From 1ffa25152f18588b381fa9260f437f07d9a04003 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Anton=20Luka=20=C5=A0ijanec?= <anton@sijanec.eu>
Date: Mon, 15 Apr 2024 10:33:24 +0200
Subject: =?UTF-8?q?=C5=A1ola?=
MIME-Version: 1.0
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Content-Transfer-Encoding: 8bit

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+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\begin_preamble
+\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);
+    }%
+}%
+\end_preamble
+\use_default_options true
+\begin_modules
+enumitem
+theorems-ams
+\end_modules
+\maintain_unincluded_children false
+\language slovene
+\language_package default
+\inputencoding auto
+\fontencoding global
+\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_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\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_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\leftmargin 1cm
+\topmargin 0cm
+\rightmargin 1cm
+\bottommargin 2cm
+\headheight 1cm
+\headsep 1cm
+\footskip 1cm
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style german
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Title
+Rešitev šeste domače naloge Linearne Algebre
+\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
+Za boljšo preglednost sem svoje rešitve domače naloge prepisal na računalnik.
+ Dokumentu sledi še rokopis.
+ Naloge je izdelala asistentka Ajda Lemut.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+newcommand
+\backslash
+euler{e}
+\backslash
+newcommand
+\backslash
+rang{
+\backslash
+text{rang}}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Naj bosta 
+\begin_inset Formula $V,W$
+\end_inset
+
+ vektorska prostora.
+ Pokaži, da je množica vseh linearnih preslikav 
+\begin_inset Formula $\mathcal{L}\left(V,W\right)=\left\{ A:V\to W:A\text{ linearna}\right\} $
+\end_inset
+
+ vektorski prostor.
+\end_layout
+
+\begin_deeper
+\begin_layout Paragraph
+Rešitev
+\end_layout
+
+\begin_layout Standard
+Definirali smo, da za linearno preslikavo velja aditivnost 
+\begin_inset Formula $L\left(v_{1}+v_{2}\right)=Lv_{1}+Lv_{2}$
+\end_inset
+
+ in homogenost 
+\begin_inset Formula $L\alpha v=\alpha Lv$
+\end_inset
+
+, skupaj 
+\begin_inset Formula $L\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)=\alpha_{1}Lv_{2}+\alpha_{2}Lv_{2}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Vektorski prostor pa smo definirali kot urejeno trojico 
+\begin_inset Formula $\left(V,+,\cdot\right)\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(V,+\right)$
+\end_inset
+
+ je Abelova grupa: komutativnost, asociativnost, inverzi, enota, notranjost
+\end_layout
+
+\begin_layout Enumerate
+aksiomi množenja s skalarjem iz polja 
+\begin_inset Formula $F$
+\end_inset
+
+: 
+\begin_inset Formula $\forall\alpha,\beta\in F\forall a,b\in V:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\alpha\cdot\left(a+b\right)=\alpha\cdot a+\alpha\cdot b$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\left(\alpha+\beta\right)\cdot a=\alpha\cdot a+\beta\cdot a$
+\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
+
+, kjer je 
+\begin_inset Formula $1$
+\end_inset
+
+ enota 
+\begin_inset Formula $F$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+Da linearne preslikave 
+\begin_inset Formula $L:V\to W$
+\end_inset
+
+ sploh obstajajo, privzemam, da sta 
+\begin_inset Formula $V$
+\end_inset
+
+ in 
+\begin_inset Formula $W$
+\end_inset
+
+ vektorska prostora nad istim poljem.
+\end_layout
+
+\begin_layout Standard
+Treba je definirati 
+\begin_inset Formula $+$
+\end_inset
+
+, 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ ter dokazati, da je pri izbranih 
+\begin_inset Formula $+$
+\end_inset
+
+, 
+\begin_inset Formula $F$
+\end_inset
+
+ in 
+\begin_inset Formula $\cdot$
+\end_inset
+
+ 
+\begin_inset Formula $\left(\mathcal{L},+,\cdot\right)$
+\end_inset
+
+ vektorski prostor po tej definiciji.
+ Vzemimo za 
+\begin_inset Formula $+$
+\end_inset
+
+ operacijo 
+\begin_inset Formula $+$
+\end_inset
+
+ iz vektorskega prostora 
+\begin_inset Formula $W$
+\end_inset
+
+ in definirajmo operacijo na 
+\begin_inset Formula $\mathcal{L}$
+\end_inset
+
+: 
+\begin_inset Formula $\forall L_{1},L_{2}\in\mathcal{L}:\quad\left(L_{1}+L_{2}\right)v\coloneqq L_{1}v+L_{2}v$
+\end_inset
+
+.
+ Dokažimo, da je 
+\begin_inset Formula $\left(\mathcal{L},+\right)$
+\end_inset
+
+ abelova grupa:
+\end_layout
+
+\begin_layout Enumerate
+Notranjost operacije: Trdimo, da je 
+\begin_inset Formula $L_{1}+L_{2}$
+\end_inset
+
+ linearna transformacija.
+ Dokaz: 
+\begin_inset Formula $\forall v\in V:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+Aditivnost: 
+\begin_inset Formula $\left(L_{1}+L_{2}\right)\left(v_{1}+v_{2}\right)\overset{\text{def}+}{=}L_{1}\left(v_{1}+v_{2}\right)+L_{2}\left(v_{1}+v_{2}\right)\overset{\text{aditivnost}}{=}L_{1}v_{1}+L_{1}v_{2}+L_{2}v_{1}+L_{2}v_{2}\overset{W\text{V.P.}}{=}L_{1}v_{1}+L_{2}v_{1}+L_{1}v_{2}+L_{2}v_{2}\overset{def+}{=}\left(L_{1}+L_{2}\right)v_{1}+\left(L_{1}+L_{2}\right)v_{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Homogenost: 
+\begin_inset Formula $\alpha\left(L_{1}+L_{2}\right)v\overset{\text{def}+}{=}\alpha\left(L_{1}v+L_{2}v\right)\overset{W\text{V.P.}}{=}\alpha L_{1}v+\alpha L_{2}v\overset{\text{homogenost}}{=}L_{1}\alpha v+L_{2}\alpha v\overset{\text{def}+}{=}\left(L_{1}+L_{2}\right)\left(\alpha v\right)$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Enote: Enota naj bo tista linearna preslikava 
+\begin_inset Formula $L_{0}$
+\end_inset
+
+, ki slika ves 
+\begin_inset Formula $V$
+\end_inset
+
+ v 
+\begin_inset Formula $0\in W$
+\end_inset
+
+.
+ Dokaz: 
+\begin_inset Formula $\forall L\in\mathcal{L}:\quad$
+\end_inset
+
+
+\begin_inset Formula $Lv+L_{0}v\text{\ensuremath{\overset{\text{def}L_{0}}{=}}}Lv+0\ensuremath{\overset{\left(W,+\right)\text{abelova grupa}}{=}}Lv$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Inverzi
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:Inverzi:-Ker-je"
+
+\end_inset
+
+: Ker je 
+\begin_inset Formula $W$
+\end_inset
+
+ V.
+ P., 
+\begin_inset Formula $\forall w\in W\exists!-w\in W\ni:w+\left(-w\right)=0$
+\end_inset
+
+, zato 
+\begin_inset Formula $\forall L\in\mathcal{L}\exists-L\in\mathcal{L}\ni:-L+L=L_{0}$
+\end_inset
+
+ s predpisom 
+\begin_inset Formula $-L$
+\end_inset
+
+ slika element 
+\begin_inset Formula $v\in V$
+\end_inset
+
+ v tisti 
+\begin_inset Formula $w\in W$
+\end_inset
+
+, ki je inverz 
+\begin_inset Formula $Lv\in W$
+\end_inset
+
+.
+ 
+\begin_inset Formula $-L$
+\end_inset
+
+ je res 
+\begin_inset Formula $\in\mathcal{L}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $-L\coloneqq$
+\end_inset
+
+
+\begin_inset Formula $\left(-1\right)\cdot L$
+\end_inset
+
+, kjer je 
+\begin_inset Formula $-1$
+\end_inset
+
+ inverz enote polja, ki ga izberemo kasneje.
+ 
+\begin_inset Formula $\forall v\in V,L\in\mathcal{L}:\left(\left(-1\right)\cdot L\right)v\overset{\text{def}\cdot\text{sledi}}{=}\left(-1\right)\left(Lv\right)\overset{\text{def\ensuremath{\cdot},homogenost}}{=}L\left(-1v\right)\overset{\text{karakteristika F}\not=0}{L\left(-v\right)}$
+\end_inset
+
+.
+ Ta dokaz se sklicuje na določitev polja in skalarnega množenja, ki ga podam
+ kasneje.
+\end_layout
+
+\begin_layout Enumerate
+Asociativnost: 
+\begin_inset Formula $\forall L_{1},L_{2},L_{3}\in\mathcal{L}:L_{1}+\left(L_{2}+L_{3}\right)=\left(L_{1}+L_{2}\right)+L_{3}$
+\end_inset
+
+ velja očitno iz definicije 
+\begin_inset Formula $+$
+\end_inset
+
+, saj je 
+\begin_inset Formula $W$
+\end_inset
+
+ vektorski prostor.
+ Komutativnost spet iz istih razlogov.
+\end_layout
+
+\begin_layout Standard
+Določiti moramo še polje in množenje s skalarjem.
+ Vzemimo za 
+\begin_inset Formula $F$
+\end_inset
+
+ polje vektorskega prostora 
+\begin_inset Formula $W$
+\end_inset
+
+ in množenje s skalarjem definirajmo takole: 
+\begin_inset Formula $\forall v\in V,\alpha\in F:\left(\alpha L\right)v\coloneqq\alpha\left(Lv\right)$
+\end_inset
+
+.
+ Zopet za vsak slučaj dokažimo še linearnost dobljene preslikave 
+\begin_inset Formula $\forall\alpha,\beta\in F\forall L\in\mathcal{L}\forall v_{1},v_{2}\in V:$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Aditivnost: 
+\begin_inset Formula $\left(\alpha L\right)\left(v_{1}+v_{2}\right)\overset{\text{def}\cdot}{=}\alpha\left(L\left(v_{1}+v_{2}\right)\right)\overset{\text{aditivnost}}{=}\alpha\left(Lv_{1}+Lv_{2}\right)\overset{W\text{V.P.}}{=}\alpha\left(Lv_{1}\right)+\alpha\left(Lv_{2}\right)\overset{\text{def}\cdot}{=}\left(\alpha L\right)v_{1}+\left(\alpha L\right)v_{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+Homogenost: 
+\begin_inset Formula $\left(\alpha L\right)\left(\beta v\right)\overset{\text{def}\cdot}{=}\alpha\left(L\left(\beta v\right)\right)\overset{\text{homogenost}}{=}\alpha$
+\end_inset
+
+
+\begin_inset Formula $\beta Lv\overset{\text{F\text{polje}}}{=}\beta\alpha\left(Lv\right)\overset{\text{def}\cdot}{=}\beta\left(\alpha L\right)v$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Iz tega dokaza sledi tudi obstoj inverzov (
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:Inverzi:-Ker-je"
+plural "false"
+caps "false"
+noprefix "false"
+
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+Sedaj lahko dokažemo še štiri aksiome vektorskih prostorov za množenje s
+ skalarjem.
+ 
+\begin_inset Formula $\forall\alpha,\beta\in F\forall L_{1},L_{2}\in V:$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+alpha
+\backslash
+left(L_1+L_2
+\backslash
+right)
+\backslash
+overset{?}{=}
+\backslash
+alpha L_1+
+\backslash
+alpha L_2$}
+\end_layout
+
+\end_inset
+
+: 
+\begin_inset Formula $\left(\alpha\left(L_{1}+L_{2}\right)\right)v\overset{\text{def}+\cdot}{=}\alpha\left(L_{1}v+L_{2}v\right)\overset{W\text{V.P.}}{=}\alpha L_{1}+\alpha L_{2}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+left(
+\backslash
+alpha+
+\backslash
+beta
+\backslash
+right)L_1=
+\backslash
+alpha L_1+
+\backslash
+beta L_1$}
+\end_layout
+
+\end_inset
+
+: Po definiciji našega 
+\begin_inset Formula $\cdot$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$
+\backslash
+alpha
+\backslash
+left(
+\backslash
+beta L_1
+\backslash
+right)=
+\backslash
+left(
+\backslash
+alpha
+\backslash
+beta
+\backslash
+right)L_1$}
+\end_layout
+
+\end_inset
+
+: Po definiciji našega 
+\begin_inset Formula $\cdot$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+udensdash{$1
+\backslash
+cdot L_1=L_1$}
+\end_layout
+
+\end_inset
+
+: 
+\begin_inset Formula $\left(1\cdot L_{1}\right)v\overset{\text{def}\cdot}{=}1\cdot\left(L_{1}v\right)\overset{W\text{V.P.}}{=}L_{1}v$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Naj bo 
+\begin_inset Formula $Z:\mathbb{R}^{3}\to\mathbb{R}^{3}$
+\end_inset
+
+ zrcaljenje preko ravnine 
+\begin_inset Formula $x+y+z=0$
+\end_inset
+
+.
+ Določi matriko 
+\begin_inset Formula $Z$
+\end_inset
+
+ v standardni bazi.
+\end_layout
+
+\begin_deeper
+\begin_layout Paragraph
+Rešitev
+\end_layout
+
+\begin_layout Standard
+Tri točke na taki ravnini so 
+\begin_inset Formula $\left(0,0,0\right)$
+\end_inset
+
+, 
+\begin_inset Formula $\left(1,0,-1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(0,1,-1\right)$
+\end_inset
+
+.
+ Normala ravnine je 
+\begin_inset Formula $\left(1,0,-1\right)\times\left(0,1,-1\right)=\left(1,1,1\right)$
+\end_inset
+
+.
+ Parametrično to ravnino zapišemo kot 
+\begin_inset Formula $\left\{ s\vec{r}+p\vec{q};s,p\in\mathbb{R}\right\} $
+\end_inset
+
+, kjer 
+\begin_inset Formula $\vec{r}=\left(1,0,-1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{q}=\left(0,1,-1\right)$
+\end_inset
+
+.
+ Za določitev matrike linearne preslikave 
+\begin_inset Formula $Z$
+\end_inset
+
+ bomo zrcalili vektorje standardne baze 
+\begin_inset Formula $\left(1,0,0\right)$
+\end_inset
+
+, 
+\begin_inset Formula $\left(0,1,0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(0,0,1\right)$
+\end_inset
+
+ čez to ravnino.
+ Zrcaljenje 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ v 
+\begin_inset Formula $Z\vec{t}$
+\end_inset
+
+ čez ravnino je opisano z enačbo 
+\begin_inset Formula $Z\vec{t}=\vec{t}+2\left(\hat{t}-\vec{t}\right)=2\hat{t}-\vec{t}$
+\end_inset
+
+, kjer s 
+\begin_inset Formula $\hat{t}$
+\end_inset
+
+ označim pravokotno projekcijo točke 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ na ravnino.
+ Torej najprej tako projicirajmo standardno bazo na ravnino.
+\begin_inset Formula 
+\[
+\langle\hat{t}-\vec{t},\vec{q}\rangle=0=\langle\hat{t}-\vec{t},\vec{r}\rangle\quad\text{(pravokotna projekcija)}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\langle s\vec{r}+p\vec{q}-\vec{t},\vec{q}\rangle=0=\langle s\vec{r}+p\vec{q}-\vec{t},\vec{r}\rangle\quad\text{(parametrični zapis \ensuremath{\hat{t}})}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s\langle\vec{r},\vec{q}\rangle+p\langle\text{\ensuremath{\vec{q},\vec{q}\rangle-\langle\vec{t},\vec{q}\rangle=0=s\langle\vec{r},\vec{r}\rangle+p\langle\vec{q},\vec{r}\rangle-\langle\vec{t},\vec{r}\rangle}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+begin{align*}
+\end_layout
+
+\begin_layout Plain Layout
+
+s
+\backslash
+langle
+\backslash
+vec{r},
+\backslash
+vec{q}
+\backslash
+rangle+p
+\backslash
+langle
+\backslash
+vec{q},
+\backslash
+vec{q}
+\backslash
+rangle&=
+\backslash
+langle
+\backslash
+vec{t},
+\backslash
+vec{q}
+\backslash
+rangle
+\backslash
+
+\backslash
+
+\end_layout
+
+\begin_layout Plain Layout
+
+s
+\backslash
+langle
+\backslash
+vec{r},
+\backslash
+vec{r}
+\backslash
+rangle+p
+\backslash
+langle
+\backslash
+vec{q},
+\backslash
+vec{r}
+\backslash
+rangle&=
+\backslash
+langle
+\backslash
+vec{t},
+\backslash
+vec{r}
+\backslash
+rangle
+\end_layout
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{align*}
+\end_layout
+
+\end_inset
+
+Dobimo sistem enačb z neznankama 
+\begin_inset Formula $s$
+\end_inset
+
+ in 
+\begin_inset Formula $p$
+\end_inset
+
+, parametroma projekcije.
+ Vstavimo 
+\begin_inset Formula $\vec{r}=\left(1,0,-1\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\vec{q}=\left(0,1,-1\right)$
+\end_inset
+
+ ter za 
+\begin_inset Formula $\vec{t}$
+\end_inset
+
+ posamično vse tri točke standardne baze in izračunajmo njihove projekcije.
+\begin_inset Formula 
+\[
+s\cdot1+p\cdot2=\langle\vec{t},\left(0,1,-1\right)\rangle
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s\cdot2+p\cdot1=\langle\vec{t},\left(1,0,-1\right)\rangle
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Nato izračunamo še njihovo zrcaljenje iz projekcij po enačbi za zrcaljenje
+ zgoraj.
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+setlength{
+\backslash
+columnseprule}{0.2pt}
+\backslash
+begin{multicols}{3}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\vec{b_{1}}=\left(1,0,0\right)$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s+2p=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+2s+p=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s=-2p
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p-4p=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p=-\frac{1}{3},\quad s=\frac{2}{3}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\hat{t}=\left(\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Z\vec{b_{1}}=\left(\frac{1}{3},-\frac{2}{3},-\frac{2}{3}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\vec{b_{2}}=\left(0,1,0\right)$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s+2p=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+2s+p=0
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p=-2s
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s-4s=1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p=\frac{2}{3},\quad s=-\frac{1}{3}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\hat{b_{2}}=\left(-\frac{1}{3},\frac{2}{3},-\frac{1}{3}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Z\vec{b_{2}}=\left(-\frac{2}{3},\frac{1}{3},-\frac{2}{3}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $\vec{b_{3}}=\left(0,0,1\right)$
+\end_inset
+
+
+\begin_inset Formula 
+\[
+s+2p=-1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+2s+p=-1
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+s=-2p-1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+2\left(-2p-1\right)+p=-1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+p=-\frac{1}{3},\quad s=-\frac{1}{3}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\hat{b_{2}}=\left(-\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+Z\vec{b_{3}}=\left(-\frac{2}{3},-\frac{2}{3},\frac{1}{3}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+end{multicols}
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Dobljene z 
+\begin_inset Formula $Z$
+\end_inset
+
+ preslikane (čez ravnino zrcaljene) vektorje po stolpcih zložimo v matriko
+ 
+\begin_inset Formula $Z$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+Z=\left[\begin{array}{ccc}
+1/3 & -2/3 & -2/3\\
+-2/3 & 1/3 & -2/3\\
+-2/3 & -2/3 & 1/3
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Določi rang matrike
+\begin_inset Formula 
+\[
+B=\left[\begin{array}{cccc}
+-2-t & 4 & 5+t & 4\\
+1 & -1 & -2 & 1\\
+-t & 3 & 1+t & 4+t
+\end{array}\right]
+\]
+
+\end_inset
+
+v odvisnosti od parametra 
+\begin_inset Formula $t$
+\end_inset
+
+.
+\end_layout
+
+\begin_deeper
+\begin_layout Paragraph
+Rešitev
+\end_layout
+
+\begin_layout Standard
+Za 
+\begin_inset Formula $A:V\to U$
+\end_inset
+
+ smo definirali 
+\begin_inset Formula $\rang A\coloneqq\dim\text{Im}A$
+\end_inset
+
+, kjer je 
+\begin_inset Formula $\text{Im}A\coloneqq\left\{ Av;v\in V\right\} $
+\end_inset
+
+.
+ Dokazali smo, da je rang matrike enak številu linearno neodvisnih vrstic
+ matrike in da velja 
+\begin_inset Formula $\rang A=\rang A^{T}$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\rang\left[\begin{array}{cccc}
+-2-t & 4 & 5+t & 4\\
+1 & -1 & -2 & 1\\
+-t & 3 & 1+t & 4+t
+\end{array}\right]=\rang\left[\begin{array}{ccc}
+-2-t & 1 & -t\\
+4 & -1 & 3\\
+5+t & -2 & 1+t\\
+4 & 1 & 4+t
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\rang\left[\begin{array}{ccc}
+4 & 1 & -3\\
+-2-t & 1 & -t\\
+5+t & -2 & 1+t\\
+4 & 1 & 4+t
+\end{array}\right]=\rang\left[\begin{array}{ccc}
+4 & 1 & -3\\
+3 & -1 & 1\\
+5+t & -2 & 1+t\\
+4 & 1 & 4+t
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\rang\left[\begin{array}{ccc}
+4 & 1 & -3\\
+3 & -1 & 1\\
+1+t & -3 & -3\\
+4 & 1 & 4+t
+\end{array}\right]=\rang\left[\begin{array}{ccc}
+1 & 2 & -4\\
+3 & -1 & 1\\
+1+t & -3 & -3\\
+4 & 1 & 4+t
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\rang\left[\begin{array}{ccc}
+1 & 2 & -4\\
+0 & -7 & 13\\
+0 & -5-t & 1+4t\\
+0 & -7 & 20+t
+\end{array}\right]=\rang\left[\begin{array}{ccc}
+1 & 2 & -4\\
+0 & -7 & 13\\
+0 & 0 & \frac{-58+15t}{7}\\
+0 & 0 & 7+t
+\end{array}\right]
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Rang je vsaj 2, ker sta 
+\begin_inset Formula $\left(1,2,-4\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(0,-7,13\right)$
+\end_inset
+
+ linearno neodvisna.
+ Rang je kvečjemu 3, ker je manjša izmed stranic matrike dolžine 3.
+ Rang ne more biti 2, ker sistem 
+\begin_inset Formula $\frac{-58+15t}{7}=7+t=0$
+\end_inset
+
+ nima rešitve.
+ 
+\begin_inset Formula $\rang B=3$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Enumerate
+Poišči karakteristični in minimalni polinom matrike
+\begin_inset Formula 
+\[
+A=\left[\begin{array}{ccc}
+4 & -5 & 3\\
+2 & -3 & 2\\
+-1 & 1 & 0
+\end{array}\right]
+\]
+
+\end_inset
+
+in s pomočjo Cayley-Hamiltonovega izreka določi njen inverz.
+\end_layout
+
+\begin_deeper
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\nabla_{P}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left|\begin{array}{ccc}
+4-\lambda & -5 & 3\\
+2 & -3-\lambda & 2\\
+-1 & 1 & -\lambda
+\end{array}\right|=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=-3\left(3+\lambda\right)-2\left(4-\lambda\right)-10\lambda+\lambda\left(3+\lambda\right)\left(4-\lambda\right)+10+6=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+-9-3\lambda-8+2\lambda-10\lambda+\left(3\lambda+\lambda^{2}\right)\left(4-\lambda\right)+16=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=-17+16-11\lambda+12\lambda-3\lambda^{2}+4\lambda^{2}-\lambda^{3}=-\lambda^{3}+\lambda^{2}+\lambda-1
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Eno ničlo uganemo (
+\begin_inset Formula $\lambda_{1}=1$
+\end_inset
+
+), nato 
+\begin_inset Formula $-\lambda^{3}+\lambda^{2}+\lambda-1:\lambda-1=-\lambda^{2}+1=\left(1+\lambda\right)\left(1-\lambda\right)$
+\end_inset
+
+.
+ 1 je torej dvojna ničla, 
+\begin_inset Formula $\lambda_{2}=-1$
+\end_inset
+
+ pa enojna.
+ Ker 
+\begin_inset Formula $m_{A}\left(\lambda\right)|\nabla_{A}\left(\lambda\right)$
+\end_inset
+
+, je kandidat za 
+\begin_inset Formula $m_{A}\left(\lambda\right)$
+\end_inset
+
+ poleg 
+\begin_inset Formula $-\nabla_{A}\left(\lambda\right)$
+\end_inset
+
+ še 
+\begin_inset Formula $p\left(\lambda\right)=\left(\lambda-x\right)\left(\lambda+1\right)=1-\lambda^{2}$
+\end_inset
+
+.
+ Po Cayley-Hamiltonovem izreku 
+\begin_inset Formula $m_{A}\left(A\right)=0=\nabla_{A}\left(A\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $I-A^{2}\not=0$
+\end_inset
+
+, je 
+\begin_inset Formula $m_{A}\left(\lambda\right)=-\nabla_{A}\left(\lambda\right)=\lambda^{3}-\lambda^{2}-\lambda+1$
+\end_inset
+
+.
+ Izračunajmo inverz:
+\begin_inset Formula 
+\[
+m_{A}\left(A\right)=A^{3}-A^{2}-A+I=0\quad\quad/-I
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A^{3}-A^{2}-A=-I\quad\quad/\cdot A^{-1}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+A^{3}A^{-1}-A^{2}A^{-1}-AA^{-1}=-IA^{-1}\quad\quad/\cdot\left(-I\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+-A^{2}+A+I=A^{1}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=-\left[\begin{array}{ccc}
+4 & -5 & 3\\
+2 & -3 & 2\\
+-1 & 1 & 0
+\end{array}\right]\left[\begin{array}{ccc}
+4 & -5 & 3\\
+2 & -3 & 2\\
+-1 & 1 & 0
+\end{array}\right]+\left[\begin{array}{ccc}
+4 & -5 & 3\\
+2 & -3 & 2\\
+-1 & 1 & 0
+\end{array}\right]+\left[\begin{array}{ccc}
+1 & 0 & 0\\
+0 & 1 & 0\\
+0 & 0 & 1
+\end{array}\right]=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\left[\begin{array}{ccc}
+2 & -3 & 1\\
+2 & -3 & 2\\
+1 & -1 & 2
+\end{array}\right]\text{, kar je res \ensuremath{A^{-1}.}}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Rokopisi, ki sledijo, naj služijo le kot dokaz samostojnega reševanja.
+ Zavedam se namreč njihovega neličnega izgleda.
+\end_layout
+
+\begin_layout Standard
+\begin_inset External
+	template PDFPages
+	filename /mnt/slu/shramba/upload/www/d/LADN6FMF1.pdf
+	extra LaTeX "pages=-"
+
+\end_inset
+
+
+\begin_inset External
+	template PDFPages
+	filename /mnt/slu/shramba/upload/www/d/LADN6FMF2.pdf
+	extra LaTeX "pages=-"
+
+\end_inset
+
+
+\end_layout
+
+\end_body
+\end_document
-- 
cgit v1.2.3