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author | Anton Luka Šijanec <anton@sijanec.eu> | 2024-04-15 10:33:24 +0200 |
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committer | Anton Luka Šijanec <anton@sijanec.eu> | 2024-04-15 10:33:24 +0200 |
commit | 1ffa25152f18588b381fa9260f437f07d9a04003 (patch) | |
tree | 99b688314ca245b80185798e12f1e3450eb2a667 /šola/la | |
parent | dn05p2 (diff) | |
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-rw-r--r-- | šola/la/dn6/dokument.lyx | 1514 |
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diff --git a/šola/la/dn6/dokument.lyx b/šola/la/dn6/dokument.lyx new file mode 100644 index 0000000..a75e37f --- /dev/null +++ b/šola/la/dn6/dokument.lyx @@ -0,0 +1,1514 @@ +#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 |