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\begin_body

\begin_layout Title
Teorija linearne algebre za ustni izpit —
 IŠRM 2023/24
\end_layout

\begin_layout Author

\noun on
Anton Luka Šijanec
\end_layout

\begin_layout Date
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
today
\end_layout

\end_inset


\end_layout

\begin_layout Abstract
Povzeto po zapiskih s predavanj prof.
 Cimpriča.
\end_layout

\begin_layout Part
Teorija
\end_layout

\begin_layout Section
Prvi semester
\end_layout

\begin_layout Subsection
Vektorji v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset


\end_layout

\begin_layout Standard
Identificaramo 
\begin_inset Formula $n-$
\end_inset

terice realnih števil,
 točke v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

,
 množice paroma enakih geometrijskih vektorjev.
\end_layout

\begin_layout Standard
Osnovne operacije z vektorji:
 Vsota (po komponentah) in množenje s skalarjem (po komponentah),
 kjer je skalar realno število.
\end_layout

\begin_layout Standard
Lastnosti teh računskih operacij:
 asociativnost in komutativnost vsote,
 aditivna enota,
 
\begin_inset Formula $-\vec{a}=\left(-1\right)\cdot\vec{a}$
\end_inset

,
 leva in desna distributivnost,
 homogenost,
 multiplikativna enota.
\end_layout

\begin_layout Subsubsection
Linearna kombinacija vektorjev
\end_layout

\begin_layout Definition*
Linearna kombinacija vektorjev 
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset

 je izraz oblike 
\begin_inset Formula $\alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}}$
\end_inset

,
 kjer so 
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
\end_inset

 skalarji.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Množico vseh linearnih kombinacij vektorjev 
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset

 označimo z 
\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} $
\end_inset

 in ji pravimo linearna ogrinjača (angl.
 span).
 
\begin_inset Formula $\Lin\left\{ \vec{v_{1}},\dots,\vec{v_{n}}\right\} =\left\{ \alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}};\forall\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\right\} $
\end_inset


\end_layout

\begin_layout Subsubsection
Linearna neodvisnost vektorjev
\end_layout

\begin_layout Paragraph*
Ideja
\end_layout

\begin_layout Standard
En vektor je linearno neodvisen,
 če ni enak 
\begin_inset Formula $\vec{0}$
\end_inset

.
 Dva,
 če ne ležita na isti premici.
 Trije,
 če ne ležijo na isti ravnini.
\end_layout

\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:odvisni"

\end_inset

Vektorji 
\begin_inset Formula $\vec{v_{1}},\dots,\vec{v_{n}}$
\end_inset

 so linearno odvisni,
 če se da enega izmed njih izraziti z linearno kombinacijo preostalih 
\begin_inset Formula $n-1$
\end_inset

 vektorjev.
 Vektorji so linearno neodvisni,
 če niso linearno odvisni (in obratno).
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:vsi0"

\end_inset

Vektorji 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 so linearno neodvisni,
 če za vsake skalarje,
 ki zadoščajo 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset

,
 velja 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset

.
 ZDB poleg 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset

 ne obstajajo nobeni drugi 
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
\end_inset

,
 kjer bi veljalo 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:kvečjemu1"

\end_inset


\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 so linearno neodvisni,
 če se da vsak vektor na kvečjemu en način izraziti kot linearno kombinacijo 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

.
\end_layout

\begin_layout Theorem*
Te tri definicije so ekvivalentne.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:odvisni}\Rightarrow\ref{def:vsi0}\right)$
\end_inset

 Recimo,
 da so 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 linearno odvisni v smislu 
\begin_inset CommandInset ref
LatexCommand ref
reference "def:odvisni"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
 Dokažimo,
 da so tedaj linearno odvisni tudi v smislu 
\begin_inset Formula $\ref{def:vsi0}$
\end_inset

.
 Obstaja tak 
\begin_inset Formula $i$
\end_inset

,
 da lahko 
\begin_inset Formula $v_{i}$
\end_inset

 izrazimo z linearno kombinacijo preostalih,
 torej 
\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset

 za neke 
\begin_inset Formula $\alpha$
\end_inset

.
 Sledi 
\begin_inset Formula $0=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\left(-1\right)v_{i}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset

,
 kar pomeni,
 da obstaja linearna kombinacija,
 ki je enaka 0,
 toda niso vsi koeficienti 0 (že koeficient pred 
\begin_inset Formula $v_{i}$
\end_inset

 je 
\begin_inset Formula $-1$
\end_inset

),
 tedaj so vektorji po definiciji 
\begin_inset CommandInset ref
LatexCommand ref
reference "def:vsi0"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 linearno odvisni.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:vsi0}\Rightarrow\ref{def:odvisni}\right)$
\end_inset

 Recimo,
 da so 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 linearno odvisno v smislu 
\begin_inset Formula $\ref{def:vsi0}$
\end_inset

.
 Tedaj obstajajo 
\begin_inset Formula $\alpha$
\end_inset

,
 ki niso vse 0,
 da velja 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset

.
 Tedaj 
\begin_inset Formula $\exists i\ni:\alpha_{i}\not=0$
\end_inset

 in velja
\begin_inset Formula 
\[
\alpha_{i}v_{i}=-\alpha_{1}v_{1}-\cdots-\alpha_{i-1}v_{i-1}-\alpha_{i+1}v_{i+1}-\cdots-\alpha_{n}v_{n}\quad\quad\quad\quad/:\alpha_{i}
\]

\end_inset


\begin_inset Formula 
\[
v_{i}=-\frac{\alpha_{1}}{\alpha_{i}}v_{i}-\cdots-\frac{\alpha_{i-1}}{\alpha_{i}}v_{i-1}-\frac{\alpha_{i+1}}{\alpha_{i}}v_{i+1}-\cdots-\frac{\alpha_{n}}{\alpha_{i}}v_{n}\text{,}
\]

\end_inset

s čimer smo 
\begin_inset Formula $v_{i}$
\end_inset

 izrazili kot linearno kombinacijo preostalih vektorjev.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\ref{def:vsi0}\Leftrightarrow\ref{def:kvečjemu1}\right)$
\end_inset

 Naj bodo 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 LN.
 Recimo,
 da obstaja 
\begin_inset Formula $v$
\end_inset

,
 ki se ga da na dva načina izraziti kot linearno kombinacijo 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

.
 Naj bo 
\begin_inset Formula $v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{n}v_{n}$
\end_inset

.
 Sledi 
\begin_inset Formula $0=\left(\alpha_{1}-\beta_{1}\right)v_{1}+\cdots+\left(\alpha_{n}-\beta_{n}\right)v_{n}$
\end_inset

.
 Po definiciji 
\begin_inset CommandInset ref
LatexCommand ref
reference "def:vsi0"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 velja 
\begin_inset Formula $\forall i:\alpha_{i}-\beta_{i}=0\Leftrightarrow\alpha_{i}=\beta_{i}$
\end_inset

,
 torej sta načina,
 s katerima izrazimo 
\begin_inset Formula $v$
\end_inset

,
 enaka,
 torej lahko 
\begin_inset Formula $v$
\end_inset

 izrazimo na kvečjemu en način z 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

,
 kar ustreza definiciji 
\begin_inset CommandInset ref
LatexCommand ref
reference "def:kvečjemu1"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsubsection
Ogrodje in baza
\end_layout

\begin_layout Definition*
Vektorji 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 so ogrodje (angl.
 span),
 če 
\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} =\mathbb{R}^{n}\Leftrightarrow\forall v\in\mathbb{R}^{n}\exists\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Vektorji 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 so baza,
 če so LN in ogrodje 
\begin_inset Formula $\Leftrightarrow\forall v\in\mathbb{R}^{n}:\exists!\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}\ni:v=\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}$
\end_inset

 ZDB vsak vektor 
\begin_inset Formula $\in\mathbb{R}^{n}$
\end_inset

 se da na natanko en način izraziti kot LK 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

.
\end_layout

\begin_layout Example*
Primer baze je standardna baza 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

:
 
\begin_inset Formula $\left\{ \left(1,0,0,\dots,0\right),\left(0,1,0,\dots,0\right),\left(0,0,1,\dots,0\right),\left(0,0,0,\dots,1\right)\right\} $
\end_inset

.
 To pa ni edina baza.
 Primer nestandardne baze v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset

 je 
\begin_inset Formula $\left\{ \left(1,1,1\right),\left(0,1,1\right),\left(0,0,1\right)\right\} $
\end_inset

.
\end_layout

\begin_layout Subsubsection
Norma in skalarni produkt
\end_layout

\begin_layout Definition*
Norma vektorja 
\begin_inset Formula $v=\left(\alpha_{1},\dots,\alpha_{n}\right)$
\end_inset

 je definirana z 
\begin_inset Formula $\left|\left|v\right|\right|=\sqrt{\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}}$
\end_inset

.
 Geometrijski pomen norme je dolžina krajevnega vektorja z glavo v 
\begin_inset Formula $v$
\end_inset

.
\end_layout

\begin_layout Standard
Osnovne lastnosti norme:
 
\begin_inset Formula $\left|\left|v\right|\right|\geq0$
\end_inset

,
 
\begin_inset Formula $\left|\left|v\right|\right|=0\Rightarrow v=\vec{0}$
\end_inset

,
 
\begin_inset Formula $\left|\left|\alpha v\right|\right|=\left|\alpha\right|\cdot\left|\left|v\right|\right|$
\end_inset

,
 
\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
\end_inset

 (trikotniška neenakost)
\end_layout

\begin_layout Definition*
Skalarni produkt 
\begin_inset Formula $u=\left(\alpha_{1},\dots,\alpha_{n}\right),v=\left(\beta_{1},\dots,\beta_{n}\right)$
\end_inset

 označimo z 
\begin_inset Formula $\left\langle u,v\right\rangle \coloneqq\alpha_{1}\beta_{1}+\cdots+\alpha_{n}\beta_{n}$
\end_inset

.
 Obstaja tudi druga oznaka in pripadajoča drugačna definicija 
\begin_inset Formula $u\cdot v\coloneqq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
\end_inset

,
 kjer je 
\begin_inset Formula $\varphi$
\end_inset

 kot med 
\begin_inset Formula $u,v$
\end_inset

.
\end_layout

\begin_layout Claim*
Velja 
\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v$
\end_inset

.
\end_layout

\begin_layout Proof
Uporabimo kosinusni izrek,
 ki pravi,
 da v trikotniku s stranicami dolžin 
\begin_inset Formula $a,b,c$
\end_inset

 velja 
\begin_inset Formula $c^{2}=a^{2}+b^{2}-2ab\cos\varphi$
\end_inset

,
 kjer je 
\begin_inset Formula $\varphi$
\end_inset

 kot med 
\begin_inset Formula $b$
\end_inset

 in 
\begin_inset Formula $c$
\end_inset

.
 Za vektorja 
\begin_inset Formula $v$
\end_inset

 in 
\begin_inset Formula $u$
\end_inset

 z vmesnim kotom 
\begin_inset Formula $\varphi$
\end_inset

 torej velja 
\begin_inset Formula 
\[
\left|\left|u-v\right|\right|^{2}=\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]

\end_inset

Obenem velja 
\begin_inset Formula $\left|\left|u\right|\right|^{2}=\alpha_{1}^{2}+\cdots+\alpha_{n}^{2}=\left\langle u,u\right\rangle $
\end_inset

,
 torej lahko zgornjo enačbo prepišemo v
\begin_inset Formula 
\[
\left\langle u-v,u-v\right\rangle =\left\langle u,u\right\rangle +\left\langle v,v\right\rangle -2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]

\end_inset

Naj bo 
\begin_inset Formula $w=u,v$
\end_inset

.
 Iz prihodnosti si izposodimo obe linearnosti in simetričnost.
 
\begin_inset Formula 
\[
\left\langle u-v,u-v\right\rangle =\left\langle u-v,w\right\rangle =\left\langle u,w\right\rangle -\left\langle v,w\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
\]

\end_inset

 Prišli smo do enačbe
\begin_inset Formula 
\[
\cancel{\left\langle u,u\right\rangle }-2\left\langle u,v\right\rangle +\cancel{\left\langle v,v\right\rangle }=\cancel{\left\langle u,u\right\rangle }+\cancel{\left\langle v,v\right\rangle }-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\quad\quad\quad\quad/:-2
\]

\end_inset


\begin_inset Formula 
\[
\left\langle u,v\right\rangle =\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi.
\]

\end_inset


\end_layout

\begin_layout Claim*
Paralelogramska identiteta.
 
\begin_inset Formula $\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}$
\end_inset

 ZDB vsota kvadratov dolžin obeh diagonal je enota vsoti kvadratov dolžin vseh štirih stranic.
\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\left|\left|u+v\right|\right|^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u+v\right\rangle +\left\langle v,u+v\right\rangle =\left\langle u,u\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
\]

\end_inset

 
\begin_inset Formula 
\[
\left|\left|u-v\right|\right|^{2}=\left\langle u-v,u-v\right\rangle =\left\langle u,u-v\right\rangle -\left\langle v,u-v\right\rangle =\left\langle u,u\right\rangle -\left\langle u,v\right\rangle -\left\langle v,u\right\rangle +\left\langle v,v\right\rangle 
\]

\end_inset


\begin_inset Formula 
\[
\left|\left|u+v\right|\right|^{2}+\left|\left|u-v\right|\right|^{2}=2\left\langle u,u\right\rangle +2\left\langle v,v\right\rangle =2\left|\left|u\right|\right|^{2}+2\left|\left|v\right|\right|^{2}
\]

\end_inset


\end_layout

\begin_layout Claim*
Cauchy-Schwarzova neenakost.
 
\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $\left|\left\langle u,v\right\rangle \right|=\left|\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi\right|=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\left|\cos\varphi\right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|$
\end_inset

,
 kajti 
\begin_inset Formula $\left|\cos\varphi\right|\in\left[0,1\right]$
\end_inset

.
\end_layout

\begin_layout Claim*
Trikotniška neenakost.
 
\begin_inset Formula $\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|$
\end_inset


\end_layout

\begin_layout Proof
Sledi iz Cauchy-Schwarzove.
 Velja 
\begin_inset Formula 
\[
-\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq\left|\left\langle u,v\right\rangle \right|\leq\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/\cdot2
\]

\end_inset


\begin_inset Formula 
\[
-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\leq2\left|\left\langle u,v\right\rangle \right|\leq2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\quad\quad\quad\quad/+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
\]

\end_inset


\begin_inset Formula 
\[
-2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq\cancel{2\left|\left\langle u,v\right\rangle \right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}\leq}2\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|+\left|\left|u\right|\right|^{2}+\left|\left|v\right|\right|^{2}
\]

\end_inset

uporabimo kosinusni izrek na levi strani enačbe,
 desno pa zložimo v kvadrat:
\begin_inset Formula 
\[
\left|\left|u+v\right|\right|^{2}\leq\left(\left|\left|u\right|\right|+\left|\left|v\right|\right|\right)^{2}\quad\quad\quad\quad/\sqrt{}
\]

\end_inset


\begin_inset Formula 
\[
\left|\left|u+v\right|\right|\leq\left|\left|u\right|\right|+\left|\left|v\right|\right|
\]

\end_inset


\end_layout

\begin_layout Claim*
Za neničelna vektorja velja 
\begin_inset Formula $u\perp v\Leftrightarrow\left\langle u,v\right\rangle =0$
\end_inset

.
\end_layout

\begin_layout Proof
\begin_inset Formula $\left\langle u,v\right\rangle =u\cdot v=\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\cos\varphi$
\end_inset

,
 kar je 0 
\begin_inset Formula $\Leftrightarrow\varphi=\pi=90°$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Vektorski in mešani produkt
\end_layout

\begin_layout Standard
Definirana sta le za vektorje v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset

.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $u=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right),v=\left(\beta_{1},\beta_{2},\beta_{3}\right)$
\end_inset

.
 
\begin_inset Formula $u\times v=\left(\alpha_{2}\beta_{3}-\alpha_{3}\beta_{2},\alpha_{3}\beta_{1}-\alpha_{1}\beta_{3},\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}\right)$
\end_inset

.
\end_layout

\begin_layout Paragraph
Geometrijski pomen
\end_layout

\begin_layout Standard
Vektor 
\begin_inset Formula $u\times v$
\end_inset

 je pravokoten na 
\begin_inset Formula $u$
\end_inset

 in 
\begin_inset Formula $v$
\end_inset

,
 njegova dolžina je 
\begin_inset Formula $\left|\left|u\right|\right|\cdot\left|\left|v\right|\right|\cdot\sin\varphi$
\end_inset

,
 kar je ploščina paralelograma,
 ki ga oklepata 
\begin_inset Formula $u$
\end_inset

 in 
\begin_inset Formula $v$
\end_inset

.
\end_layout

\begin_layout Standard
Pravilo desnega vijaka nam je v pomoč pri doložanju usmeritve vektorskega produkta.
 Če iztegnjen kazalec desne roke predstavlja 
\begin_inset Formula $u$
\end_inset

 in iztegnjen sredinec 
\begin_inset Formula $v$
\end_inset

,
 iztegnjen palec kaže v smeri 
\begin_inset Formula $u\times v$
\end_inset

.
\end_layout

\begin_layout Claim*
Lagrangeva identiteta.
 
\begin_inset Formula $\left|\left|u\times v\right|\right|+\left\langle u,v\right\rangle ^{2}=\left|\left|u\right|\right|^{2}\cdot\left|\left|v\right|\right|^{2}$
\end_inset


\begin_inset Note Note
status open

\begin_layout Plain Layout
DOKAZ???????
\end_layout

\end_inset


\end_layout

\begin_layout Definition*
Mešani produkt vektorjev 
\begin_inset Formula $u,v,w$
\end_inset

 je skalar 
\begin_inset Formula $\left\langle u\times v,w\right\rangle $
\end_inset

.
 Oznaka:
 
\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle $
\end_inset

.
\end_layout

\begin_layout Paragraph*
Geometrijski pomen
\end_layout

\begin_layout Standard
Volumen paralelpipeda,
 ki ga določajo 
\begin_inset Formula $u,v,w$
\end_inset

.
 Razlaga:
 
\begin_inset Formula $\left[u,v,w\right]=\left\langle u\times v,w\right\rangle =\left|\left|u\times v\right|\right|\cdot\left|\left|w\right|\right|\cdot\cos\varphi$
\end_inset

;
 
\begin_inset Formula $\left|\left|u\times v\right|\right|$
\end_inset

 je namreč ploščina osnovne ploskve,
 
\begin_inset Formula $\left|\left|w\right|\right|\cdot\cos\varphi$
\end_inset

 pa je višina paralelpipeda.
\end_layout

\begin_layout Claim*
Osnovne lastnosti vektorskega produkta so 
\begin_inset Formula $u\times u=0$
\end_inset

,
 
\begin_inset Formula $u\times v=-\left(v\times u\right)$
\end_inset

,
 
\begin_inset Formula $\left(\alpha u+\beta v\right)\times w=\alpha\left(u\times w\right)+\beta\left(v\times w\right)$
\end_inset

 (linearnost)
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Claim*
Osnovne lastnosti mešanega produkta so linearnost v vsakem faktorju,
 menjava dveh faktorjev spremeni predznak (
\begin_inset Formula $\left[u,v,w\right]=-\left[v,u,w\right]$
\end_inset

),
 cikličen pomik ne spremeni vrednosti (
\begin_inset Formula $\left[u,v,w\right]=\left[v,w,u\right]=\left[w,u,v\right]$
\end_inset

).
\end_layout

\begin_layout Subsubsection
Premica v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset


\end_layout

\begin_layout Standard
Premico lahko podamo z
\end_layout

\begin_layout Itemize
dvema različnima točkama
\end_layout

\begin_layout Itemize
s točko 
\begin_inset Formula $\vec{r_{0}}$
\end_inset

 in neničelnim smernim vektorjem 
\begin_inset Formula $\vec{p}$
\end_inset

.
 Premica je tako množica točk 
\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+t\vec{p};\forall t\in\mathbb{R}\right\} $
\end_inset

.
 Taki enačbi premice rečemo parametrična.
\end_layout

\begin_layout Itemize
s točko in normalo (v 
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset

;
 v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

 potrebujemo točko in 
\begin_inset Formula $n-1$
\end_inset

 normal)
\end_layout

\begin_layout Standard
Nadaljujmo s parametričnim zapisom 
\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
\end_inset

.
 Če točke zapišemo po komponentah,
 dobimo parametrično enačbo premice po komponentah:
 
\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+t\left(p_{1},p_{2},p_{3}\right)$
\end_inset

.
\begin_inset Formula 
\[
x=x_{0}+tp_{1}
\]

\end_inset


\begin_inset Formula 
\[
y=y_{0}+tp_{2}
\]

\end_inset


\begin_inset Formula 
\[
z=z_{0}+tp_{3}
\]

\end_inset


\end_layout

\begin_layout Standard
Sedaj lahko iz vsake enačbe izrazimo 
\begin_inset Formula $t$
\end_inset

 in dobimo normalno enačbo premice v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

:
\begin_inset Formula 
\[
t=\frac{x-x_{0}}{p_{1}}=\frac{y-y_{0}}{p_{2}}=\frac{z-z_{0}}{p_{3}}\text{, oziroma v splošnem za premico v \ensuremath{\mathbb{R}^{n}}: }t=\frac{x_{1_{0}}-x_{1}}{p_{1}}=\cdots=\frac{x_{n_{0}}-x_{n}}{p_{n}}
\]

\end_inset


\end_layout

\begin_layout Standard
Osnovne naloge s premicami so projekcija točke na premico,
 zrcaljenje točke čez premico in razdalja med točko in premico.
\end_layout

\begin_layout Paragraph*
Iskanje projekcije dane točke na dano premico
\end_layout

\begin_layout Standard
(skica prepuščena bralcu) 
\begin_inset Formula $\vec{r_{1}}$
\end_inset

 projiciramo na 
\begin_inset Formula $\vec{r}=\vec{r_{0}}+t\vec{p}$
\end_inset

 in dobimo 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

.
 Za 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

 vemo,
 da leži na premici,
 torej 
\begin_inset Formula $\exists t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}$
\end_inset

.
 Poleg tega vemo,
 da je 
\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
\end_inset

 pravokoten na premico oz.
 njen smerni vektor 
\begin_inset Formula $\vec{p}$
\end_inset

,
 torej 
\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0$
\end_inset

.
 Ti dve enačbi združimo,
 da dobimo 
\begin_inset Formula $t$
\end_inset

,
 ki ga nato vstavimo v prvo enačbo:
\begin_inset Formula 
\[
\left\langle \vec{r_{0}}+t\vec{p}-\vec{r_{1},}\vec{p}\right\rangle =0\Longrightarrow\left\langle \vec{r_{0}},\vec{p}\right\rangle +t\left\langle \vec{p},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0\Longrightarrow t=\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }
\]

\end_inset


\begin_inset Formula 
\[
\vec{r_{1}'}=\vec{r_{0}}+t\vec{p}=\vec{r_{0}}+\frac{\left\langle \vec{r_{1}},\vec{p}\right\rangle -\left\langle \vec{r_{0}},\vec{p}\right\rangle }{\left\langle \vec{p},\vec{p}\right\rangle }\vec{p}
\]

\end_inset


\end_layout

\begin_layout Standard
Spotoma si lahko izpišemo obrazec za oddaljenost točke od premice:
 
\begin_inset Formula $a=\left|\left|\vec{r_{1}'}-\vec{r_{1}}\right|\right|$
\end_inset

 in obrazec za zrcalno sliko (
\begin_inset Formula $\vec{r_{1}''}$
\end_inset

):
 
\begin_inset Formula $\vec{r_{1}'}=\frac{\vec{r_{1}''}+\vec{r_{1}}}{2}\Longrightarrow\vec{r_{1}''}=2\vec{r_{1}'}-\vec{r_{1}}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Ravnine v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset


\end_layout

\begin_layout Standard
Ravnino lahko podamo
\end_layout

\begin_layout Itemize
s tremi nekolinearnimi točkami
\end_layout

\begin_layout Itemize
s točko na ravnini in dvema neničelnima smernima vektorjema,
 ki sta linarno neodvisna.
 Ravnina je tako množica točk 
\begin_inset Formula $\left\{ \vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q};\forall s,t\in\mathbb{R}\right\} $
\end_inset

.
 Taki enačbi ravnine rečemo parametrična.
\end_layout

\begin_layout Itemize
s točko in na ravnini in normalo (v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset

;
 v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

 poleg točke potrebujemo 
\begin_inset Formula $n-2$
\end_inset

 normal)
\end_layout

\begin_layout Standard
Nadaljujmo s parametričnim zapisom 
\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset

.
 Če točke zapišemo po komponentah,
 dobimo parametrično enačbo ravnine po komponentah:
 
\begin_inset Formula $\left(x,y,z\right)=\left(x_{0},y_{0},z_{0}\right)+s\left(p_{1},p_{2},p_{3}\right)+t\left(q_{1},q_{2},q_{3}\right)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
x=x_{0}+sp_{1}+tq_{1}
\]

\end_inset


\begin_inset Formula 
\[
y=y_{0}+sp_{2}+tq_{2}
\]

\end_inset


\begin_inset Formula 
\[
z=y_{0}+sp_{3}+tq_{3}
\]

\end_inset


\end_layout

\begin_layout Paragraph
Normalna enačba ravnine v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset


\end_layout

\begin_layout Standard
(skica prepuščena bralcu) Vemo,
 da je 
\begin_inset Formula $\vec{n}$
\end_inset

 (normala) pravokotna na vse vektorje v ravnini,
 tudi na 
\begin_inset Formula $\vec{r}-\vec{r_{0}}$
\end_inset

 za poljuben 
\begin_inset Formula $\vec{r}$
\end_inset

 na ravnini.
 Velja torej normalna enačba ravnine:
 
\begin_inset Formula $\left\langle \vec{r}-\vec{r_{0}},\vec{n}\right\rangle =0$
\end_inset

.
 Razpišimo jo po komponentah,
 da na koncu dobimo normalno enačbo ravnine po komponentah:
\begin_inset Formula 
\[
\left\langle \left(x,y,z\right)-\left(x_{0},y_{0},z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle =0=\left\langle \left(x-x_{0},y-y_{0},z-z_{0}\right),\left(n_{1},n_{2},n_{3}\right)\right\rangle 
\]

\end_inset


\begin_inset Formula 
\[
n_{1}\left(x-x_{0}\right)+n_{2}\left(y-y_{0}\right)+n_{3}\left(z-z_{0}\right)=0=n_{1}x-n_{1}x_{0}+n_{2}y-n_{2}y_{0}+n_{3}z-n_{3}z_{0}=0
\]

\end_inset


\begin_inset Formula 
\[
n_{1}x+n_{2}y+n_{3}z=n_{1}x_{0}+n_{2}y_{0}+n_{3}z_{0}=d
\]

\end_inset


\end_layout

\begin_layout Paragraph
Iskanje pravokotne projekcije dane točke na dano ravnino
\end_layout

\begin_layout Standard
(skica prepuščena bralcu) Projicirati želimo 
\begin_inset Formula $\vec{r_{1}}$
\end_inset

 v 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

 na ravnini 
\begin_inset Formula $\vec{r}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset

.
 Vemo,
 da 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

 leži na ravnini,
 zato 
\begin_inset Formula $\exists s,t\in\mathbb{R}\ni:\vec{r_{1}'}=\vec{r_{0}}+s\vec{p}+t\vec{q}$
\end_inset

.
 Poleg tega vemo,
 da je 
\begin_inset Formula $\vec{r_{1}'}-\vec{r_{1}}$
\end_inset

 pravokoten na ravnino oz.
 na 
\begin_inset Formula $\vec{p}$
\end_inset

 in na 
\begin_inset Formula $\vec{q}$
\end_inset

 hkrati,
 torej 
\begin_inset Formula $\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{1}'}-\vec{r_{1}},\vec{q}\right\rangle $
\end_inset

.
 Vstavimo 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

 iz prve enačbe v drugo in dobimo
\begin_inset Formula 
\[
\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}}+s\vec{p}+t\vec{q}-\vec{r_{1}},\vec{q}\right\rangle 
\]

\end_inset


\begin_inset Formula 
\[
\left\langle \vec{r_{0}},\vec{p}\right\rangle +s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle -\left\langle \vec{r_{1}},\vec{p}\right\rangle =0=\left\langle \vec{r_{0}},\vec{q}\right\rangle +s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle -\left\langle \vec{r_{1}},\vec{q}\right\rangle 
\]

\end_inset

dobimo sistem dveh enačb
\begin_inset Formula 
\[
s\left\langle \vec{p},\vec{p}\right\rangle +t\left\langle \vec{q},\vec{p}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{p}\right\rangle 
\]

\end_inset


\begin_inset Formula 
\[
s\left\langle \vec{p},\vec{q}\right\rangle +t\left\langle \vec{q},\vec{q}\right\rangle =\left\langle \vec{r_{1}}-\vec{r_{0}},\vec{q}\right\rangle 
\]

\end_inset

sistem rešimo in dobljena 
\begin_inset Formula $s,t$
\end_inset

 vstavimo v prvo enačbo zgoraj,
 da dobimo 
\begin_inset Formula $\vec{r_{1}'}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Regresijska premica
\end_layout

\begin_layout Standard
Regresijska premica je primer uporabe zgornje naloge.
 V ravnini je danih 
\begin_inset Formula $n$
\end_inset

 točk 
\begin_inset Formula $\left(x_{1},y_{1}\right),\dots,\left(x_{n},y_{n}\right)$
\end_inset

.
 Iščemo tako premico 
\begin_inset Formula $y=ax+b$
\end_inset

,
 ki se najbolj prilega tem točkam.
 Prileganje premice točkam merimo z metodo najmanjših kvadratov:
 naj bo 
\begin_inset Formula $d_{i}$
\end_inset

 navpična razdalja med 
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset

 in premico 
\begin_inset Formula $y=ax+b$
\end_inset

,
 torej razdalja med točkama 
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset

 in 
\begin_inset Formula $\left(x_{i},ax_{i}+b\right)$
\end_inset

,
 kar je 
\begin_inset Formula $\left|y_{i}-ax_{i}-b\right|$
\end_inset

.
 Minimizirati želimo vsoto kvadratov navpičnih razdalj,
 torej izraz 
\begin_inset Formula $d_{1}^{2}+\cdots+d_{n}^{2}=\left(y_{1}-ax_{1}-b\right)^{2}+\cdots+\left(y_{n}-ax_{n}-b\right)^{2}=\left|\left|\left(y_{1}-ax_{1}-b,\dots,y_{n}-ax_{n}-b\right)\right|\right|^{2}=\left|\left|\left(y_{1},\dots,y_{n}\right)-a\left(x_{1},\dots,x_{n}\right)-b\left(1,\dots,1\right)\right|\right|^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
Če je torej 
\begin_inset Formula $\vec{r}=\vec{0}+a\left(x_{1},\dots,x_{n}\right)+b\left(1,\dots,1\right)$
\end_inset

 hiperravnina v 
\begin_inset Formula $n-$
\end_inset

dimenzionalnem prostoru,
 bo norma,
 ki jo želimo minimizirati,
 najmanjša tedaj,
 ko 
\begin_inset Formula $a,b$
\end_inset

 izberemo tako,
 da najdemo projekcijo 
\begin_inset Formula $\left(y_{1},\dots,y_{n}\right)$
\end_inset

 na to hiperravnino (skica prepuščena bralcu).
 Rešimo sedaj nalogo projekcije točke na ravnino:
\end_layout

\begin_layout Standard
Označimo 
\begin_inset Formula $\vec{y}\coloneqq\left(y_{1},\dots,y_{n}\right)$
\end_inset

,
 
\begin_inset Formula $\vec{x}\coloneqq\left(x_{1},\dots,x_{n}\right)$
\end_inset

,
 
\begin_inset Formula $\vec{1}=\left(1,\dots,1\right)$
\end_inset

.
 Vemo,
 da 
\begin_inset Formula $\vec{y}-a\vec{x}-b\vec{1}\perp\vec{x},\vec{1}$
\end_inset

,
 torej 
\begin_inset Formula $\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{x}\right\rangle =0=\left\langle \vec{y}-a\vec{x}-b\vec{1},\vec{1}\right\rangle $
\end_inset

 in dobimo sistem enačb
\begin_inset Formula 
\[
\left\langle \vec{y},\vec{x}\right\rangle =a\left\langle \vec{x},\vec{x}\right\rangle +b\left\langle \vec{1},\vec{x}\right\rangle 
\]

\end_inset


\begin_inset Formula 
\[
\left\langle \vec{y},\vec{1}\right\rangle =a\left\langle \vec{x},\vec{1}\right\rangle +b\left\langle \vec{1},\vec{1}\right\rangle .
\]

\end_inset

V sistem sedaj vstavimo definicije točk 
\begin_inset Formula $\left(x_{i},y_{i}\right)$
\end_inset

 in ga nato delimo s številom točk,
 da dobimo sistem s povprečji,
 ki ga nato rešimo (izluščimo 
\begin_inset Formula $a,b$
\end_inset

):
\begin_inset Formula 
\[
\sum_{i=1}^{n}y_{i}x_{i}=a\sum_{i=i}^{n}x_{i}^{2}+b\sum_{i=1}^{n}x_{i}\quad\quad\quad\quad/:n
\]

\end_inset


\begin_inset Formula 
\[
\sum_{i=1}^{n}y_{i}=a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}1=a\sum_{i=1}^{n}x_{i}+bn\quad\quad\quad\quad/:n
\]

\end_inset


\begin_inset Formula 
\[
\overline{yx}=a\overline{x^{2}}+b\overline{x}
\]

\end_inset


\begin_inset Formula 
\[
\overline{y}=a\overline{x}+b\Longrightarrow\overline{y}-a\overline{x}=b
\]

\end_inset


\begin_inset Formula 
\[
\overline{yx}=a\overline{x^{2}}+\left(\overline{y}-a\overline{x}\right)\overline{x}=a\overline{x^{2}}+\overline{y}\cdot\overline{x}-a\overline{x}^{2}\Longrightarrow a\left(\overline{x^{2}}-\overline{x}^{2}\right)=\overline{yx}-\overline{y}\cdot\overline{x}\Longrightarrow a=\frac{\overline{yx}-\overline{y}\cdot\overline{x}}{\overline{x^{2}}-\overline{x}^{2}}
\]

\end_inset


\end_layout

\begin_layout Subsection
Sistemi linearnih enačb
\end_layout

\begin_layout Standard
Ta sekcija,
 z izjemo prve podsekcije,
 je precej dobesedno povzeta po profesorjevi beamer skripti.
\end_layout

\begin_layout Subsubsection
Linearna enačba
\end_layout

\begin_layout Definition*
\begin_inset Formula $\sim$
\end_inset

 je enačba oblike 
\begin_inset Formula $a_{1}x_{1}+\cdots+a_{n}x_{n}=b$
\end_inset

 in vsebuje koeficiente,
 spremenljivke in desno stran.
 Množica rešitev so vse 
\begin_inset Formula $n-$
\end_inset

terice realnih števil,
 ki zadoščajo enačbi 
\begin_inset Formula $R=\left\{ \left(x_{1},\dots,x_{n}\right)\in\mathbb{R}^{n};a_{1}x_{1}+\cdots+a_{n}x_{n}=b\right\} $
\end_inset

.
 Če so vsi koeficienti 0,
 pravimo,
 da je enačba trivialna,
 sicer (torej čim je en koeficient neničeln) je netrivialna.
\end_layout

\begin_layout Remark*
Za trivialno enačbo velja 
\begin_inset Formula $R=\begin{cases}
\emptyset & ;b\not=0\\
\mathbb{R}^{n} & ;b=0
\end{cases}$
\end_inset

.
 Za netrivialno enačbo pa velja 
\begin_inset Formula $a_{i}\not=0$
\end_inset

,
 torej:
\begin_inset Formula 
\[
a_{1}x_{1}+\cdots+a_{i}x_{i}+\cdots+a_{n}x_{n}=b
\]

\end_inset


\begin_inset Formula 
\[
a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=b-a_{i}x_{i}=-a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)
\]

\end_inset


\begin_inset Formula 
\[
a_{1}x_{1}+\cdots+a_{i-1}x_{i-1}+a_{i}\left(x_{i}-\frac{b}{a_{i}}\right)+a_{i+1}x_{i+1}+\cdots+a_{n}x_{n}=0
\]

\end_inset


\begin_inset Formula 
\[
\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i-1},x_{i}-\frac{b}{a_{i}},x_{i+1},\dots,x_{n}\right)\right\rangle =0=\left\langle \left(a_{i},\dots,a_{n}\right),\left(x_{1},\dots,x_{i},\dots,x_{n}\right)-\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)\right\rangle 
\]

\end_inset

Tu lahko označimo 
\begin_inset Formula $\vec{n}\coloneqq\left(a_{i},\dots,a_{n}\right)$
\end_inset

,
 
\begin_inset Formula $\vec{r}=\left(x_{1},\dots,x_{i},\dots,x_{n}\right)$
\end_inset

,
 
\begin_inset Formula $\vec{r_{0}}=\left(0,\dots,0,\frac{b}{a},0,\dots,0\right)$
\end_inset

 in dobimo 
\begin_inset Formula $\left\langle \vec{n},\vec{r}-\vec{r_{0}}\right\rangle $
\end_inset

,
 kar je normalna enačba premice v 
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset

,
 normalna enačba ravnine v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset

 oziroma normalna enačba hiperravnine v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Sistem linearnih enačb
\end_layout

\begin_layout Definition*
Sistem 
\begin_inset Formula $m$
\end_inset

 linearnih enačb z 
\begin_inset Formula $n$
\end_inset

 spremenljivkami je sistem enačb oblike
\begin_inset Formula 
\[
\begin{array}{ccccccc}
a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
\vdots &  &  &  & \vdots &  & \vdots\\
a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
\end{array}.
\]

\end_inset


\end_layout

\begin_layout Fact*
Množica rešitev sistema je 
\begin_inset Formula $\mathbb{R}^{n}\Leftrightarrow\forall i,j:a_{i,j}=b_{i}=0$
\end_inset

.
 Sicer je množica rešitev presek hiperravnin v 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

 —
 rešitev posameznih enačb.
 To vključuje tudi primer prazne množice rešitev,
 saj je takšna na primer presek dveh vzporednih hiperravnin.
\end_layout

\begin_layout Example*
Množica rešitev 
\begin_inset Formula $2\times2$
\end_inset

 sistema je lahko
\end_layout

\begin_layout Itemize
cela ravnina
\end_layout

\begin_layout Itemize
premica v ravnini
\end_layout

\begin_layout Itemize
točka v ravnini
\end_layout

\begin_layout Itemize
prazna množica
\end_layout

\begin_layout Remark*
Enako velja za množico rešitev 
\begin_inset Formula $3\times2$
\end_inset

 sistema.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
Množica rešitev sistema 
\begin_inset Formula $2\times3$
\end_inset

 pa ne more biti točka v prostoru,
 lahko pa je cel prostor,
 ravnina v prostoru,
 premica v prostoru ali prazna množica.
\end_layout

\begin_layout Paragraph*
Algebraičen pomen rešitev sistema
\end_layout

\begin_layout Standard
Rešitve sistema
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{array}{ccccccc}
a_{1,1}x_{1} & + & \cdots & + & a_{1,n}x_{n} & = & b_{1}\\
\vdots &  &  &  & \vdots &  & \vdots\\
a_{m,1}x_{1} & + & \cdots & + & a_{m,n}x_{n} & = & b_{m}
\end{array}
\]

\end_inset


\end_layout

\begin_layout Standard
lahko zapišemo kot linearno kombinacijo stolpecv sistema in spremenljivk:
\begin_inset Formula 
\[
\left(b_{1},\dots,b_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)=x_{1}\left(a_{1,1},\dots,a_{m,1}\right)+\cdots+x_{n}\left(a_{1,n},\dots,a_{m,n}\right)
\]

\end_inset


\begin_inset Formula 
\[
\vec{b}=x_{1}\vec{a_{1}}+\cdots+x_{n}\vec{a_{n}}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Klasifikacija sistemov linearnih enačb
\end_layout

\begin_layout Standard
Sisteme linearnih enačb delimo glede na velikost na
\end_layout

\begin_layout Itemize
kvadratne (toliko enačb kot spremenljivk),
\end_layout

\begin_layout Itemize
poddoločene (več spremenljivk kot enačb),
\end_layout

\begin_layout Itemize
predoločene (več enačb kot spremenljivk);
\end_layout

\begin_layout Standard
glede na rešljivost na
\end_layout

\begin_layout Itemize
nerešljive (prazna množica rešitev),
\end_layout

\begin_layout Itemize
enolično rešljive (množica rešitev je singleton),
\end_layout

\begin_layout Itemize
neenolično rešljive (moč množice rešitev je več kot 1);
\end_layout

\begin_layout Standard
glede na obliko desnih strani na
\end_layout

\begin_layout Itemize
homogene (vektor desnih stani je ničeln)
\end_layout

\begin_layout Itemize
nehomogene (vektor desnih strani je neničen)
\end_layout

\begin_layout Remark*
Če sta 
\begin_inset Formula $\vec{x}$
\end_inset

 in 
\begin_inset Formula $\vec{y}$
\end_inset

 dve različni rešitvi sistema,
 je rešitev sistema tudi 
\begin_inset Formula $\left(1-t\right)\vec{x}+t\vec{y}$
\end_inset

 za vsak realen 
\begin_inset Formula $t$
\end_inset

,
 torej ima vsak neenolično rešljiv sistem neskončno rešitev.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
Pogosto (a nikakor ne vedno) se zgodi,
 da je kvadraten sistem enolično rešljiv,
 predoločen sistem nerešljiv,
 poddoločen sistem pa neenolično rešljiv.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
Homogen sistem je vedno rešljiv,
 saj obstaja trivialna rešitev 
\begin_inset Formula $\vec{0}$
\end_inset

.
 Vprašanje pri homogenih sistemih je torej,
 kdaj je enolično in kdaj neenolično rešljiv.
 Dokazali bomo,
 da je vsak poddoločen homogen sistem linearnih enačb neenolično rešljiv.
\end_layout

\begin_layout Subsubsection
Reševanje sistema
\end_layout

\begin_layout Standard
Sisteme lahko rešujemo z izločanjem spremenljivk.
 Iz ene enačbe izrazimo spremenljivko in jo vstavimo v druge enačbe,
 da izrazimo zopet nove spremenljivke,
 ki jih spet vstavimo v nove enačbe,
 iz katerih spremenljivk še nismo izražali in tako naprej,
 vse dokler ne pridemo do zadnjega možnega izražanja (dodatno branje prepuščeno bralcu).
\end_layout

\begin_layout Standard
Sisteme pa lahko rešujemo tudi z Gaussovo metodo.
 Trdimo,
 da se rešitev sistema ne spremeni,
 če na njem uporabimo naslednje elementarne vrstične transformacije:
\end_layout

\begin_layout Itemize
menjava vrstnega reda enačb,
\end_layout

\begin_layout Itemize
množenje enačbe z neničelno konstanto,
\end_layout

\begin_layout Itemize
prištevanje večkratnika ene enačbe k drugi.
\end_layout

\begin_layout Standard
Z Gaussovo metodo (dodatno branje prepuščeno bralcu) mrcvarimo razširjeno matriko sistema,
 dokler ne dobimo reducirane kvadratne stopničaste forme (angl.
 row echelon),
 ki izgleda takole (
\begin_inset Formula $\times$
\end_inset

 reprezentira poljubno realno številko,
 
\begin_inset Formula $0$
\end_inset

 ničlo in 
\begin_inset Formula $1$
\end_inset

 enico):
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\left[\begin{array}{ccccccccccccccccc|c}
0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
\vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \times & \cdots & \times & 0 & \cdots & \times\\
 &  &  & \vdots & \vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \times & \cdots & \times & 0 & \cdots & \times\\
 &  &  &  &  &  &  & \vdots & \vdots &  & \vdots & 0 & 0 & \cdots & 0 & 1 & \cdots & \times\\
 &  &  &  &  &  &  &  &  &  &  & \vdots & \vdots &  & \vdots & 0 & \cdots & \vdots\\
\vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots & \vdots &  & \vdots & \vdots &  & \vdots\\
0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & \cdots & \times
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Homogeni sistemi
\end_layout

\begin_layout Definition*
Sistem je homogen,
 če je vektor desnih strani ničeln.
\end_layout

\begin_layout Claim*
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

 Že dokazano zgoraj.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(1\Rightarrow3\right)$
\end_inset

 Že dokazano zgoraj.
\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)
\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.
 
\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 
\[
\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det AB=\det B=\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A\det B
\]

\end_inset


\begin_inset Formula 
\[
\det AB=\det A\det B
\]

\end_inset


\end_layout

\begin_deeper
\begin_layout Remark*
\begin_inset Formula $\exists A^{-1}\Leftrightarrow\det A\not=0$
\end_inset


\end_layout

\begin_deeper
\begin_layout Proof
Predpostavimo 
\begin_inset Formula $A$
\end_inset

 je obrnljiva.
 Tedaj 
\begin_inset Formula $\exists A^{-1}=B\ni:AB=I\overset{\circ\det}{\Longrightarrow}\det\left(AB\right)=\det I$
\end_inset

.
 PDDRAA 
\begin_inset Formula $\det A\not=0$
\end_inset

,
 tedaj 
\begin_inset Formula $\det AB=\det B\det A=0\not=\det I=1$
\end_inset

.
 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset


\end_layout

\begin_layout Proof
Predpostavimo sedaj 
\begin_inset Formula $A$
\end_inset

 ni obrnljiva.
 Tedaj 
\begin_inset Formula $\nexists A^{-1}\Rightarrow\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=R$
\end_inset

 ima ničelno vrstico.
 Uporabimo isti razmislek kot spodaj,
 torej 
\begin_inset Formula $\det R=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
\end_inset

.
 Ker so determinante elementarnih matrik vse neničelne,
 mora biti 
\begin_inset Formula $\det A$
\end_inset

 ničeln,
 da je produkt ničeln.
\end_layout

\end_deeper
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $R$
\end_inset

 ima ničelno vrstico.
 Tedaj 
\begin_inset Formula $\det\left(R\right)=0\Rightarrow0=\det R=\det E_{n}\cdots\det E_{1}\det A$
\end_inset

.
 Ker so determinante elementarnih matrik vse neničelne,
 mora biti 
\begin_inset Formula $\det A$
\end_inset

 ničeln,
 da je produkt ničeln.
\end_layout

\end_deeper
\begin_layout Enumerate
Dokazujemo 
\begin_inset Formula $\det A^{T}=\det A$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Enumerate
Če je 
\begin_inset Formula $A$
\end_inset

 elementarna matrika,
 to drži:
 
\begin_inset Formula $\det P_{ij}=-1=\det P_{ij}^{T}=\det P_{ij}$
\end_inset

,
 
\begin_inset Formula $\det E_{i}\left(\alpha\right)=\alpha=\det E_{i}\left(\alpha\right)^{T}=\det E_{i}\left(\alpha\right)$
\end_inset

,
 
\begin_inset Formula $\det E_{ij}\left(\alpha\right)=1=\det E_{ji}\left(\alpha\right)^{T}=\det E_{ij}\left(\alpha\right)^{T}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Če ima 
\begin_inset Formula $A$
\end_inset

 ničelno vrstico,
 to drži,
 saj ima tedaj 
\begin_inset Formula $A^{T}$
\end_inset

 ničeln stolpec in 
\begin_inset Formula $\det A=0=\det A^{T}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Splošen primer:
 Po Gaussovi metodi 
\begin_inset Formula $\exists E_{n},\dots,E_{1}\ni:E_{n}\cdots E_{1}A=\text{RVSO}\left(A\right)=R$
\end_inset

.
 Zopet ločimo dva primera:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $R=I$
\end_inset

.
 
\begin_inset Formula $\det R=\det R^{T}=1$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $R$
\end_inset

 ima ničelno vrstico.
 
\begin_inset Formula $\det R=\det R^{T}=0$
\end_inset


\end_layout

\begin_layout Standard
Sedaj vemo,
 da 
\begin_inset Formula $\det R=\det R^{T}$
\end_inset

.
 Računajmo:
\begin_inset Formula 
\[
\det R=\det R^{T}
\]

\end_inset


\begin_inset Formula 
\[
\det\left(E_{n}\cdots E_{1}A\right)=\det\left(E_{n}\cdots E_{1}A\right)^{T}
\]

\end_inset


\begin_inset Formula 
\[
\det\left(E_{n}\cdots E_{1}A\right)=\det\left(A^{T}E_{1}^{T}\cdots E_{n}^{T}\right)
\]

\end_inset


\begin_inset Formula 
\[
\cancel{\det E_{n}}\cdots\cancel{\det E_{1}}\det A=\det A^{T}\cancel{\det E_{1}^{T}}\cdots\cancel{\det E_{n}^{T}}
\]

\end_inset


\begin_inset Formula 
\[
\det A=\det A^{T}
\]

\end_inset


\end_layout

\end_deeper
\end_deeper
\begin_layout Enumerate
Dokazujemo 
\begin_inset Formula $\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det A\det C$
\end_inset

.
 Levi izraz v enačbi vsebuje t.
 i.
 bločno matriko.
 Upoštevamo poprej dokazano multiplikativnost determinante in opazimo,
 da pri bločnem množenju matrik velja
\begin_inset Formula 
\[
\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\left[\begin{array}{cc}
I & 0\\
0 & C
\end{array}\right]\left[\begin{array}{cc}
A & B\\
0 & I
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det\left[\begin{array}{cc}
I & 0\\
0 & C
\end{array}\right]\det\left[\begin{array}{cc}
A & B\\
0 & I
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
\det\left[\begin{array}{cc}
A & B\\
0 & C
\end{array}\right]=\det C\det A
\]

\end_inset


\end_layout

\begin_deeper
\begin_layout Standard
Pojasnilo:
 Za 
\begin_inset Formula $\det C$
\end_inset

 si razpišemo bločno matriko,
 za 
\begin_inset Formula $\det A$
\end_inset

 si zopet razpišemo bločno matriko in nato z Gaussovimi transformacijami z enicami iz spodnjega desnega bloka izničimo zgornji desni blok (
\begin_inset Formula $B$
\end_inset

).
\end_layout

\end_deeper
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Subsubsection
Cramerjevo pravilo —
 eksplicitna formula za rešitve kvadratnega sistema linearnih enačb
\end_layout

\begin_layout Standard
Radi bi dobili eksplicitne formule za komponente rešitve 
\begin_inset Formula $x_{i}$
\end_inset

 kvadratnega sistema linearnih enačb 
\begin_inset Formula $A\vec{x}=\vec{b}$
\end_inset

.
 Izpeljimo torej eksplicitno formulo .
 Druga/srednja matrika je identična,
 v kateri smo 
\begin_inset Formula $i-$
\end_inset

ti stolpec zamenjali z vektorjem spremenljivk 
\begin_inset Formula $\vec{x}$
\end_inset

 (to označimo z 
\begin_inset Formula $I_{i}\left(\vec{x}\right)$
\end_inset

),
 tretja/desna matrika pa je matrika koeficientov v kateri smo 
\begin_inset Formula $i-$
\end_inset

ti stolpec zamenjali z vektorjem desnih strani 
\begin_inset Formula $b$
\end_inset

 (to označimo z 
\begin_inset Formula $A_{i}\left(\vec{x}\right)$
\end_inset

).
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots &  & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]\left[\begin{array}{ccccccc}
1 &  & 0 & x_{1} &  &  & 0\\
 & \ddots &  & \vdots\\
 &  & 1 & x_{i-1}\\
 &  &  & x_{i}\\
 &  &  & x_{i+1} & 1\\
 &  &  & \vdots &  & \ddots\\
0 &  &  & x_{n} & 0 &  & 1
\end{array}\right]=\left[\begin{array}{ccccc}
a_{11} & \cdots & b_{1} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots & \iddots & \vdots\\
a_{i1} &  & b_{i} &  & a_{in}\\
\vdots & \iddots & \vdots & \ddots & \vdots\\
a_{n1} & \cdots & b_{n} & \cdots & a_{nn}
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots &  & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{array}\right]\left[\begin{array}{ccccccc}
1 &  & 0 & x_{1} &  &  & 0\\
 & \ddots &  & \vdots\\
 &  & 1 & x_{i-1}\\
 &  &  & x_{i}\\
 &  &  & x_{i+1} & 1\\
 &  &  & \vdots &  & \ddots\\
0 &  &  & x_{n} & 0 &  & 1
\end{array}\right]=\left[\begin{array}{ccccc}
a_{11} & \cdots & a_{11}x_{1}+\cdots+a_{1n}x_{n} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots & \iddots & \vdots\\
a_{i1} &  & a_{i1}x_{1}+\cdots+a_{in}x_{n} &  & a_{in}\\
\vdots & \iddots & \vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{n1}x_{1}+\cdots+a_{nn}x_{n} & \cdots & a_{nn}
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
AI_{i}\left(\vec{x}\right)=A_{i}\left(\vec{b}\right)\quad\quad\quad\quad/\det
\]

\end_inset


\begin_inset Formula 
\[
\det\left(AI_{i}\left(\vec{x}\right)\right)=\det A_{i}\left(\vec{b}\right)
\]

\end_inset


\begin_inset Formula 
\[
\det A\det I_{i}\left(\vec{x}\right)=\det A_{i}\left(\vec{b}\right)
\]

\end_inset

Izračunamo 
\begin_inset Formula $\det I_{i}\left(\vec{x}\right)$
\end_inset

 z razvojem po 
\begin_inset Formula $i-$
\end_inset

ti vrstici.
\begin_inset Formula 
\[
\det A\cdot x_{i}=\det A_{i}\left(\vec{b}\right)
\]

\end_inset


\begin_inset Formula 
\[
x_{i}=\frac{\det A_{i}\left(\vec{b}\right)}{\det A}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Formula-za-inverz-matrike"

\end_inset

Formula za inverz matrike
\end_layout

\begin_layout Standard
Za dano obrnljivo 
\begin_inset Formula $A_{n\times n}$
\end_inset

 iščemo eksplicitno formulo za celice 
\begin_inset Formula $X$
\end_inset

,
 da velja 
\begin_inset Formula $AX=I$
\end_inset

.
 Ideja:
 najprej bomo problem prevedli na reševanje sistemov linearnih enačb in uporabili Cramerjevo pravilo ter končno poenostavili formule.
 Naj bodo 
\begin_inset Formula $\vec{x_{1}},\dots,\vec{x_{n}}$
\end_inset

 stolpci 
\begin_inset Formula $X$
\end_inset

 in 
\begin_inset Formula $\vec{i_{1}},\dots,\vec{i_{n}}$
\end_inset

.
 Potemtakem je 
\begin_inset Formula $\left[\begin{array}{ccc}
A\vec{x_{1}} & \cdots & A\vec{x_{n}}\end{array}\right]=A\left[\begin{array}{ccc}
\vec{x_{1}} & \cdots & \vec{x_{n}}\end{array}\right]=AX=I=\left[\begin{array}{ccc}
\vec{i_{1}} & \cdots & \vec{i_{n}}\end{array}\right]$
\end_inset

.
 Primerjajmo sedaj stolpce na obeh straneh:
 
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :A\vec{x_{i}}=\vec{i_{i}}$
\end_inset

.
 ZDB za vsak stolpec 
\begin_inset Formula $X$
\end_inset

 smo dobili sistem 
\begin_inset Formula $n\times n$
\end_inset

 linearnih enačb.
 Te sisteme 
\begin_inset Formula $A\vec{x_{j}}=\vec{i_{j}}$
\end_inset


\begin_inset Foot
status open

\begin_layout Plain Layout
Tokrat uporabimo indeks 
\begin_inset Formula $j$
\end_inset

,
 ker z njim reprezentiramo stolpec in ponavadi,
 ko govorimo o elementu 
\begin_inset Formula $x_{ij}$
\end_inset

 matrike 
\begin_inset Formula $X$
\end_inset

,
 z 
\begin_inset Formula $i$
\end_inset

 označimo vrstico.
\end_layout

\end_inset

 bomo rešili s Cramerjevim pravilom.
\begin_inset Formula 
\[
x_{ij}=\left(\vec{x}_{j}\right)_{i}=\frac{\det A_{i}\left(\vec{i_{j}}\right)}{\det A}\overset{\text{razvoj po \ensuremath{j-}ti vrstici}}{=}\frac{\det A_{ji}\cdot\left(-1\right)^{j+i}}{\det A}
\]

\end_inset


\begin_inset Formula 
\[
X=A^{-1}=\left[\begin{array}{ccc}
\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\
\vdots &  & \vdots\\
\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A}
\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc}
\det A_{11}\cdot\left(-1\right)^{1+1} & \cdots & \det A_{1n}\cdot\left(-1\right)^{1+1}\\
\vdots &  & \vdots\\
\det A_{n1}\cdot\left(-1\right)^{n+1} & \cdots & \det A_{nn}\cdot\left(-1\right)^{n+n}
\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T},
\]

\end_inset


\end_layout

\begin_layout Standard
kjer 
\begin_inset Formula $\tilde{A}$
\end_inset

 pravimo kofaktorska matrika.
\end_layout

\begin_layout Subsection
Algebrske strukture
\end_layout

\begin_layout Subsubsection
Uvod
\end_layout

\begin_layout Standard
Naj bo 
\begin_inset Formula $M$
\end_inset

 neprazna množica.
 Operacija na 
\begin_inset Formula $M$
\end_inset

 pove,
 kako iz dveh elementov 
\begin_inset Formula $M$
\end_inset

 dobimo nov element 
\begin_inset Formula $M$
\end_inset

.
 Na primer,
 če 
\begin_inset Formula $a,b\in M$
\end_inset

,
 je 
\begin_inset Formula $a\circ b$
\end_inset

 nov element 
\begin_inset Formula $M$
\end_inset

.
\end_layout

\begin_layout Definition*
Operacija na 
\begin_inset Formula $M$
\end_inset

 je funkcija 
\begin_inset Formula $\circ:M\times M\to M$
\end_inset

,
 kjer je 
\begin_inset Formula $M\times M$
\end_inset

 kartezični produkt (urejeni pari).
 
\begin_inset Formula $\left(a,b\right)\mapsto\circ\left(a,b\right)$
\end_inset

,
 slednje pa označimo z 
\begin_inset Formula $\circ\left(a,b\right)=a\circ b$
\end_inset

.
\end_layout

\begin_layout Standard
Na isti množici imamo lahko več različno definiranih operacij.
 Ločimo jih tako,
 da uvedemo pojem grupoida.
\end_layout

\begin_layout Definition*
Grupoid je 
\begin_inset Formula $\left(\text{neprazna množica},\text{izbrana operacija }\circ:M\times M\to M\right)$
\end_inset

.
 Na primer 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

.
\end_layout

\begin_layout Standard
Še posebej nas zanimajo operacije z lepimi lastnostmi,
 denimo asociativnost,
 komutativnost,
 obstoj enot,
 inverzov.
\end_layout

\begin_layout Definition*
Grupoid,
 katerega 
\begin_inset Formula $\circ$
\end_inset

 je asociativna 
\begin_inset Formula $\Leftrightarrow\forall a,b,c\in M:\left(a\circ b\right)\circ c=a\circ\left(b\circ c\right)$
\end_inset

,
 je polgrupa.
 Tedaj skladnja dopušča pisanje brez oklepajev:
 
\begin_inset Formula $a\circ b\circ c\circ d$
\end_inset

 je nedvoumen/veljaven izraz,
 ko je 
\begin_inset Formula $\circ$
\end_inset

 asociativna.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Komutativnost:
 
\begin_inset Formula $\circ$
\end_inset

 je komutativna 
\begin_inset Formula $\Leftrightarrow\forall a,b\in M:a\circ b=b\circ a$
\end_inset

.
 Grupoidom s komutativno operacijo pravimo,
 da so komutativni.
\end_layout

\begin_layout Example*
Asociativni in komutativni grupoidi (komutativne polgrupe):
 
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{Q},\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset

 —
 številske operacije.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Asociativni,
 a ne komutativni grupoidi (nekomutativne polgrupe):
 
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset

 —
 množenje matrik.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Komutativni,
 a ne asociativni grupoidi:
 Jordanski produkt matrik:
 
\begin_inset Formula $A\circ B=\frac{1}{2}\left(AB+BA\right)$
\end_inset

\SpecialChar endofsentence

\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Niti komutativni niti asociativni grupoidi:
 Vektorski produkt v 
\begin_inset Formula $\mathbb{R}^{3}$
\end_inset

:
 
\begin_inset Formula $\left(\mathbb{R}^{3},\times\right)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $M\not=\emptyset$
\end_inset

.
 
\begin_inset Formula $F$
\end_inset

 naj bodo vse funkcije 
\begin_inset Formula $M\to M$
\end_inset

,
 
\begin_inset Formula $\circ$
\end_inset

 pa kompozitum dveh funkcij.
 Izkaže se,
 da:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(F,\circ\right)$
\end_inset

 je vedno polgrupa.
\end_layout

\begin_deeper
\begin_layout Proof
Definicija kompozituma:
 
\begin_inset Formula $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
\end_inset

.
\begin_inset Formula 
\[
\left(f\circ g\right)\circ h\overset{?}{=}f\circ\left(g\circ h\right)
\]

\end_inset


\begin_inset Formula 
\[
\forall x:\left(\left(f\circ g\right)\circ h\right)\left(x\right)=\left(f\circ g\right)\left(h\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
\]

\end_inset


\begin_inset Formula 
\[
\forall x:\left(f\circ\left(g\circ h\right)\right)\left(x\right)=f\left(\left(g\circ h\right)\left(x\right)\right)=f\left(g\left(h\left(x\right)\right)\right)
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Itemize
Čim ima 
\begin_inset Formula $M$
\end_inset

 vsaj tri elemente,
 
\begin_inset Formula $\left(F,\circ\right)$
\end_inset

 ni komutativna.
\end_layout

\end_deeper
\begin_layout Definition*
Naj bo 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 grupoid.
 Element 
\begin_inset Formula $e\in M$
\end_inset

 je enota,
 če 
\begin_inset Formula $\forall a\in M:e\circ a=a\wedge a\circ e=a$
\end_inset

.
 Če velja le eno v konjunkciji,
 je 
\begin_inset Formula $e$
\end_inset

 bodisi leva bodisi desna enota (respectively) in v takem primeru 
\begin_inset Formula $e$
\end_inset

 ni enota.
\end_layout

\begin_layout Example*
Ali spodnji grupoidi imajo enoto in kakšna je?
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{R},+\right)$
\end_inset

:
 enota je 0.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset

:
 enota je 1.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset

:
 ni enote,
 kajti 
\begin_inset Formula $0\not\in\mathbb{N}$
\end_inset

.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset

:
 enota je 
\begin_inset Formula $I_{n\times n}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Claim*
Vsak grupoid ima kvečjemu eno enoto.
 Dve enoti v istem grupoidu sta enaki.
 Še več:
 vsaka leva enota je enaka vsaki desni enoti.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $e$
\end_inset

 leva enota in 
\begin_inset Formula $f$
\end_inset

 desna enota,
 torej 
\begin_inset Formula $\forall a:e\circ a=a\wedge a\circ f=a$
\end_inset

.
 Tedaj 
\begin_inset Formula $e\circ f=f$
\end_inset

 in 
\begin_inset Formula $e\circ f=e$
\end_inset

.
 Ker je vsaka leva enota vsaki desni,
 sta poljubni enoti enaki.
 Enota je,
 če obstaja,
 ena sama in je obenem edina leva in edina desna enota.
\end_layout

\begin_layout Example*
Lahko se zgodi,
 da obstaja poljubno različnih levih,
 a nobene desne enote.
 Primer so vse matrike oblike 
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right]$
\end_inset

.
 Račun 
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right]\cdot\left[\begin{array}{cc}
c & d\\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
ac & ad\\
0 & 0
\end{array}\right]$
\end_inset

 pokaže,
 da so vsi elementi 
\begin_inset Formula $\left[\begin{array}{cc}
1 & \times\\
0 & 0
\end{array}\right]$
\end_inset

 leve enote.
 Iz dejstva,
 da je več (tu celo neskončno) levih enot,
 sledi dejstvo,
 da ni desnih.
\end_layout

\begin_layout Definition*
Polgrupi z enoto pravimo monoid.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 monoid z enoto 
\begin_inset Formula $e$
\end_inset

.
 Inverz elementa 
\begin_inset Formula $a\in M$
\end_inset

 je tak 
\begin_inset Formula $b\in M\ni:b\circ a=e\wedge a\circ b=e$
\end_inset

.
 Elementu,
 ki zadošča levi strani konjunkcije,
 pravimo levi inverz 
\begin_inset Formula $a$
\end_inset

,
 elemetu,
 ki zadošča desni strani konjunkcije,
 pa desni inverz 
\begin_inset Formula $a$
\end_inset

.
 Inverz 
\begin_inset Formula $a$
\end_inset

 je torej tak element,
 ki je hkrati levi in desni inverz 
\begin_inset Formula $a$
\end_inset

.
\end_layout

\begin_layout Remark*
Ni nujno,
 da ima vsak element monoida inverz.
 Primer je 
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset

;
 niso vse matrike obrnljive.
\end_layout

\begin_layout Claim*
Vsak element monoida ima kvečjemu en inverz.
 Vsak levi inverz je enak vsakemu desnemu.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $b$
\end_inset

 levi in 
\begin_inset Formula $c$
\end_inset

 desni inverz 
\begin_inset Formula $a$
\end_inset

,
 torej 
\begin_inset Formula $b\circ a=e=a\circ c$
\end_inset

.
 Računajmo:
 
\begin_inset Formula $b=b\circ e=b\circ\left(a\circ c\right)=\left(b\circ a\right)\circ c=e\circ c=c$
\end_inset

.
 Če obstaja,
 je torej inverz en sam,
 in ta je edini levi in edini desni inverz.
\end_layout

\begin_layout Definition*
Ker vemo,
 da je inverz enoličen,
 lahko vpeljemo oznako 
\begin_inset Formula $a^{-1}$
\end_inset

 za inverz elementa 
\begin_inset Formula $a$
\end_inset

.
\end_layout

\begin_layout Example*
Ali v spodnjih monoidih obstajajo inverzi in kakšni so?
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset

:
 inverz 
\begin_inset Formula $a$
\end_inset

 je 
\begin_inset Formula $-a$
\end_inset

.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Z},\cdot\right)$
\end_inset

:
 inverz 
\begin_inset Formula $1$
\end_inset

 je 
\begin_inset Formula $1$
\end_inset

,
 inverz 
\begin_inset Formula $-1$
\end_inset

 je 
\begin_inset Formula $-1$
\end_inset

,
 ostali elementi pa inverza nimajo.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset

:
 inverz 
\begin_inset Formula $a$
\end_inset

 je 
\begin_inset Formula $\frac{1}{a}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Remark*
Če desnega inverza ni,
 je lahko levih inverzov več.
 Primer:
 Naj bodo 
\begin_inset Formula $M$
\end_inset

 vse funkcije 
\begin_inset Formula $\mathbb{N}\to\mathbb{N}$
\end_inset

 in naj bo 
\begin_inset Formula $\circ$
\end_inset

 kompozitum funkcij.
 Tedaj velja:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset

 ima levi inverz 
\begin_inset Formula $\Leftrightarrow f$
\end_inset

 injektivna.
\end_layout

\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset

 ima desni inverz 
\begin_inset Formula $\Leftrightarrow f$
\end_inset

 surjektivna.
\end_layout

\begin_layout Itemize
\begin_inset Formula $f\in M$
\end_inset

 ima inverz 
\begin_inset Formula $\Leftrightarrow f$
\end_inset

 bijektivna.
\end_layout

\end_deeper
\begin_layout Example*
\begin_inset Formula $f\left(n\right)=n+1$
\end_inset

 je injektivna,
 a ne surjektivna.
 Vsi za komponiranje levi inverzi 
\begin_inset Formula $f$
\end_inset

 so funkcije oblike 
\begin_inset Formula $g\left(x\right)=\begin{cases}
x-1 & ;x>1\\
\times & ;x=1
\end{cases}$
\end_inset

 ZDB 
\begin_inset Formula $x$
\end_inset

 lahko slikajo v karkoli,
 pa bo 
\begin_inset Formula $\left(g\circ f\right)$
\end_inset

 še vedno funkcija identiteta.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
V 
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),\cdot\right)$
\end_inset

 je vsak levi inverz tudi desni inverz.
 To je res tudi za funkcije na končni množici,
 toda ni res v splošnem.
\end_layout

\begin_layout Definition*
Grupa je tak monoid,
 v katerem ima vsak element inverz.
 Daljše:
 grupa je taka neprazna množica 
\begin_inset Formula $G$
\end_inset

 z operacijo 
\begin_inset Formula $\circ$
\end_inset

,
 ki zadošča asociativnosti,
 obstaja enota in za vsak element obstaja njegov inverz.
 Grupi s komutativno operacijo pravimo Abelova grupa.
\end_layout

\begin_layout Example*
Nekaj abelovih grup:
 
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{Q}\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(M_{n\times n}\left(\mathbb{R}\right),+\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{R}^{n},+\right)$
\end_inset

.
 Nekaj neabelovih grup:
 
\begin_inset Formula $\left(\text{vse obrnljive matrike fiksne dimenzije},\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(\text{vse permutacije neprazne končne množice},\circ\right)$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Podstrukture
\end_layout

\begin_layout Standard
Naj bo 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 grupoid.
 Reciumi,
 da je 
\begin_inset Formula $N$
\end_inset

 neprazna podmnožica 
\begin_inset Formula $M$
\end_inset

.
 Pod temi pogoji se lahko zgodi,
 da 
\begin_inset Formula $\exists a,b\in N\ni:a\circ b\not\in N$
\end_inset

.
\end_layout

\begin_layout Example*
Oglejmo si grupoid 
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset

.
 
\begin_inset Formula $N\subseteq\mathbb{Z}$
\end_inset

 naj bodo liha cela števila.
 
\begin_inset Formula $\forall a,b\in N:a+b\not\in N\Rightarrow\exists a,b\in N\ni:a+b\not\in N$
\end_inset

,
 kajti vsota lihih števil je soda.
\end_layout

\begin_layout Definition*
Pravimo,
 da je podmnožica 
\begin_inset Formula $N\subseteq M$
\end_inset

 zaprta za 
\begin_inset Formula $\circ$
\end_inset

,
 če 
\begin_inset Formula $\forall a,b\in N:a\circ b\in N$
\end_inset

.
\end_layout

\begin_layout Example*
Oglejmo si spet grupoid 
\begin_inset Formula $\left(\mathbb{Z},+\right)$
\end_inset

.
 
\begin_inset Formula $N\subseteq\mathbb{Z}$
\end_inset

 naj bodo soda cela števila.
 
\begin_inset Formula $N$
\end_inset

 je zaprta za 
\begin_inset Formula $+$
\end_inset

.
\end_layout

\begin_layout Definition*
Takemu 
\begin_inset Formula $N$
\end_inset

,
 kjer je 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 z implicitno podedovano operacijo (
\begin_inset Formula $a\circ_{N}b=a\circ b$
\end_inset

) pravimo podgrupoid 
\begin_inset Formula $\left(N,\circ_{N}\right)$
\end_inset

.
\end_layout

\begin_layout Exercise*
Pokaži,
 da je 
\begin_inset Quotes gld
\end_inset

general linear
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A\not=0\right\} $
\end_inset

 grupa za matrično množenje.
\end_layout

\begin_deeper
\begin_layout Standard
Asociativnost je dokazana zgoraj.
 Enota je 
\begin_inset Formula $I_{n}$
\end_inset

.
 Inverzi obstajajo,
 ker so determinante neničelne in tudi inverzi imajo neničelne determinante.
 Preveriti je treba še vsebovanost,
 torej 
\begin_inset Formula $\forall A,B\in GL_{n}\left(\mathbb{R}\right):A\cdot B\in GL_{n}\left(\mathbb{R}\right)$
\end_inset

.
 Vzemimo poljubni 
\begin_inset Formula $A,B\in GL_{n}\left(\mathbb{R}\right)$
\end_inset

,
 torej 
\begin_inset Formula $\det A\not=0\wedge\det B\not=0$
\end_inset

.
 
\begin_inset Formula $\det\left(AB\right)=\det A\det B=0\Leftrightarrow\det A=0\vee\det B=0$
\end_inset

,
 toda ker noben izmed izrazov disjunkcije ne drži,
 determinanta 
\begin_inset Formula $AB$
\end_inset

 nikdar ni 0.
 Enota 
\begin_inset Formula $I$
\end_inset

 je vsebovana v 
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset

,
 saj 
\begin_inset Formula $\det I=1\not=0$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
Ali je 
\begin_inset Quotes gld
\end_inset

special linear
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)\coloneqq\left\{ A\in M_{n\times n}\left(\mathbb{R}\right);\det A=1\right\} $
\end_inset

 grupa za matrično množenje?
\end_layout

\begin_deeper
\begin_layout Standard
Vse lastnosti (razen vsebovanosti) smo preverili zgoraj.
 Preveriti je treba vsebovanost,
 torej ali 
\begin_inset Formula $\forall A,B\in SL_{n}\left(\mathbb{R}\right):A\cdot B\in SL_{n}\left(\mathbb{R}\right)$
\end_inset

.
 Vzemimo poljubni 
\begin_inset Formula $A,B\in SL_{n}\left(\mathbb{R}\right)$
\end_inset

,
 torej 
\begin_inset Formula $\det A=1\wedge\det B=1$
\end_inset

.
 
\begin_inset Formula $\det\left(AB\right)=\det A\det B=1\cdot1=1$
\end_inset

.
 Preveriti je treba še,
 da so inverzi vsebovani.
 Za poljubno 
\begin_inset Formula $A\in SL_{n}\left(\mathbb{R}\right)$
\end_inset

 je 
\begin_inset Formula $\det A^{-1}=\frac{1}{\det A}=1$
\end_inset

,
 ker je 
\begin_inset Formula $\det A=1$
\end_inset

.
 Enota 
\begin_inset Formula $I$
\end_inset

 je vsebovana v 
\begin_inset Formula $SL_{n}\left(\mathbb{R}\right)$
\end_inset

,
 saj 
\begin_inset Formula $\det I=1$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Fact*
Za podedovano operacijo 
\begin_inset Formula $\circ_{N}$
\end_inset

 v podstrukturi se asociativnost in komutativnost podedujeta,
 ni pa nujno,
 da če obstaja enota v 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

,
 obstaja enota tudi v 
\begin_inset Formula $\left(N,\circ_{N}\right)$
\end_inset

.
 Prav tako ni rečeno,
 da se podeduje obstoj inverzov.
\end_layout

\begin_layout Definition*
Če je 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 polgrupa (asociativen grupoid) in 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 pravimo,
 da je 
\begin_inset Formula $N$
\end_inset

 podpolgrupa,
 če je zaprta za 
\begin_inset Formula $\circ$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:podmonoid"

\end_inset

Če je 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 monoid (polgrupa z enoto) in 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 je 
\begin_inset Formula $N$
\end_inset

 podmonoid,
 če je zaprt za 
\begin_inset Formula $\circ$
\end_inset

 in vsebuje enoto iz 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 (prav tisto enoto,
 glej primer 
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:nxnmonoid"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 spodaj).
\end_layout

\begin_layout Example*
\begin_inset Formula $\left(\mathbb{N},\cdot\right)$
\end_inset

 je monoid.
 Soda števila so podpolgrupa (zaprta so za množenje),
 niso pa podmonoid,
 saj ne vsebujejo enice (enote).
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:nxnmonoid"

\end_inset


\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
\end_inset

 je monoid za operacijo 
\begin_inset Formula $\left(a,b\right)\circ\left(c,d\right)=\left(ac,bd\right)$
\end_inset

,
 saj je enota 
\begin_inset Formula $\left(1,1\right)$
\end_inset

.
 
\begin_inset Formula $\left(\mathbb{N}\times\left\{ 0\right\} ,\circ\right)$
\end_inset

 pa za 
\begin_inset Formula $\circ$
\end_inset

 kot prej je sicer podpolgrupa v 
\begin_inset Formula $\left(\mathbb{N}\times\mathbb{N},\circ\right)$
\end_inset

 in ima enoto 
\begin_inset Formula $\left(1,0\right)$
\end_inset

,
 vendar,
 ker 
\begin_inset Formula $\left(1,0\right)\not=\left(1,1\right)$
\end_inset

,
 to ni podmonoid.
 Enota mora torej biti,
 kot pravi definicija 
\begin_inset CommandInset ref
LatexCommand ref
reference "def:podmonoid"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

,
 ista kot enota v 
\begin_inset Quotes gld
\end_inset

starševski
\begin_inset Quotes grd
\end_inset

 strukturi.
\end_layout

\begin_layout Definition*
Če je 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 grupa in 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 pravimo,
 da je 
\begin_inset Formula $N$
\end_inset

 podgrupa 
\begin_inset Formula $\Longleftrightarrow$
\end_inset

 hkrati velja
\end_layout

\begin_deeper
\begin_layout Itemize
je zaprta za 
\begin_inset Formula $\circ$
\end_inset

,
\end_layout

\begin_layout Itemize
vsebuje isto enoto kot 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 in
\end_layout

\begin_layout Itemize
vsebuje inverz vsakega svojega elementa;
 ti inverzi pa so itak po enoličnosti enaki inverzom iz 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Example*
special linear,
 
\begin_inset Formula $SL_{n}$
\end_inset

,
 grupa vseh matrik z determinanto enako 1,
 je podgrupa 
\begin_inset Quotes gld
\end_inset

general linear
\begin_inset Quotes grd
\end_inset

,
 
\begin_inset Formula $GL_{n}$
\end_inset

,
 grupe vseh obrnljivih 
\begin_inset Formula $n\times n$
\end_inset

 matrik,
 kajti 
\begin_inset Formula $\det I=1$
\end_inset

,
 
\begin_inset Formula $\det$
\end_inset

 je multiplikativna (glej vajo zgoraj) in 
\begin_inset Formula $\det A=1\Leftrightarrow\det A^{-1}=1$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
ortogonalne matrike,
 
\begin_inset Formula $O_{n}$
\end_inset

,
 vse 
\begin_inset Formula $n\times n$
\end_inset

 matrike 
\begin_inset Formula $A$
\end_inset

,
 ki zadoščajo 
\begin_inset Formula $A^{T}A=I$
\end_inset

,
 je podgrupa 
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset

,
 kajti:
\end_layout

\begin_deeper
\begin_layout Itemize
Je zaprta:
\begin_inset Formula 
\[
A,B\in O_{n}\overset{?}{\Longrightarrow}AB\in O_{n}
\]

\end_inset


\begin_inset Formula 
\[
\left(AB\right)^{T}\left(AB\right)\overset{?}{=}I
\]

\end_inset


\begin_inset Formula 
\[
B^{T}\left(A^{T}A\right)B\overset{?}{=}I
\]

\end_inset


\begin_inset Formula 
\[
I=I
\]

\end_inset


\end_layout

\begin_layout Itemize
Vsebuje enoto 
\begin_inset Formula $I$
\end_inset

:
\begin_inset Formula 
\[
I^{T}I=I
\]

\end_inset


\end_layout

\begin_layout Itemize
Vsebuje inverze vseh svojih elementov:
 Uporabimo 
\begin_inset Formula $A^{T}A=I\Rightarrow A^{T}=A^{-1}$
\end_inset


\begin_inset Formula 
\[
A\in O_{n}\overset{?}{\Longrightarrow}A^{-1}\in O_{n}
\]

\end_inset


\begin_inset Formula 
\[
\left(A^{-1}\right)^{T}A^{-1}\overset{?}{=}I
\]

\end_inset


\begin_inset Formula 
\[
\left(A^{T}\right)^{T}A^{T}=AA^{T}=I
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Fact*
specialna ortogonalna grupa,
 
\begin_inset Formula $SO_{n}\coloneqq O_{n}\cap SL_{n}$
\end_inset

 je podgrupa 
\begin_inset Formula $GL_{n}\left(\mathbb{R}\right)$
\end_inset

.
 Dokazati je moč še bolj splošno,
 namreč,
 da je presek dveh podgrup spet podgrupa.
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
DOKAŽI?????
\end_layout

\end_inset


\end_layout

\begin_layout Claim*
Naj bo 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

 grupa in 
\begin_inset Formula $N\subseteq M$
\end_inset

 neprazna.
 Tedaj velja 
\begin_inset Formula $N$
\end_inset

 podgrupa 
\begin_inset Formula $\Leftrightarrow\forall a,b\in N:a\circ b^{-1}\in N$
\end_inset

 (zaprtost za odštevanje —
 v abelovih grupah namreč običajno operacijo označimo s 
\begin_inset Formula $+$
\end_inset

 in označimo 
\begin_inset Formula $a+b^{-1}=a-b$
\end_inset

).
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Naj bo 
\begin_inset Formula $N$
\end_inset

 podgrupa v 
\begin_inset Formula $\left(M,\circ\right)$
\end_inset

.
 Vzemimo 
\begin_inset Formula $a,b\in N$
\end_inset

.
 Upoštevamo 
\begin_inset Formula $b\in N\Rightarrow b^{-1}\in N$
\end_inset

 iz definicije podgrupe.
 Torej velja 
\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ b^{-1}\in N$
\end_inset

,
 zopet iz definicije podgrupe.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Naj 
\begin_inset Formula $\forall a,b\in N:a\circ b^{-1}\in N$
\end_inset

.
 Preverimo lastnosti iz definicije podgrupe:
\end_layout

\begin_deeper
\begin_layout Itemize
Vsebovanost enote:
 Ker je 
\begin_inset Formula $N$
\end_inset

 neprazna,
 vsebuje nek 
\begin_inset Formula $a$
\end_inset

.
 Po predpostavki je 
\begin_inset Formula $a\circ a^{-1}\in N$
\end_inset

,
 
\begin_inset Formula $a\circ a^{-1}$
\end_inset

 pa je po definiciji inverza enota.
\end_layout

\begin_layout Itemize
Vsebovanost inverzov:
 Naj bo 
\begin_inset Formula $a\in N$
\end_inset

 poljuben.
 Od prej vemo,
 da 
\begin_inset Formula $e\in N$
\end_inset

.
 Po predpostavki,
 ker 
\begin_inset Formula $e,a\in N\Rightarrow e\circ a^{-1}\in N$
\end_inset

,
 
\begin_inset Formula $e\circ a^{-1}$
\end_inset

 pa je po definiciji enote 
\begin_inset Formula $a^{-1}$
\end_inset

.
\end_layout

\begin_layout Itemize
Zaprtost:
 Naj bosta 
\begin_inset Formula $a,b\in N$
\end_inset

 poljubna.
 Od prej vemo,
 da 
\begin_inset Formula $b^{-1}\in N$
\end_inset

.
 Po predpostavki,
 ker 
\begin_inset Formula $a,b^{-1}\in N\Rightarrow a\circ\left(b^{-1}\right)^{-1}\in N$
\end_inset

,
 
\begin_inset Formula $a\circ\left(b^{-1}\right)^{-1}$
\end_inset

 pa je po definiciji inverza 
\begin_inset Formula $a\circ b$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout Subsubsection
Homomorfizmi
\end_layout

\begin_layout Standard
\begin_inset Formula $\sim$
\end_inset

 so operacije,
 ki 
\begin_inset Quotes gld
\end_inset

ohranjajo strukturo
\begin_inset Quotes grd
\end_inset

.
\end_layout

\begin_layout Definition*
Naj bosta 
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset

 in 
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset

 dva grupoida.
 Preslikava 
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset

 je homomorfizem grupoidov,
 če 
\begin_inset Formula $\forall a,b\in M_{1}:f\left(a\circ_{1}b\right)=f\left(a\right)\circ_{2}f\left(b\right)$
\end_inset

.
 Enaka definicija v polgrupah.
 Za homomorfizem monoidov zahtevamo še,
 da 
\begin_inset Formula $f\left(e_{1}\right)=e_{2}$
\end_inset

,
 kjer je 
\begin_inset Formula $e_{1}$
\end_inset

 enota 
\begin_inset Formula $M_{1}$
\end_inset

 in 
\begin_inset Formula $e_{2}$
\end_inset

 enota 
\begin_inset Formula $M_{2}$
\end_inset

.
\end_layout

\begin_layout Example*
\begin_inset Formula $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$
\end_inset

,
 ki slika 
\begin_inset Formula $a\mapsto\left(a,0\right)$
\end_inset

.
 
\begin_inset Formula $\circ_{1}$
\end_inset

 naj bo množenje,
 
\begin_inset Formula $\circ_{2}$
\end_inset

 pa 
\begin_inset Formula $\left(a,b\right)\circ_{2}\left(c,d\right)=\left(ac,bd\right)$
\end_inset

 (množenje po komponentah).
 
\begin_inset Formula $\left(1,1\right)$
\end_inset

 je enota v 
\begin_inset Formula $\mathbb{N\times\mathbb{N}}$
\end_inset

,
 
\begin_inset Formula $1$
\end_inset

 pa je enota v 
\begin_inset Formula $\mathbb{N}$
\end_inset

.
 
\begin_inset Formula $f$
\end_inset

 je homomorfizem,
 ker 
\begin_inset Formula $f\left(a\circ_{1}b\right)=\left(a\cdot b,0\right)=\left(a,0\right)\circ_{2}\left(b,0\right)=f\left(a\right)\circ_{2}f\left(b\right)$
\end_inset

,
 ni pa homomorfizem monoidov,
 saj 
\begin_inset Formula $f\left(1\right)=\left(1,0\right)\not=\left(1,1\right)$
\end_inset

.
\end_layout

\begin_layout Definition*
Za homomorfizem grup zahtevamo še,
 da 
\begin_inset Formula $f\left(a^{-1}\right)=f\left(a\right)^{-1}$
\end_inset

.
\end_layout

\begin_layout Remark*
Izkaže se,
 da ohranjanje enote in inverzov pri homomorfizmih grup sledi že iz definicije homomorfizmov grupoidov.
\end_layout

\begin_layout Claim*
Naj bosta 
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset

 in 
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset

 grupi.
 Naj bo 
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset

 preslikava,
 ki je homomorfizem grupoidov.
 Trdimo,
 da slika enoto v enoto in inverze v inverze.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $e_{1}$
\end_inset

 enota za 
\begin_inset Formula $\left(M_{1},\circ_{1}\right)$
\end_inset

 in 
\begin_inset Formula $e_{2}$
\end_inset

 enota za 
\begin_inset Formula $\left(M_{2},\circ_{2}\right)$
\end_inset

.
 Dokažimo,
 da 
\begin_inset Formula $f\left(e_{1}\right)\overset{?}{=}e_{2}$
\end_inset

.
\begin_inset Formula 
\[
f\left(e_{1}\right)=f\left(e_{1}\circ_{1}e_{1}\right)=f\left(e_{1}\right)\circ_{2}f\left(e_{1}\right)=f\left(e_{1}\right)^{-1}\circ f\left(e_{1}\right)\circ e_{2}=e_{2}\circ e_{2}=e_{2}
\]

\end_inset

Dokažimo še ohranjanje inverzov,
 se pravi 
\begin_inset Formula $b$
\end_inset

 je inverz 
\begin_inset Formula $a\overset{?}{\Longrightarrow}f\left(b\right)$
\end_inset

 je inverz 
\begin_inset Formula $f\left(a\right)$
\end_inset

.
\begin_inset Formula 
\[
a\circ_{1}b=e_{1}\overset{?}{\Longrightarrow}f\left(a\right)\circ_{2}f\left(b\right)=f\left(a\circ_{1}b\right)=f\left(e_{1}\right)=e_{2}
\]

\end_inset


\begin_inset Formula 
\[
b\circ_{1}a=e_{1}\overset{?}{\Longrightarrow}f\left(b\right)\circ_{2}f\left(a\right)=f\left(b\circ_{1}a\right)=f\left(e_{1}\right)=e_{2}
\]

\end_inset


\end_layout

\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:primeri-homomorfizmov"

\end_inset

Primeri homomorfizmov.
\end_layout

\begin_deeper
\begin_layout Enumerate
Determinanta:
 
\begin_inset Formula $M_{n}\left(\mathbb{R}\right)\to\mathbb{R}$
\end_inset

 je homomorfizem,
 ker ima multiplikativno lastnost:
 
\begin_inset Formula $\det\left(AB\right)=\det A\det B$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:permutacijska-matrika"

\end_inset


\begin_inset Formula $S_{n}$
\end_inset

 so vse permutacije množice 
\begin_inset Formula $\left\{ 1..n\right\} $
\end_inset

.
 Vsaki permutaciji 
\begin_inset Formula $\sigma\in S_{n}$
\end_inset

 priredimo permutacijsko matriko 
\begin_inset Formula $P_{\sigma}\in M_{n}\left(\mathbb{R}\right)$
\end_inset

 tako,
 da vsebuje vektorje standardne baze 
\begin_inset Formula $\mathbb{R}^{n}$
\end_inset

 kot stolpce:
\begin_inset Formula 
\[
P_{\sigma}\coloneqq\left[\begin{array}{ccc}
\vec{e_{\sigma\left(1\right)}} & \cdots & \vec{e_{\sigma\left(n\right)}}\end{array}\right]
\]

\end_inset

Imamo preslikavo 
\begin_inset Formula $S\to M_{n}\left(\mathbb{R}\right)$
\end_inset

,
 ki slika 
\begin_inset Formula $\sigma\mapsto P_{\sigma}$
\end_inset

 in trdimo,
 da je homomorfizem.
 Dokažimo,
 da je 
\begin_inset Formula $\forall\sigma,\tau\in S_{n}:P_{\sigma\circ\tau}=P_{\sigma}\cdot P_{\tau}$
\end_inset

.
 Opazimo,
 da je 
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{i}}=\vec{e_{\sigma\left(i\right)}}$
\end_inset

 (tu množimo matriko z vektorjem).
 Če namesto 
\begin_inset Formula $i$
\end_inset

 pišemo 
\begin_inset Formula $\tau\left(i\right)$
\end_inset

,
 dobimo 
\begin_inset Formula $\forall i\in\left\{ 1..n\right\} :P_{\sigma}\vec{e_{\tau\left(i\right)}}=\vec{e_{\left(\sigma\circ\tau\right)\left(i\right)}}$
\end_inset

.
 Preverimo sedaj množenje 
\begin_inset Formula $P_{\sigma}P_{\tau}=P_{\sigma}\left[\begin{array}{ccc}
\vec{e_{\tau\left(1\right)}} & \cdots & \vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
P_{\sigma}\vec{e_{\tau\left(1\right)}} & \cdots & P_{\sigma}\vec{e_{\tau\left(n\right)}}\end{array}\right]=\left[\begin{array}{ccc}
\vec{e_{\left(\sigma\circ\tau\right)\left(1\right)}} & \cdots & \vec{e_{\left(\sigma\circ\tau\right)\left(n\right)}}\end{array}\right]=P_{\sigma\circ\tau}$
\end_inset

.
 Preslikava je res homomorfizem.
\end_layout

\end_deeper
\begin_layout Claim*
Kompozitum dveh homomorfizmov je tudi sam zopet homomorfizem.
\end_layout

\begin_layout Proof
Imejmo tri grupoide in homomorfizma,
 ki slikata med njimi takole:
 
\begin_inset Formula $\left(M_{1},\circ_{1}\right)\overset{f}{\longrightarrow}\left(M_{2},\circ_{2}\right)\overset{g}{\longrightarrow}\left(M_{3},\circ_{3}\right)$
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $g\circ f$
\end_inset

 spet homorfizem.
\begin_inset Formula 
\[
\left(g\circ f\right)\left(a\circ_{1}b\right)=g\left(f\left(a\circ_{1}b\right)\right)=g\left(f\left(a\right)\circ_{2}f\left(b\right)\right)=g\left(f\left(a\right)\right)\circ_{3}g\left(f\left(b\right)\right)=\left(g\circ f\right)\left(a\right)\circ_{3}\left(g\circ f\right)\left(b\right)
\]

\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $S_{n}\overset{\sigma}{\longrightarrow}M_{n}\left(\mathbb{R}\right)\overset{\det}{\rightarrow}\mathbb{R}$
\end_inset

,
 kjer je 
\begin_inset Formula $\sigma$
\end_inset

 preslikava iz točke 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:permutacijska-matrika"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 zgleda 
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:primeri-homomorfizmov"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 zgoraj.
 
\begin_inset Formula $\sgn=\det\circ\sigma$
\end_inset

,
 kjer je 
\begin_inset Formula $\sgn$
\end_inset

 parnost permutacije.
 Preslikava 
\begin_inset Formula $\sgn$
\end_inset

 je homomorfizem,
 ker je kompozitum dveh homomorfizmov.
\end_layout

\begin_layout Definition*
Izomorfizem je preslikava,
 ki je bijektivna in je homomorfizem.
 Dve grupi sta izomorfni,
 kadar med njima obstaja izomorfizem.
\end_layout

\begin_layout Remark*
S stališča algebre sta dve izomorfni grupi v abstraktnem smislu enaki,
 saj je izomorfizem zgolj reverzibilno preimenovanje elementov.
\end_layout

\begin_layout Subsubsection
Bigrupoidi,
 polkolobarji,
 kolobarji
\end_layout

\begin_layout Definition*
Neprazni množici 
\begin_inset Formula $M$
\end_inset

 z dvema operacijama 
\begin_inset Formula $\circ_{1}$
\end_inset

 in 
\begin_inset Formula $\circ_{2}$
\end_inset

 pravimo bigrupoid in ga označimo z 
\begin_inset Formula $\left(M,\circ_{1},\circ_{2}\right)$
\end_inset

.
 Običajno operaciji označimo z 
\begin_inset Formula $+,\cdot$
\end_inset

,
 tedaj bigrupoid pišemo kot 
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset

.
\end_layout

\begin_layout Quotation
\begin_inset Quotes gld
\end_inset

Če 
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 ena z drugo nimata nobene zveze,
 je vseeno,
 če ju študiramo skupaj ali posebej.
\begin_inset Quotes grd
\end_inset


\end_layout

\begin_layout Definition*
Distributivnost je značilnost bigrupoida 
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset

.
 Ločimo levo distributivnost:
 
\begin_inset Formula $\forall a,b,c\in M:a\cdot\left(b+c\right)=a\cdot b+a\cdot c$
\end_inset

 in desno distributivnost:
 
\begin_inset Formula $\forall a,b,c\in M:\left(a+b\right)\cdot c=a\cdot c+b\cdot c$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Bigrupoid,
 ki zadošča levi in desni distributivnosti,
 je distributiven.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Distributiven bigrupoid,
 je polkolobar,
 če je 
\begin_inset Formula $\left(M,+\right)$
\end_inset

 komutativna polgrupa.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Distributiven grupoid je kolobar,
 če je 
\begin_inset Formula $\left(M,+\right)$
\end_inset

 komutativna grupa.
\end_layout

\begin_layout Example*
Primer polkolobarja,
 ki ni kolobar,
 je 
\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
\end_inset

.
 Ni enote niti inverza za 
\begin_inset Formula $+$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{N},+\right)$
\end_inset

 pa je polgrupa.
\end_layout

\begin_layout Standard
Kolobarje delimo glede na lastnosti operacije 
\begin_inset Formula $\cdot$
\end_inset

:
\end_layout

\begin_layout Definition*
Asociativen kolobar je tak,
 kjer je 
\begin_inset Formula $\cdot$
\end_inset

 asociativna operacija 
\begin_inset Formula $\sim\left(M,\cdot\right)$
\end_inset

 je polgrupa.
\end_layout

\begin_layout Example*
Primer kolobarja,
 ki ni asociativen,
 je 
\begin_inset Formula $\left(\mathbb{R}^{3},+,\times\right)$
\end_inset

,
 kjer je 
\begin_inset Formula $\times$
\end_inset

 vektorski produkt.
 Primer kolobarja,
 ki je asociativen,
 je 
\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
\end_inset

,
 kjer je 
\begin_inset Formula $\cdot$
\end_inset

 matrično množenje.
\end_layout

\begin_layout Definition*
Asociativen kolobar z enoto je tak,
 ki ima multiplikativno enoto,
 torej enoto za drugo operacijo 
\begin_inset Formula $\sim\left(M,\cdot\right)$
\end_inset

 je monoid.
 Tipično se enoto za 
\begin_inset Formula $\cdot$
\end_inset

 označi z 1,
 enoto za 
\begin_inset Formula $+$
\end_inset

 pa z 0.
\end_layout

\begin_layout Example*
Primer asociativnega kolobarja brez enote je 
\begin_inset Formula $\left(\text{soda }\mathbb{N},+,\cdot\right)$
\end_inset

.
 Primer asociativnega kolobarja z enoto je 
\begin_inset Formula $\left(\mathbb{N},+,\cdot\right)$
\end_inset

.
\end_layout

\begin_layout Definition*
\begin_inset Formula $b$
\end_inset

 je inverz 
\begin_inset Formula $a$
\end_inset

,
 če 
\begin_inset Formula $b\cdot a=e$
\end_inset

 in 
\begin_inset Formula $a\cdot b=e$
\end_inset

,
 kjer je 
\begin_inset Formula $e$
\end_inset

 multiplikativna enota kolobarja.
\end_layout

\begin_layout Remark*
Element 0 nima nikoli inverza,
 ker 
\begin_inset Formula $\forall a\in M:0\cdot a=0$
\end_inset

.
\end_layout

\begin_layout Proof
\begin_inset Formula $\cancel{0\cdot a}=\left(0+0\right)\cdot a=0\cdot a+\cancel{0\cdot a}$
\end_inset

 (dokaz velja za kolobarje,
 ne pa polkolobarje,
 ker imamo pravilo krajšanja
\begin_inset Foot
status open

\begin_layout Plain Layout
Dokaz v mojih Odgovorih na vprašanja za ustni izpit Diskretnih struktur 2 IŠRM
\end_layout

\end_inset

 le,
 kadar je 
\begin_inset Formula $\left(M,+\right)$
\end_inset

 grupa).
\end_layout

\begin_layout Definition*
Asociativen kolobar z enoto,
 v katerem ima vsak neničen element inverz,
 je obseg.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Kolobar je komutativen,
 če je 
\begin_inset Formula $\cdot$
\end_inset

 komutativna operacija (
\begin_inset Formula $+$
\end_inset

 je itak po definiciji že komutativna).
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Komutativen obseg je polje.
\end_layout

\begin_layout Example*
Primeri polj:
 
\begin_inset Formula $\left(\mathbb{Q},+,\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
\end_inset

,
 
\begin_inset Formula $\left(F\left[\mathbb{R}\right],+,\cdot\right)$
\end_inset

,
 kjer je 
\begin_inset Formula $F\left[\mathbb{R}\right]$
\end_inset

 polje racionalnih funkcij.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Primer obsega,
 ki ni polje:
 
\begin_inset Formula $\left(\mathbb{H},+,\cdot\right)$
\end_inset

.
\end_layout

\begin_layout Definition*
Kvaternioni so 
\begin_inset Formula $M_{2\times2}\left(\mathbb{R}\right)$
\end_inset

 take oblike:
 za 
\begin_inset Formula $\alpha,\beta\in\mathbb{C}$
\end_inset

 je 
\begin_inset Formula $\mathbb{H}\coloneqq\left[\begin{array}{cc}
\alpha & \beta\\
-\overline{\beta} & \overline{\alpha}
\end{array}\right]=\left[\begin{array}{cc}
a+bi & c+di\\
-c+di & a-bi
\end{array}\right]=\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]a+\left[\begin{array}{cc}
i & 0\\
0 & -i
\end{array}\right]b+\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]c+\left[\begin{array}{cc}
0 & i\\
i & 0
\end{array}\right]d=1a+bi+cj+dk$
\end_inset

 za 
\begin_inset Formula $a,b,c,d\in\mathbb{R}$
\end_inset

 in dimenzije 
\begin_inset Formula $1,i,j,k$
\end_inset

.
\end_layout

\begin_layout Example*
Primer kolobarja:
 Naj bo 
\begin_inset Formula $X$
\end_inset

 neprazna množica in 
\begin_inset Formula $R$
\end_inset

 kolobar.
 
\begin_inset Formula $R^{X}$
\end_inset

 so vse funkcije 
\begin_inset Formula $X\to R$
\end_inset

.
 Naj bosta 
\begin_inset Formula $f,g\in R^{X}$
\end_inset

.
 Definirajmo operaciji:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $+$
\end_inset

 
\begin_inset Formula $f+g\coloneqq\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
\end_inset


\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\cdot$
\end_inset

 
\begin_inset Formula $f\cdot g\coloneqq\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$
\end_inset


\end_layout

\end_deeper
\begin_layout Subsubsection
Podkolobarji
\end_layout

\begin_layout Definition*
Podbigrupoid od 
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset

 je taka podmnožica 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 ki je zaprta za 
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 ZDB 
\begin_inset Formula $N\subseteq M$
\end_inset

 je podgrupoid v 
\begin_inset Formula $\left(M,+\right)$
\end_inset

 in 
\begin_inset Formula $\left(M,\cdot\right)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Podkolobar kolobarja 
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset

 je taka podmnožica 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 da je 
\begin_inset Formula $N$
\end_inset

 podgrupa v 
\begin_inset Formula $\left(M,+\right)$
\end_inset

 in 
\begin_inset Formula $N$
\end_inset

 podgrupoid v 
\begin_inset Formula $\left(N,\cdot\right)\Leftrightarrow N$
\end_inset

 zaprta za 
\begin_inset Formula $\cdot$
\end_inset

.
 Skrajšana definicija je torej,
 da je 
\begin_inset Formula $\forall a,b\in N:a+b^{-1}\in N\wedge a\cdot b\in N$
\end_inset

,
 torej zaprtost za odštevanje in množenje.
\end_layout

\begin_layout Example*
Primeri podkolobarjev
\end_layout

\begin_deeper
\begin_layout Itemize
v 
\begin_inset Formula $\left(M_{n}\left(\mathbb{R}\right),+,\cdot\right)$
\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
zgornjetrikotne matrike
\end_layout

\begin_layout Itemize
diagonalne matrike
\end_layout

\begin_layout Itemize
matrike s spodnjo vrstico ničelno
\end_layout

\begin_layout Itemize
matrike z ničelnim 
\begin_inset Formula $i-$
\end_inset

tim stolpcem
\end_layout

\begin_layout Itemize
\begin_inset Formula $M_{n}\left(\mathbb{Z}\right)$
\end_inset

,
 
\begin_inset Formula $M_{n}\left(\mathbb{Q}\right)$
\end_inset


\end_layout

\begin_layout Itemize
matrike oblike 
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
b & a
\end{array}\right]$
\end_inset


\end_layout

\end_deeper
\begin_layout Itemize
v 
\begin_inset Formula $\left(\mathbb{R}^{[a,b]},+,\cdot\right)$
\end_inset

 (vse funkcije 
\begin_inset Formula $\left[a,b\right]\to\mathbb{R}$
\end_inset

 za seštevanje in množenje)
\end_layout

\begin_deeper
\begin_layout Itemize
vse omejene funkcije
\end_layout

\begin_layout Itemize
vse zvezne funkcije
\end_layout

\begin_layout Itemize
vse odvedljive funkcije
\end_layout

\end_deeper
\end_deeper
\begin_layout Definition*
Podobseg obsega 
\begin_inset Formula $\left(M,+,\cdot\right)$
\end_inset

 je taka 
\begin_inset Formula $N\subseteq M$
\end_inset

,
 da velja:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $N$
\end_inset

 podgrupa v 
\begin_inset Formula $\left(M,+\right)$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $N\setminus\left\{ 0\right\} $
\end_inset

 podgrupa v 
\begin_inset Formula $\left(M\setminus\left\{ 0\right\} ,\cdot\right)$
\end_inset


\end_layout

\begin_layout Standard
ZDB:
 
\begin_inset Formula $N$
\end_inset

 je zaprta za odštevanje (seštevanje z aditivnim inverzom) in za deljenje (množenje z multiplikativnim inverzom) z neničelnimi elementi.
\end_layout

\end_deeper
\begin_layout Example*
Primeri podobsegov:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\mathbb{R}$
\end_inset

 je podobseg v 
\begin_inset Formula $\left(\mathbb{C},+,\cdot\right)$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $\mathbb{Q}$
\end_inset

 je podobseg v 
\begin_inset Formula $\left(\mathbb{R},+,\cdot\right)$
\end_inset


\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Izkaže se,
 da je najmanjše podpolje v 
\begin_inset Formula $\mathbb{R}$
\end_inset

,
 ki vsebuje 
\begin_inset Formula $\mathbb{Q}$
\end_inset

 in 
\begin_inset Formula $\sqrt{3}$
\end_inset

 množica 
\begin_inset Formula $\left\{ a+b\sqrt{3};\forall a,b\in\mathbb{Q}\right\} $
\end_inset

.
 Očitno je zaprt za odštevanje.
 Za deljenje?
\end_layout

\begin_deeper
\begin_layout Standard
\begin_inset Formula 
\[
\frac{a+b\sqrt{3}}{c+d\sqrt{3}}=\frac{\left(a+b\sqrt{3}\right)\left(c-d\sqrt{3}\right)}{\left(c+d\sqrt{3}\right)\left(c-d\sqrt{3}\right)}=\frac{ac-ad\sqrt{3}+bc\sqrt{3}-3bd}{c^{2}-3d^{2}}=\frac{ac-3bd+\left(bc-ad\right)\sqrt{3}}{c^{2}-3d^{2}}=
\]

\end_inset


\begin_inset Formula 
\[
=\frac{ac-3bd}{c^{2}-3d^{2}}+\frac{bc-ad}{c^{2}-3d^{2}}\sqrt{3}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Subsubsection
Homomorfizmi kolobarjev
\end_layout

\begin_layout Definition*
Naj bosta 
\begin_inset Formula $\left(M_{1},+_{1},\cdot_{1}\right)$
\end_inset

 in 
\begin_inset Formula $\left(M_{2},+_{2},\cdot_{2}\right)$
\end_inset

 kolobarja.
 
\begin_inset Formula $f:M_{1}\to M_{2}$
\end_inset

 je homomorfizem kolobarjev 
\begin_inset Formula $\Leftrightarrow\forall a,b\in M_{1}:f\left(a+_{1}b\right)=f\left(a\right)+_{2}f\left(b\right)\wedge f\left(a\cdot_{1}b\right)=f\left(a\right)\cdot_{2}f\left(b\right)$
\end_inset

.
 ZDB 
\begin_inset Formula $f$
\end_inset

 mora biti homomorfizem grupoidov 
\begin_inset Formula $\left(M_{1},+_{1}\right)\to\left(M_{2},+_{2}\right)$
\end_inset

 in 
\begin_inset Formula $\left(M_{1},\cdot_{1}\right)\to\left(M_{2},\cdot_{2}\right)$
\end_inset

.
 Za homomorfizem kolobarjev z enoto zahtevamo še 
\begin_inset Formula $f\left(1_{1}\right)=1_{2}$
\end_inset

.
\end_layout

\begin_layout Example*
\begin_inset Formula $f:M_{2}\left(\mathbb{R}\right)\to M_{3}\left(\mathbb{R}\right)$
\end_inset

 s predpisom 
\begin_inset Formula $\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]\mapsto\left[\begin{array}{ccc}
a & b & 0\\
c & d & 0\\
0 & 0 & 0
\end{array}\right]$
\end_inset

 je homomorfizem kolobarjev,
 ni pa homomorfizem kolobarjev z enoto,
 kajti 
\begin_inset Formula $f\left(\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]\right)=\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{array}\right]$
\end_inset

,
 kar ni enota v 
\begin_inset Formula $M_{3}\left(\mathbb{R}\right)$
\end_inset

 (
\begin_inset Formula $I_{3}$
\end_inset

) za implicitni operaciji 
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $g:M_{n}\left(\mathbb{R}\right)\to M_{n}\left(\mathbb{R}\right)$
\end_inset

 ki slika 
\begin_inset Formula $A\mapsto S^{-1}AS$
\end_inset

,
 kjer je 
\begin_inset Formula $S$
\end_inset

 neka fiksna obrnljiva matrika v 
\begin_inset Formula $M_{n}\left(\mathbb{R}\right)$
\end_inset

.
 Uporabimo implicitni operaciji 
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 za matrike.
 Računa 
\begin_inset Formula $g\left(A+B\right)=S^{-1}\left(A+B\right)S=S^{-1}AS+S^{-1}BS=g\left(A\right)+g\left(B\right)$
\end_inset

 in 
\begin_inset Formula $g\left(AB\right)=S^{-1}ABS=S^{-1}AIBS=S^{-1}ASS^{-1}BS=g\left(A\right)g\left(B\right)$
\end_inset

 pokažeta,
 da je 
\begin_inset Formula $g$
\end_inset

 homomorfizem kolobarjev,
 celo z enoto,
 kajti 
\begin_inset Formula $g\left(I\right)=S^{-1}IS=S^{-1}S=I$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $h:\mathbb{C}\to M_{n}\left(\mathbb{R}\right)$
\end_inset

 s predpisom 
\begin_inset Formula $\alpha+\beta i\to\left[\begin{array}{cc}
\alpha & \beta\\
-\beta & \alpha
\end{array}\right]$
\end_inset

 je homomorfizem kolobarjev z enoto.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Kolobar ostankov 
\begin_inset Formula $\mathbb{Z}_{n}\coloneqq\left\{ 0..\left(n-1\right)\right\} $
\end_inset

 je asociativni kolobar z enoto.
 Če je 
\begin_inset Formula $p$
\end_inset

 praštevilo,
 pa je celo 
\begin_inset Formula $\mathbb{Z}_{p}$
\end_inset

 polje za implicitni operaciji seštevanje in množenja po modulu.
\end_layout

\begin_layout Subsection
Vektorski prostori
\end_layout

\begin_layout Standard
Ideja:
 Vektorski prostor je Abelova grupa z dodatno strukturo —
 množenje s skalarjem.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\left(F,+,\cdot\right)$
\end_inset

 polje.
 Vektorski prostor z operacijama 
\begin_inset Formula $V+V\to V$
\end_inset

 in 
\begin_inset Formula $F\cdot V\to V$
\end_inset

 nad 
\begin_inset Formula $F$
\end_inset

 je taka 
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset

,
 da velja:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\left(V,+\right)$
\end_inset

 je abelova grupa:
 komutativnost,
 asociativnost,
 enota,
 aditivni inverzi
\end_layout

\begin_layout Enumerate
Lastnosti množenja s skalarjem.
 
\begin_inset Formula $\forall\alpha,\beta\in F,a,b\in V:$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\alpha\left(a+b\right)=\alpha a+\alpha b$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left(\alpha+\beta\right)a=\alpha a+\beta a$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left(\alpha\cdot\beta\right)\cdot a=\alpha\cdot\left(\beta\cdot a\right)$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $1\cdot a=a$
\end_inset


\end_layout

\begin_layout Standard
Alternativna abstraktna formulacija aksiomov množenja s skalarjem se glasi:
\end_layout

\begin_layout Standard
\begin_inset Formula $\forall\alpha\in F$
\end_inset

 priredimo preslikavo 
\begin_inset Formula $\varphi_{\alpha}:V\to V$
\end_inset

,
 ki pošlje 
\begin_inset Formula $v\mapsto\alpha v$
\end_inset

.
 Štiri zgornje aksiome množenja s skalarjem sedaj označimo z abstraktnimi formulacijami:
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha}\left(a+b\right)\overset{\text{def.}}{=}\alpha\left(a+b\right)=\alpha b+\alpha b\overset{\text{def.}}{=}\varphi_{\alpha}\left(a\right)+\varphi_{\alpha}\left(b\right)$
\end_inset

 —
 vidimo,
 da je 
\begin_inset Formula $\varphi_{\alpha}$
\end_inset

 homomorfizem iz 
\begin_inset Formula $\left(V,+\right)$
\end_inset

 v 
\begin_inset Formula $\left(V,+\right)$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha+\beta}\left(a\right)\overset{\text{def.}}{=}\left(\alpha+\beta\right)a=\alpha a+\beta a\overset{\text{def.}}{=}\varphi_{\alpha}a+\varphi_{\beta}a$
\end_inset

 —
 torej 
\begin_inset Formula $\varphi_{\alpha+\beta}=\varphi_{\alpha}+\varphi_{\beta}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\varphi_{\alpha\beta}a\overset{\text{def.}}{=}\left(\alpha\beta\right)a=\alpha\left(\beta a\right)\overset{\text{def.}}{=}\varphi_{\alpha}\left(\varphi_{\beta}\left(a\right)\right)=\left(\varphi_{\alpha}\circ\varphi_{\beta}\right)\left(a\right)$
\end_inset

 —
 torej 
\begin_inset Formula $\varphi_{\alpha\beta}=\varphi_{\alpha}\circ\varphi_{\beta}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\varphi_{1}a\overset{\text{def.}}{=}1a=a$
\end_inset

 —
 torej 
\begin_inset Formula $\varphi_{1}=id$
\end_inset

.
\end_layout

\begin_layout Paragraph
\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO ALTERNATIVNA DEFINICIJA VEKTORSKEGA PROSTORA Z GRUPO ENDOMORFIZMOV
\end_layout

\end_inset


\end_layout

\end_deeper
\end_deeper
\begin_layout Remark*
Če v definiciji vektorskega prostora zamenjamo polje 
\begin_inset Formula $F$
\end_inset

 s kolobarjem 
\begin_inset Formula $F$
\end_inset

,
 dobimo definicijo 
\series bold
modula 
\series default
nad 
\begin_inset Formula $F$
\end_inset

.
\end_layout

\begin_layout Example*
Primeri vektorskih prostorov:
\end_layout

\begin_deeper
\begin_layout Itemize
standarden primer:
 naj bo 
\begin_inset Formula $F$
\end_inset

 pojle in 
\begin_inset Formula $n\in\mathbb{N}$
\end_inset

.
 Naj bo 
\begin_inset Formula $V=F^{n}$
\end_inset

,
 
\begin_inset Formula $+$
\end_inset

 seštevanje po komponentah in 
\begin_inset Formula $\cdot$
\end_inset

 množenje s skalarjem po komponentah.
 Pod temi pogoji je 
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset

 vektorski prostor —
 ustreza vsem osmim aksiomom.
\end_layout

\begin_layout Itemize
Naj bo 
\begin_inset Formula $F$
\end_inset

 polje in 
\begin_inset Formula $n,m\in\mathbb{N}$
\end_inset

.
 Naj bo 
\begin_inset Formula $V\coloneqq M_{m,n}\left(\mathbb{F}\right)=m\times n$
\end_inset

 matrike nad 
\begin_inset Formula $F$
\end_inset

.
 
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 definiramo kot pri matrikah.
\end_layout

\begin_layout Itemize
Naj bo 
\begin_inset Formula $F$
\end_inset

 polje,
 
\begin_inset Formula $S\not=\emptyset$
\end_inset

 množica.
 Naj bo 
\begin_inset Formula $V\coloneqq F^{S}$
\end_inset

 (vse funkcije 
\begin_inset Formula $S\to F$
\end_inset

).
 Naj bosta 
\begin_inset Formula $\varphi,\tau:S\to F$
\end_inset

.
 Definirajmo 
\begin_inset Formula $\forall s\in S$
\end_inset

 operaciji 
\begin_inset Formula $\left(\varphi+\tau\right)\left(s\right)=\varphi\left(s\right)+\tau\left(s\right)$
\end_inset

 in 
\begin_inset Formula $\left(\varphi\cdot\tau\right)\left(s\right)=\varphi\left(s\right)\cdot\tau\left(s\right)$
\end_inset

.
 Tedaj je 
\begin_inset Formula $V$
\end_inset

 vektorski prostor.
 Ta definicija je podobna kot definiciji z 
\begin_inset Formula $n-$
\end_inset

terico elementov polja,
 saj lahko 
\begin_inset Formula $n-$
\end_inset

terico identificiramo s funkcijo 
\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} \to F$
\end_inset

,
 toda ta primer dovoli neskončno razsežne vektorske prostore,
 saj 
\begin_inset Formula $S$
\end_inset

 ni nujno končna,
 
\begin_inset Formula $n-$
\end_inset

terica pa nekako implicitno je,
 saj 
\begin_inset Formula $n\in\mathbb{N}$
\end_inset

.
\end_layout

\begin_layout Itemize
Polinomi.
 Naj bo 
\begin_inset Formula $V\coloneqq F\left[x\right]$
\end_inset

 (polinomi v spremenljivki 
\begin_inset Formula $x$
\end_inset

 s koeficienti v 
\begin_inset Formula $F$
\end_inset

).
 Seštevanje definirajmo po komponentah:
 
\begin_inset Formula $\left(\alpha+\beta x+\gamma x^{2}\right)+\left(\pi+\tau x\right)=\left(\alpha+\pi+\left(\beta+\tau\right)x+\gamma x^{2}\right)$
\end_inset

,
 množenje s skalarjem pa takole:
 
\begin_inset Formula $\alpha\left(a+bx+cx^{2}\right)=\alpha a+\alpha bx+\alpha cx^{2}$
\end_inset

.
\end_layout

\begin_layout Itemize
Naj bosta 
\begin_inset Formula $V_{1}$
\end_inset

 in 
\begin_inset Formula $V_{2}$
\end_inset

 dva vektorska prostora nad istim poljem 
\begin_inset Formula $F$
\end_inset

.
 Tvorimo nov vektorski prostor nad 
\begin_inset Formula $F$
\end_inset

,
 ki mu pravimo 
\begin_inset Quotes gld
\end_inset

direktna vsota
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $V_{1}$
\end_inset

 in 
\begin_inset Formula $V_{2}$
\end_inset

 in ga označimo z 
\begin_inset Formula $V_{1}\oplus V_{2}\coloneqq$
\end_inset


\begin_inset Formula $\left\{ \left(v_{1},v_{2}\right);\forall v_{1}\in V_{1},v_{2}\in V:2\right\} $
\end_inset

.
 Seštevamo po komponentah:
 
\begin_inset Formula $\left(v_{1},v_{2}\right)+\left(v_{1}',v_{2}'\right)=\left(v_{1}+v_{1}',v_{2}+v_{2}'\right)$
\end_inset

,
 s skalarjem pa množimo prvi komponento:
 
\begin_inset Formula $\forall\alpha\in F:\alpha\left(v_{1},v_{2}\right)=\left(\alpha v_{1},v_{2}\right)$
\end_inset

.
 Definicijo lahko posplošimo na 
\begin_inset Formula $n$
\end_inset

 vektorskih prostorov.
 Tedaj so elementi prostora urejene 
\begin_inset Formula $n-$
\end_inset

terice.
\end_layout

\end_deeper
\begin_layout Subsubsection
Podprostori vekrorskih prostorov —
 vektorski podprostori
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\left(V,+,\cdot\right)$
\end_inset

 vektorski prostor nad 
\begin_inset Formula $F$
\end_inset

.
 Vektorski podprostor je taka neprazna podmnožica 
\begin_inset Formula $V$
\end_inset

,
 ki je zaprta za seštevanje in množenje s skalarjem.
 Natančneje:
 
\begin_inset Formula $\left(W,+,\cdot\right)$
\end_inset

 je vektorski podprostor 
\begin_inset Formula $\left(V,+,\cdot\right)\Longleftrightarrow$
\end_inset

 velja hkrati:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $W\subseteq V$
\end_inset

 in 
\begin_inset Formula $W\not=\emptyset$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:zaprtost+"

\end_inset


\begin_inset Formula $\forall a,b\in W:a+b\in W$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:zaprtostskalar"

\end_inset


\begin_inset Formula $\forall a\in W,\alpha\in F:\alpha a\in W$
\end_inset


\end_layout

\begin_layout Standard
Lastnosti 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtost+"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 in 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtostskalar"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 je moč združiti v eno:
 
\begin_inset Formula $\forall a_{i},a_{2}\in W,\alpha_{1},\alpha_{2}\in F:\alpha_{1}a_{1}+\alpha_{2}a_{2}\in W$
\end_inset

.
\end_layout

\begin_layout Standard
Z drugimi besedami je vektorski podprostor taka podmnožica,
 ki vsebuje vse linearne kombinacije svojih elementov.
 Odštevanje 
\begin_inset Formula $a-b$
\end_inset

 je poseben primer linearne kombinacije,
 kajti 
\begin_inset Formula $a_{1}-a_{2}=1a_{1}+\left(-1\right)a_{2}$
\end_inset

.
 Sledi,
 da mora biti 
\begin_inset Formula $\left(W,+\right)$
\end_inset

 podgrupa 
\begin_inset Formula $\left(V,+\right)$
\end_inset

,
 torej taka podmnožica 
\begin_inset Formula $V$
\end_inset

,
 ki je zaprta za odštevanje.
\end_layout

\end_deeper
\begin_layout Example*
Primeri vektorskih podprostorov:
\end_layout

\begin_deeper
\begin_layout Itemize
Naj bo 
\begin_inset Formula $V=\mathbb{R}^{2}$
\end_inset

 (ravnina).
 Vsi vektorski podprostori 
\begin_inset Formula $V$
\end_inset

 so premice,
 ki gredo skozi izhodišče,
 izhodišče samo in cela ravnina.
 Slednja sta t.
 i.
 trivialna podprostora.
\end_layout

\end_deeper
\begin_layout Remark*
\begin_inset Formula $\forall\left(V,+,\cdot\right)$
\end_inset

 vektorski prostor 
\begin_inset Formula $:\left\{ 0\right\} ,V$
\end_inset

 sta vektorska podprostora.
 Imenujemo ju trivialna vektorska podprostora.
\end_layout

\begin_layout Claim*
Vsak podprostor vsebuje aditivno enoto 0.
\end_layout

\begin_layout Proof
Po definiciji je vsak vektorski podprostor neprazen,
 torej 
\begin_inset Formula $\exists w\in W$
\end_inset

.
 Polje gotovo vsebuje aditivno enoto 0,
 torej po aksiomu 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:zaprtostskalar"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 za podprostore sledi 
\begin_inset Formula $0\cdot w\in W$
\end_inset

.
 Dokažimo 
\begin_inset Formula $0\cdot w\overset{?}{=}0$
\end_inset

:
 
\begin_inset Formula $\cancel{0\cdot w}=\left(0+0\right)\cdot w=0\cdot w+\cancel{0\cdot w}$
\end_inset

 (pravilo krajšanja v grupi),
 torej 
\begin_inset Formula $0=0\cdot w$
\end_inset

.
\end_layout

\begin_layout Claim*
Množica rešitev homogene (desna stran je 0) linearne enačbe je vselej vektorski podprostor.
\end_layout

\begin_layout Proof
Imamo 
\begin_inset Formula $\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n}=0$
\end_inset

.
 Če sta 
\begin_inset Formula $\vec{a}=\left(a_{1},\dots,a_{n}\right)$
\end_inset

 in 
\begin_inset Formula $\vec{b}=\left(b_{1},\dots,b_{n}\right)$
\end_inset

 rešitvi,
 velja 
\begin_inset Formula $\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}=0$
\end_inset

 in 
\begin_inset Formula $\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}=0$
\end_inset

.
 Vzemimo poljubna 
\begin_inset Formula $\alpha,\beta\in F$
\end_inset

 in si oglejmo 
\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}$
\end_inset

:
\begin_inset Formula 
\[
\alpha\left(\alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}\right)+\beta\left(\alpha_{1}b_{1}+\cdots+\alpha_{n}b_{n}\right)=0
\]

\end_inset


\begin_inset Formula 
\[
\alpha_{1}\left(\alpha a_{1}+\beta b_{1}\right)+\cdots+\alpha_{n}\left(\alpha a_{n}+\beta b_{n}\right)=0
\]

\end_inset

Vzemimo koeficiente v oklepajih pred 
\begin_inset Formula $\alpha_{i}$
\end_inset

 v enačbi pred to vrstico in jih zložimo v vektor.
 Tedaj je 
\begin_inset Formula $\alpha\vec{a}+\beta\vec{b}=\left(\alpha a_{1}+\beta b_{1},\dots,\alpha a_{n}+\beta b_{n}\right)$
\end_inset

 spet rešitev homogene linearne enačbe.
 Ker je linearna kombinacija elementov vektorskega podprostora spet element vektorskega podprostora,
 je po definiciji množica rešitev homogene linearne enačbe res vselej vektorski podprostor.
\end_layout

\begin_layout Remark*
Podoben računa velja tudi za množico rešitev sistema linearnih enačb,
 kar sicer sledi tudi iz naslednje trditve.
\end_layout

\begin_layout Claim*
Presek dveh podprostorov je tudi sam spet podprostor.
\end_layout

\begin_layout Proof
Naj bosta 
\begin_inset Formula $W_{1},W_{2}$
\end_inset

 podprostora v 
\begin_inset Formula $V$
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $W_{1}\cap V_{2}$
\end_inset

 spet podprostor.
 Vzemimo poljubna 
\begin_inset Formula $a,b\in W_{1}\cap W_{2}$
\end_inset

 in poljubna 
\begin_inset Formula $\alpha,\beta\in F$
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
\end_inset

.
 Vemo,
 da 
\begin_inset Formula $a,b\in W_{1}$
\end_inset

 in 
\begin_inset Formula $a,b\in W_{2}$
\end_inset

.
 Ker je podprostor po definiciji zaprt za linearne kombinacije svojih elementov,
 je 
\begin_inset Formula $\alpha a+\beta b\in W_{1}$
\end_inset

 in 
\begin_inset Formula $\alpha a+\beta b\in W_{2}$
\end_inset

,
 torej 
\begin_inset Formula $\alpha a+\beta b\in W_{1}\cap W_{2}$
\end_inset

,
 torej je presek podprostorov res zaprt za LK svojih elementov in je s tem tudi sam podprostor.
\end_layout

\begin_deeper
\begin_layout Remark*
Slednji dokaz lahko očitno posplošimo na več podprostorov.
 Presek nikdar ni prazen,
 saj vsi podprostori vsebujejo aditivno enoto 0 (dokaz za to je malce višje).
\end_layout

\end_deeper
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Vsota-podprostorov"

\end_inset

Vsota podprostorov
\end_layout

\begin_layout Definition*
Naj bosta 
\begin_inset Formula $W_{1}$
\end_inset

 in 
\begin_inset Formula $W_{2}$
\end_inset

 podprostora v 
\begin_inset Formula $V$
\end_inset

.
 Vsoto podprostorov 
\begin_inset Formula $W_{1}$
\end_inset

 in 
\begin_inset Formula $W_{2}$
\end_inset

 označimo z 
\begin_inset Formula $W_{1}+W_{2}=\left\{ w_{1}+w_{w};\forall w_{1}\in W_{1},w_{2}\in W_{2}\right\} $
\end_inset

.
\end_layout

\begin_layout Claim*
Vsota podprostorov je tudi sama spet podprostor.
\end_layout

\begin_layout Proof
Naj bosta 
\begin_inset Formula $a,b\in W_{1}+W_{2}$
\end_inset

 poljubna.
 Tedaj po definiciji 
\begin_inset Formula $a=a_{1}+a_{2}$
\end_inset

,
 kjer 
\begin_inset Formula $a_{1}\in W_{1}$
\end_inset

 in 
\begin_inset Formula $a_{2}\in W_{2}$
\end_inset

,
 in 
\begin_inset Formula $b=b_{1}+b_{2}$
\end_inset

,
 kjer 
\begin_inset Formula $b_{1}\in W_{1}$
\end_inset

 in 
\begin_inset Formula $b_{2}\in W_{2}$
\end_inset

.
 
\begin_inset Formula $\forall\alpha,\beta\in F$
\end_inset

:
\end_layout

\begin_layout Proof
\begin_inset Formula 
\[
\alpha a+\beta b=\alpha\left(a_{1}+a_{2}\right)+\beta\left(b_{1}+b_{2}\right)=\alpha a_{1}+\alpha a_{2}+\beta b_{1}+\beta b_{2}=\left(\alpha a_{1}+\beta b_{1}\right)+\left(\alpha a_{2}+\beta b_{2}\right)\in W_{1}+W_{2},
\]

\end_inset

kajti 
\begin_inset Formula $\left(\alpha a_{1}+\beta b_{1}\right)\in W_{1}$
\end_inset

 in 
\begin_inset Formula $\left(\alpha a_{2}+\beta b_{2}\right)\in W_{2}$
\end_inset

,
 saj sta to linearni kombinaciji elementov prostorov.
 Njuna vsota pa je element 
\begin_inset Formula $W_{1}+W_{2}$
\end_inset

 po definiciji vsote podprostorov.
\end_layout

\begin_layout Subsubsection
Baze
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $V$
\end_inset

 vektorski prostor nad poljem 
\begin_inset Formula $F$
\end_inset

.
 Množica 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 je baza,
 če je LN in če je ogrodje.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 je LN,
 če za vsake 
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
\end_inset

,
 ki zadoščajo 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n}=0$
\end_inset

 velja 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{n}=0$
\end_inset

.
 Ekvivalentni definiciji LN:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 je LN 
\begin_inset Formula $\Leftrightarrow\forall v\in V$
\end_inset

 se da kvečjemu na en način izraziti kot linearno kombinacijo 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

.
\end_layout

\begin_layout Itemize
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 je LN 
\begin_inset Formula $\Leftrightarrow\nexists v\in\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

,
 da bi se ga dalo izraziti kot LK preostalih elementov.
\end_layout

\begin_layout Standard
Dokaz ekvivalentnosti teh definicij je enak tistemu za 
\begin_inset Formula $V=\mathbb{R}^{n}$
\end_inset

 višje.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 je ogrodje 
\begin_inset Formula $\Leftrightarrow\forall v\in V$
\end_inset

 se da na vsaj en način izraziti kot LK te množice 
\begin_inset Formula $\Leftrightarrow\Lin\left\{ v_{1},\dots,v_{n}\right\} =V$
\end_inset

.
\end_layout

\begin_layout Example*
Primeri baz:
\end_layout

\begin_deeper
\begin_layout Itemize
standardna baza:
 Naj bo 
\begin_inset Formula $V=F^{n}$
\end_inset

.
 
\begin_inset Formula $v_{1}=\left(1,0,0,\dots,0,0\right)$
\end_inset

,
 
\begin_inset Formula $v_{2}=\left(0,1,0,\dots,0,0\right)$
\end_inset

,
 ...,
 
\begin_inset Formula $v_{n}=\left(0,0,0,\dots,0,1\right)$
\end_inset

.
 Da je 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} \subseteq F^{n}$
\end_inset

 res baza,
 preverimo z determinanto (
\begin_inset Formula $\det A\not=0\Leftrightarrow\exists A^{-1}\Leftrightarrow$
\end_inset

 stolpci so baza prostora):
\begin_inset Formula 
\[
\det\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]=0\Leftrightarrow\left\{ v_{1},\dots,v_{n}\right\} \text{ \textbf{ni} baza}
\]

\end_inset


\end_layout

\begin_layout Itemize
baze v 
\begin_inset Formula $F\left[x\right]_{<n}$
\end_inset

 (polinomi stopnje,
 manjše od 
\begin_inset Formula $n$
\end_inset

)
\end_layout

\begin_deeper
\begin_layout Itemize
standardna baza:
 
\begin_inset Formula $\left\{ 1,x,x^{2},x^{3},\dots,x^{n-1}\right\} $
\end_inset


\end_layout

\begin_layout Itemize
vzemimo paroma različne 
\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}\in F$
\end_inset

 in definirajmo 
\begin_inset Formula $p_{i}\left(x\right)=\left(x-\alpha_{1}\right)\cdots\left(x-\alpha_{i-1}\right)\left(x-\alpha_{i+1}\right)\cdots\left(x-\alpha_{n}\right)$
\end_inset

 za vsak 
\begin_inset Formula $i\in\left\{ 1..n\right\} $
\end_inset

,
 kar je polimom stopnje 
\begin_inset Formula $n-1$
\end_inset

.
\begin_inset Formula $\left\{ \alpha_{1}p_{1}\left(x\right),\dots,\alpha_{n}p_{n}\left(x\right)\right\} $
\end_inset

 je baza za 
\begin_inset Formula $F\left[x\right]_{<n}$
\end_inset

.
 
\end_layout

\begin_deeper
\begin_layout Proof
Dokazujemo,
 da so LN in ogrodje:
\end_layout

\begin_layout Itemize
LN:
 
\begin_inset Formula $\beta_{1}p_{i}\left(x\right)+\cdots+\beta_{n}p_{n}\left(x\right)=0\overset{?}{\Longrightarrow}\beta_{1}=\cdots=\beta_{n}=0$
\end_inset

.
 Opazimo,
 da 
\begin_inset Formula $p_{i}\left(\alpha_{j}\right)=0\Leftrightarrow i=j$
\end_inset

.
 Torej če za 
\begin_inset Formula $x$
\end_inset

 vstavimo katerikoli 
\begin_inset Formula $\alpha_{i}$
\end_inset

,
 bodo vsi členi 0,
 razen 
\begin_inset Formula $\beta_{i}p_{i}\left(x\right)$
\end_inset

.
 Ker pa 
\begin_inset Formula $\alpha_{i}$
\end_inset

 ni ničla 
\begin_inset Formula $p_{i}\left(x\right)$
\end_inset

,
 je 
\begin_inset Formula $\beta_{i}=0$
\end_inset

,
 čim je 
\begin_inset Formula $\beta_{i}p_{i}\left(x\right)=0$
\end_inset

.
 Preverjati je treba le 
\begin_inset Formula $\alpha_{i}$
\end_inset

,
 ker je dovolj najti eno vrednost spremenljivke,
 v kateri se vrednosti polinomov ne ujemajo,
 da lahko rečemo,
 da polinomi niso isti.
\end_layout

\begin_layout Itemize
ogrodje:
 Trdimo,
 da za vsak polimom velja formula 
\begin_inset Formula $f\left(x\right)=\frac{f\left(\alpha_{1}\right)}{p_{1}\left(\alpha_{1}\right)}p_{1}\left(x\right)+\cdots+\frac{f\left(\alpha_{n}\right)}{p_{n}\left(\alpha_{n}\right)}p_{n}\left(x\right)$
\end_inset

.
 Obe strani enačbe imata stopnjo največ 
\begin_inset Formula $n-1$
\end_inset

 in se ujemata v 
\begin_inset Formula $n$
\end_inset

 različnih točkah.
 Če za 
\begin_inset Formula $x$
\end_inset

 vstavimo 
\begin_inset Formula $\alpha_{i}$
\end_inset

 za vsak 
\begin_inset Formula $i$
\end_inset

,
 dobimo 0 v vseh členih,
 razen v 
\begin_inset Formula $i-$
\end_inset

tem,
 kjer se vrednost 
\begin_inset Formula $f\left(x\right)$
\end_inset

 ujema z vrednostjo 
\begin_inset Formula $f\left(\alpha_{1}\right)$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\end_deeper
\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Obstoj-baze"

\end_inset

Obstoj baze
\end_layout

\begin_layout Standard
Omejimo se na končno razsežne vektorske prostore.
\end_layout

\begin_layout Definition*
Vektorski prostor je končno razsežen,
 če ima končno ogrodje:
 
\begin_inset Formula $\exists n\in\mathbb{N}\exists v_{1},\dots,v_{n}\ni:V=\Lin\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

.
\end_layout

\begin_layout Theorem*
obstoj baze.
 Vsak končno razsežen vektorski prostor ima vsaj eno bazo.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $V$
\end_inset

 KRVP in naj bo 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 njegovo ogrodje.
 Ker ogrodje ni nujno LN,
 naj bo 
\begin_inset Formula $S$
\end_inset

 minimalna/najmanjša podmnožica 
\begin_inset Formula $\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

,
 ki je še ogrodje za 
\begin_inset Formula $V$
\end_inset

.
 Trdimo,
 da je 
\begin_inset Formula $S$
\end_inset

 baza za 
\begin_inset Formula $V$
\end_inset

.
 Po konstrukciji je ogrodje,
 dokažimo še,
 da je LN:
 PDDRAA 
\begin_inset Formula $S$
\end_inset

 je linearno odvisna.
 Tedaj 
\begin_inset Formula $\exists v_{i}\in S\ni:v_{i}$
\end_inset

 je LK 
\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $S\setminus\left\{ v_{i}\right\} $
\end_inset

 ogrodje manjše moči,
 kar bi bilo v protislovju s predpostavko.
 Tedaj obstajajo koeficienti,
 da velja 
\begin_inset Formula $v_{i}=\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $v\in V$
\end_inset

.
 Ker je 
\begin_inset Formula $S$
\end_inset

 ogrodje 
\begin_inset Formula $V$
\end_inset

,
 obstajajo neki koeficienti 
\begin_inset Formula $\beta_{1},\dots,\beta_{n}$
\end_inset

,
 da velja 
\begin_inset Formula 
\[
v=\beta_{1}v_{1}+\cdots+\beta_{i}v_{i}+\cdots+\beta_{n}v_{n}=\beta_{1}v_{1}+\cdots+\beta_{i}\left(\alpha_{1}v_{1}+\cdots+\alpha_{i-1}v_{i-1}+\alpha_{i+1}v_{i+1}+\cdots+\alpha_{n}v_{n}\right)+\cdots+\beta_{n}v_{n}=
\]

\end_inset


\begin_inset Formula 
\[
=\left(\beta_{1}+\beta_{i}\alpha_{1}\right)v_{1}+\cdots+\left(\beta_{i-1}+\beta_{i}\alpha_{i-1}\right)v_{i-1}+\left(\beta_{i+1}+\beta_{i}\alpha_{i+1}\right)v_{i+1}+\cdots+\left(\beta_{n}+\beta_{i}\alpha_{n}\right)v_{n}
\]

\end_inset

To pa je 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

,
 saj je bilo rečeno,
 da je 
\begin_inset Formula $S$
\end_inset

 najmanjše ogrodje,
 mi pa smo razvili poljuben 
\begin_inset Formula $v$
\end_inset

 po manjšem ogrodju.
 Torej ima vsak KRVP bazo in vsako ogrodje ima podmnožico,
 ki je baza.
\end_layout

\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "enoličnost-moči-baze."

\end_inset

enoličnost moči baze.
 Naj bo 
\begin_inset Formula $V$
\end_inset

 KRVP z 
\begin_inset Formula $n-$
\end_inset

elementno bazo.
 Tedaj velja vse to:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset

 LN množica 
\begin_inset Formula $A$
\end_inset

 v 
\begin_inset Formula $V$
\end_inset

 ima 
\begin_inset Formula $\leq n$
\end_inset

 elementov
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset

 ogrodje v 
\begin_inset Formula $V$
\end_inset

 ima 
\begin_inset Formula $\geq n$
\end_inset

 elementov
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset

 baza v 
\begin_inset Formula $V$
\end_inset

 ima 
\begin_inset Formula $n$
\end_inset

 elementov
\end_layout

\end_deeper
\begin_layout Proof
Dokaz je dolg.
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"

\end_inset


\end_layout

\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:Vsak-poddoločen-homogen"

\end_inset

Vsak poddoločen homogen sistem linearnih enačb ima netrivialno rešitev.
\end_layout

\begin_deeper
\begin_layout Proof
Dokaz se nahaja pod identično trditvijo 
\begin_inset CommandInset ref
LatexCommand vref
reference "claim:Vpoddol-hom-sist-ima-ne0-reš"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:ln<=ogr"

\end_inset

Če je 
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset

 LN množica v 
\begin_inset Formula $V$
\end_inset

 in 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 ogrodje za 
\begin_inset Formula $V$
\end_inset

,
 je 
\begin_inset Formula $m\leq n$
\end_inset

.
 ZDB moč katerekoli LN množice je manjša ali enaka od kateregakoli ogrodja v 
\begin_inset Formula $V$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Proof
RAAPDD 
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset

 je LN,
 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 je ogrodje in 
\begin_inset Formula $m>n$
\end_inset

.
 Iščemo protislovje.
 Vsakega od 
\begin_inset Formula $u_{i}$
\end_inset

 lahko razvijemo po 
\begin_inset Formula $v$
\end_inset

.
\begin_inset Formula 
\[
\begin{array}{ccccccc}
u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
\vdots &  & \vdots &  &  &  & \vdots\\
u_{m} & = & \alpha_{m1}v_{1} & + & \cdots & + & \alpha_{mn}v_{n}
\end{array}
\]

\end_inset


\begin_inset Formula $\forall i\in\left\{ 1..m\right\} $
\end_inset

 pomnožimo 
\begin_inset Formula $i-$
\end_inset

to enačbo s skalarjem 
\begin_inset Formula $x_{i}$
\end_inset

 in jih seštejmo.
 
\begin_inset Formula $\vec{x}$
\end_inset

 so abstraktne spremenljivke.
 Tedaj:
\begin_inset Formula 
\[
x_{1}u_{1}+\cdots+x_{m}u_{m}=x_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+x_{m}\left(\alpha_{m1}v_{1}+\cdots+\alpha_{mn}v_{n}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=v_{1}\left(\alpha_{11}x_{1}+\cdots+\alpha_{m1}x_{m}\right)+\cdots+v_{n}\left(\alpha_{1n}x_{1}+\cdots+\alpha_{mn}x_{m}\right)
\]

\end_inset


\end_layout

\begin_layout Proof
Izenačimo koeficiente za 
\begin_inset Formula $v_{i}$
\end_inset

 z 0 in dobimo poddoločen homogen sistem enačb (ima 
\begin_inset Formula $n$
\end_inset

 enačb in 
\begin_inset Formula $m$
\end_inset

 spremenljivk,
 po predpostavki pa velja 
\begin_inset Formula $m>n$
\end_inset

):
\begin_inset Formula 
\[
\begin{array}{ccccccc}
\alpha_{11}x_{1} & + & \cdots & + & \alpha_{m1}x_{m} & = & 0\\
\vdots &  &  &  & \vdots &  & \vdots\\
\alpha_{1n}x_{1} & + & \cdots & + & \alpha_{mn}x_{m} & = & 0
\end{array}
\]

\end_inset

Po lemi 
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:Vsak-poddoločen-homogen"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 ima ta sistem netrivialno rešitev,
 recimo 
\begin_inset Formula $\left(\mu_{1},\dots,\mu_{m}\right)$
\end_inset

.
 Če to rešitev vstavimo v 
\begin_inset Formula $u_{1}x_{1}+\cdots+u_{m}x_{m}$
\end_inset

,
 dobimo 
\begin_inset Formula $u_{1}\mu_{1}+\cdots+u_{m}\mu_{m}=0$
\end_inset

.
 Ker so 
\begin_inset Formula $u_{1},\dots,u_{m}$
\end_inset

 LN,
 so 
\begin_inset Formula $\mu_{1}=\cdots=\mu_{m}=0$
\end_inset

,
 kar je v 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

 s predpostavko.
\end_layout

\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset

 baza je ogrodje 
\begin_inset Formula $\Rightarrow$
\end_inset

 po lemi 
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:ln<=ogr"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 velja,
 da ima vsaka LN množica manj ali enako elementov kot vsako ogrodje,
 torej tudi manj ali enako kot 
\begin_inset Formula $n$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall$
\end_inset

 baza je LN 
\begin_inset Formula $\Rightarrow$
\end_inset

 po lemi 
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:ln<=ogr"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 velja,
 da ima vsako ogrodje več ali enako elementov kot vsaka LN,
 torej tudi več ali enako kot 
\begin_inset Formula $n$
\end_inset

.
\end_layout

\begin_layout Enumerate
Sledi iz zgornjih dveh točk,
 saj je baza tako ogrodje kot LN hkrati.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $V$
\end_inset

 KRVP.
 Njegova dimenzija,
 
\begin_inset Formula $\dim V$
\end_inset

,
 je moč baze v 
\begin_inset Formula $V$
\end_inset

.
\end_layout

\begin_layout Example*
\begin_inset Formula $\dim F^{n}=n$
\end_inset

,
 
\begin_inset Formula $\dim M_{m\times n}\left(\mathbb{F}\right)=m\cdot n$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Dopolnitev LN množice do baze
\end_layout

\begin_layout Claim*
Naj bo 
\begin_inset Formula $V$
\end_inset

 vektorski prostor z dimenzijo 
\begin_inset Formula $n$
\end_inset

.
 Trdimo,
 da
\end_layout

\begin_deeper
\begin_layout Enumerate
ima vsaka LN množica 
\begin_inset Formula $\leq n$
\end_inset

 elementov,
\end_layout

\begin_layout Enumerate
je vsaka LN množica v 
\begin_inset Formula $V$
\end_inset

 z 
\begin_inset Formula $n$
\end_inset

 elementi baza,
\end_layout

\begin_layout Enumerate
lahko vsako LN množico v 
\begin_inset Formula $V$
\end_inset

 dopolnimo do baze.
\end_layout

\end_deeper
\begin_layout Proof
Dokaz je dolg
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"

\end_inset


\end_layout

\begin_deeper
\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:večja-ln"

\end_inset

Če so 
\begin_inset Formula $v_{1},\dots,v_{m}\in V$
\end_inset

 LN in če 
\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

,
 potem so tudi 
\begin_inset Formula $v_{1},\dots,v_{m},v_{m+1}$
\end_inset

 LN.
\end_layout

\begin_deeper
\begin_layout Proof
Naj velja 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0$
\end_inset

 za nek 
\begin_inset Formula $\vec{\alpha}\in F^{m+1}$
\end_inset

.
 Dokažimo 
\begin_inset Formula $\vec{a}=\vec{0}$
\end_inset

.
 Če 
\begin_inset Formula $\alpha_{m+1}=0$
\end_inset

,
 sledi 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{m}v_{m}=0$
\end_inset

,
 ker pa so po predpostavki 
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset

 LN,
 je 
\begin_inset Formula $\vec{\alpha}=\vec{0}$
\end_inset

.
 Sicer pa,
 če PDDRAA 
\begin_inset Formula $\alpha_{m+1}\not=0$
\end_inset

,
 lahko z 
\begin_inset Formula $a_{m+1}$
\end_inset

 delimo:
\begin_inset Formula 
\[
\alpha_{1}v_{1}+\cdots+\alpha_{m+1}v_{m+1}=0
\]

\end_inset


\begin_inset Formula 
\[
\alpha_{m+1}v_{m+1}=-\alpha_{1}v_{1}-\cdots-\alpha_{m}v_{m}
\]

\end_inset


\begin_inset Formula 
\[
v_{m+1}=\frac{-\alpha_{1}}{\alpha_{m+1}}v_{1}+\cdots+\frac{-\alpha_{m}}{\alpha_{m+1}}v_{m}
\]

\end_inset

Tedaj pridemo do 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

,
 saj smo 
\begin_inset Formula $v_{m+1}$
\end_inset

 izrazili kot LK 
\begin_inset Formula $\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

,
 po predpostavki pa je vendar 
\begin_inset Formula $v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
že dokazano z dokazom trditve 
\begin_inset CommandInset ref
LatexCommand vref
reference "enoličnost-moči-baze."
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 v razdelku 
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Obstoj-baze"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{Vsaka LN množica v 
\backslash
ensuremath{V} z 
\backslash
ensuremath{n} elementi je baza.}
\end_layout

\end_inset

 PDDRAA 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 je LN,
 ki ni baza.
 Tedaj 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

 ni ogrodje.
 Tedaj 
\begin_inset Formula $\Lin\left\{ v_{1},\dots,v_{n}\right\} \not=V$
\end_inset

.
 Zatorej 
\begin_inset Formula $\exists v_{n+1}\in V\ni:\left\{ v_{1},\dots,v_{n},v_{n+1}\right\} $
\end_inset

 je LN,
 kar je v 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

 s trditvijo,
 da ima vsaka 
\begin_inset Formula $LN$
\end_inset

 množica v 
\begin_inset Formula $V$
\end_inset

 kvečjemu 
\begin_inset Formula $n$
\end_inset

 elementov.
\end_layout

\begin_layout Enumerate
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{Vsako LN množico v $V$ z $n$ elementi lahko dopolnimo do baze.}
\end_layout

\end_inset

 Naj bo 
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset

 LN množica v 
\begin_inset Formula $V$
\end_inset

.
 Vemo,
 da je 
\begin_inset Formula $m\leq n$
\end_inset

.
 Če 
\begin_inset Formula $m=n$
\end_inset

,
 je 
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset

 baza po zgornji trditvi.
 Sicer pa je 
\begin_inset Formula $m<n$
\end_inset

:
 Tedaj 
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset

 ni ogrodje,
 sicer bi imeli neko LN množico z več elementi kot neko ogrodje,
 saj ima po popraj dokazanem vsako ogrodje vsaj toliko elementov kot vsaka LN množica.
 Ker 
\begin_inset Formula $v_{1},\dots,v_{m}$
\end_inset

 ni ogrodje,
 
\begin_inset Formula $\exists v_{m+1}\not\in\Lin\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

.
 Po lemi 
\begin_inset CommandInset ref
LatexCommand ref
reference "lem:večja-ln"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 je torej 
\begin_inset Formula $v_{1},\dots,v_{m+1}$
\end_inset

 LN množica.
 Če je 
\begin_inset Formula $m+1=n$
\end_inset

,
 je to že baza,
 sicer ponavljamo dodajanje elementov,
 dokler ne dodamo 
\begin_inset Formula $k$
\end_inset

 elementov in dosežemo 
\begin_inset Formula $m+k=n$
\end_inset

.
 Tedaj je to baza.
 Naredili smo 
\begin_inset Formula $k=m-n$
\end_inset

 korakov.
\end_layout

\end_deeper
\begin_layout Proof
Uporabna vrednost tega izreka sta dva nova izreka o dimenzijah podprostorov:
\end_layout

\begin_layout Claim*
Če je 
\begin_inset Formula $V$
\end_inset

 je KRVP in 
\begin_inset Formula $W$
\end_inset

 njegov podprostor,
 je 
\begin_inset Formula $\dim W\leq\dim V$
\end_inset

.
\end_layout

\begin_layout Proof
PDDRAA 
\begin_inset Formula $\dim W>\dim V$
\end_inset

.
 Čim ima baza 
\begin_inset Formula $W$
\end_inset

 večjo moč kot baza 
\begin_inset Formula $V$
\end_inset

,
 obstaja v 
\begin_inset Formula $W$
\end_inset

 LN množica z večjo močjo kot baza 
\begin_inset Formula $V$
\end_inset

.
 Toda ker je ta LN množica LN tudi v 
\begin_inset Formula $V$
\end_inset

,
 obstaja v 
\begin_inset Formula $V$
\end_inset

 LN množica z več elementi kot baza 
\begin_inset Formula $V$
\end_inset

,
 kar je v 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

 s trditvijo 
\begin_inset CommandInset ref
LatexCommand vref
reference "enoličnost-moči-baze."
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 v razdelku 
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Obstoj-baze"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Claim*
dimenzijska formula za podprostore.
 Naj bo 
\begin_inset Formula $V$
\end_inset

 KRVP in 
\begin_inset Formula $W_{1},W_{2}$
\end_inset

 podprostora v 
\begin_inset Formula $V$
\end_inset

.
 Velja 
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}-\dim\left(W_{1}\cap W_{2}\right)$
\end_inset

.
 Vsota vektorskih podprostorov je definirana v razdelku 
\begin_inset CommandInset ref
LatexCommand vref
reference "subsec:Vsota-podprostorov"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Proof
Izberimo bazo 
\begin_inset Formula $w_{1},\dots,w_{m}$
\end_inset

 za 
\begin_inset Formula $W_{1}\cap W_{2}$
\end_inset

.
 Naj bo 
\begin_inset Formula $u_{1},\dots,u_{k}$
\end_inset

 njena dopolnitev do baze 
\begin_inset Formula $W_{1}$
\end_inset

 in 
\begin_inset Formula $v_{1},\dots,v_{l}$
\end_inset

 njena dopolnitev do baze 
\begin_inset Formula $W_{2}$
\end_inset

.
 Trdimo,
 da je 
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
\end_inset

 baza za 
\begin_inset Formula $W_{1}+W_{2}$
\end_inset

.
 Tedaj bi namreč veljalo 
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=m+k+l$
\end_inset

,
 
\begin_inset Formula $\dim\left(W_{1}\cap W_{2}\right)=m$
\end_inset

,
 
\begin_inset Formula $\dim\left(W_{1}\right)=m+k$
\end_inset

 in 
\begin_inset Formula $\dim\left(W_{2}\right)=m+l$
\end_inset

.
 Treba je dokazati še,
 da je 
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}$
\end_inset

 baza za 
\begin_inset Formula $W_{1}+W_{2}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Je ogrodje?
 Vzemimo poljuben 
\begin_inset Formula $v\in W_{1}+W_{2}$
\end_inset

.
 Po definiciji 
\begin_inset Formula $W_{1}+W_{2}\exists z_{1}\in W_{1},z_{2}\in W_{2}\ni:v=z_{1}+z_{2}$
\end_inset

.
 Razvijmo 
\begin_inset Formula $z_{1}$
\end_inset

 po bazi 
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
\end_inset

 za 
\begin_inset Formula $W_{1}$
\end_inset

 in 
\begin_inset Formula $z_{2}$
\end_inset

 po bazi 
\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{k}$
\end_inset

 za 
\begin_inset Formula $W_{2}$
\end_inset

.
 Takole:
 
\begin_inset Formula $z_{1}=\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}$
\end_inset

 in 
\begin_inset Formula $z_{2}=\gamma_{1}w_{1}+\cdots\gamma_{m}w_{m}+\delta_{1}v_{1}+\cdots\delta_{l}v_{l}$
\end_inset

.
 Torej 
\begin_inset Formula $v=z_{1}+z_{2}=\left(\alpha_{1}+\gamma_{1}\right)w_{1}+\cdots+\left(\alpha_{m}+\gamma_{m}\right)w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\delta_{1}v_{1}+\cdots+\delta_{l}v_{l}\in\Lin\left\{ w_{1},\dots,w_{m},u_{1},\dots,u_{k},v_{1},\dots,v_{l}\right\} $
\end_inset

.
 Je ogrodje.
\end_layout

\begin_layout Itemize
Je LN?
 Naj bo 
\begin_inset Formula 
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
\]

\end_inset


\begin_inset Formula 
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
\]

\end_inset

Leva stran enačbe je 
\begin_inset Formula $\in W_{1}$
\end_inset

,
 desna pa 
\begin_inset Formula $\in W_{2}$
\end_inset

,
 zatorej je element,
 ki ga izraza na obeh straneh enačbe opisujeta,
 
\begin_inset Formula $\in W_{1}\cap W_{2}$
\end_inset

.
 Torej je 
\begin_inset Formula $v_{1},\dots,v_{l}$
\end_inset

 baza za 
\begin_inset Formula $W_{1}\cap W_{1}$
\end_inset

.
 Toda baza od 
\begin_inset Formula $W_{1}\cap W_{2}$
\end_inset

 je tudi 
\begin_inset Formula $w_{1},\dots,w_{m}$
\end_inset

,
 zatorej lahko ta element razpišemo po njej:
\begin_inset Formula 
\[
\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}=\delta_{1}w_{1}+\cdots+\delta_{m}v_{m}
\]

\end_inset


\begin_inset Formula 
\[
\delta_{1}w_{1}+\cdots+\delta_{m}w_{m}+\gamma_{1}v_{1}+\cdots+\gamma_{l}v_{l}=0
\]

\end_inset

Toda 
\begin_inset Formula $w_{1},\dots,w_{m},v_{1},\dots,v_{l}$
\end_inset

 je baza za 
\begin_inset Formula $W_{2}$
\end_inset

 po naši prejšnji definiciji,
 torej je LN množica,
 zato 
\begin_inset Formula $\delta_{1}=\cdots=\delta_{m}=\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset

.
 Ker 
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset

,
 se lahko vrnemo k drugi enačbi te točke in to ugotovitev upoštevamo:
\begin_inset Formula 
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=\left(-\gamma_{1}\right)v_{1}+\cdots+\left(-\gamma_{l}\right)v_{l}
\]

\end_inset


\begin_inset Formula 
\[
\alpha_{1}w_{1}+\cdots+\alpha_{m}w_{m}+\beta_{1}u_{1}+\cdots+\beta_{k}u_{k}=0
\]

\end_inset

Toda 
\begin_inset Formula $w_{1},\dots,w_{m},u_{1},\dots,u_{k}$
\end_inset

 je baza za 
\begin_inset Formula $W_{1}$
\end_inset

 po naši prejšnji definiciji,
 torej je LN množica,
 zato 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=0$
\end_inset

.
 Torej velja 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{m}=\beta_{1}=\cdots=\beta_{k}=\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset

,
 torej je ta množica res LN.
\end_layout

\end_deeper
\begin_layout Corollary*
Velja torej 
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim\left(W_{1}\right)+\dim\left(W_{2}\right)$
\end_inset

.
 Enačaj velja 
\begin_inset Formula $\Leftrightarrow W_{1}\cap W_{2}=\left\{ 0\right\} $
\end_inset

,
 kajti 
\begin_inset Formula $\dim\left(\left\{ 0\right\} \right)=0$
\end_inset

.
\end_layout

\begin_layout Definition
\begin_inset CommandInset label
LatexCommand label
name "def:vsota-je-direktna"

\end_inset

Pravimo,
 da je vsota 
\begin_inset Formula $W_{1}+W_{2}$
\end_inset

 direktna,
 če velja 
\begin_inset Formula $W_{1}\cap W_{2}=\left\{ 0\right\} $
\end_inset

 oziroma ekvivalentno če je 
\begin_inset Formula $\dim\left(W_{1}+W_{2}\right)=\dim W_{1}+\dim W_{2}$
\end_inset

 oziroma ekvivalentno 
\begin_inset Formula $\forall w_{1}\in W_{1},w_{2}\in W_{2}:w_{1}+w_{2}=0\Rightarrow w_{1}=w_{2}=0$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Prehod na novo bazo
\end_layout

\begin_layout Standard
Naj bo 
\begin_inset Formula $V$
\end_inset

 vektorski prostor dimenzije 
\begin_inset Formula $n$
\end_inset

.
 Recimo,
 da imamo dve bazi v 
\begin_inset Formula $V$
\end_inset

.
 
\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

 naj bo 
\begin_inset Quotes gld
\end_inset

stara baza
\begin_inset Quotes grd
\end_inset

,
 
\begin_inset Formula $C=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 pa naj bo 
\begin_inset Quotes gld
\end_inset

nova baza
\begin_inset Quotes grd
\end_inset

.
 
\begin_inset Formula $\forall v\in V$
\end_inset

 lahko razvijemo po 
\begin_inset Formula $B$
\end_inset

 in po 
\begin_inset Formula $C$
\end_inset

.
 Razvoj po 
\begin_inset Formula $B$
\end_inset

:
 
\begin_inset Formula $v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
\end_inset

,
 razvoj po 
\begin_inset Formula $C$
\end_inset

:
 
\begin_inset Formula $v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}$
\end_inset

.
 Kakšna je zveza med 
\begin_inset Formula $\vec{\beta}$
\end_inset

 in 
\begin_inset Formula $\vec{\gamma}$
\end_inset

 v obeh razvojih?
\end_layout

\begin_layout Standard
Uvedimo oznako 
\begin_inset Formula $\left[v\right]_{B}$
\end_inset

,
 to naj bodo koeficienti vektorja 
\begin_inset Formula $v$
\end_inset

 pri razvoju po 
\begin_inset Formula $B$
\end_inset

.
 
\begin_inset Formula $\left[v\right]_{B}=\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]$
\end_inset

.
\end_layout

\begin_layout Standard
Vsak vektor stare baze razvijmo po novi bazi,
 kjer 
\begin_inset Formula $\left[u_{i}\right]_{C}=\left[\begin{array}{c}
\alpha_{i1}\\
\vdots\\
\alpha_{in}
\end{array}\right]$
\end_inset

:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\begin{array}{ccccccc}
u_{1} & = & \alpha_{11}v_{1} & + & \cdots & + & \alpha_{1n}v_{n}\\
\vdots &  & \vdots &  &  &  & \vdots\\
u_{n} & = & a_{n1}v_{1} & + & \cdots & + & a_{nn}v_{n}
\end{array}
\]

\end_inset

Koeficiente 
\begin_inset Formula $\alpha$
\end_inset

 zložimo v tako imenovanj prehodno matriko 
\begin_inset Formula $P_{C\leftarrow B}$
\end_inset

:
\begin_inset Formula 
\[
P_{C\leftarrow B}=\left[\begin{array}{ccc}
\left[u_{1}\right]_{C} & \cdots & \left[u_{n}\right]_{C}\end{array}\right]=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{n1}\\
\vdots &  & \vdots\\
a_{1n} & \cdots & a_{nn}
\end{array}\right]
\]

\end_inset

Sledi 
\begin_inset Formula 
\[
v=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}=\beta_{1}\left(\alpha_{11}v_{1}+\cdots+\alpha_{1n}v_{n}\right)+\cdots+\beta_{n}\left(\alpha_{n1}v_{1}+\cdots+\alpha_{nn}v_{n}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=v_{1}\left(\beta_{1}\alpha_{11}+\beta_{2}\alpha_{21}+\cdots+\beta_{n}\alpha_{n1}\right)+\cdots+v_{n}\left(\beta_{1}\alpha_{1n}+\beta_{2}\alpha_{2n}+\cdots+\beta_{n}\alpha_{nn}\right)=
\]

\end_inset

po drugi strani je 
\begin_inset Formula $v$
\end_inset

 tudi lahko razvit po novi bazi:
\begin_inset Formula 
\[
=v=\gamma_{1}v_{1}+\cdots+\gamma_{n}v_{n}
\]

\end_inset

Iz česar,
 ker je razvoj po bazi enoličen,
 sledi
\begin_inset Formula 
\[
\begin{array}{ccccccc}
\gamma_{1} & = & \beta_{1}\alpha_{11} & + & \cdots & + & \beta_{n}\alpha_{n1}\\
\vdots &  & \vdots &  &  &  & \vdots\\
\gamma_{n} & = & \beta_{1}a_{1n} & + & \cdots & + & \beta_{n}\alpha_{nn}
\end{array},
\]

\end_inset

kar v matrični obliki zapišemo
\begin_inset Formula 
\[
\left[\begin{array}{ccc}
\alpha_{11} & \cdots & \alpha_{n1}\\
\vdots &  & \vdots\\
\alpha_{1n} & \cdots & \alpha_{nn}
\end{array}\right]\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]=\left[\begin{array}{c}
\gamma_{1}\\
\vdots\\
\gamma_{n}
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
P_{C\leftarrow B}\left[v\right]_{B}=\left[v\right]_{C}.
\]

\end_inset


\end_layout

\begin_layout Remark*
Velja:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $P_{B\leftarrow B}=I$
\end_inset

.
\end_layout

\begin_layout Itemize
Naj bodo prehodi med bazami takšnile:
 
\begin_inset Formula $B\overset{P_{C\leftarrow B}}{\longrightarrow}C\overset{P_{D\leftarrow C}}{\longrightarrow}D$
\end_inset

.
 Potem je 
\begin_inset Formula $P_{D\leftarrow B}=P_{C\leftarrow B}\cdot P_{C\leftarrow D}$
\end_inset

.
\end_layout

\begin_layout Itemize
\begin_inset Formula $P_{B\leftarrow C}\cdot P_{C\leftarrow B}=I$
\end_inset

,
 
\begin_inset Formula $\left(P_{B\leftarrow C}\right)^{-1}=P_{C\leftarrow B}$
\end_inset

.
\end_layout

\begin_layout Itemize
Naj bo 
\begin_inset Formula $v\in F^{n}$
\end_inset

 in 
\begin_inset Formula $S$
\end_inset

 standardna baza za 
\begin_inset Formula $F^{n}$
\end_inset

.
 Potem 
\begin_inset Formula $\left[v\right]_{S}=\left[\begin{array}{c}
\alpha_{1}\\
\vdots\\
\alpha_{n}
\end{array}\right]=v$
\end_inset

.
 Sledi 
\begin_inset Formula $P_{S\leftarrow B}=\left[\begin{array}{ccc}
\left[u_{1}\right]_{S} & \cdots & \left[u_{n}\right]_{S}\end{array}\right]=\left[\begin{array}{ccc}
u_{1} & \cdots & u_{n}\end{array}\right]$
\end_inset

 za 
\begin_inset Formula $B=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

.
 Sledi tudi 
\begin_inset Formula $P_{S\leftarrow C}=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]$
\end_inset

,
 kjer so 
\begin_inset Formula $v,u,B,C$
\end_inset

 kot prej (kot definirano na začetku tega razdelka).
\end_layout

\begin_layout Itemize
\begin_inset Formula $P_{C\leftarrow B}=P_{C\leftarrow S}\cdot P_{S\leftarrow B}$
\end_inset

 (slednji dve točki veljata samo v 
\begin_inset Formula $F^{n}$
\end_inset

,
 kjer je standardna baza lepa in zapisljiva kot elementi v matriki)
\end_layout

\end_deeper
\begin_layout Section
Drugi semester
\end_layout

\begin_layout Subsection
Linearne preslikave
\end_layout

\begin_layout Standard
Radi bi definirali homomorfizem vektorskih prostorov.
 Homomorfizem za abelove grupe smo že definirali,
 vektorski prostor pa je le abelova grupa z dodatno strukturo (množenje s skalarjem).
\end_layout

\begin_layout Definition*
Preslikava 
\begin_inset Formula $f:V_{1}\to V_{2}$
\end_inset

 je homomorfizem vektorskih prostorov nad istim poljem oziroma linearna preslikava,
 če je aditivna (homomorfizem) (
\begin_inset Formula $\forall u,v\in V_{1}:f\left(u+_{1}v\right)=fu+_{2}fv$
\end_inset

) in če je homogena:
 
\begin_inset Formula $\forall u\in V_{1},\alpha\in F:f\left(\alpha u\right)=\alpha f\left(u\right)$
\end_inset

.
\end_layout

\begin_layout Remark*
Ekvivalentno je preverjati oba pogoja hkrati.
 Če za 
\begin_inset Formula $L:U\to V$
\end_inset

 velja 
\begin_inset Formula $\forall\alpha_{1},\alpha_{2}\in F,u_{1},u_{2}\in U:L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}$
\end_inset

,
 je 
\begin_inset Formula $L$
\end_inset

 linearna preslikava.
\end_layout

\begin_layout Example*
Vrtež za kot 
\begin_inset Formula $\tau$
\end_inset

 v ravnini:
\begin_inset Formula 
\[
\left[\begin{array}{c}
x\\
y
\end{array}\right]\to\left[\begin{array}{cc}
\cos\tau & -\sin\tau\\
\sin\tau & \cos\tau
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Linearna funkcija iz analize ni linearna preslikava.
 Premik za vektor 
\begin_inset Formula $w$
\end_inset

 ni linearna preslikava.
 Odvajanje in integriranje sta linearni preslikavi.
\end_layout

\begin_layout Fact*
Vsaka linearna preslikava slika 0 v 0.
\end_layout

\begin_layout Definition*
Bijektivni linearni preslikavi pravimo linearni izomorfizem.
\end_layout

\begin_layout Claim
\begin_inset CommandInset label
LatexCommand label
name "claim:invLinIzJeLinIz"

\end_inset

Inverz linearnega izomorfizma je zopet linearni izomorfizem.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $L:U\to V$
\end_inset

 bijektivna linearna preslikava med vektorskima prostoroma nad istim poljem 
\begin_inset Formula $F$
\end_inset

.
 Dokazati je treba,
 da je 
\begin_inset Formula $L^{-1}:V\to U$
\end_inset

 spet linearna preslikava.
 Ker je 
\begin_inset Formula $L$
\end_inset

 linearna,
 velja 
\begin_inset Formula 
\[
\forall\alpha_{1},\alpha_{2}\in F,v_{1},v_{2}\in V:L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=\alpha_{1}LL^{-1}v_{1}+\alpha_{2}LL^{-1}v_{2}=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)
\]

\end_inset

Ker je 
\begin_inset Formula $L$
\end_inset

 injektivna,
 iz 
\begin_inset Formula $L\left(\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}\right)=LL^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
\end_inset

 sledi 
\begin_inset Formula $\alpha_{1}L^{-1}v_{1}+\alpha_{2}L^{-1}v_{2}=L^{-1}\left(\alpha_{1}v_{1}+\alpha_{2}v_{2}\right)$
\end_inset

.
\end_layout

\begin_layout Subsubsection
\begin_inset Formula $F^{n}$
\end_inset

 je linearno izomorfen 
\begin_inset Formula $n-$
\end_inset

razsežnem 
\begin_inset Formula $V$
\end_inset

 nad 
\begin_inset Formula $F$
\end_inset


\end_layout

\begin_layout Claim*
Vsak 
\begin_inset Formula $n-$
\end_inset

razsežen vektorski prostor nad 
\begin_inset Formula $F$
\end_inset

 je linearno izomorfen 
\begin_inset Formula $F^{n}$
\end_inset

.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $V$
\end_inset

 
\begin_inset Formula $n-$
\end_inset

razsežen vektorski prostor nad 
\begin_inset Formula $F$
\end_inset

 in 
\begin_inset Formula $B=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 baza za 
\begin_inset Formula $V$
\end_inset

.
 Definirajmo preslikavo 
\begin_inset Formula $\phi_{B}:F^{n}\to V$
\end_inset

 s predpisom 
\begin_inset Formula $\left(x_{1},\cdots,x_{1}\right)\mapsto x_{1}v_{1}+\cdots+x_{n}v_{n}$
\end_inset

.
 Ker je 
\begin_inset Formula $B$
\end_inset

 ogrodje,
 je 
\begin_inset Formula $\phi_{B}$
\end_inset

 surjektivna.
 Ker je 
\begin_inset Formula $B$
\end_inset

 linearno neodvisna,
 je 
\begin_inset Formula $\phi_{B}$
\end_inset

 injektivna.
 Pokažimo še,
 da je linearna presikava:
\begin_inset Formula 
\[
\phi_{B}\left(\alpha\left(x_{1},\dots,x_{n}\right)+\beta\left(y_{1},\dots,y_{n}\right)\right)=\phi_{B}\left(\alpha x_{1}+\beta y_{1},\dots,\alpha x_{n}+\beta x_{n}\right)=v_{1}\left(\alpha x_{1}+\beta y_{1}\right)+\cdots+v_{n}\left(\alpha x_{n}+\beta x_{n}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=\alpha\left(v_{1}x_{1}+\cdots+v_{n}x_{n}\right)+\cdots+\beta\left(v_{1}y_{1}+\cdots+v_{n}y_{n}\right)=\alpha\phi_{B}\left(x_{1},\dots,x_{n}\right)+\beta\phi_{B}\left(y_{1},\dots y_{n}\right)
\]

\end_inset


\end_layout

\begin_layout Subsubsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Matrika-linearne-preslikave"

\end_inset

Matrika linearne preslikave —
 linearni izomorfizem 
\begin_inset Formula $M_{m,n}\left(F\right)\to\mathcal{L}\left(F^{n},F^{m}\right)$
\end_inset


\end_layout

\begin_layout Claim*
Naj bo 
\begin_inset Formula $F$
\end_inset

 polje in 
\begin_inset Formula $m,n\in F$
\end_inset

.
 
\begin_inset Formula $\mathcal{L}\left(F^{n},F^{m}\right)$
\end_inset

 je vektorski prostor linearnih preslikav iz 
\begin_inset Formula $F^{n}\to F^{m}$
\end_inset

.
 Seštevanje definiramo z 
\begin_inset Formula $\left(L_{1}+L_{2}\right)u\coloneqq L_{1}u+L_{2}u$
\end_inset

,
 množenje s skalarjem pa 
\begin_inset Formula $\left(\alpha L\right)u=\alpha\left(Lu\right)$
\end_inset

.
 Naj bo 
\begin_inset Formula $M_{m,n}\left(F\right)$
\end_inset

 vektorski prostor vseh 
\begin_inset Formula $m\times n$
\end_inset

 matrik nad 
\begin_inset Formula $F$
\end_inset

 z znanim seštevanjem in množenjem.
 Obstaja linearni izomorfizem med tema dvema prostoroma.
\end_layout

\begin_layout Proof
Oznake kot v trditvi.
 Za vsako 
\begin_inset Formula $m\times n$
\end_inset

 matriko 
\begin_inset Formula $A=\left[a_{i,j}\right]$
\end_inset

 definirajmo preslikavo 
\begin_inset Formula $L_{A}$
\end_inset

 iz 
\begin_inset Formula $F^{n}$
\end_inset

 v 
\begin_inset Formula $F^{m}$
\end_inset

 takole:
 
\begin_inset Formula $L_{A}\left(x_{1},\dots,x_{n}\right)=\left(a_{1,1}x_{1}+\cdots+a_{1,n}x_{n},\dots,a_{m,1}x_{1}+\cdots+a_{m,n}x_{n}\right)$
\end_inset

.
 Po definiciji matričnega množenja ta preslikava ustreza 
\begin_inset Formula $L_{A}\vec{x}=A\vec{x}$
\end_inset

.
 Dokažimo,
 da je linearni izomorfizem.
\end_layout

\begin_deeper
\begin_layout Itemize
Linearnost:
 
\begin_inset Formula $L_{\alpha A+\beta B}\vec{x}=\left(\alpha A+\beta B\right)\vec{x}=\alpha A\vec{x}+\beta B\vec{x}=\alpha L_{A}\vec{x}+\beta L_{B}\vec{x}=\left(\alpha L_{A}+\beta L_{B}\right)\vec{x}$
\end_inset


\end_layout

\begin_layout Itemize
Bijektivnost:
 Konstruirajmo inverzno preslikavo (iz trditve 
\begin_inset CommandInset ref
LatexCommand ref
reference "claim:invLinIzJeLinIz"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 vemo,
 da bo linearna).
 Vsaki linearni presikavi 
\begin_inset Formula $L:F^{n}\to F^{m}$
\end_inset

 priredimo 
\begin_inset Formula $m\times n$
\end_inset

 matriko 
\begin_inset Formula $\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]$
\end_inset

,
 kjer je 
\begin_inset Formula $e_{1},\dots,e_{n}$
\end_inset

 standardna baza za 
\begin_inset Formula $F^{n}$
\end_inset

.
 Pokažimo,
 da je ta preslikava res inverz,
 torej preverimo,
 da je kompozitum 
\begin_inset Formula $A\mapsto L_{A}\mapsto\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]$
\end_inset

 identiteta in da je 
\begin_inset Formula $\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]\mapsto L_{A}\mapsto A$
\end_inset

 tudi identiteta.
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$A
\backslash
mapsto L_{A}
\backslash
mapsto
\backslash
left[
\backslash
begin{array}{ccc}
Le_{1} & 
\backslash
cdots & Le_{n}
\backslash
end{array}
\backslash
right]
\backslash
overset{?}{=}id$}
\end_layout

\end_inset

:
 
\begin_inset Formula 
\[
\left[\begin{array}{ccc}
L_{A}e_{1} & \cdots & L_{A}e_{n}\end{array}\right]=\left[\begin{array}{ccc}
Ae_{1} & \cdots & Ae_{n}\end{array}\right]=A\left[\begin{array}{ccc}
e_{1} & \cdots & e_{n}\end{array}\right]=AI=A
\]

\end_inset


\end_layout

\begin_layout Itemize
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
udensdash{$
\backslash
left[
\backslash
begin{array}{ccc}
L_{A}e_{1} & 
\backslash
cdots & L_{A}e_{n}
\backslash
end{array}
\backslash
right]
\backslash
mapsto L_{A}
\backslash
mapsto A
\backslash
overset{?}{=}id$}
\end_layout

\end_inset

:
\begin_inset Formula 
\[
\forall x:L_{\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]}x=\left[\begin{array}{ccc}
Le_{1} & \cdots & Le_{n}\end{array}\right]\left[\begin{array}{c}
x_{1}\\
\vdots\\
x_{n}
\end{array}\right]=x_{1}Le_{1}+\cdots+x_{n}Le_{n}=L\left(x_{1}e_{1}+\cdots+x_{n}e_{n}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=L\left(x_{1}\left[\begin{array}{c}
1\\
0\\
\vdots\\
0
\end{array}\right]+\cdots+x_{n}\left[\begin{array}{c}
0\\
\vdots\\
0\\
1
\end{array}\right]\right)=Lx
\]

\end_inset


\end_layout

\end_deeper
\end_deeper
\begin_layout Proof
Vsaki linearni preslikavi med dvema vektorskima prostoroma sedaj lahko priredimo matriko.
 Prirejanje je odvisno od izbire baz v obeh vektorskih prostorih.
 Matrika namreč preslika koeficiente iz polja 
\begin_inset Formula $F$
\end_inset

,
 s katerimi je dan vektor,
 ki ga z leve množimo z matriko,
 razvit po 
\begin_inset Quotes gld
\end_inset

vhodni
\begin_inset Quotes grd
\end_inset

 bazi,
 v koeficiente iz istega polja,
 s katerimi je rezultantni vektor razvit po 
\begin_inset Quotes gld
\end_inset

izhodni
\begin_inset Quotes grd
\end_inset

 bazi.
\end_layout

\begin_layout Proof
Naj bosta 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $V$
\end_inset

 vektorska prostora nad istim poljem 
\begin_inset Formula $F$
\end_inset

 in naj bo 
\begin_inset Formula $L:U\to V$
\end_inset

 linearna preslikava.
 Izberimo bazo 
\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

 za 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

 za 
\begin_inset Formula $V$
\end_inset

.
 Razvijmo vektorje 
\begin_inset Formula $Lu_{1},\dots,Lu_{n}$
\end_inset

 po bazi 
\begin_inset Formula $\mathcal{C}$
\end_inset

:
\begin_inset Formula 
\[
\begin{array}{ccccccc}
Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
\vdots &  & \vdots &  &  &  & \vdots\\
Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
\end{array}
\]

\end_inset

Skalarje 
\begin_inset Formula $\alpha_{i,j}$
\end_inset

 sedaj zložimo v spodnjo matriko,
 ki ji pravimo 
\series bold
matrika linearne preslikave 
\begin_inset Formula $L$
\end_inset

 glede na bazi 
\begin_inset Formula $\mathcal{B}$
\end_inset

 in 
\begin_inset Formula $\mathcal{C}$
\end_inset


\series default
.
 
\begin_inset Formula 
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[\begin{array}{ccc}
a_{1,1} & \cdots & \alpha_{n,1}\\
\vdots &  & \vdots\\
\alpha_{1,m} & \cdots & \alpha_{n,m}
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $L:\mathbb{R}\left[x\right]_{\leq3}\to\mathbb{R}\left[x\right]_{\leq2}$
\end_inset

 —
 linearna preslikava iz realnih polinomov stopnje kvečjemu 3 v realne polinome stopnje kvečjemu 2,
 ki predstavlja odvajanje polinomov.
 Bazi sta 
\begin_inset Formula $\mathcal{B}=\left\{ 1,x,x^{2},x^{2}\right\} $
\end_inset

 in 
\begin_inset Formula $\mathcal{C}=\left\{ 1,x,x^{2}\right\} $
\end_inset

.
\begin_inset Formula 
\[
\begin{array}{ccccccc}
L\left(1\right) & = & 0 & + & 0x & + & 0x^{2}\\
L\left(x\right) & = & 1 & + & 0x & + & 0x^{2}\\
L\left(x^{2}\right) & = & 0 & + & 2x & + & 0x^{2}\\
L\left(x^{3}\right) & = & 0 & + & 0x & + & 3x^{2}
\end{array}
\]

\end_inset

Zapišimo matriko te linearne preslikave:
\begin_inset Formula 
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cccc}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Prehodna matrika je poseben primer matrike linearne preslikave.
 
\begin_inset Formula $L:V\to V$
\end_inset

 z bazama 
\begin_inset Formula $\mathcal{B}=\left\{ v_{1},\dots,v_{n}\right\} $
\end_inset

 in 
\begin_inset Formula $\mathcal{C}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

,
 kjer 
\begin_inset Formula $L=id$
\end_inset

.
\begin_inset Formula 
\[
\begin{array}{ccccccc}
id\left(u_{1}\right) & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,n}v_{n}\\
\vdots &  & \vdots &  &  &  & \vdots\\
id\left(u_{n}\right) & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,n}v_{n}
\end{array}
\]

\end_inset

Zapišimo matriko te linearne preslikave:
 
\begin_inset Formula $\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}=P_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Lastnosti matrik linearnih preslikav
\end_layout

\begin_layout Standard
\begin_inset CommandInset counter
LatexCommand set
counter "theorem"
value "0"
lyxonly "false"

\end_inset


\end_layout

\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:osnovna-formula"

\end_inset

osnovna formula.
 Posplošitev formule 
\begin_inset Formula $\left[u\right]_{\mathcal{C}}=P_{\mathcal{C}\leftarrow\mathcal{B}}\cdot\left[u\right]_{\mathcal{B}}$
\end_inset

 se glasi 
\begin_inset Formula $\left[Lu\right]_{\mathcal{C}}=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}$
\end_inset

 za linearno preslikavo 
\begin_inset Formula $L:U\to V$
\end_inset

,
 
\begin_inset Formula $u\in U$
\end_inset

,
 kjer je 
\begin_inset Formula $\mathcal{\mathcal{B}}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

 baza za 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $\mathcal{C}=\left\{ v_{1},\dots,v_{m}\right\} $
\end_inset

 baza za 
\begin_inset Formula $V$
\end_inset

.
\end_layout

\begin_layout Proof
Po korakih:
\end_layout

\begin_deeper
\begin_layout Enumerate
Razvijmo 
\begin_inset Formula $u$
\end_inset

 po bazi 
\begin_inset Formula $\mathcal{B}$
\end_inset

:
 
\begin_inset Formula $u=\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:Uporabimo-L-na"

\end_inset

Uporabimo 
\begin_inset Formula $L$
\end_inset

 na obeh straneh:
 
\begin_inset Formula $Lu=L\left(\beta_{1}u_{1}+\cdots+\beta_{n}u_{n}\right)=\beta_{1}Lu_{1}+\cdots+\beta_{n}Lu_{n}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Razvijmo bazo 
\begin_inset Formula $\mathcal{B}$
\end_inset

,
 preslikano z 
\begin_inset Formula $L$
\end_inset

,
 po bazi 
\begin_inset Formula $\mathcal{C}$
\end_inset

:
\begin_inset Formula 
\[
\begin{array}{ccccccc}
Lu_{1} & = & \alpha_{1,1}v_{1} & + & \cdots & + & \alpha_{1,m}v_{m}\\
\vdots &  & \vdots &  &  &  & \vdots\\
Lu_{n} & = & \alpha_{n,1}v_{1} & + & \cdots & + & \alpha_{n,m}v_{m}
\end{array}
\]

\end_inset


\end_layout

\begin_layout Enumerate
Razvoj vstavimo v enačbo iz koraka 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:Uporabimo-L-na"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 in uredimo:
\begin_inset Formula 
\[
Lu=\beta_{1}\left(\alpha_{1,1}v_{1}+\cdots+\alpha_{1,m}v_{m}\right)+\cdots+\beta_{n}\left(\alpha_{n,1}v_{1}+\cdots+\alpha_{n,m}v_{m}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=v_{1}\left(\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\right)+\cdots+v_{m}\left(\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}\right)
\]

\end_inset


\end_layout

\begin_layout Enumerate
Odtod sledi:
\begin_inset Formula 
\[
\left[Lu\right]_{\mathcal{C}}=\left[\begin{array}{c}
\beta_{1}\alpha_{1,1}+\cdots+\beta_{n}\alpha_{n,1}\\
\vdots\\
\beta_{1}\alpha_{1,m}+\cdots+\beta_{n}\alpha_{n,m}v_{m}
\end{array}\right]=\left[\begin{array}{ccc}
\alpha_{1,1} & \cdots & \alpha_{n,1}\\
\vdots &  & \vdots\\
\alpha_{1,m} & \cdots & \alpha_{n,m}
\end{array}\right]\left[\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{n}
\end{array}\right]=\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[u\right]_{\mathcal{B}}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:matrika-kompozituma-linearnih"

\end_inset

matrika kompozituma linearnih preslikav.
 Posplošitev formule 
\begin_inset Formula $P_{\mathcal{\mathcal{D}\leftarrow\mathcal{B}}}=P_{\mathcal{D\leftarrow C}}\cdot P_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset

 se glasi 
\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset

.
 Trdimo,
 da je kompozitum linearnih preslikav spet linearna preslikava in da enačba velja.
\end_layout

\begin_layout Proof
Najprej dokažimo,
 da je kompozitum linearnih preslikav spet linearna preslikava.
\begin_inset Formula 
\[
\left(K\circ L\right)\left(\alpha u+\beta v\right)=K\left(L\left(\alpha u+\beta v\right)\right)=K\left(\alpha Lu+\beta Lv\right)=\alpha KLu+\beta KLv=\alpha\left(K\circ L\right)u+\beta\left(K\circ L\right)v
\]

\end_inset

Sedaj pa dokažimo še enačbo 
\begin_inset Formula $\left[K\circ L\right]_{\mathcal{D\leftarrow B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\cdot\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset

.
 Naj bosta 
\begin_inset Formula $L:U\to V$
\end_inset

 in 
\begin_inset Formula $K:V\to W$
\end_inset

 linearni preslikavi,
 
\begin_inset Formula $\mathcal{B}=\left\{ u_{1},\dots,u_{n}\right\} $
\end_inset

 baza 
\begin_inset Formula $U$
\end_inset

,
 
\begin_inset Formula $\mathcal{C}$
\end_inset

 baza 
\begin_inset Formula $V$
\end_inset

 in 
\begin_inset Formula $\mathcal{D}$
\end_inset

 baza 
\begin_inset Formula $W$
\end_inset

.
 Od prej vemo,
 da:
\begin_inset Formula 
\[
\left[L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{D}} & \cdots & \left[Lu_{n}\right]_{\mathcal{D}}\end{array}\right],
\]

\end_inset

zato pišimo
\begin_inset Formula 
\[
\left[K\circ L\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[\begin{array}{ccc}
\left[\left(K\circ L\right)u_{1}\right]_{\mathcal{D}} & \cdots & \left[\left(K\circ L\right)u_{n}\right]_{\mathcal{D}}\end{array}\right]=\left[\begin{array}{ccc}
\left[KLu_{1}\right]_{\mathcal{D}} & \cdots & \left[KLu_{n}\right]_{\mathcal{D}}\end{array}\right]\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}
\]

\end_inset


\begin_inset Formula 
\[
\overset{\text{izrek \ref{thm:osnovna-formula}}}{=}\left[\begin{array}{ccc}
\left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[K\right]_{\mathcal{D}\leftarrow C}\left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[\begin{array}{ccc}
\left[Lu_{1}\right]_{\mathcal{C}} & \cdots & \left[Lu_{n}\right]_{\mathcal{C}}\end{array}\right]=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Jedro in slika linearne preslikave
\end_layout

\begin_layout Definition*
Naj bosta 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $V$
\end_inset

 vektorska prostora nad istim poljem 
\begin_inset Formula $F$
\end_inset

 in 
\begin_inset Formula $L:U\to V$
\end_inset

 linearna preslikava.
 Jedro 
\begin_inset Formula $L$
\end_inset

 naj bo 
\begin_inset Formula $\Ker L\coloneqq\left\{ u\in U;Lu=0\right\} $
\end_inset

 (angl.
 kernel/null space) in Slika/zaloga vrednosti 
\begin_inset Formula $L$
\end_inset

 naj bo 
\begin_inset Formula $\Slika L\coloneqq\left\{ Lu;\forall u\in U\right\} $
\end_inset

 (angl.
 image/range).
\end_layout

\begin_layout Claim*
Trdimo naslednje:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\Ker L$
\end_inset

 je vektorski podprostor v 
\begin_inset Formula $U$
\end_inset

 (če vsebuje 
\begin_inset Formula $\vec{a}$
\end_inset

 in 
\begin_inset Formula $\vec{b}$
\end_inset

,
 vsebuje tudi vse LK 
\begin_inset Formula $\vec{a}$
\end_inset

 in 
\begin_inset Formula $\vec{b}$
\end_inset

)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\Slika L$
\end_inset

 je vektorski podprostor v 
\begin_inset Formula $V$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Proof
Dokazujemo dve trditvi:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall u_{1},u_{2}\in\Ker L,\alpha_{1},\alpha_{2}\in F\overset{?}{\Longrightarrow}\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
\end_inset

.
 Po predpostavki velja 
\begin_inset Formula $Lu_{1}=0$
\end_inset

 in 
\begin_inset Formula $Lu_{2}=0$
\end_inset

,
 torej 
\begin_inset Formula $\alpha_{1}Lu_{1}+\alpha_{2}Lu_{2}=0$
\end_inset

.
 Iz linearnosti 
\begin_inset Formula $L$
\end_inset

 sledi 
\begin_inset Formula $L\left(\alpha_{1}u_{1}+\alpha_{2}u_{2}\right)=0$
\end_inset

,
 torej 
\begin_inset Formula $\alpha_{1}u_{1}+\alpha_{2}u_{2}\in\Ker L$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall v_{1},v_{2}\in\Slika L,\beta_{1},\beta_{2}\in F\overset{?}{\Longrightarrow}\beta_{1}v_{1}+\beta_{2}v_{2}\in\Slika L$
\end_inset

.
 Po predpostavki velja 
\begin_inset Formula $\exists u_{1},u_{2}\in U\ni:v_{1}=Lu_{1}\wedge v_{2}=Lu_{2}$
\end_inset

.
 Velja torej 
\begin_inset Formula $\beta_{1}v_{1}+\beta_{2}v_{2}=\beta_{1}Lu_{1}+\beta_{2}Lu_{2}\overset{\text{linearnost}}{=}L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)$
\end_inset

 in 
\begin_inset Formula $\beta_{1}u_{1}+\beta_{2}u_{2}\in U$
\end_inset

,
 torej je 
\begin_inset Formula $L\left(\beta_{1}u_{1}+\beta_{2}u_{2}\right)\in\Slika L$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Definition*
Ničnost 
\begin_inset Formula $L$
\end_inset

 je 
\begin_inset Formula $\n\left(L\right)\coloneqq\dim\Ker L$
\end_inset

 (angl.
 nullity) in rang 
\begin_inset Formula $L$
\end_inset

 je 
\begin_inset Formula $\rang\left(L\right)=\dim\Slika L$
\end_inset

 (angl.
 rank).
\end_layout

\begin_layout Remark*
Jedro in sliko smo definirali za linearne preslikave,
 vendar ju lahko definiramo tudi za poljubno matriko 
\begin_inset Formula $A$
\end_inset

 nad poljem 
\begin_inset Formula $F$
\end_inset

,
 saj smo v 
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Matrika-linearne-preslikave"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 dokazali linearni izomorfizem med 
\begin_inset Formula $m\times n$
\end_inset

 matrikami nad 
\begin_inset Formula $F$
\end_inset

 in linearnimi preslikavami 
\begin_inset Formula $F^{n}\to F^{m}$
\end_inset

.
\begin_inset Formula 
\[
Au=\left[\begin{array}{ccc}
a_{11} & \cdots & a_{1n}\\
\vdots &  & \vdots\\
a_{m1} & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{c}
u_{1}\\
\vdots\\
u_{n}
\end{array}\right]=\left[\begin{array}{c}
a_{11}u_{1}+\cdots+a_{1n}u_{n}\\
\vdots\\
a_{m1}u_{1}+\cdots+a_{mn}u_{n}
\end{array}\right]=\left[\begin{array}{c}
a_{11}\\
\vdots\\
a_{m1}
\end{array}\right]u_{1}+\cdots+\left[\begin{array}{c}
a_{1n}\\
\vdots\\
a_{mn}
\end{array}\right]u_{n}
\]

\end_inset

Iz tega je razvidno,
 da je 
\begin_inset Formula $\Slika A$
\end_inset

 torej linearna ogrinjača stolpcev matrike 
\begin_inset Formula $A$
\end_inset

.
 Pravimo tudi,
 da je 
\begin_inset Formula $\Slika A$
\end_inset

 stolpični prostor 
\begin_inset Formula $A$
\end_inset

 oziroma 
\begin_inset Formula $\Col A$
\end_inset

 (angl.
 column space).
 
\begin_inset Formula $\rang A=\dim\Slika A$
\end_inset

 je torej največje število linearno neodvisnih stolpcev 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Claim*
Linearna preslikava 
\begin_inset Formula $L$
\end_inset

 je injektivna (
\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
\end_inset

) 
\begin_inset Formula $\Leftrightarrow\Ker L=\left\{ 0\right\} $
\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Predpostavimo,
 da je 
\begin_inset Formula $L$
\end_inset

 injektivna,
 torej 
\begin_inset Formula $Lu_{1}=Lu_{2}\Rightarrow u_{1}=u_{2}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $u\in\Ker L$
\end_inset

.
 Zanj velja 
\begin_inset Formula $Lu=0=L0\Rightarrow u=0$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Predpostavimo,
 
\begin_inset Formula $\Ker L=\left\{ 0\right\} $
\end_inset

.
 Računajmo:
 
\begin_inset Formula $Lu_{1}=Lu_{2}\Longrightarrow Lu_{1}-Lu_{2}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(u_{1}-u_{2}\right)=0\Longrightarrow u_{1}-u_{2}=0\Longrightarrow u_{1}=u_{2}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Theorem*
osnovna formula.
 Naj bo 
\begin_inset Formula $L:U\to V$
\end_inset

 linearna preslikava.
 Tedaj je 
\begin_inset Formula $\dim\Ker L+\dim\Slika L=\dim U$
\end_inset

,
 torej 
\begin_inset Formula $\n L+\rang L=\dim U$
\end_inset

.
 Za matrike torej trdimo 
\begin_inset Formula $\n A+\rang A=\dim F^{n}=n$
\end_inset

 za 
\begin_inset Formula $m\times n$
\end_inset

 matriko 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Proof
Vemo,
 da sta jedro in slika podprostora.
 Naj bo 
\begin_inset Formula $w_{1},\dots,w_{k}$
\end_inset

 baza jedra in 
\begin_inset Formula $u_{1},\dots,u_{l}$
\end_inset

 njena dopolnitev do baze 
\begin_inset Formula $U$
\end_inset

.
 Torej 
\begin_inset Formula $\dim U=k+l=\n L+l$
\end_inset

.
 Treba je še dokazati,
 da je 
\begin_inset Formula $l=\rang A$
\end_inset

.
 Konstruirajmo bazo za 
\begin_inset Formula $\Slika L$
\end_inset

,
 ki ima 
\begin_inset Formula $l$
\end_inset

 elementov in dokažimo,
 da so 
\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
\end_inset

 baza za 
\begin_inset Formula $\Slika L$
\end_inset

:
\end_layout

\begin_deeper
\begin_layout Itemize
Je ogrodje?
 Vzemimo poljuben 
\begin_inset Formula $v\in\Slika L$
\end_inset

.
 Zanj obstaja nek 
\begin_inset Formula $u\in U\ni:Lu=v$
\end_inset

,
 ki ga lahko razvijemo po bazi 
\begin_inset Formula $U$
\end_inset

 takole 
\begin_inset Formula $u=\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}$
\end_inset

.
 Sedaj na obeh straneh uporabimo 
\begin_inset Formula $L$
\end_inset

 in upoštevamo linearnost:
\begin_inset Formula 
\[
v=Lu=L\left(\alpha_{1}w_{1}+\cdots+\alpha_{k}w_{k}+\beta_{1}u_{1}+\cdots+\beta_{l}u_{l}\right)=\alpha_{1}Lw_{1}+\cdots+\alpha_{k}Lw_{k}+\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
\]

\end_inset


\begin_inset Formula 
\[
=\beta_{1}Lu_{1}+\cdots+\beta_{l}Lu_{l}
\]

\end_inset

Ker so 
\begin_inset Formula $w_{i}$
\end_inset

 baza 
\begin_inset Formula $\Ker L$
\end_inset

,
 so elementi 
\begin_inset Formula $\Ker L$
\end_inset

,
 torej je 
\begin_inset Formula $Lw_{i}=0$
\end_inset

 za vsak 
\begin_inset Formula $i$
\end_inset

.
 Tako poljuben 
\begin_inset Formula $v\in\Slika L$
\end_inset

 razpišemo z bazo velikosti 
\begin_inset Formula $l$
\end_inset

.
\end_layout

\begin_layout Itemize
Je LN?
 Računajmo:
 
\begin_inset Formula $\gamma_{1}Lu_{1}+\cdots+\gamma_{l}Lu_{l}=0\overset{\text{linearnost }}{\Longrightarrow}L\left(\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\right)=0\Longrightarrow\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}\in\Ker L$
\end_inset

,
 kar pomeni,
 da ga je moč razviti po bazi 
\begin_inset Formula $\Ker L$
\end_inset

:
\begin_inset Formula 
\[
\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}=\delta_{1}w_{1}+\cdots+\delta_{k}w_{k}
\]

\end_inset


\begin_inset Formula 
\[
\gamma_{1}u_{1}+\cdots+\gamma_{l}u_{l}-\delta_{1}w_{1}-\cdots-\delta_{k}w_{k}=0
\]

\end_inset

Ker je 
\begin_inset Formula $w_{1},\dots,w_{k},u_{1},\dots,u_{l}$
\end_inset

 baza 
\begin_inset Formula $U$
\end_inset

,
 je LN,
 zato velja 
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=w_{1}=\cdots=w_{k}=0$
\end_inset

,
 kar pomeni,
 da očitno velja 
\begin_inset Formula $\gamma_{1}=\cdots=\gamma_{l}=0$
\end_inset

,
 torej je res LN.
\end_layout

\end_deeper
\begin_layout Remark*
Bralcu prav pride skica s 3.
 strani zapiskov predavanja 
\begin_inset Quotes gld
\end_inset

LA1P FMF 2024-02-28
\begin_inset Quotes grd
\end_inset

.
\end_layout

\begin_layout Paragraph*
Do preproste matrike preslikave z ustreznimi bazami
\end_layout

\begin_layout Standard
Imenujmo sedaj 
\begin_inset Formula $\mathcal{B}=\left\{ w_{1},\dots,w_{k},u_{1},\dots,u_{l}\right\} $
\end_inset

 bazo za 
\begin_inset Formula $U$
\end_inset

,
 in 
\begin_inset Formula $\mathcal{C}=\left\{ Lu_{1},\dots,Lu_{l},z_{1},\dots,z_{m}\right\} $
\end_inset

 baza za 
\begin_inset Formula $V$
\end_inset

,
 kjer je 
\begin_inset Formula $z_{1},\dots,z_{m}$
\end_inset

 dopolnitev 
\begin_inset Formula $Lu_{1},\dots,Lu_{l}$
\end_inset

 do baze 
\begin_inset Formula $V$
\end_inset

,
 kajti 
\begin_inset Formula $V$
\end_inset

 je lahko večji kot samo 
\begin_inset Formula $\Slika L$
\end_inset

,
 in si oglejmo matriko naše preslikave 
\begin_inset Formula $L:U\to V$
\end_inset

,
 ki slika iz baze 
\begin_inset Formula $\mathcal{B}$
\end_inset

 v bazo 
\begin_inset Formula $\mathcal{C}$
\end_inset

.
 Najprej razpišimo preslikane elemente baze 
\begin_inset Formula $\mathcal{B}$
\end_inset

 po bazi 
\begin_inset Formula $\mathcal{C}$
\end_inset

:
\begin_inset Formula 
\[
\begin{array}{ccccccccccccc}
Lu_{1} & = & 1\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
\vdots &  & \vdots &  &  &  & \vdots &  & \vdots &  &  &  & \vdots\\
Lu_{l} & = & 0\cdot Lu_{1} & + & \cdots & + & 1\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
Lw_{1} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}\\
\vdots &  & \vdots &  &  &  & \vdots &  & \vdots &  &  &  & \vdots\\
Lw_{k} & = & 0\cdot Lu_{1} & + & \cdots & + & 0\cdot Lu_{l} & + & 0\cdot z_{1} & + & \cdots & + & 0\cdot z_{m}
\end{array}
\]

\end_inset


\begin_inset Formula 
\[
\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}=\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]
\]

\end_inset

S primerno izbiro baz 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $V$
\end_inset

 je torej matrika preslikave precej preprosta,
 zgolj bločna matrika z identiteto,
 veliko 
\begin_inset Formula $\rang L$
\end_inset

 in ničlami,
 ki ustrezajo dimenzijam 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $V$
\end_inset

.
\end_layout

\begin_layout Standard
Kaj pa,
 če je 
\begin_inset Formula $L$
\end_inset

 matrika?
 Recimo ji 
\begin_inset Formula $A$
\end_inset

,
 da je 
\begin_inset Formula $L_{A}=L$
\end_inset

 od prej.
 Tedaj 
\begin_inset Formula $A\in M_{p,n}\left(F\right)$
\end_inset

.
 Naj bo 
\begin_inset Formula $P=\left[\begin{array}{cccccc}
u_{1} & \cdots & u_{l} & w_{1} & \cdots & w_{k}\end{array}\right]$
\end_inset

 matrika,
 katere stolpci so baza 
\begin_inset Formula $U$
\end_inset

 in 
\begin_inset Formula $Q=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]$
\end_inset

 matrika,
 katere stolpci so baza 
\begin_inset Formula $V$
\end_inset

.
 Po karakterizaciji obrnljivih matrik sta obrnljivi.
 Tedaj
\begin_inset Formula 
\[
AP=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & Aw_{1} & \cdots & Aw_{k}\end{array}\right]\overset{\text{jedro}}{=}\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
Q\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & z_{1} & \cdots & z_{m}\end{array}\right]\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]=\left[\begin{array}{cccccc}
Au_{1} & \cdots & Au_{l} & 0 & \cdots & 0\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
AP=Q\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]\Longrightarrow Q^{-1}AP=\left[\begin{array}{cc}
I_{l} & 0\\
0 & 0
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Ekvivalentnost matrik
\end_layout

\begin_layout Definition*
Matriki 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 sta ekvivalentni (oznaka 
\begin_inset Formula $A\sim B$
\end_inset


\begin_inset Foot
status open

\begin_layout Plain Layout
Isto oznako uporabljamo tudi za podobne matrike,
 vendar podobnost ni enako kot ekvivalentnost.
\end_layout

\end_inset

) 
\begin_inset Formula $\Leftrightarrow\exists$
\end_inset

 obrnljivi 
\begin_inset Formula $P,Q\ni:B=PAQ$
\end_inset

.
\end_layout

\begin_layout Example*
Dokazali smo,
 da je vsaka matrika 
\begin_inset Formula $A$
\end_inset

 ekvivalentna matriki 
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset

,
 kjer je 
\begin_inset Formula $r=\rang A$
\end_inset

.
\end_layout

\begin_layout Proof
Dokažimo,
 da je relacija 
\begin_inset Formula $\sim$
\end_inset

 ekvivalenčna:
\end_layout

\begin_deeper
\begin_layout Itemize
refleksivnost:
 
\begin_inset Formula $A\sim A$
\end_inset

 velja.
 Naj bo 
\begin_inset Formula $A\in M_{m,n}\left(F\right)$
\end_inset

.
 Tedaj 
\begin_inset Formula $A=I_{m}AI_{n}$
\end_inset

.
\end_layout

\begin_layout Itemize
simetričnost:
 
\begin_inset Formula $A\sim B\Rightarrow B\sim A$
\end_inset

,
 kajti če velja 
\begin_inset Formula $B=PAQ$
\end_inset

 in sta 
\begin_inset Formula $P$
\end_inset

 in 
\begin_inset Formula $Q$
\end_inset

 obrnljivi,
 velja 
\begin_inset Formula $P^{-1}BQ^{-1}=A$
\end_inset

.
\end_layout

\begin_layout Itemize
tranzitivnost:
 
\begin_inset Formula $A\sim B\wedge B\sim C\Rightarrow A\sim C$
\end_inset

,
 kajti,
 če velja 
\begin_inset Formula $B=PAQ$
\end_inset

 in 
\begin_inset Formula $C=SBT$
\end_inset

 in so 
\begin_inset Formula $P,Q,S,T$
\end_inset

 obrnljive,
 velja 
\begin_inset Formula $C=\left(SP\right)A\left(QT\right)$
\end_inset

 in produkt obrnljivih matrik je obrnljiva matrika.
\end_layout

\end_deeper
\begin_layout Theorem*
Dve matriki sta ekvivalentni natanko tedaj,
 ko imata enako velikost in enak rang.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Po predpostavki imata 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 enako velikost in enak rang 
\begin_inset Formula $r$
\end_inset

.
 Od prej vemo,
 da sta obe ekvivalentni 
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset

,
 ker pa je relacija ekvivalentnosti ekvivalenčna,
 sta 
\begin_inset Formula $A\sim B$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Po predpostavki 
\begin_inset Formula $A\sim B$
\end_inset

,
 torej 
\begin_inset Formula $\exists P,Q\ni:B=PAQ$
\end_inset

.
 Če je 
\begin_inset Formula $A$
\end_inset

 
\begin_inset Formula $m\times n$
\end_inset

,
 je 
\begin_inset Formula $P$
\end_inset

 
\begin_inset Formula $m\times m$
\end_inset

 in 
\begin_inset Formula $Q$
\end_inset

 
\begin_inset Formula $n\times n$
\end_inset

,
 zatorej je po definiciji matričnega množenja 
\begin_inset Formula $B$
\end_inset

 
\begin_inset Formula $m\times n$
\end_inset

.
 Dokazati je treba še 
\begin_inset Formula $\rang A=\rang B$
\end_inset

.
\begin_inset Formula 
\[
\rang B=\rang PAQ\overset{?}{=}\rang PA=\overset{?}{=}\rang A
\]

\end_inset

Dokažimo najprej 
\begin_inset Formula $\rang PAQ=\rang PA$
\end_inset

 oziroma 
\begin_inset Formula $\rang CQ=\rang C$
\end_inset

 za obrnljivo 
\begin_inset Formula $Q$
\end_inset

 in poljubno C.
 Dokažemo lahko celo 
\begin_inset Formula $\Slika CQ=\Slika C$
\end_inset

:
 
\begin_inset Formula 
\[
\forall u:u\in\Slika CQ\Leftrightarrow\exists v\ni:u=\left(CQ\right)v\Leftrightarrow\exists v'\ni:u=Cv'\Leftrightarrow u\in\Slika C.
\]

\end_inset

Sedaj dokažimo še 
\begin_inset Formula $\rang\left(PA\right)=\rang\left(A\right)$
\end_inset

.
 Zadošča dokazati,
 da je 
\begin_inset Formula $\Ker\left(PA\right)=\Ker A$
\end_inset

,
 kajti tedaj bi iz enakosti izrazov
\begin_inset Formula 
\[
\dim\Slika A+\dim\Ker A=\dim F^{n}=n
\]

\end_inset


\begin_inset Formula 
\[
\dim\Slika PA+\dim\Ker PA=\dim F^{n}=n
\]

\end_inset

dobili 
\begin_inset Formula $\dim\Slika PA=\dim\Slika A$
\end_inset

.
 Dokažimo torej 
\begin_inset Formula $\Ker PA=\Ker A$
\end_inset

:
\begin_inset Formula 
\[
\forall u:u\in\Ker PA\Leftrightarrow PAu=0\overset{P\text{ obrnljiva}}{\Longleftrightarrow}Au=0\Leftrightarrow u\in\Ker A.
\]

\end_inset

Torej je res 
\begin_inset Formula $\Ker PA=\Ker A$
\end_inset

,
 torej je res 
\begin_inset Formula $\rang PA=\rang A$
\end_inset

,
 torej je res 
\begin_inset Formula $\rang A=\rang B$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsubsection
Podobnost matrik
\end_layout

\begin_layout Definition*
Kvadratni matriki 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $B$
\end_inset

 sta podobni,
 če 
\begin_inset Formula $\exists$
\end_inset

 taka obrnljiva matrika 
\begin_inset Formula $P\ni:B=PAP^{-1}$
\end_inset

.ž
\end_layout

\begin_layout Claim*
Podobnost je ekvivalenčna relacija.
\end_layout

\begin_layout Proof
Dokazujemo,
 da je relacija ekvivalenčna,
 torej:
\end_layout

\begin_deeper
\begin_layout Itemize
refleksivna:
 
\begin_inset Formula $A=IAI^{-1}=IAI=A$
\end_inset


\end_layout

\begin_layout Itemize
simetrična:
 
\begin_inset Formula $B=PAP^{-1}\Rightarrow P^{-1}BP=A$
\end_inset


\end_layout

\begin_layout Itemize
tranzitivna:
 
\begin_inset Formula $B=PAP^{-1}\wedge C=QBQ^{-1}\Rightarrow C=QPAP^{-1}Q^{-1}=\left(QP\right)A\left(QP\right)^{-1}$
\end_inset


\end_layout

\end_deeper
\begin_layout Remark
\begin_inset CommandInset label
LatexCommand label
name "rem:nista-podobni"

\end_inset

Očitno velja podobnost 
\begin_inset Formula $\Rightarrow$
\end_inset

 ekvivalentnost,
 toda obrat ne velja vedno.
 Na primer 
\begin_inset Formula $\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right]$
\end_inset

 in 
\begin_inset Formula $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset

 sta ekvivalentni (sta enake velikosti in ranga),
 toda nista podobni (dokaz kasneje).
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Od prej vemo,
 da je vsaka matrika ekvivalentna matriki 
\begin_inset Formula $\left[\begin{array}{cc}
I_{r} & 0\\
0 & 0
\end{array}\right]$
\end_inset

,
 kjer je 
\begin_inset Formula $r$
\end_inset

 njen rang.
 A je vsaka kvadratna matrika podobna kakšni lepi matriki?
 Ja.
 Vsaka matrika je podobna zgornjetrikotni matriki in jordanski kanonični formi (več o tem kasneje).
 Toda a je vsaka kvadratna matrika podobna diagonalni matriki?
 Ne.
\end_layout

\begin_layout Definition*
Matrika 
\begin_inset Formula $D$
\end_inset

 je diagonalna 
\begin_inset Formula $\sim d_{ij}\not=0\Rightarrow i=j$
\end_inset

.
\end_layout

\begin_layout Standard
Kdaj je matrika 
\begin_inset Formula $A$
\end_inset

 podobna neki diagonalni matriki?
 Kdaj 
\begin_inset Formula $\exists$
\end_inset

 diagonalna 
\begin_inset Formula $D$
\end_inset

 in obrnljiva 
\begin_inset Formula $P\ni:A=PDP^{-1}$
\end_inset

?
 Izpeljimo iz nastavka.
 
\begin_inset Formula $D=\left[\begin{array}{ccc}
\lambda_{1} &  & 0\\
 & \ddots\\
0 &  & \lambda n
\end{array}\right]$
\end_inset

 in 
\begin_inset Formula $P=\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]$
\end_inset

,
 kjer sta 
\begin_inset Formula $D$
\end_inset

 in 
\begin_inset Formula $P$
\end_inset

 neznani.
 Ker mora biti 
\begin_inset Formula $P$
\end_inset

 obrnljiva,
 so njeni stolpični vektorji LN.
\begin_inset Formula 
\[
A=PDP^{-1}\Leftrightarrow AP=PD\Leftrightarrow A\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
\vec{v_{1}} & \cdots & \vec{v_{n}}\end{array}\right]\left[\begin{array}{ccc}
\lambda_{1} &  & 0\\
 & \ddots\\
0 &  & \lambda_{n}
\end{array}\right]\text{ in }P\text{ obrnljiva}
\]

\end_inset


\begin_inset Formula 
\[
\left[\begin{array}{ccc}
A\vec{v_{1}} & \cdots & A\vec{v_{n}}\end{array}\right]=\left[\begin{array}{ccc}
\lambda_{1}\vec{v_{1}} & \cdots & \lambda_{n}\vec{v_{n}}\end{array}\right]\text{ in }v_{i}\text{ so LN}
\]

\end_inset


\begin_inset Formula 
\[
A\vec{v_{1}}=\lambda_{1}\vec{v_{1}},\dots,A\vec{v_{n}}=\lambda_{n}v_{n}\text{ in }\forall i:v_{i}\not=0
\]

\end_inset


\end_layout

\begin_layout Standard
Porodi se naloga,
 imenovana 
\begin_inset Quotes gld
\end_inset

Lastni problem
\begin_inset Quotes grd
\end_inset

.
 Iščemo pare 
\begin_inset Formula $\left(\lambda,\vec{v}\right)$
\end_inset

,
 ki zadoščajo enačbi 
\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
\end_inset

.
 
\end_layout

\begin_layout Definition*
Pravimo,
 da je 
\begin_inset Formula $\lambda$
\end_inset

 je lastna vrednost matrike 
\begin_inset Formula $A$
\end_inset

,
 če obstaja tak 
\begin_inset Formula $\vec{v}\not=0$
\end_inset

,
 da je 
\begin_inset Formula $A\vec{v}=\lambda\vec{v}$
\end_inset

.
 V tem primeru pravimo,
 da je 
\begin_inset Formula $\vec{v}$
\end_inset

 lastni vektor,
 ki pripada lastni vrednosti 
\begin_inset Formula $\lambda$
\end_inset

.
 Paru 
\begin_inset Formula $\left(\lambda,\vec{v}\right)$
\end_inset

,
 ki zadošča enačbi,
 pravimo lastni par matrike 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Standard
Nalogo 
\begin_inset Quotes gld
\end_inset

Lastni problem
\begin_inset Quotes grd
\end_inset

 rešujemo v dveh korakih.
 Najprej najdemo vse 
\begin_inset Formula $\lambda$
\end_inset

,
 nato za vsako poiščemo pripadajoče 
\begin_inset Formula $\vec{v}$
\end_inset

,
 ki za lastno vrednost obstajajo po definiciji.
\end_layout

\begin_layout Standard
Za nek 
\begin_inset Formula $v\not=0$
\end_inset

 pišimo 
\begin_inset Formula $Av=\lambda v=\lambda Iv\Leftrightarrow Av-\lambda Iv=0\Leftrightarrow\left(A-\lambda I\right)v=0$
\end_inset

 za nek 
\begin_inset Formula $v\not=0\Leftrightarrow\Ker\left(A-\lambda I\right)\not=\left\{ 0\right\} \overset{\text{K.O.M.}}{\Longleftrightarrow}A-\lambda I$
\end_inset

 ni obrnljiva 
\begin_inset Formula $\Leftrightarrow\det\left(A-\lambda I\right)=0$
\end_inset

.
\end_layout

\begin_layout Definition*
Polinom 
\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-xI\right)$
\end_inset

 je karakteristični polinom matrike 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Definition*
Premislek zgoraj nam pove,
 da so lastne vrednosti 
\begin_inset Formula $A$
\end_inset

 ničle 
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset

.
\end_layout

\begin_layout Remark*
Karakteristični polinom lahko nima nobene ničle:
 
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]$
\end_inset

,
 
\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\det\left[\begin{array}{cc}
-\lambda & 1\\
-1 & -\lambda
\end{array}\right]=x^{2}+1$
\end_inset

,
 katerega ničli sta 
\begin_inset Formula $\lambda_{1}=i$
\end_inset

 in 
\begin_inset Formula $\lambda_{2}=-i$
\end_inset

,
 ki nista realni števili.
 V nadaljevanju se zato omejimo na kompleksne matrike in kompleksne lastne vrednosti,
 saj ima po Osnovnem izreku Algebre polinom s kompleksnimi koeficienti vedno vsaj kompleksne ničle.
\end_layout

\begin_layout Standard
Kako pa iščemo lastne vektorje za lastno vrednost 
\begin_inset Formula $\lambda$
\end_inset

?
 Spomnimo se na 
\begin_inset Formula $Av=\lambda v\Leftrightarrow v\in\Ker\left(A-\lambda I\right)$
\end_inset

.
 Rešiti moramo homogen sistem linearnih enačb.
 Po definiciji so lastni vektorji neničelni,
 zato nas trivialna rešitev ne zanima.
\end_layout

\begin_layout Definition*
Množici 
\begin_inset Formula $\Ker\left(A-\lambda I\right)$
\end_inset

 pravimo lastni podprostor matrike 
\begin_inset Formula $A$
\end_inset

,
 ki pripada 
\begin_inset Formula $\lambda$
\end_inset

.
 Slednji vsebuje 
\begin_inset Formula $\vec{0}$
\end_inset

 in množico vektorjev,
 ki so vsi lastni vektorji 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Exercise*
Izračunaj lastne vrednosti od 
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right]$
\end_inset

.
 Od prej vemo,
 da 
\begin_inset Formula $\lambda_{1}=i$
\end_inset

,
 
\begin_inset Formula $\lambda_{2}=-i$
\end_inset

.
 Izračunajmo 
\begin_inset Formula $\Ker\left(A-iI\right)$
\end_inset

 in 
\begin_inset Formula $\Ker\left(A+iI\right)$
\end_inset

:
\begin_inset Formula 
\[
\Ker\left(A-iI\right):\quad\left[\begin{array}{cc}
-i & 1\\
-1 & -i
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=0\quad\Longrightarrow\quad-ix+y=0,-x-iy=0\quad\Longrightarrow\quad y=ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
1\\
i
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
\Ker\left(A+iI\right):\quad\left[\begin{array}{cc}
i & 1\\
-1 & i
\end{array}\right]\left[\begin{array}{c}
x\\
y
\end{array}\right]=0\quad\Longrightarrow\quad ix+y=0,-x+y=0\quad\Longrightarrow\quad y=-ix\quad\Longrightarrow\quad v=x\left[\begin{array}{c}
1\\
-i
\end{array}\right]
\]

\end_inset


\begin_inset Formula 
\[
\Ker\left(A-iI\right)=\Lin\left\{ \left[\begin{array}{c}
1\\
i
\end{array}\right]\right\} ,\quad\Ker\left(A+iI\right)=\Lin\left\{ \left[\begin{array}{c}
1\\
-i
\end{array}\right]\right\} 
\]

\end_inset


\end_layout

\begin_layout Exercise*
Vstavimo lastna vektorja v 
\begin_inset Formula $P$
\end_inset

 in lastne vrednosti v 
\begin_inset Formula $D$
\end_inset

 na pripadajoči mesti.
 Dobimo obrnljivo 
\begin_inset Formula $P$
\end_inset

 in velja 
\begin_inset Formula $A=PDP^{-1}$
\end_inset


\begin_inset Formula 
\[
P=\left[\begin{array}{cc}
1 & 1\\
i & -i
\end{array}\right],\quad D=\left[\begin{array}{cc}
i & 0\\
0 & -i
\end{array}\right]
\]

\end_inset


\end_layout

\begin_layout Exercise*
Temu početju pravimo 
\begin_inset Quotes gld
\end_inset

diagonalizacija matrike 
\begin_inset Formula $A$
\end_inset


\begin_inset Quotes grd
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Primer matrike,
 ki ni diagonalizabilna:
 
\begin_inset Formula $A=\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset

.
 
\begin_inset Formula $\det\left(A-\lambda I\right)=\left[\begin{array}{cc}
-\lambda & 1\\
0 & -\lambda
\end{array}\right]=\lambda^{2}$
\end_inset

.
 Ničli/lastni vrednosti sta 
\begin_inset Formula $\lambda_{1}=0$
\end_inset

 in 
\begin_inset Formula $\lambda_{2}=0$
\end_inset

.
 Toda 
\begin_inset Formula $\Ker\left(A-0I\right)=\Ker A=\Lin\left\{ \left[\begin{array}{c}
1\\
0
\end{array}\right]\right\} $
\end_inset

 in 
\begin_inset Formula $P=\left[\begin{array}{cc}
1 & 1\\
0 & 0
\end{array}\right]$
\end_inset

 ni obrnljiva.
 S tem dokažemo trditev v primeru 
\begin_inset CommandInset ref
LatexCommand ref
reference "rem:nista-podobni"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
 
\begin_inset Formula $\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right]$
\end_inset

 in 
\begin_inset Formula $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$
\end_inset

 nista podobni,
 ker je prva diagonalna,
 druga pa ni podobna diagonalni matriki (ne da se je diagonalizirati).
\end_layout

\begin_layout Standard
Lastne vrednosti lahko definiramo tudi za linearne preslikave,
 saj so linearne preslikave linearno izomorfne matrikam.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $V$
\end_inset

 vektorski prostor nad 
\begin_inset Formula $F=\mathbb{C}$
\end_inset

 in 
\begin_inset Formula $L:V\to V$
\end_inset

 linearna preslikava.
 Število 
\begin_inset Formula $\lambda\in F$
\end_inset

 je lastna vrednost 
\begin_inset Formula $L$
\end_inset

,
 le obstaja tak neničelni 
\begin_inset Formula $v\in V$
\end_inset

,
 da velja 
\begin_inset Formula $Lv=\lambda v$
\end_inset

.
\end_layout

\begin_layout Standard
Kako pa rešujemo 
\begin_inset Quotes gld
\end_inset

Lastni problem
\begin_inset Quotes grd
\end_inset

 za linearne preslikave?
 
\begin_inset Formula $Lv=\lambda v\Leftrightarrow Lv-\lambda\left(id\right)v=0\Leftrightarrow\left(L-\lambda\left(id\right)\right)v=0\Leftrightarrow v\in\Ker\left(L-\lambda\left(id\right)\right)\overset{v\not=0}{\Longleftrightarrow}\det\left(L-\lambda\left(id\right)\right)=0$
\end_inset

.
 Toda determinante linearne preslikave nismo definirali.
 Lahko pa determinanto izračunamo na matriki,
 ki pripada tej linearni preslikavi.
 Toda dvem različnim bazam pripadata različni matriki linearne preslikave.
 Dokazati je treba,
 da sta determinanti dveh matrik,
 pripadajočih eni linearni preslikavi,
 enaki,
 četudi sta matriki v različnih bazah.
\end_layout

\begin_layout Lemma
\begin_inset CommandInset label
LatexCommand label
name "lem:Podobni-matriki-imata"

\end_inset

Podobni matriki imata isto determinanto.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $B=PAP^{-1}$
\end_inset

 za neko obrnljivo 
\begin_inset Formula $P$
\end_inset

.
 Tedaj 
\begin_inset Formula $\det B=\det PAP^{-1}=\det P\det A\det P^{-1}=\det P\det P^{-1}\det A=\det PP^{-1}\det A=\det I\det A=1\cdot\det A=\det A$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
\begin_inset Formula $L:V\to V$
\end_inset

 naj bo linearna preslikava,
 
\begin_inset Formula $V$
\end_inset

 prostor nad 
\begin_inset Formula $F=\mathbb{C}$
\end_inset

,
 
\begin_inset Formula $\mathcal{B}$
\end_inset

 in 
\begin_inset Formula $\mathcal{C}$
\end_inset

 pa bazi 
\begin_inset Formula $V$
\end_inset

.
 Priredimo matriki 
\begin_inset Formula $L_{\mathcal{B}\leftarrow\mathcal{B}}$
\end_inset

 in 
\begin_inset Formula $L_{\mathcal{C}\leftarrow\mathcal{C}}$
\end_inset

.
 Spomnimo se izreka 
\begin_inset CommandInset ref
LatexCommand vref
reference "thm:matrika-kompozituma-linearnih"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

:
 
\begin_inset Formula $\left[KL\right]_{\mathcal{D}\leftarrow\mathcal{B}}=\left[K\right]_{\mathcal{D}\leftarrow\mathcal{C}}\left[L\right]_{\mathcal{C}\leftarrow\mathcal{B}}$
\end_inset

.
 
\begin_inset Formula $L=\left[id\circ L\circ id\right]$
\end_inset

,
 zato 
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\circ L\circ id\right]_{\mathcal{C}\leftarrow\mathcal{C}}=\left[id\right]_{\mathcal{C}\leftarrow\mathcal{B}}\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}\left[id\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{C}}}}=P\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}P^{-1}$
\end_inset

 za neko obrnljivo 
\begin_inset Formula $P$
\end_inset

.
 Torej sta matriki 
\begin_inset Formula $\left[L\right]_{\mathcal{B}\leftarrow\text{\ensuremath{\mathcal{B}}}}$
\end_inset

 in 
\begin_inset Formula $\left[L\right]_{\mathcal{C}\leftarrow\mathcal{C}}$
\end_inset

 podobni,
 torej imata po lemi 
\begin_inset CommandInset ref
LatexCommand vref
reference "lem:Podobni-matriki-imata"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 isto determinanto.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
Alternativen dokaz,
 da imata podobni matriki iste lastne vrednosti:
 
\begin_inset Formula $A$
\end_inset

 podobna 
\begin_inset Formula 
\[
B\Rightarrow B=PAP^{-1}\Rightarrow B-xI=P\left(A-xI\right)P^{-1}\Rightarrow\det\left(B-xI\right)=\det\left(A-xI\right)\Rightarrow p_{A}=p_{B},
\]

\end_inset

torej so lastne vrednosti enake.
 Kaj pa lastni vektorji?
 Naj bo 
\begin_inset Formula $v$
\end_inset

 lastni vektor 
\begin_inset Formula $A$
\end_inset

,
 torej
\begin_inset Formula 
\[
Av=\lambda v\Rightarrow PAv=\lambda Pv\Rightarrow PAP^{-1}Pv=\lambda Pv\Rightarrow BPv=\lambda Pv,
\]

\end_inset

torej za 
\begin_inset Formula $v$
\end_inset

 lastni vektor 
\begin_inset Formula $A$
\end_inset

 sledi,
 da je 
\begin_inset Formula $Pv$
\end_inset

 lastni vektor 
\begin_inset Formula $B$
\end_inset

.
\end_layout

\begin_layout Standard
Linearni transformaciji torej priredimo tako matriko,
 ki ima v začetnem in končnem prostoru isto bazo.
 Tedaj lahko izračunamo lastne pare na tej matriki.
\end_layout

\begin_layout Theorem*
Schurov izrek.
 Vsaka kompleksna kvadratna matrika je podobna zgornjetrikotni matriki.
\end_layout

\begin_layout Proof
Indukcija po velikosti matrike.
\end_layout

\begin_deeper
\begin_layout Itemize
Baza:
 
\begin_inset Formula $A_{1\times1}$
\end_inset

 je zgornjetrikotna.
\end_layout

\begin_layout Itemize
Korak:
 Po I.
 P.
 trdimo,
 da je vsaka 
\begin_inset Formula $A_{\left(n-1\right)\times\left(n-1\right)}$
\end_inset

 podobna kaki zgornjetrikotni matriki.
 Dokažimo še za poljubno 
\begin_inset Formula $A_{n\times n}$
\end_inset

.
 Naj bo 
\begin_inset Formula $\lambda$
\end_inset

 lastna vrednost 
\begin_inset Formula $A$
\end_inset

 in 
\begin_inset Formula $v_{1}$
\end_inset

 pripadajoči lastni vektor ter 
\begin_inset Formula $v_{2},\dots,v_{n}$
\end_inset

 dopolnitev 
\begin_inset Formula $v_{1}$
\end_inset

 do baze 
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset

.
 Potem je matrika 
\begin_inset Formula $P=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]$
\end_inset

 obrnljiva.
\begin_inset Formula 
\[
AP=\left[\begin{array}{ccc}
Av_{1} & \cdots & Av_{n}\end{array}\right]=\left[\begin{array}{ccc}
v_{1} & \cdots & v_{n}\end{array}\right]\left[\begin{array}{cccc}
\lambda & a_{1,2} & \cdots & a_{1,n}\\
0 & \vdots &  & \vdots\\
\vdots & \vdots &  & \vdots\\
0 & a_{m,n} & \cdots & a_{m.n}
\end{array}\right]=P\left[\begin{array}{cc}
\lambda & B\\
0 & C
\end{array}\right]
\]

\end_inset

Po I.
 P.
 obstaja taka zgornjetrikotna 
\begin_inset Formula $T$
\end_inset

 in obrnljiva 
\begin_inset Formula $Q$
\end_inset

,
 da 
\begin_inset Formula $C=QTQ^{-1}$
\end_inset

.
\begin_inset Formula 
\[
\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]^{-1}P^{-1}AP\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]^{-1}\left[\begin{array}{cc}
\lambda & B\\
0 & C
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=
\]

\end_inset


\begin_inset Formula 
\[
=\left[\begin{array}{cc}
\lambda & C\\
0 & Q^{-1}B
\end{array}\right]\left[\begin{array}{cc}
1 & 0\\
0 & Q
\end{array}\right]=\left[\begin{array}{cc}
\lambda & CQ\\
0 & B
\end{array}\right]
\]

\end_inset


\begin_inset Formula $A$
\end_inset

 je torej podobna 
\begin_inset Formula $\left[\begin{array}{cc}
\lambda & CQ\\
0 & B
\end{array}\right]$
\end_inset

,
 ki je zgornjetrikotna.
\end_layout

\end_deeper
\begin_layout Proof
\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO karakterizacija linearnih preslikav 
\begin_inset Quotes gld
\end_inset

LA1V FMF 2024-03-12
\begin_inset Quotes grd
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Zadosten pogoj za diagonalizabilnost
\end_layout

\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:lave-razl-lavr-so-LN"

\end_inset

Lastni vektorji,
 ki pripadajo različnim lastnim vrednostim,
 so linearno neodvisni.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $A_{n\times n}$
\end_inset

 matrika,
 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 njene lastne vrednosti in 
\begin_inset Formula $v_{1},\dots,v_{k}$
\end_inset

 njim pripadajoči lastni vektorji.
 Dokazujemo 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 paroma različni 
\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
\end_inset

 LN.
 Dokaz z indukcijo po 
\begin_inset Formula $k$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Baza 
\begin_inset Formula $k=1$
\end_inset

:
 Elementi 
\begin_inset Formula $\left\{ \lambda_{1}\right\} $
\end_inset

 so trivialno paroma različni in 
\begin_inset Formula $v_{1}$
\end_inset

 je kot neničen vektor LN.
\end_layout

\begin_layout Itemize
Korak:
 Dokazujemo 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k+1}$
\end_inset

 so paroma različne 
\begin_inset Formula $\Rightarrow v_{1},\dots,v_{k}$
\end_inset

 so LN,
 vedoč I.
 P.
 Denimo,
 da 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
\end_inset

.
 Množimo z 
\begin_inset Formula $A$
\end_inset

:
\begin_inset Formula 
\[
A\left(\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}\right)=\alpha_{1}Av_{1}+\cdots+\alpha_{k+1}Av_{k+1}=\alpha_{1}\lambda_{1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
\]

\end_inset

Množimo začetno enačbo z 
\begin_inset Formula $\lambda_{k+1}$
\end_inset

 (namesto z 
\begin_inset Formula $A$
\end_inset

,
 kot smo to storili zgoraj):
\begin_inset Formula 
\[
\alpha_{1}\lambda_{k+1}v_{1}+\cdots+\alpha_{k+1}\lambda_{k+1}v_{k+1}=0
\]

\end_inset

Odštejmo eno enačbo od druge,
 dobiti moramo 0,
 saj odštevamo 0 od 0:
\begin_inset Formula 
\[
\alpha_{1}\left(\lambda_{1}-\lambda_{k+1}\right)v_{1}+\cdots+\alpha_{k}\left(\lambda_{k}-\lambda_{k+1}\right)v_{k}+\cancel{\alpha_{k+1}\left(\lambda_{k+1}-\lambda_{k+1}\right)v_{k+1}}=0
\]

\end_inset

Ker so lastne vrednosti paroma različne (
\begin_inset Formula $\lambda_{i}=\lambda_{j}\Rightarrow i=j$
\end_inset

),
 so njihove razlike neničelne.
 Ker so 
\begin_inset Formula $v_{1},\dots,v_{k}$
\end_inset

 po predpostavki LN,
 sledi 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=0$
\end_inset

.
 Vstavimo te konstante v 
\begin_inset Formula $\alpha_{1}v_{1}+\cdots+\alpha_{k+1}v_{k+1}=0$
\end_inset

 in dobimo 
\begin_inset Formula $\alpha_{k+1}v_{k+1}=0$
\end_inset

.
 Ker je 
\begin_inset Formula $v_{k+1}$
\end_inset

 neničeln (je namreč lastni vektor),
 sledi 
\begin_inset Formula $\alpha_{k+1}=0$
\end_inset

,
 torej 
\begin_inset Formula $\alpha_{1}=\cdots=\alpha_{k}=\alpha_{k+1}=0$
\end_inset

,
 zatorej so 
\begin_inset Formula $v_{1},\dots,v_{k+1}$
\end_inset

 res LN.
\end_layout

\end_deeper
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:vsota-lastnih-podpr-direktna"

\end_inset

Vsota vseh lastnih podprostorov matrike je direktna (definicija 
\begin_inset CommandInset ref
LatexCommand vref
reference "def:vsota-je-direktna"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

).
\end_layout

\begin_layout Proof
Naj bodo 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 vse paroma različne lastne vrednosti matrike 
\begin_inset Formula $A\in M_{n}\left(\mathbb{C}\right)$
\end_inset

.
 Pripadajoči lastni podprostori so torej 
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :V_{i}=\Ker\left(A-\lambda_{i}I\right)$
\end_inset

.
 Trdimo,
 da je vsota teh podprostorov direktna,
 torej 
\begin_inset Formula $\forall v_{1}\in V_{1},\dots,v_{k}\in V_{k}:v_{1}+\cdots+v_{k}=0\Rightarrow v_{1}=\cdots=v_{k}=0$
\end_inset

.
 To sledi iz izreka 
\begin_inset CommandInset ref
LatexCommand vref
reference "thm:lave-razl-lavr-so-LN"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Corollary*
Če ima 
\begin_inset Formula $n\times n$
\end_inset

 matrika 
\begin_inset Formula $n$
\end_inset

 paroma različnih lastnih vrednosti,
 je podobna diagonalni matriki.
\end_layout

\begin_layout Proof
Po posledici 
\begin_inset CommandInset ref
LatexCommand vref
reference "cor:vsota-lastnih-podpr-direktna"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 je vsota lastnih podprostorov matrike 
\begin_inset Formula $A_{n\times n}$
\end_inset

 direktna.
 Če je torej lastnih podprostorov 
\begin_inset Formula $n$
\end_inset

,
 je njihova vsota cel prostor 
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset

.
 Matriko se da diagonalizirati,
 kadar je vsota vseh lastnih podprostorov enaka podprostoru 
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset

 (tedaj so namreč stolpci matrike 
\begin_inset Formula $P$
\end_inset

 linearno neodvisni,
 zato je 
\begin_inset Formula $P$
\end_inset

 obrnljiva).
\end_layout

\begin_layout Subsubsection
Algebraične in geometrijske vekčratnosti
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $A_{n\times n}$
\end_inset

 matrika.
 
\begin_inset Formula $p_{A}\left(\lambda\right)=\det\left(A-\lambda I\right)=\left(-1\right)^{n}\left(\lambda-\lambda_{1}\right)^{n_{1}}\cdots\left(\lambda-\lambda_{k}\right)^{n_{k}}$
\end_inset

,
 kjer so 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 vse paroma različne lastne vrednosti 
\begin_inset Formula $A$
\end_inset

.
 Stopnji ničle —
 
\begin_inset Formula $n_{i}$
\end_inset

 —
 rečemo algebraična večkratnost lastne vrednosti 
\begin_inset Formula $\lambda_{i}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Geometrijska večkratnost lastne vrednosti 
\begin_inset Formula $\lambda_{i}$
\end_inset

 je 
\begin_inset Formula $\dim\Ker\left(A-\lambda_{i}I\right)=\n$
\end_inset


\begin_inset Formula $\left(A-\lambda_{i}I\right)=m_{i}$
\end_inset

.
\end_layout

\begin_layout Standard
Algebraično večkratnost 
\begin_inset Formula $\lambda_{i}$
\end_inset

 označimo z 
\begin_inset Formula $n_{i}$
\end_inset

 in je večkratnost ničle 
\begin_inset Formula $\lambda_{i}$
\end_inset

 v 
\begin_inset Formula $p_{A}\left(\lambda\right)$
\end_inset

 (karakterističnem polinomu).
 Geometrijsko večkratnost 
\begin_inset Formula $\lambda_{i}$
\end_inset

 pa označimo z 
\begin_inset Formula $m_{i}$
\end_inset

 in je dimenzija lastnega podprostora za 
\begin_inset Formula $\lambda_{i}$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
\end_inset

 —
 geometrijska večkratnost lastne vrednosti je kvečjemu tolikšna,
 kot je algebraična večkratnost te lastne vrednosti.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $v_{1},\dots,v_{m_{i}}$
\end_inset

 baza za lastni podprostor 
\begin_inset Formula $V_{i}=\Ker\left(A-\lambda_{i}I\right)$
\end_inset

 in naj bo 
\begin_inset Formula $v_{m_{i}+1},\dots,v_{n}$
\end_inset

 njena dopolnitev do baze 
\begin_inset Formula $\mathbb{C}^{n}$
\end_inset

.
 Tedaj velja:
 
\begin_inset Formula $Av_{1}=\lambda_{1}v_{1}$
\end_inset

,
 ...,
 
\begin_inset Formula $Av_{m_{i}}=\lambda_{m_{i}}v_{m_{i}}$
\end_inset

,
 
\begin_inset Formula $Av_{m_{i}+1}=$
\end_inset

 linearna kombinacija 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

,
 ...,
 
\begin_inset Formula $Av_{n}=$
\end_inset

 linearna kombinacija 
\begin_inset Formula $v_{1},\dots,v_{n}$
\end_inset

.
 Naj bo 
\begin_inset Formula $P=\left[\begin{array}{cccccc}
v_{1} & \cdots & v_{m_{i}} & v_{m_{i}+1} & \cdots & v_{n}\end{array}\right]$
\end_inset

,
 ki je obrnljiva.
\begin_inset Formula 
\[
P^{-1}AP=\cdots=\left[\begin{array}{cc}
\lambda_{i}I_{m_{i}} & B\\
0 & C
\end{array}\right]
\]

\end_inset


\series bold
Dokaza ne razumem.
 Obupam.
\end_layout

\begin_layout Claim*
Matriko s paroma različnimi lastnimi vrednostmi 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 je moč diagonalizirati 
\begin_inset Formula $\Leftrightarrow\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
\end_inset

.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $V_{i}$
\end_inset

 lastno podprostor lastne vrednosti 
\begin_inset Formula $\lambda_{i}$
\end_inset

.
 Vemo,
 da se da 
\begin_inset Formula $A_{n\times n}$
\end_inset

 diagonalizirati 
\begin_inset Formula $\Leftrightarrow A$
\end_inset

 ima 
\begin_inset Formula $n$
\end_inset

 LN stolpičnih vektorjev 
\begin_inset Formula $\Leftrightarrow\Ker\left(A-\lambda_{1}I\right)+\cdots+\Ker\left(A-\lambda_{k}I\right)=\mathbb{C}^{n}\Leftrightarrow\dim\left(V_{i}+\cdots+V_{k}\right)=\dim V_{i}+\cdots+\dim V_{k}\Leftrightarrow$
\end_inset

 vsota lastnih podprostorov je direktna 
\begin_inset Formula $\Leftrightarrow\dim\left(V_{1}+\cdots+V_{n}\right)=n\Leftrightarrow\dim V_{1}+\cdots+\dim V_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n\Leftrightarrow m_{1}+\cdots+m_{k}=n_{1}+\cdots+n_{m}$
\end_inset

.
 Toda ker po prejšnjem izreku 
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}\leq n_{i}$
\end_inset

,
 mora veljati 
\begin_inset Formula $\forall i\in\left\{ 1..k\right\} :m_{i}=n_{i}$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Minimalni polinom matrike
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+\cdots+c_{n}x^{n}\in\mathbb{C}\left[x\right]$
\end_inset

 polinom in 
\begin_inset Formula $A$
\end_inset

 matrika.
 
\begin_inset Formula $p\left(A\right)\coloneqq c_{0}A^{0}+\cdots+c_{n}A^{n}=c_{0}I+\cdots+c_{n}A^{n}$
\end_inset

.
 Če je 
\begin_inset Formula $p\left(A\right)=0$
\end_inset

 (ničelna matrika),
 pravimo,
 da polinom 
\begin_inset Formula $p$
\end_inset

 anhilira/uniči matriko 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Fact*
\begin_inset Formula $p\left(A\right)=0\Rightarrow p\left(P^{-1}AP\right)=0$
\end_inset

.
\end_layout

\begin_layout Standard
Izkaže se,
 da karakteristični polimom anhilira matriko —
 
\begin_inset Formula $p_{A}\left(A\right)=0$
\end_inset

.
 Dokaz kasneje.
\end_layout

\begin_layout Definition*
Polinom 
\begin_inset Formula $m\left(x\right)$
\end_inset

 je minimalen polinom 
\begin_inset Formula $A$
\end_inset

,
 če velja:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $m\left(A\right)=0$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $m$
\end_inset

 ima vodilni koeficient 1
\end_layout

\begin_layout Enumerate
med vsemi polinomi,
 ki zadoščajo prvi in drugi zahtevi,
 ima 
\begin_inset Formula $m$
\end_inset

 najnižjo stopnjo
\end_layout

\end_deeper
\begin_layout Claim*
eksistenca minimalnega polinoma —
 Minimalni polinom obstaja.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $A_{n\times n}$
\end_inset

 matrika.
 Očitno je 
\begin_inset Formula $M_{n\times n}\left(\mathbb{C}\right)$
\end_inset

 vektorski prostor dimenzije 
\begin_inset Formula $n^{2}$
\end_inset

.
 Matrike 
\begin_inset Formula $\left\{ I,A,A^{2},\dots,A^{n^{2}}\right\} $
\end_inset

 so linearno odvisne,
 ker je moč te množice za 1 večja od moči vektorskega prostora.
 Torej 
\begin_inset Formula $\exists c_{0},\cdots,c_{n^{2}}\in\mathbb{C}$
\end_inset

,
 ki niso vse 0 
\begin_inset Formula $\ni:c_{0}I+c_{1}A+c_{2}A^{2}+\cdots+c_{n^{2}}A^{n^{2}}=0$
\end_inset

.
 Torej polinom 
\begin_inset Formula $p\left(x\right)=c_{0}x^{0}+c_{1}x^{1}+c_{2}x^{2}+\cdots+c_{n^{2}}x^{n^{2}}$
\end_inset

 anhilira 
\begin_inset Formula $A$
\end_inset

.
 Če ta polinom delimo z njegovim vodilnim koeficientom,
 dobimo polinom,
 ki ustreza prvima dvema zahevama za minimalni polinom.
 Če med vsemi takimi izberemo takega z najnižjo stopnjo,
 le-ta ustreza še tretji zahtevi.
\end_layout

\begin_layout Theorem*
Če je 
\begin_inset Formula $m\left(x\right)$
\end_inset

 minimalni polinom za 
\begin_inset Formula $A$
\end_inset

 in če 
\begin_inset Formula $p\left(x\right)$
\end_inset

 anhilira 
\begin_inset Formula $A$
\end_inset

,
 potem 
\begin_inset Formula $m\left(x\right)\vert p\left(x\right)$
\end_inset

 (
\begin_inset Formula $m\left(x\right)$
\end_inset

 deli 
\begin_inset Formula $p\left(x\right)$
\end_inset

).
\end_layout

\begin_layout Proof
Delimo 
\begin_inset Formula $p$
\end_inset

 z 
\begin_inset Formula $m$
\end_inset

:
 
\begin_inset Formula $\exists k\left(x\right),r\left(x\right)\ni:p\left(x\right)=k\left(x\right)m\left(x\right)+r\left(x\right)\wedge\deg r\left(x\right)<\deg m\left(x\right)$
\end_inset

.
 Vstavimo 
\begin_inset Formula $A$
\end_inset

 na obe strani:
\begin_inset Formula 
\[
0=p\left(A\right)=k\left(A\right)m\left(A\right)+r\left(A\right)=k\left(A\right)\cdot0+r\left(A\right)=0+r\left(A\right)=r\left(A\right)=0
\]

\end_inset

Sledi 
\begin_inset Formula $r\left(x\right)=0$
\end_inset

,
 kajti če 
\begin_inset Formula $r$
\end_inset

 ne bi bil ničeln polinom,
 bi ga lahko delili z vodilnim koeficientom in po predpostavki 
\begin_inset Formula $\deg r\left(x\right)<\deg m\left(x\right)$
\end_inset

 bi imel manjšo stopnjo kot 
\begin_inset Formula $m\left(x\right)$
\end_inset

,
 torej bi ustrezal zahtevam 1 in 2 za minimalni polinom in bi imel manjšo stopnjo od 
\begin_inset Formula $m$
\end_inset

,
 torej 
\begin_inset Formula $m$
\end_inset

 ne bi bil minimalni polinom,
 kar bi vodilo v protislovje.
\end_layout

\begin_layout Corollary*
enoličnost minimalnega polinoma.
 Naj bosta 
\begin_inset Formula $m_{1}$
\end_inset

 in 
\begin_inset Formula $m_{2}$
\end_inset

 minimalna polinoma matrike 
\begin_inset Formula $A$
\end_inset

.
 Ker 
\begin_inset Formula $m$
\end_inset

 po definiciji anhilira 
\begin_inset Formula $A$
\end_inset

,
 iz prejšnje trditve sledi,
 če vstavimo 
\begin_inset Formula $m=m_{1}$
\end_inset

 in 
\begin_inset Formula $p=m_{2}$
\end_inset

,
 
\begin_inset Formula $m_{1}\vert m_{2}$
\end_inset

.
 Toda če vstavimo 
\begin_inset Formula $m=m_{2}$
\end_inset

 in 
\begin_inset Formula $p=m_{1}$
\end_inset

,
 
\begin_inset Formula $m_{2}\vert m_{1}$
\end_inset

.
 Iz 
\begin_inset Formula $m_{1}\vert m_{2}\wedge m_{2}\vert m_{1}$
\end_inset

 sledi,
 da se 
\begin_inset Formula $m_{1}$
\end_inset

 in 
\begin_inset Formula $m_{2}$
\end_inset

 razlikujeta le za konstanten faktor,
 ki pa je po definiciji minimalnega polinoma 1,
 torej 
\begin_inset Formula $m_{1}=m_{2}$
\end_inset

.
 Zaradi enoličnosti lahko označimo minimalni polinom 
\begin_inset Formula $A$
\end_inset

 z 
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Ničle minimalnega polinoma
\end_layout

\begin_layout Claim*
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset

 in 
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset

 imata iste ničle 
\begin_inset Formula $\sim$
\end_inset

 ničle 
\begin_inset Formula $m_{A}\left(x\right)$
\end_inset

 so lastne vrednosti 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Proof
Ker je 
\begin_inset Formula $p_{A}\left(x\right)$
\end_inset

 (dokaz kasneje),
 velja po trditvi v dokazu enoličnosti,
 da 
\begin_inset Formula $m_{A}\vert p_{A}$
\end_inset

,
 torej je vsaka ničla 
\begin_inset Formula $m_{A}$
\end_inset

 tudi ničla 
\begin_inset Formula $p_{A}$
\end_inset

.
 Treba je dokazati še,
 da je vsaka ničla 
\begin_inset Formula $p_{A}$
\end_inset

 tudi ničla 
\begin_inset Formula $m_{A}$
\end_inset

,
 natančneje:
 Treba je dokazati,
 da če je 
\begin_inset Formula $\lambda$
\end_inset

 lastna vrednost matrike 
\begin_inset Formula $A$
\end_inset

,
 je 
\begin_inset Formula $m_{A}\left(\lambda\right)=0$
\end_inset

.
 Naj bo 
\begin_inset Formula $v\not=0$
\end_inset

 lastni vektor za 
\begin_inset Formula $\lambda$
\end_inset

.
 Tedaj 
\begin_inset Formula $Av=\lambda v$
\end_inset

.
 Potem velja 
\begin_inset Formula $A^{2}v=AAv=A\lambda v=\lambda Av=\lambda\lambda v=\lambda^{2}v$
\end_inset

 in splošneje 
\begin_inset Formula $A^{n}v=\lambda^{n}v$
\end_inset

.
 Sedaj recimo,
 da je 
\begin_inset Formula $m_{A}\left(x\right)=d_{0}x^{0}+\cdots+d_{r}x^{r}$
\end_inset

.
 Potem je,
 ker minimalni polinom anhilira 
\begin_inset Formula $A$
\end_inset

,
 
\begin_inset Formula 
\[
m_{A}\left(\lambda\right)v=\left(d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+\cdots+d_{r}\lambda^{r}\right)v=d_{0}v+d_{1}\lambda v+d_{2}\lambda^{2}v+\cdots+d_{r}\lambda^{r}v=
\]

\end_inset


\begin_inset Formula 
\[
=d_{0}v+d_{1}Av+d_{2}A^{2}v+\cdots+d_{r}A^{r}v=\left(d_{0}+d_{1}A+d_{2}A^{2}+\cdots+d_{r}A^{r}\right)v=m_{A}\left(A\right)v=0v=0
\]

\end_inset

Ker 
\begin_inset Formula $m_{A}\left(\lambda\right)v=0$
\end_inset

 in 
\begin_inset Formula $v\not=0$
\end_inset

 (je namreč lastni vektor),
 velja 
\begin_inset Formula $m_{A}\left(\lambda\right)=0$
\end_inset

.
\end_layout

\begin_layout Paragraph*
Lastnosti
\end_layout

\begin_layout Standard
Ker je 
\begin_inset Formula $p_{A}\left(x\right)=\left(-1\right)^{n}\left(x-\lambda_{1}\right)^{n_{1}}\cdots\left(x-\lambda_{k}\right)^{n_{k}}$
\end_inset

 in 
\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{r_{1}}\cdots\left(x-\lambda_{k}\right)^{r_{k}}$
\end_inset

,
 sledi iz 
\begin_inset Formula $m_{A}\vert p_{A}\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\leq n_{i}$
\end_inset

.
 Poleg tega,
 ker 
\begin_inset Formula $m_{A}\left(\lambda_{1}\right)=0\Rightarrow\forall i\in\left\{ 1..k\right\} :r_{i}\geq1$
\end_inset

.
 Toda pozor:
 
\series bold
Ni
\series default
 res,
 da 
\begin_inset Formula $r_{i}=m_{i}$
\end_inset

 (stropnja lastnega podprostora).
\end_layout

\begin_layout Theorem*
Cayley-Hamilton.
 
\begin_inset Formula $p_{A}\left(A\right)=0$
\end_inset

 —
 karakteristični polinom matrike 
\begin_inset Formula $A$
\end_inset

 anhilira matriko 
\begin_inset Formula $A$
\end_inset


\end_layout

\begin_layout Proof
Spomnimo se eksplicitne formule za celico inverza matrike (razdelek 
\begin_inset CommandInset ref
LatexCommand vref
reference "subsec:Formula-za-inverz-matrike"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

),
 ki pravi 
\begin_inset Formula $B^{-1}=\frac{1}{\det B}\tilde{B}^{T}$
\end_inset

.
 Računajmo in naposled vstavimo 
\begin_inset Formula $B=A-xI$
\end_inset

:
\begin_inset Formula 
\[
B^{-1}=\frac{1}{\det B}\tilde{B}^{T}\quad\quad\quad\quad/\cdot\left(\det B\right)B
\]

\end_inset


\begin_inset Formula 
\[
\det B\cdot I=B\tilde{B}^{T}
\]

\end_inset


\begin_inset Formula 
\[
\det\left(A-xI\right)\cdot I=p_{A}\left(x\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
\]

\end_inset

Glede na definicijo 
\begin_inset Formula $\tilde{A}$
\end_inset

 je 
\begin_inset Formula $\tilde{\left(A-xI\right)}^{T}$
\end_inset

 matrika velikosti 
\begin_inset Formula $n\times n$
\end_inset

,
 ki vsebuje polinome stopnje 
\begin_inset Formula $<n$
\end_inset

,
 kajti vsebuje determinante matrik,
 katerih elementi so polinomi stopnje 
\begin_inset Formula $\leq1$
\end_inset

,
 torej takele oblike:
\begin_inset Formula 
\[
\tilde{\left(A-xI\right)}^{T}=B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}
\]

\end_inset


\begin_inset Foot
status open

\begin_layout Plain Layout
Ne razumem,
 zakaj so tu matrike 
\begin_inset Formula $B$
\end_inset

 in ne skalarji.
\end_layout

\end_inset


\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $p_{A}\left(x\right)=\det\left(A-\lambda I\right)=c_{0}+c_{1}x+\cdots+c_{n}x^{n}$
\end_inset

.
 Kot v enačbi množimo to z 
\begin_inset Formula $I$
\end_inset

:
 
\begin_inset Formula $\det\left(A-\lambda I\right)\cdot I=c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}$
\end_inset

.
 Oglejmo si še desno stran enačbe:
\begin_inset Formula 
\[
\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}=\left(A-xI\right)\left(B_{0}+B_{1}x+\cdots+B_{n-1}x^{n-1}\right)=AB_{0}+AB_{1}x+\cdots+AB_{n-1}x^{n-1}-B_{0}x-B_{1}x^{2}-\cdots-B_{n-1}x^{n}=
\]

\end_inset


\begin_inset Formula 
\[
=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}x^{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
\]

\end_inset

In primerjajmo koeficiente v polinomih pred istoležnimi spremenljivkami na obeh straneh tele enačbe:
\begin_inset Formula 
\[
\det\left(A-xI\right)\cdot I=\left(A-xI\right)\tilde{\left(A-xI\right)}^{T}
\]

\end_inset


\begin_inset Formula 
\[
c_{0}I+c_{1}Ix+\cdots+c_{n}Ix^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)x+\left(AB_{2}-B_{1}\right)x^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)x^{n-1}-B_{n-1}x^{n}
\]

\end_inset


\begin_inset Formula 
\[
\begin{array}{cccc}
1: & c_{0}I & = & AB_{0}\\
x: & c_{1}I & = & AB_{1}-B_{0}\\
x^{2}: & c_{2}I & = & AB_{2}x^{2}-B_{1}\\
\vdots\\
x^{n-1}: & c_{n-1}I & = & AB_{n-1}-B_{n-2}\\
x^{n}: & c_{n}I & = & -B_{n-1}
\end{array}
\]

\end_inset

Vstavimo sedaj 
\begin_inset Formula $A$
\end_inset

 v enačbo namesto 
\begin_inset Formula $x$
\end_inset

:
\begin_inset Formula 
\[
p_{A}\left(A\right)=c_{0}I+c_{1}IA+\cdots+c_{n}IA^{n}=AB_{0}+\left(AB_{1}-B_{0}\right)A+\left(AB_{2}-B_{1}\right)A^{2}+\cdots+\left(AB_{n-1}-B_{n-2}\right)A^{n-1}-B_{n-1}A^{n}=
\]

\end_inset


\begin_inset Formula 
\[
=AB_{0}+A^{2}B_{1}-AB_{0}+A^{2}B^{2}-B_{1}A^{2}+\cdots+A^{n}B_{n-1}-A^{n-1}B_{n-2}-B_{n-1}A^{n}=0
\]

\end_inset


\begin_inset Formula 
\[
p_{A}\left(A\right)=0
\]

\end_inset


\end_layout

\begin_layout Theorem*
Matriko 
\begin_inset Formula $A$
\end_inset

 se da diagonalizirati 
\begin_inset Formula $\Leftrightarrow m_{A}\left(x\right)$
\end_inset

 ima samo enostavne ničle (nima večkratnih —
 potence so vse 1).
 Torej 
\begin_inset Formula $m_{A}\left(x\right)=\left(x-\lambda_{1}\right)^{1}\cdots\left(x-\lambda_{k}\right)^{1}$
\end_inset

 za 
\begin_inset Formula $\lambda_{1},\dots,\lambda_{k}$
\end_inset

 vse paroma različne lastne vrednosti 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Po predpostavki je 
\begin_inset Formula $A$
\end_inset

 podobna diagonalni matriki —
 
\begin_inset Formula $A=PDP^{-1}$
\end_inset

 za diagonalno 
\begin_inset Formula $D$
\end_inset

 in obrnljivo 
\begin_inset Formula $P$
\end_inset

.
 BSŠ naj bo 
\begin_inset Formula $D=\left[\begin{array}{ccc}
\lambda_{1} & 0 & 0\\
0 & \cdots & 0\\
0 & 0 & \lambda_{k}
\end{array}\right]$
\end_inset

 in 
\begin_inset Formula $\lambda_{1}\leq\cdots\leq\lambda_{k}$
\end_inset

.
 NADALJUJ TULE AAAA 
\begin_inset Quotes gld
\end_inset

LA1P FMF 2024-03-20
\begin_inset Quotes grd
\end_inset

 stran 2.
\end_layout

\end_deeper
\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