1 files changed, 2183 insertions, 35 deletions

diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx
index d84d320..ae7dcaf 100644
--- a/šola/la/teor.lyx
+++ b/šola/la/teor.lyx
@@ -27,8 +27,11 @@
 \DeclareMathOperator{\red}{red}
 \DeclareMathOperator{\karakteristika}{char}
 \DeclareMathOperator{\Ker}{Ker}
+\DeclareMathOperator{\Slika}{Ker}
 \DeclareMathOperator{\sgn}{sgn}
 \DeclareMathOperator{\End}{End}
+\DeclareMathOperator{\n}{n}
+\DeclareMathOperator{\Col}{Col}
 \usepackage{algorithm,algpseudocode}
 \providecommand{\corollaryname}{Posledica}
 \end_preamble
@@ -5573,7 +5576,7 @@ kjer
 \end_layout
 
 \begin_layout Subsection
-Algenrske strukture
+Algebrske strukture
 \end_layout
 
 \begin_layout Subsubsection
@@ -9785,38 +9788,15 @@ LN:
 \end_inset
 
 .
-\begin_inset Foot
-status open
-
-\begin_layout Plain Layout
-Ta dokaz mi ni povsem jasen.
- Zakaj je potrebno preverjati zgolj za 
-\begin_inset Formula $\alpha_{1},\dots,\alpha_{n}$
-\end_inset
-
-,
- ne pa za vse elemente polja 
-\begin_inset Formula $F$
+ Preverjati je treba le 
+\begin_inset Formula $\alpha_{i}$
 \end_inset
 
 ,
- torej tudi tiste,
- ki niso v množici naših 
-\begin_inset Formula $\left\{ \alpha_{1},\dots,\alpha_{n}\right\} $
-\end_inset
-
-.
- Če mi bralec zna razložiti,
- naj mi piše na 
-\family typewriter
-anton@sijanec.eu
-\family default
-.
-\end_layout
-
-\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
@@ -9836,16 +9816,34 @@ ogrodje:
 \end_inset
 
  različnih točkah.
-\begin_inset Foot
-status open
+ Če za 
+\begin_inset Formula $x$
+\end_inset
 
-\begin_layout Plain Layout
-Niti tega ne razumem.
-\end_layout
+ 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
@@ -11514,6 +11512,2156 @@ v_{1} & \cdots & v_{n}\end{array}\right]$
 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
+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 povesem 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 Part
 Vaja za ustni izpit
 \end_layout