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diff --git a/šola/la/teor.lyx b/šola/la/teor.lyx index 440fc84..d84d320 100644 --- a/šola/la/teor.lyx +++ b/šola/la/teor.lyx @@ -27,6 +27,8 @@ \DeclareMathOperator{\red}{red} \DeclareMathOperator{\karakteristika}{char} \DeclareMathOperator{\Ker}{Ker} +\DeclareMathOperator{\sgn}{sgn} +\DeclareMathOperator{\End}{End} \usepackage{algorithm,algpseudocode} \providecommand{\corollaryname}{Posledica} \end_preamble @@ -1675,9 +1677,9839 @@ Označimo \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 +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 +\[ +X=A^{-1}=\left[\begin{array}{ccc} +\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A}\\ +\vdots & & \vdots\\ +\frac{\det A_{1n}\cdot\left(-1\right)^{1+n}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A} +\end{array}\right]=\frac{1}{\det A}\left[\begin{array}{ccc} +\frac{\det A_{11}\cdot\left(-1\right)^{1+1}}{\det A} & \cdots & \frac{\det A_{1n}\cdot\left(-1\right)^{1+1}}{\det A}\\ +\vdots & & \vdots\\ +\frac{\det A_{n1}\cdot\left(-1\right)^{n+1}}{\det A} & \cdots & \frac{\det A_{nn}\cdot\left(-1\right)^{n+n}}{\det A} +\end{array}\right]^{T}=\frac{1}{\det A}\tilde{A}^{T}, +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +kjer +\begin_inset Formula $\tilde{A}$ +\end_inset + + pravimo kofaktorska matrika. +\end_layout + +\begin_layout Subsection +Algenrske 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 + +. +\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 $F$ +\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 + + +\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. +\begin_inset Foot +status open + +\begin_layout Plain Layout +Niti tega ne razumem. +\end_layout + +\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* +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 + +. +\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 |