ +0 & C +\end{array}\right]\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cc} +\lambda & C\\ +0 & Q^{-1}B +\end{array}\right]\left[\begin{array}{cc} +1 & 0\\ +0 & Q +\end{array}\right]=\left[\begin{array}{cc} +\lambda & CQ\\ +0 & B +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula $A$ +\end_inset + + je torej podobna +\begin_inset Formula $\left[\begin{array}{cc} +\lambda & CQ\\ +0 & B +\end{array}\right]$ +\end_inset + +, + ki je zgornjetrikotna. +\end_layout + +\end_deeper +\begin_layout Proof +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO karakterizacija linearnih preslikav +\begin_inset Quotes gld +\end_inset + +LA1V FMF 2024-03-12 +\begin_inset Quotes grd +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Zadosten pogoj za diagonalizabilnost +\end_layout + +\begin_layout Theorem +\begin_inset CommandInset label +LatexCommand label +name "thm:lave-razl-lavr-so-LN" + +\end_inset + +Lastni vektorji, + ki pripadajo različnim lastnim vrednostim, + so linearno neodvisni. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika, + +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + njene lastne vrednosti in +\begin_inset Formula $v_{1},\dots,v_{k}$ +\end_inset + + njim pripadajoči lastni vektorji. + Dokazujemo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + paroma različni +\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$ +\end_inset + + LN. + Dokaz z indukcijo po +\begin_inset Formula $k$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza +\begin_inset Formula $k=1$ +\end_inset + +: + Elementi +\begin_inset Formula $\left\{ \lambda_{1}\right\} $ +\end_inset + + so trivialno paroma različni in +\begin_inset Formula $v_{1}$ +\end_inset + + je kot neničen vektor LN. +\end_layout + +\begin_layout Itemize +Korak: + Dokazujemo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$ +\end_inset + + so paroma različne +\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$ +\end_inset + + so LN, + vedoč I. + P. + Denimo, + da +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$ +\end_inset + +. + Množimo z +\begin_inset Formula $A$ +\end_inset + +: +\begin_inset Formula +\[ +A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0 +\] + +\end_inset + +Množimo začetno enačbo z +\begin_inset Formula $\lambda_{k+1}$ +\end_inset + + (namesto z +\begin_inset Formula $A$ +\end_inset + +, + kot smo to storili zgoraj): +\begin_inset Formula +\[ +\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0 +\] + +\end_inset + +Odštejmo eno enačbo od druge, + dobiti moramo 0, + saj odštevamo 0 od 0: +\begin_inset Formula +\[ +\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0 +\] + +\end_inset + +Ker so lastne vrednosti paroma različne ( +\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$ +\end_inset + +), + so njihove razlike neničelne. + Ker so +\begin_inset Formula $v_{1},\dots,v_{k}$ +\end_inset + + po predpostavki LN, + sledi +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$ +\end_inset + +. + Vstavimo te konstante v +\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$ +\end_inset + + in dobimo +\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$ +\end_inset + +. + Ker je +\begin_inset Formula $v_{k+1}$ +\end_inset + + neničeln (je namreč lastni vektor), + sledi +\begin_inset Formula $\alpha_{k+1}=0$ +\end_inset + +, + torej +\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$ +\end_inset + +, + zatorej so +\begin_inset Formula $v_{1},\dots,v_{k+1}$ +\end_inset + + res LN. +\end_layout + +\end_deeper +\begin_layout Corollary +\begin_inset CommandInset label +LatexCommand label +name "cor:vsota-lastnih-podpr-direktna" + +\end_inset + +Vsota vseh lastnih podprostorov matrike je direktna (definicija +\begin_inset CommandInset ref +LatexCommand vref +reference "def:vsota-je-direktna" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +). +\end_layout + +\begin_layout Proof +Naj bodo +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti matrike +\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$ +\end_inset + +. + Pripadajoči lastni podprostori so torej +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$ +\end_inset + +. + Trdimo, + da je vsota teh podprostorov direktna, + torej +\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$ +\end_inset + +. + To sledi iz izreka +\begin_inset CommandInset ref +LatexCommand vref +reference "thm:lave-razl-lavr-so-LN" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Če ima +\begin_inset Formula $n\times n$ +\end_inset + + matrika +\begin_inset Formula $n$ +\end_inset + + paroma različnih lastnih vrednosti, + je podobna diagonalni matriki. +\end_layout + +\begin_layout Proof +Po posledici +\begin_inset CommandInset ref +LatexCommand vref +reference "cor:vsota-lastnih-podpr-direktna" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + je vsota lastnih podprostorov matrike +\begin_inset Formula $A_{n\times n}$ +\end_inset + + direktna. + Če je torej lastnih podprostorov +\begin_inset Formula $n$ +\end_inset + +, + je njihova vsota cel prostor +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Matriko se da diagonalizirati, + kadar je vsota vseh lastnih podprostorov enaka podprostoru +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + (tedaj so namreč stolpci matrike +\begin_inset Formula $P$ +\end_inset + + linearno neodvisni, + zato je +\begin_inset Formula $P$ +\end_inset + + obrnljiva). +\end_layout + +\begin_layout Subsubsection +Algebraične in geometrijske vekčratnosti +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika. + +\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$ +\end_inset + +, + kjer so +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. + Stopnji ničle — + +\begin_inset Formula $n_{i}$ +\end_inset + + — + rečemo algebraična večkratnost lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Geometrijska večkratnost lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + + je +\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$ +\end_inset + + +\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Algebraično večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + označimo z +\begin_inset Formula $n_{i}$ +\end_inset + + in je večkratnost ničle +\begin_inset Formula $\lambda_{i}$ +\end_inset + + v +\begin_inset Formula $p_{A}\left(\lambda\right)$ +\end_inset + + (karakterističnem polinomu). + Geometrijsko večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + pa označimo z +\begin_inset Formula $m_{i}$ +\end_inset + + in je dimenzija lastnega podprostora za +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "claim:geom<=alg" + +\end_inset + + +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$ +\end_inset + + — + geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna, + kot je algebraična večkratnost te lastne vrednosti. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v_{1},\dots,v_{m_{i}}$ +\end_inset + + baza za lastni podprostor +\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$ +\end_inset + + in naj bo +\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$ +\end_inset + + njena dopolnitev do baze +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Tedaj velja: + +\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$ +\end_inset + +, + ..., + +\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$ +\end_inset + +, + +\begin_inset Formula $Av_{m_{i}+1}=$ +\end_inset + + linearna kombinacija +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +, + ..., + +\begin_inset Formula $Av_{n}=$ +\end_inset + + linearna kombinacija +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $P=\left[\begin{array}{cccccc} +v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + +, + ki je obrnljiva. +\begin_inset Formula +\[ +P^{-1}AP=\left[\begin{array}{cc} +\lambda_{i}I_{m_{i}} & B\\ +0 & C +\end{array}\right] +\] + +\end_inset + +Ker je karakteristični polinom neodvisen od izbire baze, + velja +\begin_inset Formula +\[ +\det\left(A-xI_{n}\right)=\det\left(\lambda_{i}I_{m_{i}}-xI\right)\det\left(C-xI_{n-m_{i}}\right)=\left(\lambda_{i}-x\right)^{m_{i}}\det\left(C-xI_{n-m_{i}}\right) +\] + +\end_inset + +Ker +\begin_inset Formula $\left(\lambda-x\right)^{m_{i}}$ +\end_inset + + deli karakteristični polinom, + je algebraična večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + + vsaj tolikšna, + kot je geometrična. +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "claim:mi=ni=>diag" + +\end_inset + +Matriko s paroma različnimi lastnimi vrednostmi +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + je moč diagonalizirati +\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $V_{i}$ +\end_inset + + lastno podprostor lastne vrednosti +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. + Vemo, + da se da +\begin_inset Formula $A_{n\times n}$ +\end_inset + + diagonalizirati +\begin_inset Formula $\Leftrightarrow A$ +\end_inset + + ima +\begin_inset Formula $n$ +\end_inset + + LN stolpičnih vektorjev +\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$ +\end_inset + + vsota lastnih podprostorov je direktna +\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$ +\end_inset + +. + Toda ker po prejšnjem izreku +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$ +\end_inset + +, + mora veljati +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Minimalni polinom matrike +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$ +\end_inset + + polinom in +\begin_inset Formula $A$ +\end_inset + + matrika. + +\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$ +\end_inset + +. + Če je +\begin_inset Formula $p\left(A\right)=0$ +\end_inset + + (ničelna matrika), + pravimo, + da polinom +\begin_inset Formula $p$ +\end_inset + + anhilira/uniči matriko +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Fact* +\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Izkaže se, + da karakteristični polimom anhilira matriko — + +\begin_inset Formula $p_{A}\left(A\right)=0$ +\end_inset + +. + Dokaz kasneje. +\end_layout + +\begin_layout Definition* +Polinom +\begin_inset Formula $m\left(x\right)$ +\end_inset + + je minimalen polinom +\begin_inset Formula $A$ +\end_inset + +, + če velja: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $m\left(A\right)=0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $m$ +\end_inset + + ima vodilni koeficient 1 +\end_layout + +\begin_layout Enumerate +med vsemi polinomi, + ki zadoščajo prvi in drugi zahtevi, + ima +\begin_inset Formula $m$ +\end_inset + + najnižjo stopnjo +\end_layout + +\end_deeper +\begin_layout Claim* +eksistenca minimalnega polinoma — + Minimalni polinom obstaja. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + matrika. + Očitno je +\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$ +\end_inset + + vektorski prostor dimenzije +\begin_inset Formula $n^{2}$ +\end_inset + +. + Matrike +\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $ +\end_inset + + so linearno odvisne, + ker je moč te množice za 1 večja od moči vektorskega prostora. + Torej +\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$ +\end_inset + +, + ki niso vse 0 +\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$ +\end_inset + +. + Torej polinom +\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$ +\end_inset + + anhilira +\begin_inset Formula $A$ +\end_inset + +. + Če ta polinom delimo z njegovim vodilnim koeficientom, + dobimo polinom, + ki ustreza prvima dvema zahevama za minimalni polinom. + Če med vsemi takimi izberemo takega z najnižjo stopnjo, + le-ta ustreza še tretji zahtevi. +\end_layout + +\begin_layout Theorem* +Če je +\begin_inset Formula $m\left(x\right)$ +\end_inset + + minimalni polinom za +\begin_inset Formula $A$ +\end_inset + + in če +\begin_inset Formula $p\left(x\right)$ +\end_inset + + anhilira +\begin_inset Formula $A$ +\end_inset + +, + potem +\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$ +\end_inset + + ( +\begin_inset Formula $m\left(x\right)$ +\end_inset + + deli +\begin_inset Formula $p\left(x\right)$ +\end_inset + +). +\end_layout + +\begin_layout Proof +Delimo +\begin_inset Formula $p$ +\end_inset + + z +\begin_inset Formula $m$ +\end_inset + +: + +\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$ +\end_inset + +. + Vstavimo +\begin_inset Formula $A$ +\end_inset + + na obe strani: +\begin_inset Formula +\[ +0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0 +\] + +\end_inset + +Sledi +\begin_inset Formula $r\left(x\right)=0$ +\end_inset + +, + kajti če +\begin_inset Formula $r$ +\end_inset + + ne bi bil ničeln polinom, + bi ga lahko delili z vodilnim koeficientom in po predpostavki +\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$ +\end_inset + + bi imel manjšo stopnjo kot +\begin_inset Formula $m\left(x\right)$ +\end_inset + +, + torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od +\begin_inset Formula $m$ +\end_inset + +, + torej +\begin_inset Formula $m$ +\end_inset + + ne bi bil minimalni polinom, + kar bi vodilo v protislovje. +\end_layout + +\begin_layout Corollary* +enoličnost minimalnega polinoma. + Naj bosta +\begin_inset Formula $m_{1}$ +\end_inset + + in +\begin_inset Formula $m_{2}$ +\end_inset + + minimalna polinoma matrike +\begin_inset Formula $A$ +\end_inset + +. + Ker +\begin_inset Formula $m$ +\end_inset + + po definiciji anhilira +\begin_inset Formula $A$ +\end_inset + +, + iz prejšnje trditve sledi, + če vstavimo +\begin_inset Formula $m=m_{1}$ +\end_inset + + in +\begin_inset Formula $p=m_{2}$ +\end_inset + +, + +\begin_inset Formula $m_{1}\vert m_{2}$ +\end_inset + +. + Toda če vstavimo +\begin_inset Formula $m=m_{2}$ +\end_inset + + in +\begin_inset Formula $p=m_{1}$ +\end_inset + +, + +\begin_inset Formula $m_{2}\vert m_{1}$ +\end_inset + +. + Iz +\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$ +\end_inset + + sledi, + da se +\begin_inset Formula $m_{1}$ +\end_inset + + in +\begin_inset Formula $m_{2}$ +\end_inset + + razlikujeta le za konstanten faktor, + ki pa je po definiciji minimalnega polinoma 1, + torej +\begin_inset Formula $m_{1}=m_{2}$ +\end_inset + +. + Zaradi enoličnosti lahko označimo minimalni polinom +\begin_inset Formula $A$ +\end_inset + + z +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Ničle minimalnega polinoma +\end_layout + +\begin_layout Claim* +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + + in +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + + imata iste ničle +\begin_inset Formula $\sim$ +\end_inset + + ničle +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Ker je +\begin_inset Formula $p_{A}\left(x\right)$ +\end_inset + + (dokaz kasneje), + velja po trditvi v dokazu enoličnosti, + da +\begin_inset Formula $m_{A}\vert p_{A}$ +\end_inset + +, + torej je vsaka ničla +\begin_inset Formula $m_{A}$ +\end_inset + + tudi ničla +\begin_inset Formula $p_{A}$ +\end_inset + +. + Treba je dokazati še, + da je vsaka ničla +\begin_inset Formula $p_{A}$ +\end_inset + + tudi ničla +\begin_inset Formula $m_{A}$ +\end_inset + +, + natančneje: + Treba je dokazati, + da če je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +, + je +\begin_inset Formula $m_{A}\left(\lambda\right)=0$ +\end_inset + +. + Naj bo +\begin_inset Formula $v\not=0$ +\end_inset + + lastni vektor za +\begin_inset Formula $\lambda$ +\end_inset + +. + Tedaj +\begin_inset Formula $Av=\lambda v$ +\end_inset + +. + Potem velja +\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$ +\end_inset + + in splošneje +\begin_inset Formula $A^{n}v=\lambda^{n}v$ +\end_inset + +. + Sedaj recimo, + da je +\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$ +\end_inset + +. + Potem je, + ker minimalni polinom anhilira +\begin_inset Formula $A$ +\end_inset + +, + +\begin_inset Formula +\[ +m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v= +\] + +\end_inset + + +\begin_inset Formula +\[ +=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0 +\] + +\end_inset + +Ker +\begin_inset Formula $m_{A}\left(\lambda\right)v=0$ +\end_inset + + in +\begin_inset Formula $v\not=0$ +\end_inset + + (je namreč lastni vektor), + velja +\begin_inset Formula $m_{A}\left(\lambda\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph* +Lastnosti +\end_layout + +\begin_layout Standard +Ker je +\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$ +\end_inset + + in +\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$ +\end_inset + +, + sledi iz +\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$ +\end_inset + +. + Poleg tega, + ker +\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$ +\end_inset + +. + Toda pozor: + +\series bold +Ni +\series default + res, + da +\begin_inset Formula $r_{i}=m_{i}$ +\end_inset + + (stropnja lastnega podprostora). +\end_layout + +\begin_layout Theorem* +Cayley-Hamilton. + +\begin_inset Formula $p_{A}\left(A\right)=0$ +\end_inset + + — + karakteristični polinom matrike +\begin_inset Formula $A$ +\end_inset + + anhilira matriko +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\begin_layout Proof +Spomnimo se eksplicitne formule za celico inverza matrike (razdelek +\begin_inset CommandInset ref +LatexCommand vref +reference "subsec:Formula-za-inverz-matrike" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +), + ki pravi +\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$ +\end_inset + +. + Računajmo in naposled vstavimo +\begin_inset Formula $B=A-xI$ +\end_inset + +: +\begin_inset Formula +\[ +B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B +\] + +\end_inset + + +\begin_inset Formula +\[ +\det B\cdot I=B\tilde{B}^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T} +\] + +\end_inset + +Glede na definicijo +\begin_inset Formula $\tilde{A}$ +\end_inset + + je +\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$ +\end_inset + + matrika velikosti +\begin_inset Formula $n\times n$ +\end_inset + +, + ki vsebuje polinome stopnje +\begin_inset Formula $<n$ +\end_inset + +, + kajti vsebuje determinante kofaktorskih matrik, + torej je takele oblike ( +\begin_inset Formula $\forall i\in\left\{ 1..\left(n-1\right)\right\} :B_{i}\in M_{n}\left(\mathbb{C}\right)$ +\end_inset + +): +\begin_inset Foot +status open + +\begin_layout Plain Layout +To si predstavljamo tako, + da iz vsake celice matrike, + ki vsebuje polinom, + izpostavimo (homogenost) spremenljivko (torej +\begin_inset Formula $x$ +\end_inset + + na fiksno potenco) in nato koeficiente v celicah pred to spremenljivvko zložimo v eno matriko. + Slednje ponovimo za vsako potenco in dobimo te matrike +\begin_inset Formula $B_{i}$ +\end_inset + +. +\end_layout + +\end_inset + + +\begin_inset Formula +\[ +\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$ +\end_inset + +. + Kot v enačbi množimo to z +\begin_inset Formula $I$ +\end_inset + +: + +\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$ +\end_inset + +. + Oglejmo si še desno stran enačbe: +\begin_inset Formula +\[ +\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n} +\] + +\end_inset + +In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe: +\begin_inset Formula +\[ +\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T} +\] + +\end_inset + + +\begin_inset Formula +\[ +c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n} +\] + +\end_inset + + +\begin_inset Formula +\[ +\begin{array}{cccc} +1: & c_{0}I & = & AB_{0}\\ +x: & c_{1}I & = & AB_{1}-B_{0}\\ +x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\ +\vdots\\ +x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\ +x^{n}: & c_{n}I & = & -B_{n-1} +\end{array} +\] + +\end_inset + +Vstavimo sedaj +\begin_inset Formula $A$ +\end_inset + + v enačbo namesto +\begin_inset Formula $x$ +\end_inset + +: +\begin_inset Formula +\[ +p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +p_{A}\left(A\right)=0 +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +Matriko +\begin_inset Formula $A$ +\end_inset + + se da diagonalizirati +\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$ +\end_inset + + ima samo enostavne ničle (nima večkratnih — + potence so vse 1). + Torej +\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$ +\end_inset + + za +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + vse paroma različne lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Po predpostavki je +\begin_inset Formula $A$ +\end_inset + + podobna diagonalni matriki — + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + za diagonalno +\begin_inset Formula $D$ +\end_inset + + in obrnljivo +\begin_inset Formula $P$ +\end_inset + +. + BSŠ naj bo +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & 0 & 0\\ +0 & \cdots & 0\\ +0 & 0 & \lambda_{k} +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$ +\end_inset + +. + Oglejmo si izraz +\begin_inset Formula +\[ +\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)=\left[\begin{array}{cccc} +0 & & & 0\\ + & \lambda_{2}-\lambda_{1}\\ + & & \ddots\\ +0 & & & \lambda_{k}-\lambda_{1} +\end{array}\right]\cdots\left[\begin{array}{cccc} +\lambda_{1}-\lambda_{k} & & & 0\\ + & \ddots\\ + & & \lambda_{k-1}-\lambda_{k}\\ +0 & & & 0 +\end{array}\right]=0 +\] + +\end_inset + +Sedaj pa še izraz +\begin_inset Formula +\[ +\left(A-\lambda_{1}I\right)\cdots\left(A-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}I\right)\cdots\left(PDP^{-1}-\lambda_{k}I\right)=\left(PDP^{-1}-\lambda_{1}PP^{-1}\right)\cdots\left(PDP^{-1}-\lambda_{k}PP^{-1}\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=P\left(D-\lambda_{1}I\right)\cancel{P^{-1}}\cdots\cancel{P}\left(D-\lambda_{k}I\right)P^{-1}=P\left(D-\lambda_{1}I\right)\cdots\left(D-\lambda_{k}I\right)P^{-1}=P0P^{-1}=0, +\] + +\end_inset + +torej ta polinom anhilira +\begin_inset Formula $A$ +\end_inset + +. + Ker deli +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + + — + vsebuje vse ničle +\begin_inset Formula $m_{A}\left(x\right)$ +\end_inset + +, + je prav to minimalen polinom +\begin_inset Formula $A$ +\end_inset + + — + ima najmanjšo stopnjo možno. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Potrebujemo nekaj lem: +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Lemma +Za vse matrike +\begin_inset Formula $A,B$ +\end_inset + + velja +\begin_inset Formula $\n\left(AB\right)\leq\n\left(A\right)+\n\left(B\right)\sim\dim\Ker\left(AB\right)\leq\dim\Ker\left(A\right)+\dim\Ker\left(B\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Oglejmo si preslikavo +\begin_inset Formula $L:\Ker AB\to\Ker A$ +\end_inset + +, + ki slika +\begin_inset Formula $x\mapsto Bx$ +\end_inset + +. + Je dobro definirana, + kajti +\begin_inset Formula $x\in\Ker AB\Rightarrow ABx=0\Rightarrow Bx\in\Ker A$ +\end_inset + +. + Po osnovnem dimenzijskem izreku za preslikavo +\begin_inset Formula $L$ +\end_inset + + velja +\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim\Ker AB$ +\end_inset + +. + Ker velja +\begin_inset Formula $Lx=0\Rightarrow Bx=0$ +\end_inset + +, + velja +\begin_inset Formula $\Ker L\subseteq\Ker B$ +\end_inset + + in zato +\begin_inset Formula $\dim\Ker L\leq\dim\Ker B$ +\end_inset + +. + Poleg tega iz definicije velja +\begin_inset Formula $\Slika L\subseteq\Ker A$ +\end_inset + + in zato +\begin_inset Formula $\dim\Slika L\leq\dim\Ker A$ +\end_inset + +. + Vstavimo te neenakosti v enačbo iz dimenzijskega izreka: +\begin_inset Formula +\[ +\dim\Ker L+\dim\Slika L=\dim\Ker AB +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim\Ker B+\dim\Ker A\geq\dim\Ker AB +\] + +\end_inset + +Lemo lahko posplošimo na več faktorkev, + torej +\begin_inset Formula $\n\left(A_{1}\cdots A_{k}\right)\leq\n A_{1}+\cdots+\n A_{k}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Nadaljujmo z dokazom +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + +. + Denimo, + da +\begin_inset Formula $\left(x-\lambda_{1}\right)\cdots\left(x-\lambda_{k}\right)$ +\end_inset + + anhilira +\begin_inset Formula $A$ +\end_inset + +. + Upoštevamo +\begin_inset Formula +\[ +\n\left(\left(A-\lambda_{1}\right)\cdots\left(A-\lambda_{k}\right)\right)\leq\n\left(A-\lambda_{1}\right)+\cdots+\n\left(A-\lambda_{k}\right) +\] + +\end_inset + +Členi na desni strani so geometrijske večkratnosti, + ker pa +\begin_inset Formula $A$ +\end_inset + + anhilira polinom po predpostavki, + je ta produkt ničelna preslikava in je dimenzija jedra dimenzija celega prostora. +\begin_inset Formula +\[ +\n\left(0\right)=n=n_{1}+\cdots+n_{k}\leq m_{1}+\cdots+m_{k} +\] + +\end_inset + +Ker +\begin_inset Formula $\forall i:m_{i}\leq n_{i}$ +\end_inset + + ( +\begin_inset CommandInset ref +LatexCommand vref +reference "claim:geom<=alg" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +), + velja v zgornji neenačbi enakost, + torej je matrika po +\begin_inset CommandInset ref +LatexCommand vref +reference "claim:mi=ni=>diag" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + diagonalizabilna. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsubsection +Korenski podprostori +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A\in M_{n}$ +\end_inset + + in +\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$ +\end_inset + + njen minimalni polinom. + +\begin_inset Formula $\forall i\in\left\{ 1..k\right\} $ +\end_inset + + označimo z +\begin_inset Formula $W_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$ +\end_inset + + korenski podprostor matrike +\begin_inset Formula $A$ +\end_inset + + za lastno vrednost +\begin_inset Formula $\lambda_{i}$ +\end_inset + +. + Vpeljimo še oznako +\begin_inset Formula $V_{i}\coloneqq\Ker\left(A-\lambda_{i}I\right)^{1}$ +\end_inset + + (tu potenca ni +\begin_inset Formula $r_{i}$ +\end_inset + +, + temveč je +\begin_inset Formula $1$ +\end_inset + +). +\end_layout + +\begin_layout Definition* +\begin_inset CommandInset counter +LatexCommand set +counter "theorem" +value "0" +lyxonly "false" + +\end_inset + + +\end_layout + +\begin_layout Fact +Očitno je +\begin_inset Formula $\Ker\left(A-\lambda_{i}I\right)\subseteq\Ker\left(A-\lambda_{i}\right)^{2}\subseteq\Ker\left(A-\lambda_{i}I\right)^{3}\subseteq\cdots$ +\end_inset + +, + kajti če +\begin_inset Formula $x\in\Ker\left(A-\lambda_{i}I\right)^{m}\Rightarrow\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow\left(A-\lambda_{i}I\right)\left(A-\lambda_{i}I\right)^{m}x=0\Rightarrow x\in\Ker\left(A-\lambda_{i}I\right)^{m+1}$ +\end_inset + +. + Izkaže se, + da so vse inkluzije do +\begin_inset Formula $r_{i}-te$ +\end_inset + + potence stroge, + od +\begin_inset Formula $r_{i}-$ +\end_inset + +te potence dalje pa so vse inkluzije enačaji, + torej za +\begin_inset Formula $W_{i}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}}$ +\end_inset + + velja +\begin_inset Formula +\[ +\Ker\left(A-\lambda_{i}I\right)\subset\Ker\left(A-\lambda_{i}I\right)^{2}\subset\cdots\subset\Ker\left(A-\lambda_{i}I\right)^{r_{i}}=\Ker\left(A-\lambda_{i}I\right)^{r_{i}+1}=\cdots +\] + +\end_inset + +Poleg tega se izkaže, + da je +\begin_inset Formula $\dim W_{i}=n_{i}$ +\end_inset + + (algebraična večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + +). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Fact +\begin_inset Formula $\dim V_{i}=\dim\Ker\left(A-\lambda_{i}I\right)=m_{1}$ +\end_inset + + (geometrijska večkratnost +\begin_inset Formula $\lambda_{i}$ +\end_inset + +). +\end_layout + +\begin_layout Claim +\begin_inset CommandInset label +LatexCommand label +name "claim:vsota-kor-podpr-je-vse" + +\end_inset + + +\begin_inset Formula $\mathbb{C}^{n}=W_{1}\oplus W_{2}\oplus\cdots\oplus W_{k}$ +\end_inset + + — + vsota vseh korenskih podprostorov je vse in ta vsota je direktna. + Tej vsoti pravimo +\begin_inset Quotes gld +\end_inset + +korenski razcep matrike +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Remark* +Dokazujmo: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Note Note +status open + +\begin_layout Plain Layout +\begin_inset Formula $V_{1}+\cdots+V_{k}$ +\end_inset + + je tudi direktna, + ampak ni nujno enaka +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Velja +\begin_inset Formula $\mathbb{C}^{n}=V_{1}+\cdots+V_{k}\Leftrightarrow A$ +\end_inset + + se da diagonalizirati (povedano prej). + Dokažimo trditev +\begin_inset CommandInset ref +LatexCommand vref +reference "claim:vsota-kor-podpr-je-vse" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +. + Dokažimo najprej, + da če +\begin_inset Formula $w_{1}\in W_{1},\dots,w_{k}\in W_{k}$ +\end_inset + + zadoščajo +\begin_inset Formula $w_{1}+\cdots+w_{k}=0$ +\end_inset + +, + velja +\begin_inset Formula $w_{1}=\cdots=w_{k}=0$ +\end_inset + + (direktna). + Delajmo indukcijo po številu členov: +\end_layout + +\begin_layout Itemize +Baza: + +\begin_inset Formula $w_{1}=0\Rightarrow w_{1}=0$ +\end_inset + + je očitno. +\end_layout + +\begin_deeper +\begin_layout Itemize +Indukcijska predpostavka: + +\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Korak: + Naj bodo +\begin_inset Formula $w_{1},\dots,w_{i+1}$ +\end_inset + + taki, + da +\begin_inset Formula +\[ +w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}} +\] + +\end_inset + + +\begin_inset Formula +\[ +w_{1}'+\cdots+w_{i}'+0=0, +\] + +\end_inset + +kajti +\begin_inset Formula $w_{i+1}\in\Ker\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$ +\end_inset + +. + Ker je vsak korenski prostor +\begin_inset Formula $W_{j}$ +\end_inset + + invarianten za +\begin_inset Formula $\left(A-\lambda_{i+1}I\right)^{r_{i+1}}$ +\end_inset + +, + ... + Tega dokaza ne najdem, + zato +\series bold +tega dokaza ne razumem +\series default +. + Po definiciji je +\begin_inset Formula $V$ +\end_inset + + invarianten za +\begin_inset Formula $A$ +\end_inset + +, + če za vsak +\begin_inset Formula $v\in V$ +\end_inset + + velja +\begin_inset Formula $Av\in V$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Plain Layout +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Plain Layout +Nadaljuj +\begin_inset Quotes gld +\end_inset + +LA1P FMF 2024-03-20.pdf +\begin_inset Quotes grd +\end_inset + + na strani 3. +\end_layout + +\end_inset + +Če predpostavimo, + da je vsota direktna, + je lahko dokazati, + da je vsota cel prostor. + V karakteristični polinom, + ki po Cayley-Hamiltonu anhilira +\begin_inset Formula $A$ +\end_inset + +, + vstavimo +\begin_inset Formula $A$ +\end_inset + + in dobimo +\begin_inset Formula $0=\left(-1\right)^{n}\left(A-\lambda_{1}I\right)^{r_{1}}\cdots\left(A-\lambda_{k}I\right)^{r_{k}}=A_{1}\cdots A_{k}$ +\end_inset + +. + Ker je vsota direktna, + velja +\begin_inset Formula $\n\left(A_{1}\cdots A_{n}\right)=\n\left(0\right)=n=\n A_{1}\cdots\n A_{k}=\dim\left(W_{1}+\cdots+W_{k}\right)$ +\end_inset + +, + torej +\begin_inset Formula $W_{1}+\cdots+W_{k}=\mathbb{C}^{n}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Če predpostavimo, + da je +\begin_inset Formula $W_{i}\cap W_{j}=\left\{ 0\right\} $ +\end_inset + + za +\begin_inset Formula $i\not=j$ +\end_inset + +, + lahko od tod izpeljemo direktnost vsote korenskih podprostorov. + Dokaz z indukcijo: +\end_layout + +\begin_deeper +\begin_layout Itemize +Baza: + +\begin_inset Formula $W_{1}$ +\end_inset + + je direktna vsota. + Očitno ( +\begin_inset Formula $\forall w_{1}\in W_{1}:w_{1}=0\Rightarrow w_{1}=0$ +\end_inset + +). +\end_layout + +\begin_layout Itemize +Indukcijska predpostavka: + +\begin_inset Formula $w_{1}+\cdots+w_{i}=0\Rightarrow w_{1}=\cdots=w_{i}=0$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Korak: + Naj bodo +\begin_inset Formula $w_{1},\dots,w_{i+1}$ +\end_inset + + taki, + da +\begin_inset Formula +\[ +w_{1}+\cdots+w_{i+1}=0\quad\quad\quad\quad/\cdot\left(A-\lambda_{i+1}I\right)^{r_{i+1}} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{1}+\cdots+\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{i}+0=0 +\] + +\end_inset + +Ker +\begin_inset Formula $\left(A-\lambda_{h}I\right)^{r_{h}}$ +\end_inset + + in +\begin_inset Formula $\left(A-\lambda_{k}I\right)^{r_{k}}$ +\end_inset + + za vsaka +\begin_inset Formula $h,k$ +\end_inset + + komutirata (gre namreč za polinom, + v katerega je vstavljen +\begin_inset Formula $A$ +\end_inset + +), + velja za vsak +\begin_inset Formula $j$ +\end_inset + + iz +\begin_inset Formula $\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}=0=\left(A-\lambda_{i+1}I\right)^{r_{i+1}}\left(A-\lambda_{j}I\right)^{r_{j}}w_{j}$ +\end_inset + + tudi +\begin_inset Formula +\[ +\left(A-\lambda_{j}I\right)^{r_{j}}\left(A-\lambda_{i+1}I\right)^{r_{i+1}}w_{j}=0 +\] + +\end_inset + +Ker je po I. + P. + +\begin_inset Formula $W_{1}+\cdots+W_{i}$ +\end_inset + + direktna, + velja za vsak +\begin_inset Formula $j$ +\end_inset + + +\begin_inset Formula $w_{j}\in W_{j}$ +\end_inset + +, + toda zaradi našega množenja tudi +\begin_inset Formula $w_{j}\in W_{i+1}$ +\end_inset + +. + Zaradi predpostavke +\begin_inset Formula $m=n\Rightarrow W_{m}\cup W_{n}=\left\{ 0\right\} $ +\end_inset + + velja za vsak +\begin_inset Formula $j\in\left\{ 1..i\right\} :$ +\end_inset + + +\begin_inset Formula $w_{j}=0$ +\end_inset + +. + V prvi enačbi ostane le še +\begin_inset Formula $w_{i+1}=0$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Itemize +Dokazati je treba še +\begin_inset Formula $i\not=j\Rightarrow W_{i}\cup W_{j}=\left\{ 0\right\} $ +\end_inset + +. + Dokažimo, + da je +\begin_inset Formula $W_{i}$ +\end_inset + + invarianten za +\begin_inset Formula $A$ +\end_inset + +, + t. + j. + +\begin_inset Formula $v\in W_{i}\Rightarrow Av\in W_{i}$ +\end_inset + +. + Če je +\begin_inset Formula $v\in W_{i}$ +\end_inset + +, + tedaj +\begin_inset Formula +\[ +\left(A-\lambda_{i}I\right)^{r_{i}}v=0\quad\quad\quad\quad/\cdot A +\] + +\end_inset + + +\begin_inset Formula +\[ +A\left(A-\lambda_{i}I\right)^{r_{i}}v=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +Av\in\Ker\left(A-\lambda_{i}I\right)^{r_{i}} +\] + +\end_inset + + +\begin_inset Formula +\[ +Av\in W_{i} +\] + +\end_inset + +Ker so vsi +\begin_inset Formula $W_{i}$ +\end_inset + + invariantni za +\begin_inset Formula $A$ +\end_inset + +, + so tudi njihovi preseki invariantni za +\begin_inset Formula $A$ +\end_inset + +. + Definirajmo torej linearno preslikavo +\begin_inset Formula $L:W_{i}\cap W_{j}\to W_{i}\cap W_{j}$ +\end_inset + + s predpisom +\begin_inset Formula $v\mapsto Av$ +\end_inset + +. + Vemo, + da ima +\begin_inset Formula $L$ +\end_inset + + vsaj eno lastno vrednost +\begin_inset Formula $\lambda$ +\end_inset + + in pripadajoči lastni vektor +\begin_inset Formula $w$ +\end_inset + +. + Torej +\begin_inset Formula $w\in W_{i}\cap W_{j}$ +\end_inset + + in +\begin_inset Formula $Lw=\lambda w$ +\end_inset + +, + toda +\begin_inset Formula $Lw=Aw=\lambda w$ +\end_inset + +. + Ker velja +\begin_inset Formula $Av=\lambda v\Rightarrow A^{q}v=\lambda^{q}v\Rightarrow p\left(A\right)v=p\left(\lambda\right)v$ +\end_inset + + za vsak polinom +\begin_inset Formula $p$ +\end_inset + +, + velja +\begin_inset Formula $p\left(A\right)w=p\left(\lambda\right)w$ +\end_inset + + za vsak polinom +\begin_inset Formula $p$ +\end_inset + +. + Uporabimo polinom +\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{i}\right)^{r_{i}}$ +\end_inset + + in dobimo +\begin_inset Formula $\left(A-\lambda_{i}\right)^{r_{i}}w=\left(\lambda-\lambda_{i}\right)^{r_{i}}w$ +\end_inset + +. + Leva stran enačbe je 0, + PDDRAA +\begin_inset Formula $w$ +\end_inset + + ni 0, + torej +\begin_inset Formula $\left(\lambda-\lambda_{i}\right)^{r_{i}}=0$ +\end_inset + +, + torej +\begin_inset Formula $\lambda=\lambda_{i}$ +\end_inset + +. + Vendar lahko namesto tistega polimoma uporabimo polinom +\begin_inset Formula $p\left(x\right)=\left(x-\lambda_{j}\right)^{r_{j}}$ +\end_inset + +, + kar pokaže +\begin_inset Formula $\lambda=\lambda_{j}$ +\end_inset + +, + torej +\begin_inset Formula $\lambda_{j}=\lambda_{i}$ +\end_inset + +, + kar je v +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + + s tem, + da so lastne vrednosti +\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$ +\end_inset + + paroma različne. + Torej +\begin_inset Formula $w=0$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Jordanska kanonična forma +\end_layout + +\begin_layout Standard +Vsaka kvadratna matrika je podobna posebni zgornjetrikotni matriki, + ki ji pravimo JKF. + To je bločno diagonalna matrika, + ki ima za diagonalne bloke t. + i. + +\begin_inset Quotes gld +\end_inset + +jordanske kletke +\begin_inset Quotes grd +\end_inset + +, + to so matrike oblike: +\begin_inset Formula +\[ +\left[\begin{array}{ccccc} +\lambda & 1\\ + & \lambda & 1\\ + & & \ddots & \ddots\\ + & & & \lambda & 1\\ + & & & & \lambda +\end{array}\right]. +\] + +\end_inset + +Jordanska matrika je sestavljena iz jordanskih kletk po diagonali ( +\begin_inset Formula $J_{i}$ +\end_inset + + so jordanske kletke): +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +J_{1} & & 0\\ + & \ddots\\ +0 & & J_{m} +\end{array}\right]. +\] + +\end_inset + +Običajno zahtevamo še, + da so JK, + ki imajo isto lastno vrednost, + skupaj, + ter da so JK padajoče urejene po lastni vrednosti od največje do najmanjše. +\end_layout + +\begin_layout Theorem* +Za vsako kvadratno kompleksno matriko +\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$ +\end_inset + + obstaja taka jordanska matrika +\begin_inset Formula $J$ +\end_inset + + in taka obrnljiva matrika +\begin_inset Formula $P$ +\end_inset + +, + da velja +\begin_inset Formula $A=PJP^{-1}$ +\end_inset + +. + ZDB vsaka +\begin_inset Formula $A\in M_{n\times n}\left(\mathbb{C}\right)$ +\end_inset + + je podobna neki Jordanski matriki. +\end_layout + +\begin_layout Standard +Procesu iskanja jordanske matrike pravimo +\begin_inset Quotes gld +\end_inset + +jordanifikacija +\begin_inset Quotes grd +\end_inset + +. + Kako konstruiramo +\begin_inset Formula $J$ +\end_inset + + in +\begin_inset Formula $P$ +\end_inset + +? + Izračunamo lastne vrednosti in pripadajoče korenske podprostore. + +\end_layout + +\begin_layout Itemize +Naj bo +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + +. + Za preglednost pišimo +\begin_inset Formula $N\coloneqq A-\lambda I$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Izračunamo lastne vektorje in lastni podprostor +\begin_inset Formula $\Ker N^{r}$ +\end_inset + + ter ga izrazimo z njegovo bazo, + recimo ji +\begin_inset Formula $B_{r}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Nato izberemo +\begin_inset Quotes gld +\end_inset + +pomožne baze +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula $\mathcal{B}_{1},\dots,\mathcal{B}_{r}$ +\end_inset + +, + ki pripadajo prostorom +\begin_inset Formula $\Ker N^{1},\dots,\Ker N^{r}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Pomožno bazo +\begin_inset Formula $\mathcal{B}_{r-1}$ +\end_inset + + dopolnimo do baze +\begin_inset Formula $\mathcal{B}_{r}$ +\end_inset + + z elementi +\begin_inset Formula $\mathcal{B}_{r}$ +\end_inset + + +\begin_inset Formula $u_{1},\dots,u_{k_{1}}$ +\end_inset + +. + Potem je +\begin_inset Formula $\mathcal{B}_{r-1}\cup$ +\end_inset + + +\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} $ +\end_inset + + +\begin_inset Quotes gld +\end_inset + +popravek pomožne baze +\begin_inset Formula $\mathcal{B}_{r}$ +\end_inset + + +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Itemize +Vektorje +\begin_inset Formula $\left\{ u_{1},\dots,u_{k_{1}}\right\} \in\Ker N^{r}$ +\end_inset + + pomnožimo z matriko +\begin_inset Formula $N$ +\end_inset + +, + dobljeni +\begin_inset Formula $Nu_{1},\dots,Nu_{k_{1}}$ +\end_inset + + ležijo v +\begin_inset Formula $\Ker N^{r-1}$ +\end_inset + +. + Množica +\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} $ +\end_inset + + je linearno neodvisna. + Izberemo take +\begin_inset Formula $v_{1},\dots,v_{k_{2}}\in B_{r-1}$ +\end_inset + +, + ki dopolnijo LN +\begin_inset Formula $B_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{2}\right\} $ +\end_inset + + do baze +\begin_inset Formula $\Ker N^{r-1}$ +\end_inset + +. + Potem je +\begin_inset Formula $\mathcal{B}_{r-2}\cup\left\{ Nu_{1},\dots,Nu_{k_{1}}\right\} \cup\left\{ v_{1},\dots,v_{k_{2}}\right\} $ +\end_inset + + popravek pomožne baze +\begin_inset Formula $\mathcal{B}_{r-1}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Izberemo take +\begin_inset Formula $w_{1},\dots,w_{k_{3}}\in\mathcal{B}_{r-2}$ +\end_inset + +, + ki +\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\dots,N^{2}u_{k_{1}}Nv_{1},\dots,Nv_{k_{2}}\right\} $ +\end_inset + + dopolnijo do baze +\begin_inset Formula $\Ker N^{r-2}$ +\end_inset + +. + Tedaj je +\begin_inset Formula $\mathcal{B}_{r-3}\cup\left\{ N^{2}u_{1},\cdots,N^{2}u_{k_{1}},Nv_{1},\dots,Nv_{k_{2}},w_{1},\dots,w_{k_{3}}\right\} $ +\end_inset + + popravek pomožne baze +\begin_inset Formula $B_{r-2}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Postopek ponavljamo, + dokler ne popravimo vseh možnih baz. +\end_layout + +\begin_layout Standard +Dobimo t. + i. + +\begin_inset Quotes gld +\end_inset + +jordanske verige +\begin_inset Quotes grd +\end_inset + +. + Ena jordanska veriga je +\begin_inset Formula $\left(u,Nu,N^{2}u,\dots,N^{x}u\right)$ +\end_inset + +, + torej preslikanje elementa +\begin_inset Formula $u$ +\end_inset + +, + ki začne kot dopolnitev baze korenskega podprostora +\begin_inset Formula $\Ker N^{x+1}$ +\end_inset + + in je na koncu +\begin_inset Formula $x-$ +\end_inset + +krat preslikan z +\begin_inset Formula $N$ +\end_inset + +, + torej konča v korenskem podprostoru +\begin_inset Formula $\Ker N$ +\end_inset + +. + Nekatere verige se začno v največjem korenskem podprostoru +\begin_inset Formula $\Ker N^{r}$ +\end_inset + +, + nekatere šele kasneje, + v +\begin_inset Formula $\Ker N^{1}$ +\end_inset + + ali pa +\begin_inset Formula $\Ker N^{2}$ +\end_inset + + ali pa +\begin_inset Formula $\Ker N^{3}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Imamo torej +\begin_inset Formula $k_{1}$ +\end_inset + + jordanskih verig dolžine +\begin_inset Formula $r$ +\end_inset + +, + +\begin_inset Formula $k_{2}$ +\end_inset + + jordanskih verig dolžine +\begin_inset Formula $r-1$ +\end_inset + +, + +\begin_inset Formula $k_{3}$ +\end_inset + + jordanskih verig dolžine +\begin_inset Formula $r-2$ +\end_inset + +, + ..., + +\begin_inset Formula $k_{r}$ +\end_inset + + jordanskih verig dolžine 1. + Skupaj je jordanskih verig +\begin_inset Formula $k_{1}+\cdots+k_{r}=\dim\Ker N$ +\end_inset + +. + Jordanskih verig za lastno vrednost +\begin_inset Formula $\lambda$ +\end_inset + + je torej toliko, + kot je njena geometrijska večkratnost. +\end_layout + +\begin_layout Standard +Vsaki jordanski verigi dolžine +\begin_inset Formula $k$ +\end_inset + + pripada ena jordanska kletka velikosti +\begin_inset Formula $k\times k$ +\end_inset + +. + +\begin_inset Formula $k-$ +\end_inset + +vektorjev iz verige zložimo v +\begin_inset Formula $P$ +\end_inset + + tako, + da je vektor z začetka verige (torej tisti iz popravljene baze večjega prostora) na levi strani v matriki. +\end_layout + +\begin_layout Example* +Poišči jordansko kanonično formo matrike +\begin_inset Formula +\[ +A=\left[\begin{array}{cccc} +0 & 1 & -1 & 2\\ +0 & 2 & 2 & 2\\ +0 & 0 & 2 & 0\\ +0 & 0 & 0 & 2 +\end{array}\right]. +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Najprej izračunamo karakteristični polinom: + +\begin_inset Formula $\det\left(A-\lambda I\right)=x\left(x-2\right)^{3}$ +\end_inset + +. + +\begin_inset Formula $\lambda_{1}=0$ +\end_inset + +, + +\begin_inset Formula $n_{1}=1$ +\end_inset + +, + +\begin_inset Formula $\lambda_{2}=2$ +\end_inset + +, + +\begin_inset Formula $n_{2}=3$ +\end_inset + +. + Lastni vektorji: + +\begin_inset Formula $\Ker\left(A-0I\right)=\Lin\left\{ \left(1,0,0,0\right)\right\} $ +\end_inset + +, + +\begin_inset Formula $\Ker\left(A-2I\right)=\Lin\left\{ \left(3,0,-2,2\right),\left(1,2,0,0\right)\right\} $ +\end_inset + +. + Če bi dobili 4 lastne vektorje, + bi lahko matriko diagonalizirali. + Tako je ne moremo. + Ker +\begin_inset Formula $n_{1}=1$ +\end_inset + +, + je +\begin_inset Formula $r_{1}$ +\end_inset + + največ +\begin_inset Formula $1$ +\end_inset + +, + torej +\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A-0I\right)^{2}=\cdots$ +\end_inset + +. + Izračunamo korenske podprostore +\end_layout + +\begin_deeper +\begin_layout Itemize +za lastno vrednost 0: + +\begin_inset Formula $\Ker\left(A-0I\right)=\Ker\left(A\right)=\Ker\left(A^{2}\right)$ +\end_inset + +. + Dobimo eno verigo +\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$ +\end_inset + + dolžine +\begin_inset Formula $1$ +\end_inset + + za lastno vrednost +\begin_inset Formula $0$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +za lastno vrednost 2: + +\begin_inset Formula $\Ker\left(A-2I\right)^{2}=\Lin\left\{ \left(1,0,0,2\right),\left(-1,0,1,0\right),\left(1,2,0,0\right)\right\} $ +\end_inset + +, + +\begin_inset Formula $\Ker\left(A-2I\right)^{3}=\Ker\left(A-2I\right)^{2}$ +\end_inset + +. + Opazimo, + da je +\begin_inset Formula $\left(1,0,0,2\right)$ +\end_inset + + dopolnitev baze +\begin_inset Formula $\Ker\left(A-2I\right)$ +\end_inset + + do baze +\begin_inset Formula $\Ker\left(A-2I\right)^{2}$ +\end_inset + +. + Torej je +\begin_inset Formula $\left\{ \left(1,0,0,2\right)\right\} $ +\end_inset + + +\begin_inset Quotes gld +\end_inset + +popravljena baza +\begin_inset Quotes grd +\end_inset + + +\begin_inset Formula $N^{2}$ +\end_inset + +. + Preslikamo +\begin_inset Formula $\left(A-2I\right)\left(1,0,0,2\right)=\left(2,4,0,0\right)$ +\end_inset + +, + kar tvori verigo dolžine 2 +\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$ +\end_inset + +. + Edini linearno neodvisen od +\begin_inset Formula $\left(2,4,0,0\right)$ +\end_inset + + v +\begin_inset Formula $\mathcal{B}_{1}$ +\end_inset + + je +\begin_inset Formula $\left(3,0,-2,2\right)$ +\end_inset + +, + zato je slednji začetek zadnje tretje verige dolžine 1 +\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Example* +Tri verige, + ki jih dobimo, + so +\begin_inset Formula $\left(\left(1,0,0,0\right)\right)$ +\end_inset + + za lastno vrednost +\begin_inset Formula $0$ +\end_inset + + in +\begin_inset Formula $\left(\left(1,0,0,2\right),\left(2,4,0,0\right)\right)$ +\end_inset + + ter +\begin_inset Formula $\left(\left(3,0,-2,2\right)\right)$ +\end_inset + + obe za lastno vrednost 2. + Zložimo jih v matriko +\begin_inset Formula $P$ +\end_inset + +: +\begin_inset Formula +\[ +P=\left[\begin{array}{cccc} +1 & 1 & 2 & 3\\ +0 & 0 & 4 & 0\\ +0 & 0 & 0 & -2\\ +0 & 2 & 0 & 2 +\end{array}\right] +\] + +\end_inset + +V matriko +\begin_inset Formula $J$ +\end_inset + + pa zložimo kletke pripadajočih velikosti: +\begin_inset Formula +\[ +J=\left[\begin{array}{cccc} +0 & & & 0\\ + & 2 & 1\\ + & & 2\\ +0 & & & 2 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +In velja +\begin_inset Formula $A=PJP^{-1}$ +\end_inset + + ( +\begin_inset Formula $P^{-1}$ +\end_inset + + izračunamo z Gaussom). +\end_layout + +\begin_layout Subsubsection +Funkcije matrik +\end_layout + +\begin_layout Standard +Če poznamo razcep +\begin_inset Formula $A=PJP^{-1}$ +\end_inset + +, + prevedemo računanje potenc +\begin_inset Formula $A$ +\end_inset + + na računanje potenc matrike +\begin_inset Formula $J$ +\end_inset + +, + kajti +\begin_inset Formula +\[ +A^{n}=\left(PJP^{-1}\right)\left(PJP^{-1}\right)\cdots\left(PJP^{-1}\right)=PJP^{-1}PJP^{-1}\cdots PJP^{-1}=PJ^{n}P^{-1}. +\] + +\end_inset + +Ker je +\begin_inset Formula $J$ +\end_inset + + bločno diagonalna matrika, + sestavljena iz jordanskih kletk, + se potenciranje +\begin_inset Formula $J$ +\end_inset + + prevede na potenciranje kletk, + kajti +\begin_inset Formula +\[ +J^{n}=\left[\begin{array}{ccc} +J_{1}^{n} & & 0\\ + & \ddots\\ +0 & & J_{m}^{n} +\end{array}\right]. +\] + +\end_inset + +Potenciranje jordanske kletke: +\begin_inset Formula +\[ +\left[\begin{array}{ccccc} +\lambda & 1 & & & 0\\ + & \lambda & 1\\ + & & \ddots & \ddots\\ + & & & \lambda & 1\\ +0 & & & & \lambda +\end{array}\right]^{n}=\left(\lambda I+\left[\begin{array}{ccc} +1 & & 0\\ + & \ddots\\ +0 & & 1 +\end{array}\right]\right)^{n}=\left(\lambda I+N\right)^{n}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\binom{n}{0}\left(\lambda I\right)^{n}N^{0}+\binom{n}{1}\left(\lambda N\right)^{n-1}N^{1}+\cdots+\binom{n}{n}\left(\lambda N\right)^{0}N^{n}=\binom{n}{0}\lambda^{n}+\binom{n}{1}\lambda^{n-1}N^{1}+\cdots+\binom{n}{n}N^{n} +\] + +\end_inset + +Poraja se vprašanje, + kako potencirati +\begin_inset Formula $N=\left[\begin{array}{ccccc} +0 & 1 & & & 0\\ + & \ddots & \ddots\\ + & & & \ddots\\ + & & & \ddots & 1\\ +0 & & & & 0 +\end{array}\right]$ +\end_inset + +. + Velja +\begin_inset Formula $N^{2}=\left[\begin{array}{ccccc} +0 & 0 & 1 & & 0\\ + & \ddots & \ddots & \ddots\\ + & & & \ddots & 1\\ + & & & \ddots & 0\\ +0 & & & & 0 +\end{array}\right]$ +\end_inset + + in tako dalje ( +\begin_inset Quotes gld +\end_inset + +diagonalo +\begin_inset Quotes grd +\end_inset + + enic pomikamo gor in desno). + Za +\begin_inset Formula $r\times r$ +\end_inset + + jordansko kletko, + kadar +\begin_inset Formula $n\geq r$ +\end_inset + + (sicer dobimo le prvih nekaj naddiagonal), + sledi +\begin_inset Formula +\[ +\left[\begin{array}{ccccc} +\lambda & 1 & & & 0\\ + & \lambda & 1\\ + & & \ddots & \ddots\\ + & & & \lambda & 1\\ +0 & & & & \lambda +\end{array}\right]^{n}=\left[\begin{array}{ccccc} +\lambda^{n} & n\lambda^{n-1} & \cdots & \binom{n}{r-2}\lambda^{n-r+2} & \binom{n}{r-1}\lambda^{n-r+1}\\ + & \lambda^{n} & n\lambda^{n-1} & \ddots & \binom{n}{r-2}\lambda^{n-r+2}\\ + & & \ddots & \ddots & \vdots\\ + & & & \lambda^{n} & n\lambda^{n-1}\\ +0 & & & & \lambda^{n} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Za računanje poljubne funkcije jordanske kletke pa velja predpis +\begin_inset Formula +\[ +f\left(\left[\begin{array}{ccccc} +\lambda & 1 & & & 0\\ + & \lambda & 1\\ + & & \ddots & \ddots\\ + & & & \lambda & 1\\ +0 & & & & \lambda +\end{array}\right]\right)=\left[\begin{array}{ccccc} +f\left(\lambda\right) & f'\left(\lambda\right) & \frac{f''\left(\lambda\right)}{2} & \cdots & \frac{f^{\left(k-1\right)\left(\lambda\right)}}{\left(k-1\right)!}\\ + & f\left(\lambda\right) & f'\left(\lambda\right) & \ddots & \cdots\\ + & & \ddots & \ddots & \frac{f''\left(\lambda\right)}{2}\\ + & & & f\left(\lambda\right) & f'\left(\lambda\right)\\ +0 & & & & f\left(\lambda\right) +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +In torej za računanje poljubne funkcije poljubne matrike +\begin_inset Formula $f\left(A\right)=f\left(PJP^{-1}\right)=Pf\left(J\right)P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Vektorski prostori s skalarnim produktom +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad poljem +\begin_inset Formula $\mathbb{R}$ +\end_inset + + nenujno končno razsežen. + Preslikavi +\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{R}$ +\end_inset + + pravimo skalarni produtkt, + če zadošča naslednjim lastnostim: +\end_layout + +\begin_deeper +\begin_layout Enumerate +pozitivna definitnost: + +\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle >0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +simetričnost: + +\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\left\langle u,v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Enumerate +linearnost v prvem faktorju: + +\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Corollary* +linearnost v drugem faktorju. + +\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle =\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle =\beta_{1}\left\langle v,v_{1}\right\rangle +\beta_{2}\left\langle v,v_{2}\right\rangle $ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Corollary* +Skalarni produkt z 0: + +\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$ +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Corollary* +Alternativna formulacija 1: + +\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$ +\end_inset + +. +\begin_inset Note Note +status open + +\begin_layout Plain Layout +Dokazujemo ekvivalenco: + alternativna formulacija 1 +\begin_inset Formula $\Leftrightarrow$ +\end_inset + + originalna definicija 1. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Leftarrow\right)$ +\end_inset + + Predpostavimo +\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \geq0$ +\end_inset + + in izjavo negirajmo: + +\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \leq0\Rightarrow v=0$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\Rightarrow\right)$ +\end_inset + + Predpostavimo +\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$ +\end_inset + + +\end_layout + +\end_deeper +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Primeri vektorskih prostorov s skalarnim produktom: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + s standardnim skalarnim produktom: + +\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + z nestandardnim skalarnim produktom: + Za pojubne +\begin_inset Formula $\gamma_{1}>0,\dots,\gamma_{n}>0$ +\end_inset + + definirajmo +\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\beta_{1}+\cdots+\gamma_{n}\alpha_{n}\beta_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +neskončno razsežen primer s standardnim skalarnim produktom: + +\begin_inset Formula $V=C\left[a,b\right]\sim$ +\end_inset + + zvezne +\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$ +\end_inset + +. + Definirajmo +\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle =\int_{a}^{b}f\left(x\right)g\left(x\right)dx$ +\end_inset + +. + Zveznost je potrebna za dokaz aksioma 1, + sicer za neznano neničelno funkcijo +\begin_inset Formula $f\left(x\right)=\begin{cases} +1 & ;x=0\\ +0 & ;\text{drugače} +\end{cases}$ +\end_inset + + velja +\begin_inset Formula $\int_{a}^{b}f\left(x\right)g\left(x\right)dx=0$ +\end_inset + +. + Temu pravimo standardni skalarni produkt v +\begin_inset Formula $C\left[a,b\right]$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +neskončno razsežen primer z nestandardnim skalarnim produktom: + Naj bo +\begin_inset Formula $w:\left[a,b\right]\to\mathbb{R}$ +\end_inset + + zvezna, + ki zadošča +\begin_inset Formula $\forall x\in\left[a,b\right]:w\left(x\right)>0$ +\end_inset + +. + Ostalo kot prej. + +\begin_inset Formula $\forall f,g\in V:\left\langle f,g\right\rangle _{w}=\int_{a}^{b}f\left(x\right)g\left(x\right)w\left(x\right)dx$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Remark* +Vektorski prostor s skalarnnim produktom je tak par +\begin_inset Formula $\left(V,\left\langle \cdot,\cdot\right\rangle \right)$ +\end_inset + +, + kjer je +\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle $ +\end_inset + + skalarni produkt na +\begin_inset Formula $V$ +\end_inset + +. + To je torej vektorski prostor, + za katerega izberemo in fiksiramo skalarni produkt. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad poljem +\begin_inset Formula $\mathbb{C}$ +\end_inset + + nenujno končno razsežen. + Preslikavi +\begin_inset Formula $\left\langle \cdot,\cdot\right\rangle :V\times V\to\mathbb{C}$ +\end_inset + + pravimo skalarni produtkt, + če zadošča naslednjim lastnostim: +\end_layout + +\begin_deeper +\begin_layout Enumerate +pozitivna definitnost: + +\begin_inset Formula $\forall v\in V:v\not=0\Rightarrow\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle >0$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +konjugirana simetričnost: + +\begin_inset Formula $\forall u,v\in V:\left\langle v,u\right\rangle =\overline{\left\langle u,v\right\rangle }$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +linearnost v prvem faktorju: + +\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in\mathbb{R},v_{1},v_{2}\in V:\left\langle \alpha_{1}v_{1}+\alpha_{2}v_{2},v\right\rangle =\alpha_{1}\left\langle v_{1},v\right\rangle +\alpha_{2}\left\langle v_{2},v\right\rangle $ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Corollary* +konjugirana linearnost v drugem faktorju. + +\begin_inset Formula $\left\langle u,\beta_{1}v_{1}+\beta_{2}v_{2}\right\rangle =\overline{\left\langle \beta_{1}v_{1}+\beta_{2}v_{2},v\right\rangle }=\overline{\beta_{1}\left\langle v_{1},v\right\rangle +\beta_{2}\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\overline{\left\langle v_{1},v\right\rangle }+\overline{\beta_{2}}\overline{\left\langle v_{2},v\right\rangle }=\overline{\beta_{1}}\left\langle v,v_{1}\right\rangle +\overline{\beta_{2}}\left\langle v,v_{2}\right\rangle $ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Corollary* +Skalarni produkt z 0: + +\begin_inset Formula $\left\langle 0,v\right\rangle =\left\langle 0\cdot v+0\cdot v,v\right\rangle =0\left\langle v,v\right\rangle +0\left\langle v,v\right\rangle =0\Rightarrow\left\langle v,0\right\rangle =0$ +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Corollary* +Alternativna formulacija 1: + +\begin_inset Formula $\forall v\in V:\left\langle v,v\right\rangle \in\mathbb{R}\wedge\left\langle v,v\right\rangle \geq0\wedge\left\langle v,v\right\rangle =0\Leftrightarrow v=0$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri vektorskih prostorov s skalarnim produktom: +\end_layout + +\begin_deeper +\begin_layout Itemize +standardni skalarni produkt na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +: + +\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +nestandardni skalarni produkt na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +: + Za neke +\begin_inset Formula $\gamma_{1}\in\mathbb{\mathbb{R}}^{+},\dots,\gamma_{n}\in\mathbb{R}^{+}$ +\end_inset + + definiramo +\begin_inset Formula $\left\langle \left(\alpha_{1},\dots,\alpha_{n}\right),\left(\beta_{1},\dots,\beta_{n}\right)\right\rangle =\gamma_{1}\alpha_{1}\overline{\beta_{1}}+\cdots+\gamma_{n}\alpha_{n}\overline{\beta_{n}}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +neskončno razsežen vektorski prostor na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + s standardnim skalarnim produktom: + Naj bo +\begin_inset Formula $V=C\left(\left[a,b\right],\mathbb{C}\right)$ +\end_inset + + — + +\begin_inset Formula $f=g+ih$ +\end_inset + + za +\begin_inset Formula $g,h\in C\left[a,b\right]$ +\end_inset + + (zvezni funkciji iz +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + v +\begin_inset Formula $\mathbb{R}$ +\end_inset + +). + Definiramo +\begin_inset Formula $\left\langle f_{1},f_{2}\right\rangle =\int_{a}^{b}f_{1}\left(x\right)\overline{f_{2}\left(x\right)}dx=\int_{a}^{b}\left(g_{1}+ih_{1}\right)\left(x\right)\left(g_{2}-ih_{2}\right)\left(x\right)dx=\int_{a}^{b}\left(g_{1}g_{2}+g_{1}g_{2}\right)\left(x\right)dx+i\int_{a}^{b}\left(h_{1}g_{2}-g_{1}h_{2}\right)xdx$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +neskončno razsežen vektorski prostor na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + z nestandardnim skalarnim produktom: + Isto kot zgoraj, + le da spet množimo z nekimi funkcijami, + kot pri realnem skalarnem produktu. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Norma +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor s skalarnim produktom. + +\begin_inset Formula $\forall v\in V:\left|\left|v\right|\right|=\sqrt{\left\langle v,v\right\rangle }$ +\end_inset + + je norma +\begin_inset Formula $v$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph +Osnovne lastnosti norme +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(\left|\left|v\right|\right|>0\Leftrightarrow v\not=0\right)\wedge\left|\left|0\right|\right|=0$ +\end_inset + + sledi iz prvega aksioma skalarnega produkta +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall\alpha\in F,v\in V:\left|\left|\alpha v\right|\right|=\left|\alpha\right|\left|\left|v\right|\right|$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +trikotniška neenakost: + +\begin_inset Formula $\forall u,v\in V:\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$ +\end_inset + + sledi iz Cauchy-Schwarzove neenakosti na običajen način. +\end_layout + +\begin_layout Claim* +Cauchy-Schwarz. + Za +\begin_inset Formula $V$ +\end_inset + + vektorski prostor s skalarnim produktom velja +\begin_inset Formula $\forall v\in V:\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Za +\begin_inset Formula $v=0$ +\end_inset + + očitno velja +\begin_inset Formula $0=0$ +\end_inset + +. + Za +\begin_inset Formula $v\not=0$ +\end_inset + + definirajmo +\begin_inset Formula +\[ +w=u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v +\] + +\end_inset + +po prvi lastnosti velja +\begin_inset Formula +\[ +0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle +\] + +\end_inset + +Oglejmo si +\begin_inset Formula +\[ +\left\langle w,v\right\rangle =\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\left\langle u,v\right\rangle -\left\langle \frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,v\right\rangle =\cancel{\left\langle u,v\right\rangle }-\frac{\cancel{\left\langle u,v\right\rangle }}{\cancel{\left\langle v,v\right\rangle }}\cancel{\left\langle v,v\right\rangle }=0 +\] + +\end_inset + +In se vrnimo k prejšnji enačbi: +\begin_inset Formula +\[ +0\leq\left\langle w,w\right\rangle =\left\langle w,u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v\right\rangle =\left\langle w,u\right\rangle -\frac{\overline{\left\langle u,v\right\rangle }}{\left\langle v,v\right\rangle }\left\langle w,v\right\rangle =\left\langle w,u\right\rangle -0=\left\langle w,u\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle u-\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }v,u\right\rangle =\left\langle u,u\right\rangle -\frac{\left\langle u,v\right\rangle }{\left\langle v,v\right\rangle }\left\langle v,u\right\rangle =\left|\left|u\right|\right|^{2}-\frac{\left\langle u,v\right\rangle \overline{\left\langle u,v\right\rangle }}{\left|\left|v\right|\right|^{2}}=\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}} +\] + +\end_inset + + +\begin_inset Formula +\[ +0\leq\left|\left|u\right|\right|^{2}-\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}} +\] + +\end_inset + + +\begin_inset Formula +\[ +\frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|\left|v\right|\right|^{2}}\leq\left|\left|u\right|\right|^{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left\langle u,v\right\rangle \right|^{2}\leq\left|\left|u\right|\right|^{2}\left|\left|v\right|\right|^{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right| +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Z normo lahko izrazimo skalarni produkt: +\end_layout + +\begin_deeper +\begin_layout Itemize +V +\begin_inset Formula $\mathbb{R}$ +\end_inset + +: + +\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\left(\left|\left|u+v\right|\right|^{2}-\left|\left|u-v\right|\right|^{2}\right)$ +\end_inset + + +\end_layout + +\begin_layout Itemize +V +\begin_inset Formula $\mathbb{C}$ +\end_inset + +: + +\begin_inset Formula $\left\langle u,v\right\rangle =\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Dokaz v +\begin_inset Formula $\mathbb{C}$ +\end_inset + +. + Oglejmo si +\begin_inset Formula +\[ +\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u+i^{k}v,u+i^{k}v\right\rangle =\left\langle u,u+i^{k}v\right\rangle +i^{k}\left\langle v,u+i^{k}v\right\rangle =\overline{\left\langle u+i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u+i^{k}v,v\right\rangle }= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\overline{\left\langle u,u\right\rangle }+\overline{\left\langle i^{k}v,u\right\rangle }+i^{k}\overline{\left\langle u,v\right\rangle }+i^{k}\overline{\left\langle i^{k}v,v\right\rangle }=\left\langle u,u\right\rangle +\left\langle u,i^{k}v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left\langle v,i^{k}v\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +i^{k}\left(-\left(i^{k}\right)\right)\left\langle v,v\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle +\] + +\end_inset + +Dodajmo vsoto: +\begin_inset Formula +\[ +\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\sum_{k=0}^{3}i^{k}\left(\left\langle u,u\right\rangle +\left(-i^{k}\right)\left\langle u,v\right\rangle +i^{k}\left\langle v,u\right\rangle +1\left\langle v,v\right\rangle \right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\sum_{k=0}^{3}i^{k}\left\langle u,u\right\rangle +\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +\sum_{k=0}^{3}i^{k}i^{k}\left\langle v,u\right\rangle +\sum_{k=0}^{3}i^{k}\left\langle v,v\right\rangle =0+\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle +0+0, +\] + +\end_inset + +kajti +\begin_inset Formula $\sum_{k=0}^{3}i^{k}=1+i+\left(-1\right)+\left(-i\right)=0$ +\end_inset + + in +\begin_inset Formula $\sum_{k=0}^{3}i^{2k}=1+\left(-1\right)+1+\left(-1\right)=0$ +\end_inset + +. + Nadaljujmo: +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +=\sum_{k=0}^{3}i^{k}\left(-i^{k}\right)\left\langle u,v\right\rangle =\sum_{k=0}^{3}1\left\langle u,v\right\rangle =4\left\langle u,v\right\rangle +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Ortogonalne množice in ortogonalne baze +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + VPSSP +\begin_inset Formula $\forall u,v\in V:u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +trivialne opombe. + +\begin_inset Formula $\forall v\in V:v\perp\vec{0}$ +\end_inset + +, + +\begin_inset Formula $\forall v\in V:v\not=0\Leftrightarrow v\not\perp v$ +\end_inset + + (prvi aksiom skalarnega produkta), + +\begin_inset Formula $\forall v\in V:u\perp v\Leftrightarrow v\perp u$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + VPSSP in +\begin_inset Formula $v_{1},\dots,v_{k}\in V$ +\end_inset + +. + Množica +\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $ +\end_inset + + je: +\end_layout + +\begin_deeper +\begin_layout Itemize +ortogonalna, + če +\begin_inset Formula $v_{1}\not=0\wedge\cdots\wedge v_{k}\not=0$ +\end_inset + + in +\begin_inset Formula $\forall i,j\in\left\{ 1..k\right\} :i\not=j\Rightarrow v_{i}\perp v_{j}$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +normirana, + če +\begin_inset Formula $\forall v\in\left\{ v_{1},\dots,v_{k}\right\} :\left|\left|v\right|\right|=1$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +ortonormirana, + če je ortogonalna in ortonormirana hkrati. +\end_layout + +\end_deeper +\begin_layout Remark* +Iz (ortogonalne) množice +\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $ +\end_inset + + dobimo (orto)normirano tako, + da vsak element delimo z njegovo normo. + +\begin_inset Formula $\left\{ \frac{v_{1}}{\left|\left|v_{1}\right|\right|},\dots,\frac{v_{k}}{\left|\left|v_{k}\right|\right|}\right\} $ +\end_inset + + je vedno normirana. +\end_layout + +\begin_layout Claim* +Vsaka ortogonalna množica je linearno neodvisna. +\end_layout + +\begin_layout Proof +Denimo, + da je +\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $ +\end_inset + + ortogonalna. + Vzemimo take +\begin_inset Formula $\alpha_{1},\dots,\alpha_{k}\ni:\alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k}=0$ +\end_inset + +. + +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$ +\backslash +alpha_1= +\backslash +cdots= +\backslash +alpha_k=0$} +\end_layout + +\end_inset + +. + +\begin_inset Formula +\[ +\forall i\in\left\{ 1..k\right\} :0=\left\langle 0,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{k}v_{k},v\right\rangle =\alpha_{1}\left\langle v_{1},v_{i}\right\rangle +\cdots+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cdots+\alpha_{k}\left\langle v_{k},v_{i}\right\rangle =\cdots +\] + +\end_inset + +Ker je množica ortogonalna, + je +\begin_inset Formula $\left\langle v_{l},v_{k}\right\rangle =0\Leftrightarrow l\not=k$ +\end_inset + +. + Nadaljujmo ... +\begin_inset Formula +\[ +\cdots=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\] + +\end_inset + +Ker +\begin_inset Formula $\left\langle v_{i},v_{i}\right\rangle $ +\end_inset + + ni 0, + ker je +\begin_inset Formula $v_{i}$ +\end_inset + + neničeln (da, + tudi to je del definicije ortogonalnosti), + je +\begin_inset Formula $\alpha_{i}=0$ +\end_inset + +. + In to za vsak +\begin_inset Formula $i$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Ni pa vsaka ortogonalna množica ogrodje. + Ortogonalni množici, + ki je ogrodje, + rečemo ortogonalna baza (LN sledi iz ortogonalnost). +\end_layout + +\begin_layout Subsubsection +Fourierov razvoj +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP, + +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} =\mathcal{B}$ +\end_inset + + ortogonalna baza za +\begin_inset Formula $V$ +\end_inset + + in +\begin_inset Formula $v\in V$ +\end_inset + + poljuben element. + Kako razvijemo +\begin_inset Formula $v$ +\end_inset + + po +\begin_inset Formula $\mathcal{B}$ +\end_inset + +, + vedoč, + da je ta baza ortogonalna? + Postopek imenujemo Fourierov razvoj. +\end_layout + +\begin_layout Standard +Ker je +\begin_inset Formula $\mathcal{B}$ +\end_inset + + ogrodje, + +\begin_inset Formula $\exists\alpha_{1},\dots,\alpha_{n}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + +. + Množimo skalarno z +\begin_inset Formula $v_{i}$ +\end_inset + +: +\begin_inset Formula +\[ +v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\quad\quad\quad\quad/\cdot v_{i} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle v,v_{i}\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},v_{i}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle v,v_{i}\right\rangle =\cancel{\alpha_{1}\left\langle v_{1},v_{i}\right\rangle }+\cancel{\cdots}+\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\cancel{\cdots}+\cancel{\alpha_{n}\left\langle v_{n},v_{i}\right\rangle }=\alpha_{i}\left\langle v_{i},v_{i}\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }=\alpha_{i} +\] + +\end_inset + +Torej +\begin_inset Formula $\forall v\in V$ +\end_inset + + velja +\begin_inset Formula $v=\sum_{i=1}^{n}\frac{\left\langle v,v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}$ +\end_inset + +. + Koeficientu +\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\left|\left|v_{i}\right|\right|^{2}}$ +\end_inset + + pravimo Fourierov koeficient. + Če je baza ortonormirana, + je Fourierov koeficient +\begin_inset Formula $\frac{\left\langle v,v_{i}\right\rangle }{\cancel{\left|\left|v_{i}\right|\right|^{2}}}=\left\langle v,v_{i}\right\rangle $ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Parsevalova identiteta +\end_layout + +\begin_layout Theorem* +Parsevalova identiteta. + Naj bo +\begin_inset Formula $V$ +\end_inset + + VPSSP in +\begin_inset Formula $\left\{ v_{1},\dots,v_{k}\right\} $ +\end_inset + + njegova ortogonalna baza. + Tedaj +\begin_inset Formula $\forall v\in V:$ +\end_inset + + +\begin_inset Formula +\[ +\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\frac{\left|\left\langle v,v_{i}\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle }. +\] + +\end_inset + +Če je baza ortonormirana, + se enačba očitno poenostavi v +\begin_inset Formula +\[ +\left|\left|v\right|\right|^{2}=\sum_{i=1}^{n}\left|\left\langle v,v_{i}\right\rangle \right|^{2}. +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$ +\end_inset + +. + Tedaj +\begin_inset Formula $\left|\left|v\right|\right|^{2}=\left\langle v,v\right\rangle =\left\langle \alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n},\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}\right\rangle =$ +\end_inset + + (uporabimo linearnost v 1. + in konjugirano linearnost v 2. + faktorju) +\begin_inset Formula +\[ +\begin{array}{ccccccc} += & \alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle & + & \cancel{\cdots} & + & \cancel{\alpha_{1}\overline{\alpha_{n}}\left\langle v_{1},v_{n}\right\rangle } & +\\ + & \vdots & & & & \vdots\\ ++ & \cancel{\alpha_{n}\overline{\alpha_{1}}\left\langle v_{n},v_{1}\right\rangle } & + & \cancel{\cdots} & + & \alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle & = +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\alpha_{1}\overline{\alpha_{1}}\left\langle v_{1},v_{1}\right\rangle +\cdots+\alpha_{n}\overline{\alpha_{n}}\left\langle v_{n},v_{n}\right\rangle =\left|\alpha_{1}\right|^{2}\left|\left|v_{1}\right|\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}\left|\left|v_{n}\right|\right|^{2}= +\] + +\end_inset + +Vstavimo formule za koeficiente po Fourierjevem razvoju: +\begin_inset Formula +\[ +=\left|\frac{\left\langle v,v_{1}\right\rangle }{\left|\left|v_{1}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{1}\right|\right|^{2}}+\cdots+\left|\frac{\left\langle v,v_{n}\right\rangle }{\left|\left|v_{n}\right|\right|^{\cancel{2}}}\right|^{2}\cancel{\left|\left|v_{n}\right|\right|^{2}}=\frac{\left|\left\langle v,v_{1}\right\rangle \right|^{2}}{\left|\left|v_{1}\right|\right|}+\cdots+\frac{\left|\left\langle v,v_{n}\right\rangle \right|^{2}}{\left|\left|v_{n}\right|\right|}=\sum_{i=1}^{n}\frac{\left|\left\langle v_{i},v\right\rangle \right|^{2}}{\left\langle v_{i},v_{i}\right\rangle } +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Projekcija na podprostor +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $W$ +\end_inset + + podprostor +\begin_inset Formula $V$ +\end_inset + +. + Za vsak +\begin_inset Formula $v\in V$ +\end_inset + + želimo izračunati njegovo ortogonalno projekcijo na +\begin_inset Formula $W$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Vektor +\begin_inset Formula $v'\in W$ +\end_inset + + je ortogonalna projekcija vektorja +\begin_inset Formula $v\in V$ +\end_inset + +, + če +\begin_inset Formula $\forall w\in W:\left|\left|v-v'\right|\right|\leq\left|\left|v-w\right|\right|$ +\end_inset + +. + ZDB +\begin_inset Formula $v'$ +\end_inset + + je najbližje +\begin_inset Formula $v$ +\end_inset + + izmed vseh elementov +\begin_inset Formula $W$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Zadošča preveriti, + da je +\begin_inset Formula $v-v'$ +\end_inset + + ortogonalen na vse elemente +\begin_inset Formula $W$ +\end_inset + + (pitagorov izrek), + kajti v tem primeru (če predpostavimo +\begin_inset Formula $\left(v'-w\right)\perp\left(v-v'\right)$ +\end_inset + +) velja +\begin_inset Formula +\[ +\left|\left|v-w\right|\right|^{2}=\left|\left|v-v'+v'-w\right|\right|=\left|\left|v-v'\right|\right|^{2}+\left|\left|v'-w\right|\right|^{2}\geq\left|\left|v-v'\right|\right|^{2}. +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Dokaz pitagovorega izreka: + +\begin_inset Formula $\left|\left|a+b\right|\right|^{2}=\left\langle a+b,a+b\right\rangle =\left\langle a,a\right\rangle +\cancel{\left\langle a,b\right\rangle +\left\langle b,a\right\rangle }+\left\langle b,b\right\rangle =\left|\left|a\right|\right|^{2}+\left|\left|b\right|\right|^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $\left\{ w_{1},\dots,w_{k}\right\} $ +\end_inset + + ortogonalna baza za +\begin_inset Formula $W$ +\end_inset + +. + Formula za ortogonalno projekcijo se glasi: +\begin_inset Formula +\[ +v'=\frac{\left\langle v,w_{1}\right\rangle }{\left\langle w_{1},w_{1}\right\rangle }w_{1}+\cdots+\frac{\left\langle v,w_{k}\right\rangle }{\left\langle w_{k},w_{k}\right\rangle }=\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Dokažimo, + da je +\begin_inset Formula $v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i}$ +\end_inset + + pravokoten na vse elemente +\begin_inset Formula $W$ +\end_inset + +. + Zaradi linearnosti skalarnega produkta zadošča preveriti, + da je pravokoten na bazo +\begin_inset Formula $W$ +\end_inset + +. + +\begin_inset Formula $\forall j\in\left\{ 1..k\right\} $ +\end_inset + + velja (spomnimo se, + da je +\begin_inset Formula $\left\langle w_{i},w_{j}\right\rangle =0\Leftrightarrow i\not=j$ +\end_inset + +, + zato po drugem enačaju ostane le še en člen vsote): +\begin_inset Formula +\[ +\left\langle v-\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\sum_{i=1}^{k}\frac{\left\langle v,w_{i}\right\rangle }{\left\langle w_{i},w_{i}\right\rangle }\left\langle w_{i},w_{j}\right\rangle =\left\langle v,w_{j}\right\rangle -\frac{\left\langle v,w_{j}\right\rangle }{\cancel{\left\langle w_{j},w_{j}\right\rangle }}\cancel{\left\langle w_{j},w_{j}\right\rangle }= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle v,w_{j}\right\rangle -\left\langle v,w_{j}\right\rangle =0 +\] + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Obstoj ortogonalne baze — + Gram-Schmidtova ortogonalizacija +\end_layout + +\begin_layout Standard +Radi bi dokazali, + da ima vsak KRVPSSP ortogonalno bazo in da je moč vsako ortogonalno množico dopolniti do ortogonalne baze. + Konstruktiven dokaz +\begin_inset Formula $\ddot{\smile}!$ +\end_inset + + — + postopek, + imenovan Gram-Schmidtova ortogonalizacija, + iz poljubne baze naredi ortogonalno. +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + njegova poljubna baza. + Naj bo +\begin_inset Formula $v_{1}\coloneqq u_{1}$ +\end_inset + +, +\begin_inset Formula +\[ +v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=u_{2}-u_{2}' +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }v_{2}=u_{3}-u_{3}' +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{n}\coloneqq u_{n}-\sum_{i=1}^{n-1}\frac{\left\langle u_{n},v_{i}\right\rangle }{\left\langle v_{i},v_{i}\right\rangle }v_{i}=u_{n}-u_{n}' +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Trdimo, + da je +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + ortogonalna baza za +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Opazimo, + da je +\begin_inset Formula $u_{2}'$ +\end_inset + + ortogonalna projekcija +\begin_inset Formula $u_{2}$ +\end_inset + + na +\begin_inset Formula $\Lin\left\{ v_{1}\right\} $ +\end_inset + +, + +\begin_inset Formula $u_{3}'$ +\end_inset + + ortogonalna projekcija +\begin_inset Formula $u_{3}$ +\end_inset + + na +\begin_inset Formula $\Lin\left\{ v_{1},v_{2}\right\} $ +\end_inset + +, + ..., + +\begin_inset Formula $u_{n}'$ +\end_inset + + pa ortogonalna projekcija na +\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n-1}\right\} $ +\end_inset + +, + torej +\begin_inset Formula +\[ +v_{2}=u_{2}-u_{2}'\perp\Lin\left\{ v_{1}\right\} \text{, torej }v_{2}\perp v_{1} +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{3}=u_{3}-u_{3}'\perp\Lin\left\{ v_{1},v_{2}\right\} \text{, torej }v_{3}\perp v_{1},v_{3}\perp v_{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{n}=u_{n}-u_{n}'\perp\Lin\left\{ v_{1},\dots,v_{n-1}\right\} \text{, torej }v_{n}\perp v_{1},\dots,v_{n}\perp v_{n-1}, +\] + +\end_inset + +kar pomeni, + da so +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + + paroma ortogonalni. + Toda vprašanje je, + ali so neničelni, + kajti to je, + ne boste verjeli, + prav tako pogoj za ortogonalno množico. + +\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :$ +\end_inset + + dokažimo neničelnost +\begin_inset Formula $v_{i}$ +\end_inset + +:-) +\end_layout + +\begin_layout Standard +\begin_inset Formula $v_{1}$ +\end_inset + + je neničeln, + ker je enak +\begin_inset Formula $u_{1}$ +\end_inset + +, + ki je element baze +\begin_inset Formula $V$ +\end_inset + +. + +\begin_inset Formula $v_{2}$ +\end_inset + + je neničeln, + ker je +\begin_inset Formula $v_{2}=u_{2}-\alpha v_{1}$ +\end_inset + + in +\begin_inset Formula $u_{2}\not=\alpha v_{1}$ +\end_inset + +, + ker sta linearno neodvisna, + ker tvorita ortogonalno množico. + +\begin_inset Formula $v_{3}$ +\end_inset + + je neničeln, + ker +\begin_inset Formula $v_{3}=u_{3}-\left(\beta v_{1}+\gamma v_{2}\right)$ +\end_inset + + in ker so +\begin_inset Formula $v_{1},v_{2},u_{3}$ +\end_inset + + LN, + +\begin_inset Formula $u_{3}\not=\left(\beta v_{1}+\gamma v_{2}\right)$ +\end_inset + +. + In tako dalje. +\end_layout + +\begin_layout Paragraph* +Dopolnitev ortogonalne množice do baze +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $\left\{ u_{1},\dots,u_{k}\right\} $ +\end_inset + + ortogonalna množica, + torej je linearno neodvisna, + torej jo lahko dopolnimo do baze. + +\begin_inset Formula $\left\{ u_{k+1},\dots,u_{n}\right\} $ +\end_inset + + je dopolnitev do baze. + Toda slednja še ni ortogonalna. + A nič ne de, + uporabimo lahko Gram-Schmidtovo ortogonalizacijo na +\begin_inset Formula $\left\{ u_{1},\dots,u_{k},u_{k+1},\dots,u_{n}\right\} $ +\end_inset + + in dobimo ortogonalno bazo +\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $ +\end_inset + +. + Opazimo, + da ker so po predpostavki +\begin_inset Formula $u_{1},\dots,u_{k}$ +\end_inset + + ortogonalni, + velja +\begin_inset Formula $v_{1}=u_{1},\dots,v_{k}=u_{k}$ +\end_inset + + (po GS). +\end_layout + +\begin_layout Example* +primer GS ortogonalizacije iz analize. + Naj bo +\begin_inset Formula $V=\mathbb{R}\left[x\right]_{\leq3}$ +\end_inset + +. + Baza: + +\begin_inset Formula $u_{1}=1,u_{2}=x,u_{3}=x^{2},u_{4}=x^{3}$ +\end_inset + +, + skalarni produkt naj bo +\begin_inset Formula $\left\langle p,q\right\rangle =\int_{-1}^{1}p\left(x\right)q\left(x\right)dx$ +\end_inset + +. + Konstruirajmo pripadajočo ortogonalno bazo +\begin_inset Formula $v_{1},\dots,v_{4}$ +\end_inset + +: +\begin_inset Formula +\[ +v_{1}\coloneqq u_{1}=1 +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{2}\coloneqq u_{2}-\frac{\left\langle u_{2},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}=x-\frac{\int_{-1}^{1}xdx}{\int_{-1}^{1}dx}=x-0=x=u_{2} +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{3}\coloneqq u_{3}-\frac{\left\langle u_{3},v_{1}\right\rangle }{\left\langle v_{1},v_{1}\right\rangle }v_{1}-\frac{\left\langle u_{3},v_{2}\right\rangle }{\left\langle v_{2},v_{2}\right\rangle }=x^{2}-\frac{\int_{-1}^{1}x^{2}dx}{\int_{-1}^{1}dx}-\frac{\int_{-1}^{1}x^{3}dx}{\int_{-1}^{1}x^{2}dx}x=\cdots=x²-\frac{1}{3} +\] + +\end_inset + + +\begin_inset Formula +\[ +v_{4}\coloneqq\cdots=x^{2}-\frac{3}{5}x +\] + +\end_inset + +Sklep: + +\begin_inset Formula $\left\{ 1,x,x^{2}-\frac{1}{3},x^{2}-\frac{3}{5}x\right\} $ +\end_inset + + je ortogonalna baza za ta vektorski prostor s tem skalarnim produktom. + Normirajmo jo! + Norme teh baznih vektorjev po vrsti so +\begin_inset Formula $\sqrt{2},\sqrt{\frac{2}{3}},\sqrt{\frac{8}{45}},\sqrt{\frac{8}{175}}$ +\end_inset + +. + Pripadajoča ortonormirana baza je torej +\begin_inset Formula $\left\{ \frac{1}{\sqrt{2}},\frac{x}{\sqrt{\frac{2}{3}}},\frac{x^{2}-\frac{1}{3}}{\sqrt{\frac{8}{45}}},\frac{x^{2}-\frac{3}{5}x}{\sqrt{\frac{8}{175}}}\right\} .$ +\end_inset + + Normiranje bi sicer prineslo lepše formule, + vendar bi v račune prineslo te objektivno grde konstante. +\end_layout + +\begin_layout Subsubsection +Ortogonalni komplement +\end_layout + +\begin_layout Definition +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP nad +\begin_inset Formula $F$ +\end_inset + + in +\begin_inset Formula $S\subseteq V$ +\end_inset + +. + Ortogonalni komplement +\begin_inset Formula $S$ +\end_inset + + je množica +\begin_inset Formula $S^{\perp}$ +\end_inset + +. + Vsebuje vse tiste vektorje iz +\begin_inset Formula $V$ +\end_inset + +, + ki so ortogonalni na +\begin_inset Formula $S$ +\end_inset + +. + ZDB +\begin_inset Formula $S^{\perp}\coloneqq\left\{ v\in V;\forall s\in S:v\perp s\right\} =\left\{ v\in V;v\perp S\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $\forall S\subseteq V:S^{\perp}$ +\end_inset + + je podprostor +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazati je treba +\begin_inset Formula $\forall u_{1},u_{2}\in S^{\perp},\alpha_{1},\alpha_{2}\in F:\alpha_{1}v_{1}+\alpha_{2}v_{2}\in S^{\perp}$ +\end_inset + +. + Po definiciji ortogonalnega komplementa velja +\begin_inset Formula +\[ +\forall s\in S:\left\langle u_{1},s\right\rangle =0\wedge\left\langle u_{2},s\right\rangle =0\Longrightarrow0=\alpha_{1}\left\langle u_{1},s\right\rangle +\alpha_{2}\left\langle u_{2},s\right\rangle =\left\langle \alpha_{1}u_{1}+\alpha_{2}u_{2},s\right\rangle \Longrightarrow\alpha_{1}u_{1}+\alpha_{2}u_{2}\in S^{\perp} +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +ortogonalni razcep. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $W$ +\end_inset + + vektorski podprostor +\begin_inset Formula $V$ +\end_inset + +. + Potem velja +\begin_inset Formula $V=W\oplus W^{\perp}$ +\end_inset + + (ortogonalni razcep glede na +\begin_inset Formula $W$ +\end_inset + +). +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v\in V$ +\end_inset + + pojuben, + +\begin_inset Formula $V^{\perp}$ +\end_inset + + pa ortogonalna projekcija +\begin_inset Formula $V$ +\end_inset + + na +\begin_inset Formula $W$ +\end_inset + +. + Potem velja, + da je +\begin_inset Formula $v=v-v'+v'$ +\end_inset + +, + kjer je +\begin_inset Formula $v-v'$ +\end_inset + + pravokoten na +\begin_inset Formula $W$ +\end_inset + +, + +\begin_inset Formula $v'$ +\end_inset + + pa element +\begin_inset Formula $v'$ +\end_inset + +, + torej +\begin_inset Formula $v\in W\oplus W^{\perp}$ +\end_inset + +. + Vsota je direktna, + kajti +\begin_inset Formula $\forall v\in W\cap W^{\perp}:v\perp v\Leftrightarrow v\perp v\Leftrightarrow\left\langle v,v\right\rangle =0\Leftrightarrow v=0\Longrightarrow W\cap W^{\perp}=\left\{ 0\right\} $ +\end_inset + + (po karakterizaciji direktnih vsot). +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $W$ +\end_inset + + vektorski podprostor v +\begin_inset Formula $V$ +\end_inset + +. + Velja +\begin_inset Formula $\left(W^{\perp}\right)^{\perp}=W$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Po definiciji ortogonalnega komplementa je +\begin_inset Formula $W\subseteq\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + ker +\begin_inset Formula $W\perp W^{\perp}$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\dim W=\dim\left(W^{\perp}\right)^{\perp}$ +\end_inset + +. + Ortogonalni razcep glede na +\begin_inset Formula $W$ +\end_inset + + je +\begin_inset Formula $V=W\oplus W^{\perp}\Rightarrow\dim W+\dim W^{\perp}=\dim V$ +\end_inset + +, + ortogonalni razcep glede na +\begin_inset Formula $W^{\perp}$ +\end_inset + + pa je +\begin_inset Formula $V=W^{\perp}\oplus\left(W^{\perp}\right)^{\perp}\Rightarrow\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp}=\dim V$ +\end_inset + +. +\begin_inset Formula +\[ +\dim V=\dim V +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim W+\dim W^{\perp}=\dim W^{\perp}+\dim\left(W^{\perp}\right)^{\perp} +\] + +\end_inset + + +\begin_inset Formula +\[ +\dim W^{\perp}=\dim\left(W^{\perp}\right)^{\perp} +\] + +\end_inset + + +\end_layout + +\begin_layout Proof +Alternativni dokaz: + Naj bodo +\begin_inset Formula $w_{1},\dots,w_{k}$ +\end_inset + + OB za +\begin_inset Formula $W$ +\end_inset + +. + Dopolnimo jo do OB za +\begin_inset Formula $V$ +\end_inset + + z GS z +\begin_inset Formula $w_{k+1},\dots,w_{n}$ +\end_inset + +. + Tedaj je +\begin_inset Formula $w_{k+1},\dots,w_{n}$ +\end_inset + + OB za +\begin_inset Formula $W^{\perp}$ +\end_inset + + in ker je +\begin_inset Formula $w_{1},\dots,w_{n}$ +\end_inset + + njena dopolnitev do OB +\begin_inset Formula $V$ +\end_inset + +, + je +\begin_inset Formula $w_{1},\dots,w_{k}$ +\end_inset + + OB za +\begin_inset Formula $\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + torej +\begin_inset Formula $W^{\perp}=\left(W^{\perp}\right)^{\perp}$ +\end_inset + +, + saj imata isti ortogonalni bazi. +\end_layout + +\begin_layout Subsection +Adjungirana linearna preslikava +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $V$ +\end_inset + + vektorski prostor nad +\begin_inset Formula $F$ +\end_inset + +. + Vemo, + da je +\begin_inset Formula $F$ +\end_inset + + vekrorski prostor nad +\begin_inset Formula $F$ +\end_inset + +. + Linearnim preslikavam +\begin_inset Formula $V\to F$ +\end_inset + + pravimo linearni funkcionali na +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Naj bo +\begin_inset Formula $V$ +\end_inset + + VPSSP in +\begin_inset Formula $F\in\left\{ \mathbb{R},\mathbb{C}\right\} $ +\end_inset + +. + Naj bo +\begin_inset Formula $w\in V$ +\end_inset + +. + Naj bo +\begin_inset Formula $\varphi:V\to F$ +\end_inset + + (torej je +\begin_inset Formula $\varphi$ +\end_inset + + linearni funkcional), + ki slika +\begin_inset Formula $v\mapsto\left\langle v,w\right\rangle $ +\end_inset + +. + Preslikava je po aksiomu 3 za skalarni produkt linearna. +\end_layout + +\begin_layout Theorem* +Rieszov izrek o reprezentaciji linearnih funkcionalov. + Naj bo +\begin_inset Formula $V$ +\end_inset + + KRVPSSP. + Za vsak linearen funkcional +\begin_inset Formula $\varphi$ +\end_inset + + na +\begin_inset Formula $V$ +\end_inset + + obstaja natanko en vektor +\begin_inset Formula $w\in V\ni:\forall v\in V:\varphi\left(v\right)=\left\langle v,w\right\rangle $ +\end_inset + +. + ZDB slednja konstrukcija nam da vse linearne funkcionale. +\end_layout + +\begin_layout Proof +Dokazujemo enolično eksistenco: +\end_layout + +\begin_deeper +\begin_layout Itemize +Eksistenca +\begin_inset Formula $w$ +\end_inset + +: + Vzemimo poljubno OB +\begin_inset Formula $w_{1},\dots,w_{n}$ +\end_inset + + za +\begin_inset Formula $V$ +\end_inset + +. + +\begin_inset Formula $\forall v\in V:v=\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}$ +\end_inset + +. + (fourierov razvoj po OB). + Ker je +\begin_inset Formula $\varphi$ +\end_inset + + linearna, + velja +\begin_inset Formula +\[ +\varphi\left(v\right)=\varphi\left(\left\langle v,w_{1}\right\rangle w_{1}+\cdots+\left\langle v,w_{n}\right\rangle w_{n}\right)\overset{\text{linearna}}{=}\left\langle v,w_{1}\right\rangle \varphi w_{1}+\cdots+\left\langle v,w_{n}\right\rangle \varphi w_{n}\overset{\text{konj. hom. v 2. fakt.}}{=} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle v,\overline{\varphi w_{1}}w_{1}\right\rangle +\cdots+\left\langle v,\overline{\varphi w_{n}}w_{n}\right\rangle \overset{\text{konj. ad. v 2. fakt.}}{=}\left\langle v,\left(\varphi w_{1}\right)w_{1}+\cdots+\left(\varphi w_{n}\right)w_{n}\right\rangle +\] + +\end_inset + +Za dan +\begin_inset Formula $\varphi$ +\end_inset + + smo konstruirali eksplicitno formulo za iskani +\begin_inset Formula $w$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Enoličnost +\begin_inset Formula $w$ +\end_inset + +: + PDDRAA +\begin_inset Formula $\forall v\in V:\varphi\left(v\right)=\left\langle v,w_{1}\right\rangle =\left\langle v,w_{2}\right\rangle $ +\end_inset + +. + Tedaj +\begin_inset Formula $\forall v\in V:\left\langle v,w_{1}-w_{2}\right\rangle =0$ +\end_inset + +. + Vzemimo konkreten +\begin_inset Formula $v=w_{1}-w_{2}$ +\end_inset + + in ga vstavimo v formulo +\begin_inset Formula $\left\langle v,w_{1}-w_{2}\right\rangle =0=\left\langle w_{1}-w_{2},w_{1}-w_{2}\right\rangle =0\Rightarrow w_{1}-w_{2}=0\Rightarrow w_{1}=w_{2}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Definition* +Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna. + Adjungirana linearna preslikava, + pripadajoča +\begin_inset Formula $L$ +\end_inset + +, + je taka +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +, + da velja +\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $ +\end_inset + +. + Levi skalarni produkt je tisti iz +\begin_inset Formula $V$ +\end_inset + +, + desni pa tisti iz +\begin_inset Formula $U$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Da lahko pišemo +\begin_inset Formula $L^{*}$ +\end_inset + +, + trdimo, + da je +\begin_inset Formula $L^{*}$ +\end_inset + + vedno obstaja in to vselej enolično. +\end_layout + +\begin_layout Proof +Dokazujemo enolično eksistenco: +\end_layout + +\begin_deeper +\begin_layout Itemize +Enoličnost: + Naj bosta +\begin_inset Formula $L^{*}$ +\end_inset + + in +\begin_inset Formula $L^{\circ}$ +\end_inset + + dve adjungirani linearni preslikavi za +\begin_inset Formula $L$ +\end_inset + +, + torej +\begin_inset Formula $\forall u\in U,v\in V:\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle $ +\end_inset + +. + Torej +\begin_inset Formula +\[ +\left\langle u,L^{*}v\right\rangle =\left\langle u,L^{\circ}v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +0=\left\langle u,L^{*}v-L^{\circ}v\right\rangle +\] + +\end_inset + +Za vsaka +\begin_inset Formula $u$ +\end_inset + + in +\begin_inset Formula $v$ +\end_inset + +. + Sedaj vstavimo +\begin_inset Formula $u=L^{*}v-L^{\circ}v$ +\end_inset + +: +\begin_inset Formula +\[ +0=\left\langle L^{*}v-L^{\circ}v,L^{*}v-L^{\circ}v\right\rangle \Longrightarrow L^{*}v-L^{\circ}v=0\Longrightarrow\forall v\in V:L^{*}v=L^{\circ}v +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Eksistenca: + Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in +\begin_inset Formula $L:U\to V$ +\end_inset + +. + Naj bo +\begin_inset Formula $v\in V$ +\end_inset + + poljuben. + Vpeljimo linearni funkcional +\begin_inset Formula $\varphi:U\to F$ +\end_inset + + s predpisom +\begin_inset Formula $u\mapsto\left\langle Lu,v\right\rangle $ +\end_inset + +. + Prepričajmo se, + da je ta funkcional linearna preslikava: + +\begin_inset Formula +\[ +\varphi\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\left\langle L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Lu_{1}+\alpha_{2}Lu_{2},v\right\rangle =\alpha_{1}\left\langle Lu_{1},v\right\rangle +\alpha_{2}\left\langle Lu_{2},v\right\rangle +\] + +\end_inset + +Uporabimo Rieszov izrek za funkcional +\begin_inset Formula $\varphi$ +\end_inset + +: + +\begin_inset Formula $\exists!w\in U\ni:\forall u\in U:\varphi u=\left\langle u,w\right\rangle $ +\end_inset + +. + Vpeljimo +\begin_inset Formula $L^{*}v=w$ +\end_inset + +, + s čimer za poljuben +\begin_inset Formula $v$ +\end_inset + + definiramo +\begin_inset Formula $L^{*}v$ +\end_inset + +. + Dokažimo, + da je dobljena preslikava linearna: + +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$L^{*} +\backslash +left( +\backslash +beta_{1}v_{1}+ +\backslash +beta_{2}v_{2} +\backslash +right)= +\backslash +beta_{1}L^{*}v_{1}+ +\backslash +beta_{2}L^{*}v_{2}$} +\end_layout + +\end_inset + +. + Vzemimo pojuben +\begin_inset Formula $u\in U$ +\end_inset + + in računajmo (fino bi bilo dobiti nič): +\begin_inset Formula +\[ +\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\beta_{1}L^{*}v_{1}-\beta_{2}L^{*}v_{2}\right\rangle \overset{\text{kl2f}}{=} +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left\langle u,L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle \overset{\text{lin }L^{*}}{=}\left\langle u,\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle +\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle -\overline{\beta_{1}}\left\langle u,L^{*}v_{1}\right\rangle -\overline{\beta_{2}}\left\langle u,L^{*}v_{2}\right\rangle =0 +\] + +\end_inset + +Ker to velja za vsak +\begin_inset Formula $u$ +\end_inset + +, + velja tudi za +\begin_inset Formula $u=L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)$ +\end_inset + +, + torej dobimo +\begin_inset Formula +\[ +\left\langle L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right),L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)\right\rangle =0 +\] + +\end_inset + +torej po prvem aksiomu za skalarni produtk velja linearnost: +\begin_inset Formula +\[ +L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)-\left(\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2}\right)=0 +\] + +\end_inset + + +\begin_inset Formula +\[ +L^{*}\left(\beta_{1}v_{1}+\beta_{2}v_{2}\right)=\beta_{1}L^{*}v_{1}+\beta_{2}L^{*}v_{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Example* +Naj bo +\begin_inset Formula $A\in M_{m\times n}\left(F\right)$ +\end_inset + + s pripadajočo linearno preslikavo +\begin_inset Formula $L_{A}=F^{n}\to F^{m}$ +\end_inset + +, + ki slika +\begin_inset Formula $v\mapsto Av$ +\end_inset + +. + Kako izgleda matrika +\begin_inset Formula $L_{A^{*}}$ +\end_inset + +? + Odgovor je odvisen od izbire skalarnega produkta. + Izberimo standardni skalarni produkt v +\begin_inset Formula $F^{n}$ +\end_inset + + in +\begin_inset Formula $F^{m}$ +\end_inset + + in +\begin_inset Formula $L_{A^{*}}:F^{m}\to F^{n}$ +\end_inset + + definiramo z +\begin_inset Formula $v\mapsto A^{*}v$ +\end_inset + +, + kjer je +\begin_inset Formula $A^{*}=\overline{A^{T}}$ +\end_inset + +, + torej transponiranka +\begin_inset Formula $A$ +\end_inset + + z vsemi elementi konjugiranimi. + Izkaže se, + da je potemtakem +\begin_inset Formula $L_{A^{*}}$ +\end_inset + + adjungirana linearna preslikava od +\begin_inset Formula $L_{A}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Matrika adjungirane linearne preslikave +\end_layout + +\begin_layout Standard +Naj bosta +\begin_inset Formula $U,V$ +\end_inset + + KRVPSSP in naj bo +\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $ +\end_inset + + ONB za +\begin_inset Formula $U$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $ +\end_inset + + ONB za +\begin_inset Formula $V$ +\end_inset + +. + Vzemimo linearno preslikavo +\begin_inset Formula $L:U\to V$ +\end_inset + +. + Izpeljimo zvezo med +\begin_inset Formula $L$ +\end_inset + + in +\begin_inset Formula $L^{*}$ +\end_inset + + glede na bazi +\begin_inset Formula $\mathcal{B}$ +\end_inset + + in +\begin_inset Formula $\mathcal{C}$ +\end_inset + +. + Torej +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + + za +\begin_inset Formula $L:U\to V$ +\end_inset + + in +\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$ +\end_inset + + za +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +. + Izračunajmo +\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$ +\end_inset + + tako, + da uporabimo fourierov razvoj: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +Lu_{1} & = & \left\langle Lu_{1},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{1},v_{m}\right\rangle v_{m}\\ +\vdots & & \vdots & & & & \vdots\\ +Lu_{n} & = & \left\langle Lu_{n},v_{1}\right\rangle v_{1} & + & \cdots & + & \left\langle Lu_{n},v_{m}\right\rangle v_{m} +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc} +\left\langle Lu_{1},v_{1}\right\rangle & \cdots & \left\langle Lu_{n},v_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle Lu_{1},v_{m}\right\rangle & \cdots & \left\langle Lu_{n},v_{m}\right\rangle +\end{array}\right]=\left[\begin{array}{ccc} +\left\langle u_{1},L^{*}v_{1}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle u_{1},L^{*}v_{m}\right\rangle & \cdots & \left\langle u_{n},L^{*}v_{m}\right\rangle +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Sedaj izračunajmo še +\begin_inset Formula $\left[L^{*}\right]_{\mathcal{B}\leftarrow\mathcal{C}}$ +\end_inset + + spet s fourierovim razvojem in primerjajmo istoležne koeficiente: +\begin_inset Formula +\[ +\begin{array}{ccccccc} +L^{*}v_{1} & = & \left\langle L^{*}v_{1},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle L^{*}v_{1},u_{n}\right\rangle u_{n}\\ +\vdots & & \vdots & & & & \vdots\\ +L^{*}v_{m} & = & \left\langle Lv_{m},u_{1}\right\rangle u_{1} & + & \cdots & + & \left\langle Lv_{m},u_{n}\right\rangle u_{n} +\end{array} +\] + +\end_inset + + +\begin_inset Formula +\[ +\left[L\right]_{\mathcal{B}\leftarrow\mathcal{C}}=\left[\begin{array}{ccc} +\left\langle L^{*}v_{1},u_{1}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{1}\right\rangle \\ +\vdots & & \vdots\\ +\left\langle L^{*}v_{1},u_{n}\right\rangle & \cdots & \left\langle L^{*}v_{m},u_{n}\right\rangle +\end{array}\right]=\left[\begin{array}{ccc} +\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{1},L^{*}v_{m}\right\rangle }\\ +\vdots & & \vdots\\ +\overline{\left\langle u_{n},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle } +\end{array}\right]=\left[\begin{array}{ccc} +\overline{\left\langle u_{1},L^{*}v_{1}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{1}\right\rangle }\\ +\vdots & & \vdots\\ +\overline{\left\langle u_{1},L^{*}v_{m}\right\rangle } & \cdots & \overline{\left\langle u_{n},L^{*}v_{m}\right\rangle } +\end{array}\right]^{T}= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\overline{\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}}^{T} +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Kako izgleda lastnost +\begin_inset Formula $\left\langle Lu,v\right\rangle =\left\langle u,L^{*}v\right\rangle $ +\end_inset + +? + Naj bo +\begin_inset Formula $u\in F^{n}$ +\end_inset + + in +\begin_inset Formula $v\in F^{m}$ +\end_inset + + in +\begin_inset Formula $A=m\times n$ +\end_inset + + matrika. + Ali za standardna skalarna produkta v +\begin_inset Formula $F^{n}$ +\end_inset + + in +\begin_inset Formula $F^{m}$ +\end_inset + + +\begin_inset Formula $\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle $ +\end_inset + + velja tudi za matrike, + če vzamemo +\begin_inset Formula $A^{*}=\overline{A}^{T}=\overline{A^{T}}$ +\end_inset + +? + Pa preverimo (ja, + velja): +\begin_inset Formula +\[ +\left\langle u,v\right\rangle =u_{1}\overline{v_{1}}+\cdots+u_{n}\overline{v_{n}}=\left[\begin{array}{ccc} +\overline{v_{1}} & \cdots & \overline{v_{n}}\end{array}\right]\left[\begin{array}{c} +u_{1}\\ +\vdots\\ +u_{n} +\end{array}\right]=v^{*}u +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle Au,v\right\rangle =v^{*}Au +\] + +\end_inset + + +\begin_inset Formula +\[ +\left\langle u,A^{*}v\right\rangle =\left(A^{*}v\right)^{*}u=v^{*}\left(A^{*}\right)^{*}u=v^{*}Au +\] + +\end_inset + + +\end_layout + +\begin_layout Fact* +Lastnosti adjungiranja: + +\begin_inset Formula $\left(\alpha A+\beta B\right)^{*}=\overline{\alpha}A^{*}+\overline{\beta}B^{*}$ +\end_inset + +, + +\begin_inset Formula $\left(AB\right)^{*}=B^{*}A^{*}$ +\end_inset + +, + +\begin_inset Formula $\left(A^{*}\right)^{*}=A$ +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout +TODO XXX FIXME DOKAŽI +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Jedro in slika adjungirane linearne preslikave +\end_layout + +\begin_layout Claim* +Naj bo +\begin_inset Formula $L:U\to V$ +\end_inset + + linearna. + Velja +\begin_inset Formula $\Ker\left(L^{*}\right)=\left(\Slika L\right)^{\perp}$ +\end_inset + + in +\begin_inset Formula $\Slika\left(L^{*}\right)=\left(\Ker L\right)^{\perp}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $v\in\Ker L^{*}$ +\end_inset + + za +\begin_inset Formula $L^{*}:V\to U$ +\end_inset + +. + Velja +\end_layout + +\begin_layout Proof +\begin_inset Formula +\[ +v\in\Ker\left(L^{*}\right)\Leftrightarrow L^{*}v=0\Leftrightarrow\forall u\in U:\left\langle u,L^{*}v\right\rangle =0\Leftrightarrow\forall u\in U:\left\langle Lu,v\right\rangle =0\Leftrightarrow\forall w\in\Slika L:\left\langle w,v\right\rangle =0\Leftrightarrow v\in\left(\Slika L\right)^{\perp} +\] + +\end_inset + +Velja torej +\begin_inset Formula $\Ker L^{*}=\left(\Slika L\right)^{\perp}\Rightarrow\Ker L=\left(\Slika L^{*}\right)^{\perp}\Rightarrow\left(\Ker L\right)^{\perp}=\Slika L^{*}\Rightarrow\left(\Ker L^{*}\right)^{\perp}=\Slika L$ +\end_inset + + +\end_layout + +\begin_layout Claim* +Za +\begin_inset Formula $L:U\to V$ +\end_inset + + velja +\begin_inset Formula $\Ker\left(L^{*}L\right)=\Ker L$ +\end_inset + + +\end_layout + +\begin_layout Proof +Vzemimo poljuben +\begin_inset Formula $u\in U$ +\end_inset + + in dokazujemo enakost množic (obe vsebovanosti): +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\supseteq\right)$ +\end_inset + + Če +\begin_inset Formula $u\in\Ker L\Rightarrow Lu=0\overset{\text{množimo z }L^{*}}{\Longrightarrow}L^{*}Lu=L^{*}u=0\Rightarrow u\in\Ker L^{*}L\Rightarrow\Ker L\subseteq\Ker L^{*}L$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(\subseteq\right)$ +\end_inset + + Če +\begin_inset Formula $u\in\Ker L^{*}L\Rightarrow L^{*}Lu=0\Rightarrow\left\langle u,L^{*}Lu\right\rangle =0\Rightarrow\left\langle Lu,Lu\right\rangle =0\Rightarrow Lu=0\Rightarrow u\in\Ker L\Rightarrow\Ker L^{*}L\subseteq\Ker L$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Corollary* +\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +\begin_inset Formula $\Slika\left(L^{*}L\right)=\Slika\left(L^{*}\left(L^{*}\right)^{*}\right)=\Slika\left(\left(L^{*}L\right)^{*}\right)=\left(\Ker L^{*}L\right)^{\perp}=\left(\Ker L\right)^{\perp}=\Slika L^{*}$ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Lastne vrednosti adjungirane linearne preslikave. +\end_layout + +\begin_layout Claim* +Če je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + +, + je +\begin_inset Formula $\overline{\lambda}$ +\end_inset + + lastna vrednost za +\begin_inset Formula $A^{*}$ +\end_inset + +. + ZDB +\begin_inset Formula $\det\left(A-\lambda I\right)=0\Rightarrow\det\left(A-\lambda\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $B=A-\lambda I$ +\end_inset + +. + Tedaj +\begin_inset Formula $B^{*}=A^{*}-\overline{\lambda}I^{*}=A^{*}-\overline{\lambda}I$ +\end_inset + +. + Radi bi dokazali +\begin_inset Formula $\det B=0\Rightarrow\det B^{*}=0$ +\end_inset + +. + Ker je +\begin_inset Formula $B^{*}=\overline{B}^{T}\Rightarrow\det B^{*}=\det\overline{B}^{T}=\det\overline{B}=\overline{\det B}\Rightarrow\det B=0\Rightarrow\det B^{*}=0$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Iz te formule izvemo tudi karakteristični polimom +\begin_inset Formula $A^{*}$ +\end_inset + +. + +\begin_inset Formula $p_{A^{*}}\left(x\right)=\det\left(A^{*}-xI\right)\Rightarrow p_{A}\left(\overline{x}\right)=\det\left(A-\overline{x}I\right)=\det\left(A^{*}-xI\right)^{*}=\overline{\det\left(A^{*}-xI\right)}=\overline{p_{A^{*}}\left(x\right)}$ +\end_inset + +, + torej +\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{p_{A}\left(\overline{x}\right)}$ +\end_inset + +. + Torej, + če je +\begin_inset Formula $p_{A}\left(x\right)=c_{0}x^{0}+\cdots+x_{n}x^{n}$ +\end_inset + +, + je +\begin_inset Formula $p_{A^{*}}\left(x\right)=\overline{c_{0}\overline{x^{0}}+\cdots+x_{n}\overline{x^{n}}}=\overline{c_{0}}x^{0}+\cdots+\overline{c_{n}}x^{n}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Alternativen dokaz: + Najprej dokažimo +\begin_inset Formula $\dim\Ker B^{*}=\dim\Ker B$ +\end_inset + +. + Velja +\begin_inset Formula $\dim\Ker B^{*}=\dim\left(\Slika B\right)^{\perp}=n-\dim\Slika B=\dim\Ker B$ +\end_inset + +. + Torej +\begin_inset Formula $\Ker\left(B\right)\not=0\Leftrightarrow\Ker\left(B^{*}\right)\not=0$ +\end_inset + +, + torej so lastne vrednosti +\begin_inset Formula $A^{*}$ +\end_inset + + konjugirane lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Med lastnimi vektorji +\begin_inset Formula $A$ +\end_inset + + in lastnimi vektorji +\begin_inset Formula $A^{*}$ +\end_inset + + (žal) ni posebne zveze. + Primer: + +\begin_inset Formula $A=\left[\begin{array}{cc} +1 & 2\\ +i & 1 +\end{array}\right]$ +\end_inset + + ima lastne vektorje +\begin_inset Formula $\vec{v_{1}}=\left[\begin{array}{c} +1-i\\ +-1 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\vec{v_{2}}=\left[\begin{array}{c} +1-i\\ +1 +\end{array}\right]$ +\end_inset + +, + +\begin_inset Formula $A^{*}=\left[\begin{array}{cc} +1 & 1\\ +2 & -i +\end{array}\right]$ +\end_inset + + pa lastne vektorje +\begin_inset Formula $\vec{v_{1}'}=\left[\begin{array}{c} +1-i\\ +-2 +\end{array}\right]$ +\end_inset + + in +\begin_inset Formula $\vec{v_{2}'}=\left[\begin{array}{c} +1-i\\ +2 +\end{array}\right]$ +\end_inset + +. + Med temi vektorji ni nobenih kolinearnosti. + Obstajajo pa zveze v nekaterih zanimivih primerih: +\end_layout + +\begin_layout Claim* +Če matrika +\begin_inset Formula $A$ +\end_inset + + zadošča +\begin_inset Formula $A^{*}A=AA^{A}$ +\end_inset + + (pravimo +\begin_inset Formula $A$ +\end_inset + + je normalna), + iz +\begin_inset Formula $Av=\lambda v$ +\end_inset + + sledi +\begin_inset Formula $A^{*}v=\overline{\lambda}v$ +\end_inset + +, + torej imata +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A^{*}$ +\end_inset + + iste lastne vrednosti. +\end_layout + +\begin_layout Proof +Če velja +\begin_inset Formula $Av=\lambda v$ +\end_inset + +, + velja +\begin_inset Formula $Av-\lambda v=\left(A-\lambda I\right)v=Bv=0\Rightarrow v\in\Ker B$ +\end_inset + +. + Če velja +\begin_inset Formula $A^{*}v=\overline{\lambda}v$ +\end_inset + +, + velja +\begin_inset Formula $A^{*}v-\overline{\lambda}v=\left(A^{*}-\overline{\lambda}I\right)v=B^{*}v=0\Rightarrow v\in\Ker B^{*}$ +\end_inset + +. + Dokazati je treba še +\begin_inset Formula $\Ker B=\Ker B^{*}$ +\end_inset + +: +\end_layout + +\begin_deeper +\begin_layout Enumerate +Ali velja +\begin_inset Formula $A^{*}A=AA^{*}\Rightarrow B^{*}B=BB^{*}$ +\end_inset + +? + Ja. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $B^{*}B=\left(A^{*}-\overline{\lambda}I\right)\left(A-\lambda I\right)=A^{*}A-\overline{\lambda}A-\lambda A^{*}+\overline{\lambda}\lambda I$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $BB^{*}=\left(A-\lambda I\right)\left(A^{*}-\overline{\lambda}I\right)=AA^{*}-\overline{\lambda}A-\lambda A^{*}+\lambda\overline{\lambda}I$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Enumerate +Ali velja +\begin_inset Formula $B^{*}B=BB^{*}\Rightarrow\Ker B=\Ker B^{*}$ +\end_inset + +? + Iz +\begin_inset Formula $B^{*}B=BB^{*}$ +\end_inset + + sledi +\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker\left(BB^{*}\right)\Rightarrow\Ker\left(B\right)=\Ker\left(B^{*}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\Ker B=\Ker B^{*}\Rightarrow\forall v\in V:Av=\lambda v\Leftrightarrow A^{*}v=\overline{\lambda}v$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Subsubsection +Normalne matrike +\end_layout + +\begin_layout Definition* +\begin_inset Formula $A$ +\end_inset + + je normalna +\begin_inset Formula $\Leftrightarrow A^{*}A=AA^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Dokazali smo že, + da za normalne matrike velja, + da imata +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A^{*}$ +\end_inset + + iste lastne vektorje, + kar v splošnem ne velja. +\end_layout + +\begin_layout Claim* +Lastni vektorji, + ki pripadajo različnim lastnim vrednostim normalne matrike, + so paroma ortogonalni. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A^{*}A=AA^{*}$ +\end_inset + + za neko +\begin_inset Formula $A$ +\end_inset + + in naj bo +\begin_inset Formula $Au=\lambda u$ +\end_inset + + in +\begin_inset Formula $Av=\mu v$ +\end_inset + + in +\begin_inset Formula $\mu\not=\lambda$ +\end_inset + +. + +\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$ +\end_inset + +. + Računajmo: +\begin_inset Formula +\[ +\mu\left\langle u,v\right\rangle =\left\langle u,\overline{\mu}v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle Au,v\right\rangle =\left\langle \lambda u,v\right\rangle =\lambda\left\langle u,v\right\rangle +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(\mu-\lambda\right)\left\langle u,v\right\rangle =0\wedge u\not=\lambda\Rightarrow\left\langle u,v\right\rangle =0\Leftrightarrow u\perp v +\] + +\end_inset + + +\end_layout + +\begin_layout Claim* +Vsako normalno matriko se da diagonalizirati. +\end_layout + +\begin_layout Proof +Dokažimo, + da je jordanska forma normalne matrike diagonalna +\begin_inset Formula $\Leftrightarrow$ +\end_inset + + vsi korenski podprostori so lastni. + +\begin_inset Formula $\forall m,\lambda:\Ker\left(A-I\lambda\right)^{m}=\Ker\left(A-I\lambda\right)$ +\end_inset + +. + Zadošča dokazati za +\begin_inset Formula $m=2$ +\end_inset + +. + Naj bo +\begin_inset Formula $m=2$ +\end_inset + + in +\begin_inset Formula $B=A-I\lambda$ +\end_inset + +. + Dokažimo +\begin_inset Formula $\Ker B^{2}=\Ker B$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Enumerate +Če v +\begin_inset Formula $\Ker\left(A\right)=\Ker\left(A^{*}A\right)$ +\end_inset + + vstavimo +\begin_inset Formula $A=B^{2}$ +\end_inset + +, + dobimo +\begin_inset Formula $\Ker B^{2}=\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Ker je +\begin_inset Formula $A$ +\end_inset + + normalna, + je +\begin_inset Formula $B$ +\end_inset + + normalna, + torej +\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=\left(B^{*}B\right)^{2}$ +\end_inset + +. + Torej +\begin_inset Formula $\Ker\left(\left(B^{2}\right)^{*}B^{2}\right)=\Ker\left(B^{*}B\right)^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Če v +\begin_inset Formula $\Ker A^{*}A=\Ker A$ +\end_inset + + vstavimo +\begin_inset Formula $A=B^{*}B$ +\end_inset + +, + dobimo +\begin_inset Formula $\Ker B^{*}BB^{*}B=\Ker\left(B^{*}B\right)^{2}=\Ker B^{*}B$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Zopet upoštevamo +\begin_inset Formula $\Ker A^{*}A=\Ker A$ +\end_inset + +, + torej +\begin_inset Formula $\Ker\left(B^{*}B\right)=\Ker B$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Proof +Ko dokažemo +\begin_inset Formula $B$ +\end_inset + + normalna +\begin_inset Formula $\Rightarrow B^{*}$ +\end_inset + + normalna, + bo iz +\begin_inset Formula $\Ker B^{2}=\Ker B$ +\end_inset + + sledilo +\begin_inset Formula $\Ker B^{4}=\Ker B$ +\end_inset + +. + Preverimo, + a je +\begin_inset Formula $B^{2}$ +\end_inset + + normalna, + če je +\begin_inset Formula $B$ +\end_inset + + normalna: + +\begin_inset Formula $\left(B^{2}\right)^{*}B^{2}=B^{*}B^{*}BB=B^{*}BB^{*}B=BB^{*}BB^{*}=BBB^{*}B^{*}=B^{2}\left(B^{2}\right)^{*}$ +\end_inset + +. + Sedaj vemo +\begin_inset Formula $\Ker B=\Ker B^{2}=\Ker B^{4}=\Ker B^{8}=\cdots$ +\end_inset + +. + Vemo pa tudi, + da +\begin_inset Formula +\[ +\Ker B\subseteq\Ker B^{2}\subseteq\Ker B^{3}\subseteq\Ker B^{4}\subseteq\Ker B^{5}\subseteq\Ker B^{6}\subseteq\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker B\subseteq\Ker B\subseteq\Ker B^{3}\subseteq\Ker B\subseteq\Ker B^{5}\subseteq\Ker B\subseteq\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\Ker B=\Ker B^{2}=\Ker B^{3}=\Ker B^{4}=\Ker B^{5}=\cdots +\] + +\end_inset + + +\begin_inset Formula +\[ +\forall v:\Ker B^{m}=\Ker B +\] + +\end_inset + + +\end_layout + +\begin_layout Remark* +Torej za vsako normalno matriko +\begin_inset Formula $A\exists$ +\end_inset + + diagonalna +\begin_inset Formula $D$ +\end_inset + + in obrnljiva +\begin_inset Formula $P$ +\end_inset + + z ortonormiranimi stolpci, + da velja +\begin_inset Formula $AP=PD$ +\end_inset + +, + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +. + Diagonala +\begin_inset Formula $D$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A$ +\end_inset + +, + stolpci +\begin_inset Formula $P$ +\end_inset + + pa so njeni lastni vektorji. + Lastni podprostori +\begin_inset Formula $\left(A-\lambda_{1}I\right),\dots,\left(A-\lambda_{n}I\right)$ +\end_inset + + so medsebojno pravokotni. + Izberimo ONB za vsak lasten podprostor. + Unija teh ONB je ONB za +\begin_inset Formula $F^{n}$ +\end_inset + +. + +\begin_inset Formula $F^{n}=\Ker\left(A-\lambda_{1}I\right)\oplus\cdots\oplus\Ker\left(A-\lambda_{n}I\right)$ +\end_inset + +. + Ta ONB so stolpci matrike +\begin_inset Formula $P$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +Ortogonalne/unitarne matrike +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $A$ +\end_inset + + kvadratna z ON stolpci glede na standardni skalarni produkt. + Pravimo, + da je +\begin_inset Formula $A$ +\end_inset + + unitarna (v kompleksnem primer) oziroma ortogonalna (v realnem primeru). +\end_layout + +\begin_layout Claim* +Za unitarno +\begin_inset Formula $A$ +\end_inset + + velja +\begin_inset Formula $A^{*}A=AA^{*}=I$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokazujmo za unitarno. + Za ortogonalno je dokaz podoben. + Naj bo +\begin_inset Formula $A=\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]$ +\end_inset + + unitarna. + To pomeni, + da za vsaka stolpca +\begin_inset Formula $a_{i}=\left(a_{1i},\dots,a_{ni}\right)$ +\end_inset + + in +\begin_inset Formula $a_{j}=\left(a_{1j},\dots,a_{nj}\right)$ +\end_inset + + velja za vsak +\begin_inset Formula $i,j\in\left\{ 1..n\right\} $ +\end_inset + + velja +\begin_inset Formula $\left\langle \text{\left[\begin{array}{c} +a_{1i}\\ +\text{\ensuremath{\vdots}}\\ +a_{ni} +\end{array}\right],\left[\begin{array}{c} +a_{1j}\\ +\vdots\\ +a_{nj} +\end{array}\right]}\right\rangle =a_{1i}\overline{a_{1j}}+\cdots+a_{ni}\overline{a_{nj}}=\begin{cases} +0 & ;i\not=j\\ +1 & ;i=j +\end{cases}$ +\end_inset + +. + Oglejmo si +\begin_inset Formula +\[ +A^{*}A=\left[\begin{array}{ccc} +\overline{a_{11}} & \cdots & \overline{a_{n1}}\\ +\vdots & & \vdots\\ +\overline{a_{1n}} & \cdots & \overline{a_{nn}} +\end{array}\right]\left[\begin{array}{ccc} +a_{11} & \cdots & a_{1n}\\ +\vdots & & \vdots\\ +a_{n1} & \cdots & a_{nn} +\end{array}\right]=\left[\begin{array}{ccc} +1 & & 0\\ + & \ddots\\ +0 & & 1 +\end{array}\right] +\] + +\end_inset + +Očitno je res, + ker je vsak element +\begin_inset Formula $A^{*}A$ +\end_inset + + konstruiran s skalarnim množenjem vrstice leve matrike (konjugirani stolpci +\begin_inset Formula $A$ +\end_inset + +, + ker smo poprej matriko transponiralo) in stolpca desne, + za kar predpis smo poprej že razbrali. +\end_layout + +\begin_layout Remark* +Za nekvadratne unitarne velja le +\begin_inset Formula $A^{*}A=I$ +\end_inset + +, + +\begin_inset Formula $AA^{*}=I$ +\end_inset + + pa zaradi nezmožnosti množenja zaradi nepravilnih dimenzij seveda ne velja. +\end_layout + +\begin_layout Claim* +Naslednje trditve so za +\begin_inset Formula $P$ +\end_inset + + z ortonormiranimi stolpci ekvivalentne: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $P^{*}P=I$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall$ +\end_inset + + ONB +\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $ +\end_inset + + je ON množica +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\exists$ +\end_inset + + ONB +\begin_inset Formula $\left\{ u_{1},\dots,u_{n}\right\} :\left\{ Pu_{1},\dots,Pu_{n}\right\} $ +\end_inset + + je ON množica +\end_layout + +\end_deeper +\begin_layout Proof +Dokazujemo ekvivalenco: +\end_layout + +\begin_deeper +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(1\Rightarrow2\right)$ +\end_inset + + +\begin_inset Formula $\left\langle Pu,Pv\right\rangle =\left\langle u,P^{*}Pv\right\rangle =\left\langle u,v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow3\right)$ +\end_inset + + +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left\langle Pu,Pu\right\rangle =\left\langle u,u\right\rangle =\left|\left|u\right|\right|^{2}$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow1\right)$ +\end_inset + + +\begin_inset Formula +\[ +\forall u,v:\left\langle Pu,Pv\right\rangle =\left\langle u,v\right\rangle \Rightarrow\left\langle u,P^{*}Pv\right\rangle -\left\langle u,v\right\rangle =0\Rightarrow\left\langle u,\left(P^{*}P-I\right)v\right\rangle =0 +\] + +\end_inset + + Sedaj izberimo +\begin_inset Formula $u=\left(P^{*}P-I\right)v$ +\end_inset + +: + +\begin_inset Formula $\left\langle \left(P^{*}P-I\right)v,\left(P^{*}P-I\right)v\right\rangle =0\Rightarrow P^{*}P-I=0\Rightarrow P^{*}P=0$ +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(3\Rightarrow2\right)$ +\end_inset + + Po predpostavki +\begin_inset Formula $\forall u:\left|\left|Pu\right|\right|=\left|\left|u\right|\right|$ +\end_inset + + Izrazimo skalarni produkt z normo: + +\begin_inset Formula $\left\langle u,v\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}$ +\end_inset + +, + torej +\begin_inset Formula +\[ +\left\langle Pu,Pv\right\rangle =\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|Pu+i^{k}Pv\right|\right|^{2}=\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|P\left(u+i^{k}v\right)\right|\right|^{2}\overset{\text{predpostavka}}{=}\frac{1}{4}\sum_{k=0}^{3}i^{k}\left|\left|u+i^{k}v\right|\right|^{2}=\left\langle u,v\right\rangle +\] + +\end_inset + + +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(5\Rightarrow4\right)$ +\end_inset + + Vzemimo poljuben +\begin_inset Formula $u$ +\end_inset + + in ga razvijmo po ONB +\begin_inset Formula $u_{1},\dots,u_{n}$ +\end_inset + +. + Tedaj +\begin_inset Formula $u=\alpha_{1}u_{1}+\cdots+\alpha_{n}u_{n}$ +\end_inset + +. + Ker so +\begin_inset Formula $u_{i}$ +\end_inset + + ONB, + velja +\begin_inset Formula $\left|\left|u\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots\left|\alpha_{n}\right|^{2}$ +\end_inset + +. + Ker so +\begin_inset Formula $Pu_{1}$ +\end_inset + + ONB po predpostavki, + +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\alpha_{1}\right|^{2}+\cdots+\left|\alpha_{n}\right|^{2}$ +\end_inset + +, + torej velja +\begin_inset Formula $\left|\left|Pu\right|\right|^{2}=\left|\left|u\right|\right|^{2}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(2\Rightarrow4\right)$ +\end_inset + + Ker so +\begin_inset Formula $u_{1},\dots,u_{n}$ +\end_inset + + ONM, + velja +\begin_inset Formula $\left\langle u_{i},u_{j}\right\rangle =\begin{cases} +1 & ;i=j\\ +0 & ;i\not=j +\end{cases}$ +\end_inset + +. + Tudi +\begin_inset Formula $Pu_{1},\dots,Pu_{n}$ +\end_inset + + ortonormirana, + kajti po predpostavki +\begin_inset Formula $2$ +\end_inset + + velja +\begin_inset Formula $\left\langle Pu_{i},Pu_{2}\right\rangle =\begin{cases} +1 & ;i=j\\ +0 & ;i\not=j +\end{cases}$ +\end_inset + +. +\end_layout + +\begin_layout Labeling +\labelwidthstring 00.00.0000 +\begin_inset Formula $\left(4\Rightarrow5\right)$ +\end_inset + + Očitno. +\end_layout + +\end_deeper +\begin_layout Claim* +Lastne vrednosti unitarne matrike +\begin_inset Formula $A$ +\end_inset + + se nahajajo na enotski krožnici v +\begin_inset Formula $\Im$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A$ +\end_inset + + unitarna in naj bo +\begin_inset Formula $v$ +\end_inset + + tak, + da +\begin_inset Formula $Av=\lambda v$ +\end_inset + +. + Tedaj +\begin_inset Formula $\left\langle v,v\right\rangle =\left\langle Av,Av\right\rangle =\left\langle \lambda v,\lambda v\right\rangle =\lambda\overline{\lambda}\left\langle v,v\right\rangle \Rightarrow\lambda\overline{\lambda}=1\Rightarrow\left|\lambda\right|=1\Rightarrow\lambda=e^{i\varphi}$ +\end_inset + + za nek +\begin_inset Formula $\varphi$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Iz unitarnosti sledi normalnost, + zato so lastni vektorji unitarne matrike, + ki pripadajo paroma različnim lastnim vrednostim, + pravokotni (isto, + kot pri normalnih matrikah). +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Prav tako kot pri normalnih matrikah lahko unitarne diagonalitziramo v tokrat ortogonalni bazi. + Pri unitarnih so stolpci +\begin_inset Formula $P$ +\end_inset + + še celo normirani. + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +, + kjer je +\begin_inset Formula $P$ +\end_inset + + unitarna, + torej +\begin_inset Formula $P^{*}=P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Remark* +Očitno je, + da če je +\begin_inset Formula $A$ +\end_inset + + unitarna, + velja +\begin_inset Formula $A^{*}=A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Subsubsection +\begin_inset CommandInset label +LatexCommand label +name "subsec:Simetrične/hermitske-matrike" + +\end_inset + +Simetrične/hermitske matrike +\end_layout + +\begin_layout Definition* +Matrika nad +\begin_inset Formula $\mathbb{R}$ +\end_inset + + je simetrična, + če zanjo velja +\begin_inset Formula $A^{*}=A$ +\end_inset + +. + Matrika nad +\begin_inset Formula $\mathbb{C}$ +\end_inset + + je hermitska, + če zanjo velja +\begin_inset Formula $A^{*}=A$ +\end_inset + +. + Linearni preslikavi, + pripadajoči hermitski/simetrični matriki, + pravimo sebiadjungirana. +\end_layout + +\begin_layout Fact* +Vsaka hermitska/simetrična matrika je normalna, + kajti +\begin_inset Formula $A^{*}A=AA=AA^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Lastne vrednosti hermitskih/simetričnih matrik so realne. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A=A^{*}$ +\end_inset + + in naj bo +\begin_inset Formula $Av=\lambda v$ +\end_inset + + za nek neničeln +\begin_inset Formula $v$ +\end_inset + +. + Tedaj +\begin_inset Formula $\lambda\left\langle v,v\right\rangle =\left\langle \lambda v,v\right\rangle =\left\langle Av,v\right\rangle =\left\langle v,A^{*}v\right\rangle =\left\langle v,Av\right\rangle =\left\langle v,\lambda v\right\rangle =\overline{\lambda}\left\langle v,v\right\rangle $ +\end_inset + +. + Potemtakem +\begin_inset Formula $\lambda=\overline{\lambda}\Rightarrow\lambda\in\mathbb{R}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalizacija je zopet enaka kot pri normalnih matrikah z dodatkom — + vsaka hermitska matrika je podobna realni diagonalni, + kar za normalne ni res — + normalne so lahko podobne kompleksnim diagonalnim matrikam. +\end_layout + +\begin_layout Subsubsection +Pozitivno (semi)definitne matrike +\end_layout + +\begin_layout Definition* +\begin_inset Formula $A$ +\end_inset + + je pozitivno semidefinitna +\begin_inset Formula $\sim A\geq0\Leftrightarrow A=A^{*}\wedge\forall v:\left\langle Av,v\right\rangle \geq0$ +\end_inset + +. + +\begin_inset Formula $A$ +\end_inset + + je pozitivno definitna +\begin_inset Formula $\sim A>0\Leftrightarrow A=A^{*}\wedge\forall v\not=0:\left\langle Av,v\right\rangle >0$ +\end_inset + +. + S tem ko skalarni produkt primerjamo ( +\begin_inset Formula $>,\geq$ +\end_inset + +), + implicitno zahtevamo njegovo realnost. + Primerjalni operatorji namreč na kompleksnih številih niso definirani. +\end_layout + +\begin_layout Example* +Vzemimo poljubno nenujno kvadratno +\begin_inset Formula $B$ +\end_inset + + in definirajmo +\begin_inset Formula $A=B^{*}B$ +\end_inset + +. + Potem je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna, + kajti +\begin_inset Formula $A^{*}=\left(B^{*}B\right)^{*}=B^{*}B=A$ +\end_inset + + in +\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle B^{*}Bv,v\right\rangle =\left\langle Bv,Bv\right\rangle \geq0$ +\end_inset + +. + Če pa bi bili stolpci +\begin_inset Formula $B$ +\end_inset + + linearno neodvisni, + pa bi veljalo +\begin_inset Formula $\forall v:v\not=0\Rightarrow\left\langle Av,v\right\rangle =\left\langle B^{*}Bv\right\rangle =\left\langle Bv,Bv\right\rangle >0$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $A\geq0\Rightarrow$ +\end_inset + + lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + so +\begin_inset Formula $\geq0$ +\end_inset + +. + +\begin_inset Formula $A>0\Rightarrow$ +\end_inset + + lastne vrednosti +\begin_inset Formula $A$ +\end_inset + + so +\begin_inset Formula $>0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $A\geq0$ +\end_inset + +. + Tedaj +\begin_inset Formula $Av=\lambda v$ +\end_inset + + za nek +\begin_inset Formula $v\not=0$ +\end_inset + +. + Torej +\begin_inset Formula $\left\langle Av,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda\left\langle v,v\right\rangle $ +\end_inset + +. + Toda ker +\begin_inset Formula $\left\langle Av,v\right\rangle \geq0$ +\end_inset + +, + sledi +\begin_inset Formula $\lambda\left\langle v,v\right\rangle \geq0$ +\end_inset + +. + Ker je +\begin_inset Formula $\left\langle v,v\right\rangle >0$ +\end_inset + +, + sledi +\begin_inset Formula $\lambda\geq0$ +\end_inset + +. + Analogno za +\begin_inset Formula $A>0$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalizacija je ista kot za normalna, + s tem da za diagonalno +\begin_inset Formula $D$ +\end_inset + + velja še, + da je pozitivno (semi)definitna, + ko je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna. +\end_layout + +\begin_layout Claim* +\begin_inset Formula $\forall A\geq0\exists B=B^{*},B\geq0\ni:B^{2}=A$ +\end_inset + +. + ZDB Za vsako pozitivno semidefinitno matriko +\begin_inset Formula $A$ +\end_inset + + obstajaja taka unitarna pozitivno semidefinitna +\begin_inset Formula $B$ +\end_inset + +, + da velja +\begin_inset Formula $B^{2}=A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Naj bo +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + in +\begin_inset Formula $P^{*}=P^{-1}$ +\end_inset + + in +\begin_inset Formula $D=\left[\begin{array}{ccc} +\lambda_{1} & & 0\\ + & \ddots\\ +0 & & \lambda_{n} +\end{array}\right]$ +\end_inset + +. + Definirajmo +\begin_inset Formula $E=\left[\begin{array}{ccc} +\sqrt{\lambda_{1}} & & 0\\ + & \ddots\\ +0 & & \sqrt{\lambda_{n}} +\end{array}\right]\geq0$ +\end_inset + +. + Naj bo +\begin_inset Formula $B=PEP^{-1}=PEP^{*}$ +\end_inset + +. + Opazimo +\begin_inset Formula $B^{*}=B$ +\end_inset + +, + kajti +\begin_inset Formula $\left(PEP^{-1}\right)^{*}=\left(PEP^{*}\right)^{*}=PE^{*}P^{*}=PEP^{-1}=PEP^{*}$ +\end_inset + +, + ker je +\begin_inset Formula $E^{*}=E$ +\end_inset + +, + ker je +\begin_inset Formula $\forall a\in\mathbb{R}:\sqrt{a}\in\mathbb{R}$ +\end_inset + +. + Oglejmo si +\begin_inset Formula $B^{2}=PEP^{-1}PEP^{-1}=PE^{2}P^{-1}=PDP^{-1}$ +\end_inset + +. + Tako definiramo +\begin_inset Formula $\sqrt{A}=B$ +\end_inset + + (tu +\begin_inset Formula $\sqrt{}$ +\end_inset + + ni funkcija, + kot pri JKF, + temveč nov operator). +\end_layout + +\begin_layout Claim* +Naslednje trditve so ekvivalentne (zamenjamo lahko +\begin_inset Formula $\geq$ +\end_inset + + in +\begin_inset Formula $>$ +\end_inset + +): +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $A\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=A^{*}$ +\end_inset + + in vse lastne vrednosti so +\begin_inset Formula $\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + + za nek unitaren +\begin_inset Formula $P$ +\end_inset + + in diagonalen +\begin_inset Formula $D\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=A^{*}$ +\end_inset + + in obstaja +\begin_inset Formula $\sqrt{A}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A=B^{*}B$ +\end_inset + + za neko nenujno kvadratno matriko +\begin_inset Formula $B$ +\end_inset + + (za pozitivno definitno zahtevamo, + da ima +\begin_inset Formula $B$ +\end_inset + + LN stolpce). +\end_layout + +\end_deeper +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Claim* +klasifikacija skalarnih produktov na +\begin_inset Formula $\mathbb{R}^{n}$ +\end_inset + + in +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Naj bo +\begin_inset Formula $\left\langle u,v\right\rangle $ +\end_inset + + standardni skalarni produkt na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + +\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right)$ +\end_inset + + in +\begin_inset Formula $v=\left(\beta_{1},\dots,\beta_{n}\right)$ +\end_inset + + in velja +\begin_inset Formula $\left\langle u,v\right\rangle =\alpha_{1}\overline{\beta_{1}}+\cdots+\alpha_{n}\overline{\beta_{n}}=\left[\begin{array}{ccc} +\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{c} +\alpha_{1}\\ +\vdots\\ +\alpha_{n} +\end{array}\right]=v^{*}\cdot u$ +\end_inset + +. + Za +\begin_inset Formula $A>0$ +\end_inset + + definirajmo +\begin_inset Formula $\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + spet skalarni produkt na +\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$ +\end_inset + + in da je vsak skalarni produkt v +\begin_inset Formula $\mathbb{R}^{n}/\mathbb{C}^{n}$ +\end_inset + + take oblike. +\end_layout + +\begin_layout Proof +Dokazujemo oba dela trditve: +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + je skalarni produkt +\end_layout + +\begin_deeper +\begin_layout Enumerate +pozitivna semidefinitnost: + +\begin_inset Formula $\forall u\not=0:\left[u,u\right]=\left\langle Au,u\right\rangle \geq0$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +konjutirana simetričnost: + +\begin_inset Formula $\forall u,v:\left[u,v\right]=\left\langle Au,v\right\rangle =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\overline{\left\langle Av,u\right\rangle }=\overline{\left[v,u\right]}$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Linearnost in homogenost: + +\begin_inset Formula $\forall\alpha_{1},\alpha_{2},u_{1}u_{2},v:\left[\alpha_{1}u_{1}+\alpha_{2}u_{2},v\right]=\left\langle A\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right),v\right\rangle =\left\langle \alpha_{1}Au_{1}+\alpha_{2}Au_{2},v\right\rangle =\alpha_{1}\left\langle Au_{1},v\right\rangle +\alpha_{2}\left\langle Au_{2},v\right\rangle =\alpha_{1}\left[u,v\right]+\alpha_{2}\left[u,v\right]$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Enumerate +Za vsak skalarni produkt +\begin_inset Formula $\left[\cdot,\cdot\right]$ +\end_inset + + na +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + obstaja taka pozitivno definitna matrika +\begin_inset Formula $A$ +\end_inset + +, + da velja +\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=\left\langle Au,v\right\rangle =v^{*}Au$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Standard +Naj bo +\begin_inset Formula $e_{1},\dots,e_{n}$ +\end_inset + + standardna baza za +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. + Definirajmo +\begin_inset Formula $A=\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]$ +\end_inset + +. + Velja +\begin_inset Formula $A=A^{*}$ +\end_inset + +: +\begin_inset Formula +\[ +A^{*}=\left[\begin{array}{ccc} +\overline{\left[e_{1},e_{1}\right]} & \cdots & \overline{\left[e_{1},e_{n}\right]}\\ +\vdots & & \vdots\\ +\overline{\left[e_{n},e_{1}\right]} & \cdots & \overline{\left[e_{n},e_{n}\right]} +\end{array}\right]=\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]=A +\] + +\end_inset + + Preveriti je treba še +\begin_inset Formula $\forall u,v\in\mathbb{C}^{n}:\left[u,v\right]=v^{*}Au$ +\end_inset + +. + +\begin_inset Formula $u=\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n}$ +\end_inset + + in +\begin_inset Formula $v=\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}$ +\end_inset + +. + Tedaj je +\begin_inset Formula +\[ +\left[u,v\right]=\left[\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n},\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left(\alpha_{1}\overline{\beta_{1}}\left[e_{1},e_{1}\right]+\cdots+\alpha_{1}\overline{\beta_{n}}\left[e_{1},e_{n}\right]\right)+\cdots+\left(\alpha_{n}\overline{\beta_{1}}\left[e_{n},e_{1}\right]+\cdots+\alpha_{n}\overline{\beta_{n}}\left[e_{n},e_{n}\right]\right)= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{ccc} +\overline{\beta_{1}} & \cdots & \overline{\beta_{n}}\end{array}\right]\left[\begin{array}{ccc} +\left[e_{1},e_{1}\right] & \cdots & \left[e_{n},e_{1}\right]\\ +\vdots & & \vdots\\ +\left[e_{1},e_{n}\right] & \cdots & \left[e_{n},e_{n}\right] +\end{array}\right]\left[\begin{array}{c} +\alpha_{1}\\ +\vdots\\ +\alpha_{n} +\end{array}\right]=v^{*}Au=\left\langle Au,v\right\rangle +\] + +\end_inset + +Da je +\begin_inset Formula $A$ +\end_inset + + pozitivno definitna sledi, + saj mora za vsak neničeln +\begin_inset Formula $u$ +\end_inset + + po aksiomu za pozitivno definitnost skalarnega produkta veljati +\begin_inset Formula $\left\langle Au,u\right\rangle >0$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsubsection +Singularni razcep (angl. + singular value decomposition — + SVD) +\end_layout + +\begin_layout Standard +Naj bo +\begin_inset Formula $A_{n\times n}$ +\end_inset + + neka kompleksna ali realna matrika. + Tedaj je +\begin_inset Formula $A^{*}A$ +\end_inset + + hermitska ( +\begin_inset CommandInset ref +LatexCommand ref +reference "subsec:Simetrične/hermitske-matrike" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + +) matrika dimenzij +\begin_inset Formula $n\times n$ +\end_inset + +. + Ker je +\begin_inset Formula $\forall u:\left\langle A^{*}Au,u\right\rangle =\left\langle Au,Au\right\rangle \geq0$ +\end_inset + +, + je +\begin_inset Formula $A^{*}A$ +\end_inset + + pozitivno semidefinitna, + torej so vse njene lastne vrednosti +\begin_inset Formula $\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Singularne vrednosti +\begin_inset Formula $A$ +\end_inset + + so kvadratni koreni lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Če je +\begin_inset Formula $A$ +\end_inset + + normalna in +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost +\begin_inset Formula $A$ +\end_inset + +, + obstaja tak +\begin_inset Formula $v\not=0\ni:Av=\lambda v\Rightarrow A^{*}v=\overline{\lambda}v$ +\end_inset + +. + Odtod sledi, + da je +\begin_inset Formula $A^{*}Av=A^{*}\lambda v=\lambda A^{*}v=\lambda\overline{\lambda}v$ +\end_inset + +, + torej je +\begin_inset Formula $\lambda$ +\end_inset + + lastna vrednost matrike +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Po definiciji singularne vrednosti je +\begin_inset Formula $\sqrt{\lambda\overline{\lambda}}=\sqrt{\left|\lambda\right|^{2}}=\left|\lambda\right|$ +\end_inset + + singularna vrednost matrike +\begin_inset Formula $A$ +\end_inset + +. + Potemtakem so singularne vrednostni normalnih matrik enake absolutnim vrednosti lastnih vrednosti. +\end_layout + +\begin_layout Standard +Nekatere lastne vrednosti so ničelne, + nekatere pa od nič strogo večje. + Koliko je katerih? + Število ničelnih singularnih vrednosti matrike +\begin_inset Formula $A$ +\end_inset + + je število ničelnih lastnih vrednosti matrike +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Ker je +\begin_inset Formula $A^{*}A$ +\end_inset + + hermitska, + je diagonalizabilna, + zato je algebraična večkratnost lastne vrednosti 0 enaka geometrijski večkratnosti lastne vrednosti 0, + slednja pa je definirana kot +\begin_inset Formula $\dim\Ker A^{*}A$ +\end_inset + +. + Ko upoštevamo +\begin_inset Formula $\dim\Ker A^{*}A=\dim\Ker A$ +\end_inset + +, + izvemo, + da je število ničelnih singularnih vrednosti matrike +\begin_inset Formula $A$ +\end_inset + + njena ničnost ( +\begin_inset Formula $\n A$ +\end_inset + +). + Ker je +\begin_inset Formula $A^{*}A$ +\end_inset + + velikosti +\begin_inset Formula $n\times n$ +\end_inset + +, + ima +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Formula $n$ +\end_inset + + singularnih vrednosti, + torej je število neničelnih singularnih vrednosti +\begin_inset Formula $A$ +\end_inset + + enako +\begin_inset Formula $n-\Ker A$ +\end_inset + +. + Upoštevajoč osnovni dimenzijski izrek jedra in slike, + velja +\begin_inset Formula $\n A+\rang A=n$ +\end_inset + +, + torej je neničelnih singularnih vrednosti +\begin_inset Formula $=\rang A=\dim\Slika A$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Za +\begin_inset Formula $m\times n$ +\end_inset + + matriko velja +\begin_inset Formula $\rang A\le\min\left\{ m,n\right\} $ +\end_inset + +. +\end_layout + +\begin_layout Definition* +posplošitev pojma diagonalne matrike na nekvadratne matrike. + Matrika +\begin_inset Formula $D_{m\times n}$ +\end_inset + + je diagonalna, + če velja +\begin_inset Formula $\forall i:\left\{ 1..m\right\} ,j\in\left\{ 1..n\right\} :i\not=j\Rightarrow D_{i.j}=0$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Primeri pravokotnih diagonalnih matrik: +\begin_inset Formula +\[ +\left[\begin{array}{cccc} +1 & 0 & 0 & 0\\ +0 & 2 & 0 & 0\\ +0 & 0 & 3 & 0 +\end{array}\right],\left[\begin{array}{ccc} +1 & 0 & 0\\ +0 & 2 & 0\\ +0 & 0 & 3\\ +0 & 0 & 0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Theorem* +singularni razcep. + Naj bo +\begin_inset Formula $A$ +\end_inset + + kompleksna +\begin_inset Formula $m\times n$ +\end_inset + + matrika. + Potem obstajata taki unitarni +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + in taka diagonalna +\begin_inset Formula $D$ +\end_inset + + z diagonalci +\begin_inset Formula $\geq0\ni:A=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Diagonalci +\begin_inset Formula $D$ +\end_inset + + so ravno singularne vrednosti matrike +\begin_inset Formula $A$ +\end_inset + +. + Ker je +\begin_inset Formula $Q_{2}$ +\end_inset + + unitarna, + je +\begin_inset Formula $Q_{2}^{*}=Q_{2}^{-1}\Rightarrow A=Q_{1}DQ_{2}^{*}$ +\end_inset + +. + Če +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}$ +\end_inset + +, + je +\begin_inset Formula $A^{*}=Q_{2}^{**}D^{*}Q_{1}^{*}=Q_{2}D^{*}Q_{1}^{*}$ +\end_inset + + in +\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$ +\end_inset + +, + torej je +\begin_inset Formula $A^{*}A$ +\end_inset + + podobna +\begin_inset Formula $D^{*}D$ +\end_inset + +, + diagonalci +\begin_inset Formula $D^{*}D$ +\end_inset + + so lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + in stolpci +\begin_inset Formula $Q_{2}$ +\end_inset + + so lastni vektorji +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Diagonalci +\begin_inset Formula $D$ +\end_inset + + so bodisi 0 bodisi kvadratni koreni od diagonalcev +\begin_inset Formula $D^{*}D$ +\end_inset + +, + torej kvadratni koreni lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + +, + torej singularne vrednosti od +\begin_inset Formula $A$ +\end_inset + +. +\end_layout + +\begin_layout Proof +obstoj singularnega razcepa. + Konstruirajmo +\begin_inset Formula $Q_{1},D,Q_{2}$ +\end_inset + + in dokažimo veljavnost. +\end_layout + +\begin_deeper +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $Q_{2}$ +\end_inset + +: + +\begin_inset Formula $A$ +\end_inset + + je +\begin_inset Formula $m\times n$ +\end_inset + + kompleksna. + Tvorimo +\begin_inset Formula $n\times n$ +\end_inset + + matriko +\begin_inset Formula $A^{*}A$ +\end_inset + +. + Izračunajmo lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + in jih uredimo padajoče — + +\begin_inset Formula $\lambda_{1}\geq\cdots\geq\lambda_{n}$ +\end_inset + +. + Naj bodo +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + pripadajoči lastni vektorji — + +\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}$ +\end_inset + +. + Te +\begin_inset Formula $v_{i}$ +\end_inset + + izberimo tako, + da so ortonormirani. + Lastni podprostori +\begin_inset Formula $A^{*}A$ +\end_inset + + so namreč paroma pravokotni, + saj je +\begin_inset Formula $A^{*}A$ +\end_inset + + normalna, + saj je hermitska. + V vsakem podprostoru vzamemo ONB in +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + je unija teh ON baz. + Definiramo +\begin_inset Formula $Q_{2}=\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]$ +\end_inset + +. + Ker so +\begin_inset Formula $v_{1},\dots,v_{n}$ +\end_inset + + ON, + je +\begin_inset Formula $Q_{2}$ +\end_inset + + unitarna. +\end_layout + +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $D$ +\end_inset + +: + Naj bo +\begin_inset Formula $r\coloneqq\rang A$ +\end_inset + + (število ničelnih singularnih vrednosti +\begin_inset Formula $A$ +\end_inset + +). + Oglejmo si zaporedje lastnih vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + +\begin_inset Formula $\lambda_{1}\geq\cdots>\lambda_{r+1}=\cdots=\lambda_{n}$ +\end_inset + +. + Lastne vrednosti po +\begin_inset Formula $r$ +\end_inset + + so ničelne, + ostale pa večje od 0. + Lastne vrednosti +\begin_inset Formula $A^{*}A$ +\end_inset + + v tem vrstnem redu so singularne vrednosti matrike +\begin_inset Formula $A$ +\end_inset + +: + +\begin_inset Formula $\sigma_{1}^{2}=\lambda_{1}\geq\cdots\geq\sigma_{r}^{2}=\lambda_{r}>\sigma_{r+1}^{2}=\cdots=\sigma_{n}^{2}=0$ +\end_inset + +. + Definiramo +\begin_inset Formula $D$ +\end_inset + + kot +\begin_inset Formula $m\times n$ +\end_inset + + diagonalno matriko takole: +\begin_inset Formula +\[ +D=\left[\begin{array}{cccccc} +\sigma_{1} & & & & & 0\\ + & \ddots\\ + & & \sigma_{r}\\ + & & & \sigma_{r+1}=0\\ + & & & & \ddots\\ +0 & & & & & \sigma_{n}=0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Konstrukcija +\begin_inset Formula $Q_{1}$ +\end_inset + +: + +\begin_inset Formula $\forall i\in\left\{ 1..r\right\} :u_{i}\coloneqq\frac{1}{\sigma_{1}}Av_{i}$ +\end_inset + + za +\begin_inset Formula $v_{i}$ +\end_inset + + lastne vektorje +\begin_inset Formula $A^{*}A$ +\end_inset + +, + torej +\begin_inset Formula $A^{*}Av_{i}=\lambda_{i}v_{i}=\sigma_{i}^{2}v_{i}$ +\end_inset + +. + Pokažimo, + da je +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + ON množica. + +\begin_inset Formula $\forall i,j\in\left\{ 1..r\right\} :$ +\end_inset + + +\begin_inset Formula +\[ +\left\langle u_{i},u_{j}\right\rangle =\left\langle \frac{1}{\sigma_{j}}Av_{i},\frac{1}{\sigma_{j}}Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle Av_{i},Av_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle A^{*}Av_{i},v_{j}\right\rangle =\frac{1}{\sigma_{i}\sigma_{j}}\left\langle \lambda_{i}v_{i},v_{j}\right\rangle =\frac{\lambda_{i}=\sigma_{i}^{\cancel{2}}}{\cancel{\sigma}_{i}\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle = +\] + +\end_inset + + +\begin_inset Formula +\[ +=\frac{\sigma_{i}}{\sigma_{j}}\left\langle v_{i},v_{j}\right\rangle =\begin{cases} +0 & ;i\not=j\\ +1 & ;i=j +\end{cases} +\] + +\end_inset + +Sedaj ONM +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + z +\begin_inset Formula $u_{r+1},\dots,u_{n}$ +\end_inset + + dopolnimo do ONB za +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + + (GS). + Definiramo +\begin_inset Formula $Q_{1}=\left[\begin{array}{ccc} +u_{1} & \cdots & u_{n}\end{array}\right]$ +\end_inset + +. + Ker so stolpci ONB, + je matrika unitarna. +\end_layout + +\begin_layout Itemize +Sedaj preverimo, + da velja +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}\Leftrightarrow AQ_{2}=Q_{1}D$ +\end_inset + +. +\begin_inset Formula +\[ +AQ_{2}=A\left[\begin{array}{ccc} +v_{1} & \cdots & v_{n}\end{array}\right]=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\cdots +\] + +\end_inset + +Upoštevamo, + da +\begin_inset Formula $i>r\Rightarrow\lambda_{i}=0\Rightarrow A^{*}Av_{i}=\lambda_{i}v_{i}=0\Rightarrow v_{i}\in\Ker A^{*}A\Rightarrow v_{i}\in\Ker A\Leftrightarrow Av_{i}=0$ +\end_inset + +: +\begin_inset Formula +\[ +\cdots=\left[\begin{array}{ccc} +Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{cccccc} +Av_{1} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right] +\] + +\end_inset + +Sedaj izračunajmo še +\begin_inset Formula +\[ +Q_{1}D=\left[\begin{array}{ccc} +u_{1} & \cdots & u_{n}\end{array}\right]\left[\begin{array}{cccccc} +\sigma_{1}\\ + & \ddots\\ + & & \sigma_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right]=\left[\begin{array}{cccccc} +\sigma_{1}u_{1} & \cdots & \sigma_{r}u_{r} & 0 & \cdots & 0\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cccccc} +\cancel{\sigma_{1}\frac{1}{\sigma_{1}}}Av_{i} & \cdots & \cancel{\sigma_{r}\frac{1}{\sigma_{r}}}Av_{r} & 0 & \cdots & 0\end{array}\right]=\left[\begin{array}{cccccc} +Av_{i} & \cdots & Av_{r} & 0 & \cdots & 0\end{array}\right]=AQ_{2} +\] + +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Example* +Poišči singularni razcep +\begin_inset Formula $A=\left[\begin{array}{cccc} +1 & 1 & -1 & -1\\ +-1 & 0 & 1 & 0\\ +0 & -1 & 0 & 1 +\end{array}\right]$ +\end_inset + +. + Izračunajmo +\begin_inset Formula +\[ +A^{*}A=\left[\begin{array}{cccc} +1 & 1 & -1 & -1\\ +-1 & 0 & 1 & 0\\ +0 & -1 & 0 & 1 +\end{array}\right]\left[\begin{array}{ccc} +1 & -1 & 0\\ +1 & 0 & -1\\ +-1 & 1 & 0\\ +-1 & 0 & 1 +\end{array}\right]=\cdots=\left[\begin{array}{cccc} +2 & 1 & -2 & -1\\ +1 & 2 & -1 & -2\\ +-2 & -1 & 2 & 1\\ +-1 & -2 & 1 & 2 +\end{array}\right] +\] + +\end_inset + +Izračunajmo +\begin_inset Formula $p_{A^{*}A}\left(x\right)=\det\left(A^{*}A-xI\right)=\cdots=x^{2}\left(x-2\right)\left(x-6\right)$ +\end_inset + +, + torej +\begin_inset Formula $\lambda_{1}=6$ +\end_inset + +, + +\begin_inset Formula $\lambda_{2}=2$ +\end_inset + +, + +\begin_inset Formula $\lambda_{3}=0$ +\end_inset + +, + +\begin_inset Formula $\lambda_{4}=0$ +\end_inset + +, + torej +\begin_inset Formula $\sigma_{1}=\sqrt{6}$ +\end_inset + + in +\begin_inset Formula $\sigma_{2}=\sqrt{2}$ +\end_inset + + ter +\begin_inset Formula $\sigma_{3}=\sigma_{4}=0$ +\end_inset + +. + (ujema se z dejstvom, + da je +\begin_inset Formula $\rang A=2$ +\end_inset + +). + Izračunajmo lastne vektorje +\begin_inset Formula $A^{*}A$ +\end_inset + +: +\begin_inset Formula +\[ +\lambda_{1}=6:\quad v_{1}'=\left[\begin{array}{c} +1\\ +1\\ +-1\\ +-1 +\end{array}\right],\quad\left|\left|v_{1}'\right|\right|=6 +\] + +\end_inset + + +\begin_inset Formula +\[ +\lambda_{2}=2:\quad v_{2}'=\left[\begin{array}{c} +1\\ +-1\\ +-1\\ +1 +\end{array}\right],\quad\left|\left|v_{2}'\right|\right|=2 +\] + +\end_inset + + +\begin_inset Formula +\[ +\lambda_{3}=\lambda_{4}:\quad v_{3}'=\left[\begin{array}{c} +1\\ +0\\ +1\\ +0 +\end{array}\right],v_{4}'=\left[\begin{array}{c} +0\\ +1\\ +0\\ +2 +\end{array}\right],\quad\left|\left|v_{3}'\right|\right|=\sqrt{2},\left|\left|v_{4}'\right|\right|=\sqrt{2} +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Z Gram-Schmidtom naredimo ortogonalno množico (v tem primeru so že ortogonalni) in jih normirajmo: +\begin_inset Formula +\[ +v_{1}=\frac{1}{2}\left[\begin{array}{c} +1\\ +1\\ +-1\\ +-1 +\end{array}\right],\quad v_{2}=\frac{1}{2}\left[\begin{array}{c} +1\\ +-1\\ +-1\\ +1 +\end{array}\right],\quad v_{3}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +1\\ +0\\ +1\\ +0 +\end{array}\right],\quad v_{4}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +0\\ +1\\ +0\\ +1 +\end{array}\right]. +\] + +\end_inset + +Sestavimo +\begin_inset Formula +\[ +Q_{2}=\left[\begin{array}{cccc} +\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\ +\frac{1}{2} & -\frac{1}{2} & 0 & \frac{1}{\sqrt{2}}\\ +-\frac{1}{2} & -\frac{1}{2} & \frac{1}{\sqrt{2}} & 0\\ +-\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}} +\end{array}\right],\quad D=\left[\begin{array}{cccc} +\sqrt{6} & & & 0\\ + & \sqrt{2}\\ + & & 0\\ +0 & & & 0 +\end{array}\right] +\] + +\end_inset + +Izračunamo +\begin_inset Formula $u_{1},\dots,u_{r}$ +\end_inset + + za +\begin_inset Formula $Q_{1}$ +\end_inset + +: +\begin_inset Formula +\[ +u_{1}=\frac{1}{\sigma_{1}}Av_{1}=\frac{1}{\sqrt{6}}\left[\begin{array}{c} +2\\ +-1\\ +-1 +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +u_{2}=\frac{1}{\sigma_{2}}Av_{2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} +0\\ +-1\\ +1 +\end{array}\right] +\] + +\end_inset + +Dopolnimo ju do ONB za +\begin_inset Formula $\mathbb{R}^{3}$ +\end_inset + + z Gram-Schmidtom (oz. + uganemo +\begin_inset Formula $\left[\begin{array}{c} +1\\ +1\\ +1 +\end{array}\right]$ +\end_inset + +). + Dopolnitev normiramo: + +\begin_inset Formula $u_{3}=\frac{1}{\sqrt{3}}\left[\begin{array}{c} +1\\ +1\\ +1 +\end{array}\right]$ +\end_inset + + in vektorje vstavimo v +\begin_inset Formula $Q_{1}$ +\end_inset + +: +\begin_inset Formula +\[ +Q_{1}=\left[\begin{array}{ccc} +\frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}}\\ +-\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\\ +-\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Example* +Iskani razcep je +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$ +\end_inset + + (Izračunati je potrebno še en inverz — + +\begin_inset Formula $Q_{2}^{-1}$ +\end_inset + + namreč). +\end_layout + +\begin_layout Subsubsection +Psevdoinverz — + Moore-Penroseov inverz +\end_layout + +\begin_layout Standard +Psevdoinverz je posplošitev inverza na nenujno kvadratne nenujno obrnljive matrike. + Najprej diagonalne matrike: + Njihov navaden inverz je takšen: +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +d_{11} & & 0\\ + & \ddots\\ +0 & & d_{nn} +\end{array}\right]^{-1}=\left[\begin{array}{ccc} +d_{11}^{-1} & & 0\\ + & \ddots\\ +0 & & d_{nn}^{-1} +\end{array}\right] +\] + +\end_inset + +Kadar je diagonalec ničeln, + kot element polja nima multiplikativnega inverza. + Ideja za posplošeni inverz diagonelne matrike: + take diagonalce pustimo na 0, + torej na primer: +\begin_inset Formula +\[ +\left[\begin{array}{ccc} +1 & & 0\\ + & 2\\ +0 & & 0 +\end{array}\right]^{+}=\left[\begin{array}{ccc} +1 & & 0\\ + & \frac{1}{2}\\ +0 & & 0 +\end{array}\right] +\] + +\end_inset + +Za nekvadratne diagonalne matrike pa takole: +\begin_inset Formula +\[ +\left[\begin{array}{cccc} +1 & & & 0\\ + & 2\\ +0 & & 0 +\end{array}\right]^{+}=\left[\begin{array}{ccc} +1 & & 0\\ + & \frac{1}{2}\\ + & & 0\\ +0 +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Definition* +posplošeni inverz diagonalne matrike. + Naj bo +\begin_inset Formula $D$ +\end_inset + + diagonalna +\begin_inset Formula $m\times n$ +\end_inset + + z neničelnimi diagonalci +\begin_inset Formula $d_{1},\dots d_{r}$ +\end_inset + +, + je +\begin_inset Formula $D^{+}$ +\end_inset + + diagonalna +\begin_inset Formula $n\times m$ +\end_inset + + z neničelnimi diagonalci +\begin_inset Formula $\frac{1}{d_{1}^{-1}},\dots,\frac{1}{d_{r}^{-1}}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +Za diagonalno +\begin_inset Formula $D$ +\end_inset + + opazimo +\begin_inset Formula $D^{++}=D$ +\end_inset + + in za obrnljivo diagonalno +\begin_inset Formula $D$ +\end_inset + + opazimo +\begin_inset Formula $D^{+}=D^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Sedaj bi radi pojem posplošili na nediagonalne matrike — + to storimo s pomočjo SVD. + +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}\ni:D$ +\end_inset + + diagonalna in +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + unitarni. + Tedaj velja +\begin_inset Formula $A$ +\end_inset + + obrnljiva +\begin_inset Formula $\Leftrightarrow D$ +\end_inset + + obrnljiva, + kajti +\begin_inset Formula $A^{-1}=Q_{2}D^{-1}D_{1}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Za splošen nenujno obrnljiv +\begin_inset Formula $A$ +\end_inset + + definiramo +\begin_inset Formula $A^{+}\coloneqq Q_{2}D^{+}Q_{1}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Fact* +Opazimo: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $A^{++}=\left(Q_{2}DQ_{1}^{-1}\right)^{+}=Q_{1}D^{++}Q_{2}^{-1}=Q_{1}DQ_{2}^{-1}=A$ +\end_inset + + +\end_layout + +\begin_layout Itemize +\begin_inset Formula $A$ +\end_inset + + obrnljiva: + +\begin_inset Formula $A^{+}=A^{-1}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Claim* +osnovne lastnosti psevdoinverza. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $AA^{+}A=A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(A^{+}A\right)^{*}=A^{+}A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A^{+}AA^{+}=A^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(AA^{+}\right)^{*}=AA^{+}$ +\end_inset + + +\end_layout + +\end_deeper +\begin_layout Proof +Dokažimo te 4 lastnosti najprej za +\begin_inset Formula $D$ +\end_inset + + in nato za SVD. + Pri +\begin_inset Formula $D$ +\end_inset + + predpostavimo, + da so ničle spodaj desno, + sicer obstaja permutacijska matrika, + ki je ortogonalna, + s katero lahko množimo +\begin_inset Formula $D$ +\end_inset + +, + da jo pretvorimo v željeno obliko (in potem dokaz take +\begin_inset Formula $D$ +\end_inset + + pade v primer SVD): +\end_layout + +\begin_deeper +\begin_layout Itemize +Diagonalen primer. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $DD^{+}D=$ +\end_inset + + +\begin_inset Formula +\[ +\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1}^{-1} & & & & & 0\\ + & \ddots\\ + & & d_{r}^{-1}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=D +\] + +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $D^{+}DD^{+}=\cdots=D^{+}$ +\end_inset + + na podoben način +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(DD^{+}\right)^{*}=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]^{*}=\left[\begin{array}{cccccc} +1 & & & & & 0\\ + & \ddots\\ + & & 1\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=DD^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(D^{+}D\right)^{*}=\cdots=D^{+}D$ +\end_inset + + podobno +\end_layout + +\end_deeper +\begin_layout Itemize +Splošen primer +\begin_inset Formula $A$ +\end_inset + + — + vstavimo +\begin_inset Formula $A=Q_{1}DQ_{2}^{*}=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $AA^{+}A=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{1}DD^{+}DQ_{2}^{*}=Q_{1}DQ_{2}^{*}=A$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $A^{+}AA^{+}=\cdots=A^{+}$ +\end_inset + + na podoben način +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(AA^{+}\right)^{*}=\left(Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}\right)^{*}=\left(Q_{1}DD^{+}Q_{1}^{*}\right)^{*}=Q_{1}\left(DD^{+}\right)^{*}Q_{1}^{*}=Q_{1}DD^{+}Q_{1}^{*}=Q_{1}DQ_{2}^{*}Q_{2}D^{+}Q_{1}^{*}=AA^{+}$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\left(A^{+}A\right)^{*}=\cdots=A^{+}A$ +\end_inset + + podobno +\end_layout + +\end_deeper +\end_deeper +\begin_layout Remark* +\begin_inset Formula $A$ +\end_inset + + obrnljiva +\begin_inset Formula $\Leftrightarrow D$ +\end_inset + + obrnljiva, + torej +\begin_inset Formula $A^{+}=Q_{2}D^{+}Q_{1}^{-1}=Q_{2}D^{-1}Q_{1}^{-1}=A^{-1}$ +\end_inset + +. + Potemtakem za obrnljivo +\begin_inset Formula $A$ +\end_inset + + velja +\begin_inset Formula $A^{+}=A^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Da je definicija dobra, + je treba dokazati, + da je +\begin_inset Formula $A^{+}$ +\end_inset + + enoličen ne glede na SVD, + kajti SVD za +\begin_inset Formula $A$ +\end_inset + + ni enoličen. + Naj bo +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=Q_{3}EQ_{4}^{-1}$ +\end_inset + +, + njen prvi psevsoinverz +\begin_inset Formula $B=Q_{2}D^{+}Q_{1}^{-1}$ +\end_inset + + in njen drugi psevdoinverz +\begin_inset Formula $C=Q_{4}D^{+}Q_{3}^{-1}$ +\end_inset + +. + Ali velja +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +udensdash{$B +\backslash +overset{?}{=}C$} +\end_layout + +\end_inset + +? + Velja +\begin_inset Formula +\[ +AB=\left(ACA\right)B=ACAB=\left(AC\right)^{*}\left(AB\right)^{*}=C^{*}A^{*}B^{*}A^{*}=C^{*}\left(ABA\right)^{*}=C^{*}A^{*}=\left(AC\right)^{*}=AC +\] + +\end_inset + +in +\begin_inset Formula +\[ +BA=B\left(ACA\right)=BACA=\left(BA\right)^{*}\left(CA\right)^{*}=A^{*}B^{*}A^{*}C^{*}=\left(ABA\right)^{*}C^{*}=A^{*}C^{*}=\left(CA\right)^{*}=CA +\] + +\end_inset + +ter nazadnje še +\begin_inset Formula +\[ +B=BAB=CAB=CAC=C. +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Kako izračunamo +\begin_inset Formula $A^{+}$ +\end_inset + + brez SVD? +\end_layout + +\begin_layout Standard +Če je +\begin_inset Formula $A$ +\end_inset + + pozitivno semidefinitna, + jo lahko diagonaliziramo v ortonormirani bazi: + +\begin_inset Formula $A=PDP^{-1}$ +\end_inset + +, + da ima +\begin_inset Formula $D$ +\end_inset + + pozitivne diagonalce in da je +\begin_inset Formula $P^{-1}=P^{*}$ +\end_inset + +. + Opazimo, + da je to SVD od +\begin_inset Formula $A$ +\end_inset + +, + kajti +\begin_inset Formula $Q_{1}=P,Q_{2}=P,D=D$ +\end_inset + + in tedaj +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}=PDP^{-1}$ +\end_inset + +. + Potemtakem je +\begin_inset Formula $A^{+}=PD^{+}P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Za splošno matriko +\begin_inset Formula $A$ +\end_inset + + (nenujno pozitivno semidefinitno) pa velja +\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{+}A^{*}=A^{*}\left(AA^{*}\right)^{+}$ +\end_inset + +. + +\begin_inset Formula $A^{*}A$ +\end_inset + + in +\begin_inset Formula $AA^{*}$ +\end_inset + + sta pozitivno semidefinitni. +\end_layout + +\begin_layout Proof +Najprej bomo preverili za diagonalno, + nato za SVD: +\end_layout + +\begin_deeper +\begin_layout Itemize +Diagonalna +\begin_inset Formula $D_{n\times m}$ +\end_inset + +: + +\begin_inset Formula +\[ +D=\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad D^{*}=\left[\begin{array}{cccccc} +\overline{d_{1}} & & & & & 0\\ + & \ddots\\ + & & \overline{d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right],\quad D^{*}D=\left[\begin{array}{cccccc} +\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{1}{\overline{d_{r}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right], +\] + +\end_inset + + +\begin_inset Formula +\[ +\left(D^{*}D\right)^{+}=\left[\begin{array}{cccccc} +\frac{1}{\overline{d_{1}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{1}{\overline{d_{r}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right],\quad\left(D^{*}D\right)^{+}D^{*}=\left[\begin{array}{cccccc} +\frac{\cancel{\overline{d_{1}}}}{\cancel{\overline{d_{1}}}d_{1}} & & & & & 0\\ + & \ddots\\ + & & \frac{\cancel{\overline{d_{r}}}}{\cancel{\overline{d_{r}}}d_{r}}\\ + & & & 0\\ + & & & & \ddots\\ +0 & & & & & 0 +\end{array}\right]=D^{+} +\] + +\end_inset + + +\end_layout + +\begin_layout Itemize +Za splošen +\begin_inset Formula $A$ +\end_inset + + uporabimo SVD, + da to dokažemo: + +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. + Velja +\begin_inset Formula $A^{*}A=Q_{2}D^{*}Q_{1}^{*}Q_{1}DQ_{2}^{*}=Q_{2}D^{*}DQ_{2}^{*}$ +\end_inset + + in +\begin_inset Formula $\left(A^{*}A\right)^{+}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}$ +\end_inset + +. + Torej +\begin_inset Formula $\left(A^{*}A\right)^{+}A^{*}=Q_{2}\left(D^{*}D\right)^{+}Q_{2}^{*}Q_{2}D^{*}Q_{1}^{*}=Q_{2}\left(D^{*}D\right)^{+}D^{*}Q_{1}^{*}=Q_{2}DQ_{1}^{*}=A^{+}$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Remark* +V posebnih primerih lahko poenostavljamo dalje. + Recimo, + da ima +\begin_inset Formula $A$ +\end_inset + + LN stolpce in je kvadratna +\begin_inset Formula $\Rightarrow\Ker A=\left\{ 0\right\} =\Ker A^{*}A$ +\end_inset + +, + torej +\begin_inset Formula $A^{*}A$ +\end_inset + + je obrnljiva in velja +\begin_inset Formula $\left(A^{*}A\right)^{-1}=\left(A^{*}A\right)^{+}$ +\end_inset + +. + Takrat torej velja +\begin_inset Formula $A^{+}=\left(A^{*}A\right)^{-1}A^{*}$ +\end_inset + +. +\end_layout + +\begin_layout Remark* +To smo uporabili pri iskanju posplošene rešitve predoločenega sistema: + Za sistem +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + iščemo +\begin_inset Formula $\vec{x}$ +\end_inset + +, + da je +\begin_inset Formula $\left|\left|A\vec{x}-\vec{b}\right|\right|$ +\end_inset + + minimalen, + tedaj bo tak +\begin_inset Formula $\vec{x}$ +\end_inset + + posplošena rešitev sistema. + Vemo, + da je posplošena reštev +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + + enaka rešitvi od +\begin_inset Formula $A^{*}A\vec{x}=A^{*}\vec{b}$ +\end_inset + +, + kajti, + če ima +\begin_inset Formula $A$ +\end_inset + + LN stolpce, + je +\begin_inset Formula $A^{*}A$ +\end_inset + + obrnljiva (s tem dokažemo, + da ima ta sistem vedno rešitev): +\begin_inset Formula +\[ +A^{*}A\vec{x}=A^{*}\vec{b}\quad\quad\quad\quad/\cdot\left(A^{*}A\right)^{-1} +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{x}=\left(A^{*}A\right)^{-1}A^{*}\vec{b} +\] + +\end_inset + + +\begin_inset Formula +\[ +\vec{x}=A^{+}\vec{b} +\] + +\end_inset + + +\end_layout + +\begin_layout Paragraph +Uporaba psevdoinverza +\end_layout + +\begin_layout Standard +Vemo, + kaj je posplošena rešitev sistema +\begin_inset Formula $A\vec{x}=\vec{b}$ +\end_inset + +. + Problem je, + da ima sistem lahko več posplošenih rešitev (to se lahko zgodi, + če +\begin_inset Formula $A$ +\end_inset + + nima LN stolpcev). + Med vsemi rešitvami iščemo tisto, + ki je najkrajša po normi — + +\begin_inset Formula $\left|\left|\vec{x}\right|\right|$ +\end_inset + +. +\end_layout + +\begin_layout Claim* +Najkrajša posplošena rešitev sistema +\begin_inset Formula $Ax=b$ +\end_inset + + je ravno +\begin_inset Formula $x=A^{+}b$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokažimo najprej za diagonalno matriko koeficientov, + nato pa še za splošen primer: +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $Dx=b$ +\end_inset + + +\begin_inset Formula +\[ +D_{m\times n}=\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad x=\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right],\quad b=\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +\left|\left|Dx-b\right|\right|^{2}=\left|\left|\left[\begin{array}{cccccc} +d_{1} & & & & & 0\\ + & \ddots\\ + & & d_{r}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right]\left[\begin{array}{c} +x_{1}\\ +\vdots\\ +x_{n} +\end{array}\right]-\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right]\right|\right|^{2}=\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}+b_{r+1}^{2}+\cdots+b_{m}^{2} +\] + +\end_inset + +Ta izraz doseže minimum, + ko +\begin_inset Formula $\left(d_{1}x_{1}-b_{1}\right)^{2}+\cdots+\left(d_{r}x_{r}-b_{r}\right)^{2}=0$ +\end_inset + +, + torej +\begin_inset Formula $x_{1}=\frac{b_{1}}{d_{1}},\dots,x_{r}=\frac{b_{r}}{d_{r}},x_{r+1}=\times,\dots,x_{n}=\times$ +\end_inset + +, + kjer +\begin_inset Formula $\times$ +\end_inset + + predstavlja poljubno vrednost. + Najkrajša rešitev bo torej tista, + kjer +\begin_inset Formula $x_{r+1}=\cdots=x_{n}=0$ +\end_inset + +. + Trdimo, + da je +\begin_inset Formula $\left(\frac{b_{1}}{d_{1}},\cdots,\frac{b_{r}}{d_{r}},0,\cdots,0\right)=D^{+}b$ +\end_inset + +. + Preverimo: +\begin_inset Formula +\[ +D_{n\times m}^{+}=\left[\begin{array}{cccccc} +d_{1}^{-1} & & & & & 0\\ + & \ddots\\ + & & d_{r}^{-1}\\ + & & & 0\\ + & & & & \ddots\\ + & & & & & 0 +\end{array}\right],\quad b=\left[\begin{array}{c} +b_{1}\\ +\vdots\\ +b_{m} +\end{array}\right],\quad D^{+}b=\left[\begin{array}{c} +\frac{b_{1}}{d_{1}}\\ +\vdots\\ +\frac{b_{m}}{d_{m}}\\ +0\\ +\vdots\\ +0 +\end{array}\right] +\] + +\end_inset + +Res je! +\end_layout + +\begin_layout Itemize +Splošen primer s SVD: + +\begin_inset Formula $A_{m\times n}=Q_{1}DQ_{2}^{*}$ +\end_inset + +, + kjer sta +\begin_inset Formula $Q_{1},Q_{2}$ +\end_inset + + ortogonalni in +\begin_inset Formula $D$ +\end_inset + + diagonalna. + Za tretji enačaj uporabimo dejstvo, + da množenje z ortogonalno matriko ohranja normo. +\begin_inset Foot +status open + +\begin_layout Plain Layout +\begin_inset Formula $\left|\left|Q_{2}^{*}x\right|\right|^{2}=\left\langle Q_{2}^{*}x,Q_{2}^{*}x\right\rangle =\left\langle x,Q_{2}Q_{2}^{*}x\right\rangle =\left\langle x,x\right\rangle =\left|\left|x\right|\right|^{2}$ +\end_inset + + +\end_layout + +\end_inset + + +\begin_inset Formula $\left|\left|Ax-b\right|\right|=\left|\left|Q_{1}DQ_{2}^{*}x-b\right|\right|=\left|\left|Q_{1}\left(DQ_{2}^{*}x-Q_{1}^{-1}b\right)\right|\right|=\left|\left|DQ_{2}^{*}x-Q_{1}^{-1}b\right|\right|=\left|\left|Dx'-c\right|\right|$ +\end_inset + + za +\begin_inset Formula $x'=Q_{2}^{*}x$ +\end_inset + + in +\begin_inset Formula $c=Q_{1}^{-1}b$ +\end_inset + +. + Ker je +\begin_inset Formula $Q_{2}$ +\end_inset + + obrnljiva, + velja, + da če +\begin_inset Formula $x$ +\end_inset + + preteče vse vektorje v +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +, + tudi +\begin_inset Formula $x'$ +\end_inset + + preteče vse vektorje v +\begin_inset Formula $\mathbb{C}^{n}$ +\end_inset + +. +\end_layout + +\begin_deeper +\begin_layout Standard +Potemtakem je +\begin_inset Formula $\min\left|\left|Ax-b\right|\right|=\min\left|\left|Dx'-c\right|\right|$ +\end_inset + +. + Če +\begin_inset Formula $\left|\left|Ax-b\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}$ +\end_inset + +, + potem +\begin_inset Formula $\left|\left|Dx'-c\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}'=Q_{2}^{-1}x_{0}$ +\end_inset + + in obratno, + če +\begin_inset Formula $\left|\left|Dx'-c\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}'$ +\end_inset + +, + potem +\begin_inset Formula $\left|\left|Ax-b\right|\right|$ +\end_inset + + zavzame minimum v +\begin_inset Formula $x_{0}=Q_{2}x_{0}'$ +\end_inset + +. + Torej je +\begin_inset Formula $x\mapsto Q_{2}^{-1}x$ +\end_inset + + bijektivna korespondenca med posplošenimi rešitvami +\begin_inset Formula $Ax-b$ +\end_inset + + in posplošenimi rešitvami +\begin_inset Formula $Dx'-c$ +\end_inset + +. + Opazimo, + da preslikava ohranja normo, + torej +\begin_inset Formula $\left|\left|x_{0}'\right|\right|=\left|\left|Q_{2}x_{0}'\right|\right|=\left|\left|x_{0}\right|\right|$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Od prej vemmo, + da je najkrajša posplošena rešitev +\begin_inset Formula $Dx_{0}-c$ +\end_inset + + prav +\begin_inset Formula $x_{0}=D^{+}c$ +\end_inset + +. + Po zgornjem odstavku sledi, + da je +\begin_inset Formula $x_{0}=Q_{2}x_{0}'$ +\end_inset + + najkrajša posplošena rešitev od +\begin_inset Formula $Ax=b$ +\end_inset + +. + Dobimo namreč +\begin_inset Formula $x_{0}=Q_{2}x_{0}'=Q_{2}D^{+}c=Q_{2}D^{+}Q_{1}^{-1}b=A^{+}b$ +\end_inset + +. +\end_layout + +\end_deeper +\end_deeper +\begin_layout Subsection +Kvadratne forme +\end_layout + +\begin_layout Definition* +Forma je homogen polinom, + torej tak, + v katerem imajo vsi monomi isto stopnjo. + Stopnja monoma je +\begin_inset Formula $\deg\left(\beta x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}\right)\coloneqq\alpha_{1}+\cdots+\alpha_{n}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Polinom je vsota monomov. + Stopnja polinoma je najvišja stopnja monoma v njem. +\end_layout + +\begin_layout Example* +Linearna forma v treh spremenljivkah: + +\begin_inset Formula $ax+by+cz=\left[\begin{array}{ccc} +a & b & c\end{array}\right]\left[\begin{array}{c} +x\\ +y\\ +z +\end{array}\right]$ +\end_inset + +. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Kvadratna forma je homogen polinom stopnje 2. + Primer kvadratne forme: +\begin_inset Formula +\[ +ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & a +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right] +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* +Kubična forma v treh spremenljivkah: +\begin_inset Formula +\[ +ax^{3}+by^{3}+cz^{3}+dx^{2}y+ex^{2}z+fy^{2}x+gy^{2}x+iz^{2}x+jz^{2}y+kxyz +\] + +\end_inset + + +\end_layout + +\begin_layout Definition* +Pravimo, + da sta matriki +\begin_inset Formula $A$ +\end_inset + + in +\begin_inset Formula $B$ +\end_inset + + kongruentni, + če obstaja obrnljiva +\begin_inset Formula $P\ni:B=PAP^{T}$ +\end_inset + +. + +\end_layout + +\begin_layout Standard +Radi bi naredili klasifikacijo kvadratnih form. + Če naredimo primerno linearno zamenjavo koordinat, + se kvadratna forma poenostavi v +\begin_inset Formula $ex^{2}+fy^{2}$ +\end_inset + + (mešani členi izginejo). +\begin_inset Formula +\[ +x=\alpha x'+\beta y' +\] + +\end_inset + + +\begin_inset Formula +\[ +y=\gamma x'+\delta y' +\] + +\end_inset + +zapišemo kot +\begin_inset Formula +\[ +\left[\begin{array}{c} +x\\ +y +\end{array}\right]=\left[\begin{array}{cc} +\alpha & \beta\\ +\gamma & \delta +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right],\quad\quad\overset{\text{transponiranje}}{\Longrightarrow}\quad\quad\left[\begin{array}{cc} +x & y\end{array}\right]=\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +\alpha & \gamma\\ +\beta & \delta +\end{array}\right] +\] + +\end_inset + + +\begin_inset Formula +\[ +ax^{2}+bxy+cy^{2}=\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & c +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +\alpha & \gamma\\ +\beta & \delta +\end{array}\right]\left[\begin{array}{cc} +a & b/2\\ +b/2 & a +\end{array}\right]\left[\begin{array}{cc} +\alpha & \beta\\ +\gamma & \delta +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]= +\] + +\end_inset + + +\begin_inset Formula +\[ +=\left[\begin{array}{cc} +x' & y'\end{array}\right]P^{T}AP\left[\begin{array}{c} +x'\\ +y' +\end{array}\right] +\] + +\end_inset + +Ker je +\begin_inset Formula $A$ +\end_inset + + simetrična, + lahko izberemo tako ortogonalno +\begin_inset Formula $P$ +\end_inset + +, + da je +\begin_inset Formula $P^{T}AP$ +\end_inset + + diagonalna, + recimo +\begin_inset Formula $\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]$ +\end_inset + +, + torej +\begin_inset Formula +\[ +\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=d_{1}\left(x'\right)^{2}+d_{2}\left(y'\right)^{2} +\] + +\end_inset + +Kaj vemo o +\begin_inset Formula $2\times2$ +\end_inset + + ortogonalnih matrikah? + +\begin_inset Formula $P=\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]\Rightarrow P^{T}P=\left[\begin{array}{cc} +a & c\\ +b & d +\end{array}\right]\left[\begin{array}{cc} +a & b\\ +c & d +\end{array}\right]=\left[\begin{array}{cc} +a^{2}+c^{2} & ab+cd\\ +ab+cd & b^{2}+d^{2} +\end{array}\right]$ +\end_inset + +. + Da je +\begin_inset Formula $P^{T}P=I$ +\end_inset + +, + mora veljati +\begin_inset Formula $ab+cd=0$ +\end_inset + + in +\begin_inset Formula $a^{2}+c^{2}=b^{2}+d^{2}=1$ +\end_inset + +, + torej +\begin_inset Formula $a=\cos\varphi$ +\end_inset + +, + +\begin_inset Formula $c=\sin\varphi$ +\end_inset + +, + +\begin_inset Formula $b=\cos\tau$ +\end_inset + +, + +\begin_inset Formula $d=\sin\tau$ +\end_inset + +. + Iz +\begin_inset Formula $\cos\left(\varphi+\tau\right)=0$ +\end_inset + + sledi +\begin_inset Formula $\tau=\varphi\pm\frac{\pi}{2}$ +\end_inset + +, + torej je +\begin_inset Formula $P_{1}=\left[\begin{array}{cc} +\cos\varphi & -\sin\varphi\\ +\sin\varphi & \cos\varphi +\end{array}\right]$ +\end_inset + + (vrtež za +\begin_inset Formula $\varphi$ +\end_inset + +) ali +\begin_inset Formula $P_{2}=\left[\begin{array}{cc} +\cos\varphi & \sin\varphi\\ +\sin\varphi & -\cos\varphi +\end{array}\right]$ +\end_inset + + (zrcaljenje žez +\begin_inset Formula $\varphi/2$ +\end_inset + +). + +\begin_inset Formula $\det P_{1}=1$ +\end_inset + +, + +\begin_inset Formula $\det P_{2}=-1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Če je +\begin_inset Formula $A=\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]^{-1}$ +\end_inset + +, + je tudi +\begin_inset Formula $A=\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]^{-1}$ +\end_inset + +. + Če je +\begin_inset Formula $\left[\begin{array}{cc} +v_{1} & v_{2}\end{array}\right]$ +\end_inset + + ortogonalna, + je tudi +\begin_inset Formula $\left[\begin{array}{cc} +v_{1} & -v_{2}\end{array}\right]$ +\end_inset + + ortogonalna. + Če je +\begin_inset Formula $A$ +\end_inset + + +\begin_inset Formula $2\times2$ +\end_inset + + simetrična matrika, + lahko poiščemo tak vrtež +\begin_inset Formula $P=\left[\begin{array}{cc} +\cos\varphi & -\sin\varphi\\ +\sin\varphi & \cos\varphi +\end{array}\right]$ +\end_inset + +, + da je +\begin_inset Formula $A=P\left[\begin{array}{cc} +d_{1} & 0\\ +0 & d_{2} +\end{array}\right]P^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Povzetek: + +\begin_inset Formula $ax^{2}+bxy+cy^{2}\overset{\text{vrtež}}{\longrightarrow}d_{1}x^{2}+d_{2}y^{2}$ +\end_inset + + +\end_layout + +\begin_layout Example* +Nariši krivuljo +\begin_inset Formula $4x^{2}+4xy+7y^{2}=1$ +\end_inset + +. + Pripadajoča kvadratna forma: + +\begin_inset Formula +\[ +\left[\begin{array}{cc} +x & y\end{array}\right]\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]\left[\begin{array}{c} +x\\ +y +\end{array}\right]=1=\cdots +\] + +\end_inset + +Radi bi se znebili mešanega člena: +\begin_inset Formula +\[ +\cdots=\left[\begin{array}{cc} +x' & y'\end{array}\right]P^{T}\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]P\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=1=\cdots +\] + +\end_inset + +Iščemo tak vrtež +\begin_inset Formula $P$ +\end_inset + +, + da bo +\begin_inset Formula $P^{T}AP$ +\end_inset + + diagonalna. + Izračunamo lastne vrednosti +\begin_inset Formula $A=\left[\begin{array}{cc} +4 & 2\\ +2 & 7 +\end{array}\right]$ +\end_inset + +. + Lastni vrednosti sta +\begin_inset Formula $\left\{ 3,8\right\} $ +\end_inset + +. + Izračunamo lastna vektorja: + +\begin_inset Formula $\left\{ \left[\begin{array}{c} +-2\\ +1 +\end{array}\right],\left[\begin{array}{c} +1\\ +2 +\end{array}\right]\right\} $ +\end_inset + +. + Sta že ortogonalna, + treba ju je še normirati: + +\begin_inset Formula $\left\{ \left[\begin{array}{c} +-\frac{2}{\sqrt{5}}\\ +\frac{1}{\sqrt{5}} +\end{array}\right],\left[\begin{array}{c} +\frac{1}{\sqrt{5}}\\ +\frac{2}{\sqrt{5}} +\end{array}\right]\right\} $ +\end_inset + +. + Izdelamo vrzež: + +\begin_inset Formula $P=\left[\begin{array}{cc} +\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}}\\ +\frac{2}{\sqrt{5}} & \frac{2}{\sqrt{5}} +\end{array}\right]$ +\end_inset + +. + Izračunamo kot vrteža: + +\begin_inset Formula $\frac{\sin\varphi}{\cos\varphi}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=2$ +\end_inset + +, + +\begin_inset Formula $\arctan2\approx63,4^{\circ}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Ogledamo si torej kvadratno formo +\begin_inset Formula $\left[\begin{array}{cc} +x' & y'\end{array}\right]\left[\begin{array}{cc} +8 & 0\\ +0 & 3 +\end{array}\right]\left[\begin{array}{c} +x'\\ +y' +\end{array}\right]=8x'^{2}+3y'^{2}=1$ +\end_inset + +, + kar je elipsa ( +\begin_inset Formula $\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$ +\end_inset + +) s polosema +\begin_inset Formula $\frac{1}{\sqrt{8}}$ +\end_inset + + in +\begin_inset Formula $\frac{1}{\sqrt{3}}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Elipso narišemo in jo v koordinatnem sistemu zavrtimo v negativno smer za +\begin_inset Formula $63,4^{\circ}$ +\end_inset + +. + Po zasuku je risba te krivulje risba naše prvotne kvadratne forme. +\end_layout + +\begin_layout Part +Vaja za ustni izpit +\end_layout + +\begin_layout Standard +Ustni izpit je sestavljen iz treh vprašanj. + Sekcije so zaporedna vprašanja na izpitu, + podsekcije so učiteljevi naslovi iz Primerov vprašanj, + podpodsekcije pa so dejanska vprašanja, + kot so se pojavila na dosedanjih izpitih. +\end_layout + +\begin_layout Section +Prvo vprašanje +\end_layout + +\begin_layout Standard +Prvo vprašanje je iz 1. + semestra. +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $\det AB=\det A\det B$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Baze vektorskega prostora +\end_layout + +\begin_layout Subsubsection +Linearno neodvisne množice +\end_layout + +\begin_layout Subsubsection +Ogrodje +\end_layout + +\begin_layout Subsubsection +Definicija baze +\end_layout + +\begin_layout Subsubsection +Dimenzija prostora +\end_layout + +\begin_layout Subsection +Cramerovo pravilo +\end_layout + +\begin_layout Subsubsection +Trditev in dokaz +\end_layout + +\begin_layout Subsection +Obrnljive matrike +\end_layout + +\begin_layout Subsubsection +Definicija obrnljivosti +\end_layout + +\begin_layout Subsubsection +Produkt obrnljivih matrik je obrnljiva matrika +\end_layout + +\begin_layout Subsubsection +Karakterizacija obrnljivih matrik z dokazom +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $\Ker A=\left\{ 0\right\} \Leftrightarrow A$ +\end_inset + + obrnljiva +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $A$ +\end_inset + + ima desni inverz +\begin_inset Formula $\Rightarrow A$ +\end_inset + + obrnljiva +\end_layout + +\begin_layout Subsubsection +Formula za inverz matrike z dokazom +\end_layout + +\begin_layout Subsection +Vektorski podprostori +\end_layout + +\begin_layout Subsection +Elementarne matrike +\end_layout + +\begin_layout Subsection +Pod-/predoločeni sistem +\end_layout + +\begin_layout Subsubsection +Definicija, + iskanje posplošene rešitve z izpeljavo +\end_layout + +\begin_layout Subsubsection +Moč ogrodja +\begin_inset Formula $\geq$ +\end_inset + + moč LN množice +\end_layout + +\begin_layout Subsubsection +Vsak poddoločen sistem ima netrivialno rešitev +\end_layout + +\begin_layout Standard +Posledica prejšnje trditve. +\end_layout + +\begin_layout Subsection +Regresijska premica +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsection +Vektorski/mešani produkt +\end_layout + +\begin_layout Subsection +Grupe/polgrupe +\end_layout + +\begin_layout Subsubsection +Definicija in lastnosti grupe +\end_layout + +\begin_layout Subsubsection +Definicija homomorfizma +\end_layout + +\begin_layout Subsubsection +Primeri homomorfizmov z dokazi +\end_layout + +\begin_layout Subsubsection +Definicija permutacijske grupe in dokaz, + da je grupa +\end_layout + +\begin_layout Subsubsection +Primeri grup +\end_layout + +\begin_layout Subsubsection +Dokaz, + da so ortogonalne matrike podgrupa v grupi obrnljivih matrik +\end_layout + +\begin_layout Subsubsection +Matrika permutacije +\end_layout + +\begin_layout Subsubsection +Dokaz, + da je preslikava, + ki permutaciji priredi matriko, + homomorfizem +\end_layout + +\begin_layout Subsection +Projekcija točke na premico/ravnino +\end_layout + +\begin_layout Subsection +\begin_inset Formula $\det A=\det A^{T}$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Formula za inverz +\end_layout + +\begin_layout Subsection +Homogeni sistemi enačb +\end_layout + +\begin_layout Section +Drugo vprašanje +\end_layout + +\begin_layout Standard +Drugo vprašanje zajema snov linearnih preslikav/lastnih vrednosti. +\end_layout + +\begin_layout Subsection +Diagonalizacija +\end_layout + +\begin_layout Subsubsection +Definicija, + trditve +\end_layout + +\begin_layout Subsection +Prehod na novo bazo +\end_layout + +\begin_layout Subsubsection +Prehodna matrika in njene lastnosti +\end_layout + +\begin_layout Subsubsection +Predstavitev vektorjev in linearnih preslikav z različnimi bazami +\end_layout + +\begin_layout Subsubsection +Razvoj vektorja po eni in drugi bazi (prehod vektorja na drugo bazo) +\end_layout + +\begin_layout Subsection +Matrika linearne preslikave +\end_layout + +\begin_layout Subsection +Rang matrike +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsubsection +Dokaz, + da je rang število LN stolpcev +\end_layout + +\begin_layout Subsubsection +Dimenzijska formula za podprostore +\end_layout + +\begin_layout Subsection +\begin_inset Formula $\rang A=\rang A^{T}$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Ekvivalentnost matrik +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsubsection +Dokaz, + da je relacija ekvivalenčna +\end_layout + +\begin_layout Subsubsection +Dokaz, + da je vsaka matrika ekvivalentna matriki +\begin_inset Formula $I_{r}$ +\end_inset + +, + t. + j. + bločni matriki, + katere zgornji levi blok je +\begin_inset Formula $I$ +\end_inset + + dimenzije +\begin_inset Formula $r$ +\end_inset + +, + drugi trije bloki pa so ničelne matrike. +\end_layout + +\begin_layout Subsection +Jedro/slika +\end_layout + +\begin_layout Subsection +Minimalni poinom +\end_layout + +\begin_layout Subsubsection +Definicija karakterističnega in minimalnega polinoma +\end_layout + +\begin_layout Subsection +Cayley-Hamiltonov izrek +\end_layout + +\begin_layout Subsubsection +Trditev in dokaz +\end_layout + +\begin_layout Subsection +Korenski razcep +\end_layout + +\begin_layout Subsubsection +Definicija korenskih podprostorov +\end_layout + +\begin_layout Subsubsection +Presek različnih korenskih podprostorov je trivialen +\end_layout + +\begin_layout Subsubsection +Vsota korenskih podprostorov je direktna (se sklicuje na zgornjo trditev) +\end_layout + +\begin_layout Subsection +Osnovna formula rang +\begin_inset Formula $+$ +\end_inset + + ničnost +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsection +Funkcije matrik +\end_layout + +\begin_layout Section +Tretje vprašanje +\end_layout + +\begin_layout Standard +Tretje vprašanje zajema naslednje snovi: +\end_layout + +\begin_layout Itemize +vektorski prostori s skalarnim produktom, +\end_layout + +\begin_layout Itemize +adjungirana preslikava, +\end_layout + +\begin_layout Itemize +singularni razcep, +\end_layout + +\begin_layout Itemize +kvadratne forme. +\end_layout + +\begin_layout Subsubsection +Singularni razcep: + Konstrukcija +\begin_inset Formula $Q_{1},Q_{2},D$ +\end_inset + + in dokaz +\begin_inset Formula $A=Q_{1}DQ_{2}^{-1}$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Ortogonalne/unitarne matrike +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsubsection +Dokaz +\begin_inset Formula $AA^{*}=I$ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Lastne vrednosti +\end_layout + +\begin_layout Subsubsection +Prehodna matrika iz ONB v drugo ONB ima ortogonalne stolpce (dokaz) +\end_layout + +\begin_layout Subsection +Kvadratne krivulje +\end_layout + +\begin_layout Subsection +Psevdoinverz +\end_layout + +\begin_layout Subsubsection +Definicija +\end_layout + +\begin_layout Subsection +Najkrajša posplošena rešitev sistema +\end_layout + +\begin_layout Subsubsection +Definicija, + trditev in dokaz +\end_layout + +\begin_layout Subsection +Simetrične matrike +\end_layout + +\begin_layout Subsubsection +Vse o simetričnih matrikah +\end_layout + +\begin_layout Subsection +Adjungirana linearna preslikava +\end_layout + +\begin_layout Subsubsection +Definicija in celotna formulacija +\end_layout + +\begin_layout Subsubsection +Rieszov izrek +\end_layout + +\begin_layout Subsubsection +Dokaz obstoja in enoličnosti kot posledica Rieszovega izreka +\end_layout + +\begin_layout Subsubsection +Formula za matriko linearne preslikave in +\begin_inset Formula $\left\langle Au,v\right\rangle =v^{*}Au=\left\langle u,A^{*}v\right\rangle $ +\end_inset + + +\end_layout + +\begin_layout Subsubsection +Lastne vrednosti adjungirane matrike +\end_layout + +\begin_layout Subsection +Klasifikacija skalarnih produktov +\end_layout + +\begin_layout Subsection +Normalne matrike +\end_layout + +\begin_layout Subsubsection +Definicija, + lastnosti, + izreki, + dokazi +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $A$ +\end_inset + + normalna +\begin_inset Formula $\Rightarrow A$ +\end_inset + + in +\begin_inset Formula $A^{*}$ +\end_inset + + imata isto množico lastnih vrednosti +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $\Ker\left(A-xI\right)=\Ker\left(A-\overline{x}I\right)$ +\end_inset + + za normalno +\begin_inset Formula $A$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Ortogonalni komplement +\end_layout + +\begin_layout Subsubsection +Formula za ortogonalno projekcijo +\end_layout + +\begin_layout Subsection +Izrek o reprezentaciji linearnih funkcionalov +\end_layout + +\begin_layout Subsection +Pozitivno semidefinitne matrike +\end_layout + +\begin_layout Subsubsection +Definicija, + lastnosti. +\end_layout + +\begin_layout Subsubsection +Dokaz, + da imajo nenegativne lastne vrednosti. +\end_layout + +\begin_layout Subsubsection +Kvadratni koren pozitivno semidefinitne matrike. +\end_layout + +\begin_layout Subsubsection +\begin_inset Formula $A\geq0\Rightarrow A$ +\end_inset + + sebiadjungirana +\end_layout + +\begin_layout Subsection +Ortogonalne in ortonormirane baze/Gram-Schmidt +\end_layout + +\end_body +\end_document |