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| -rw-r--r-- | šola/la/teor.lyx | 2218 |
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 |