diff --git a/beviser.lyx b/beviser.lyx index ca49d6b..cf0d2bd 100644 --- a/beviser.lyx +++ b/beviser.lyx @@ -946,5 +946,1308 @@ A & =A\cdot{\rm I}_{n}\\ Beviset er nu afsluttet. \end_layout +\begin_layout Standard +\begin_inset Newpage newpage +\end_inset + + +\end_layout + +\begin_layout Section +Vektorrum, underrum og dimension +\end_layout + +\begin_layout Subsection +Definition 5.1 (Vektorrum) +\end_layout + +\begin_layout Standard +En mængde +\begin_inset Formula $V$ +\end_inset + +samt to afbildninger +\begin_inset Formula $\mathcal{A}$ +\end_inset + + og +\begin_inset Formula $\mathcal{S}$ +\end_inset + +, der opfylder identiteterne +\end_layout + +\begin_layout Enumerate +Den +\emph on +kommutative +\emph default + lov +\end_layout + +\begin_layout Enumerate +Den +\emph on +associative +\emph default + lov +\end_layout + +\begin_layout Enumerate +Eksistens af +\emph on +neutralelement +\end_layout + +\begin_layout Enumerate +Eksistens af +\emph on +inverse +\emph default + elementer +\end_layout + +\begin_layout Enumerate +Distributiv lov 1 +\end_layout + +\begin_layout Enumerate +Distributiv lov 2 +\end_layout + +\begin_layout Enumerate +Associativitet for skalarmultiplikation ( +\begin_inset Formula $\alpha\cdot(\beta\cdot\boldsymbol{u})=(\alpha\beta)\cdot\boldsymbol{u}$ +\end_inset + +) +\end_layout + +\begin_layout Subsection +Definition 5.7 (Underrum) +\end_layout + +\begin_layout Standard +Et underrum af et vektorrum +\begin_inset Formula $V$ +\end_inset + + er +\emph on +en delmængde +\emph default + +\begin_inset Formula $S$ +\end_inset + + af vektorrummet +\begin_inset Formula $V$ +\end_inset + +, der opfylder følgende betingelser: +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\boldsymbol{0}\in S$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall\boldsymbol{u},\boldsymbol{v}\in S:\quad\boldsymbol{u}+\boldsymbol{v}\in S$ +\end_inset + + +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall\boldsymbol{u}\in S,\:\alpha\in\mathbb{F}:\quad\alpha\cdot\boldsymbol{u}\in S$ +\end_inset + + +\end_layout + +\begin_layout Standard +Punkt 2 og 3 kaldes +\emph on +stabilitet overfor addition +\emph default + hhv. + +\emph on +multiplikation +\emph default + +\end_layout + +\begin_layout Subsection +Definition 5.9 (Linearkombination) +\end_layout + +\begin_layout Standard +En vektor +\begin_inset Formula $\boldsymbol{v}$ +\end_inset + + i vektorrummet +\begin_inset Formula $V$ +\end_inset + + kaldes en +\emph on +linearkombination +\emph default + af +\begin_inset Formula $\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}$ +\end_inset + +, hvis der eksisterer skalarer +\begin_inset Formula $\alpha_{1},\alpha_{2},\dots,\alpha_{n}\in\mathbb{F}$ +\end_inset + +, så +\begin_inset Formula +\[ +\boldsymbol{v}=\alpha_{1}\cdot\boldsymbol{v}_{1}+\alpha_{2}\cdot\boldsymbol{v}_{2}+\cdots+\alpha_{n}\cdot\boldsymbol{v}_{n} +\] + +\end_inset + +og +\begin_inset Formula $\boldsymbol{v}\in V$ +\end_inset + +. + Dette kan oplagt også noteres som +\begin_inset Formula $\boldsymbol{v=\sum_{i=1}^{n}}\alpha_{i}\cdot\boldsymbol{v}_{i}$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Definition 5.11 (Span) +\end_layout + +\begin_layout Standard +Mængden af alle linearkombination af +\begin_inset Formula $\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}$ +\end_inset + + kaldes for +\emph on +spannet +\emph default + af elementerne +\begin_inset Formula $\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}$ +\end_inset + + og betegnes med +\begin_inset Formula +\[ +{\rm Span}(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}) +\] + +\end_inset + +Se også Lemma 5.12 for et interessant resultat. +\end_layout + +\begin_layout Subsection +Definition 5.14 (Dimension) +\end_layout + +\begin_layout Standard +\begin_inset Formula $V$ +\end_inset + +er et +\begin_inset Formula $\mathbb{F}$ +\end_inset + +-vektorrum. + Vi definerer da: +\end_layout + +\begin_layout Itemize +Hvis +\begin_inset Formula $V=\{\boldsymbol{0}\}$ +\end_inset + + har +\begin_inset Formula $V$ +\end_inset + + dimension +\begin_inset Formula $0$ +\end_inset + + +\end_layout + +\begin_layout Itemize +Hvis +\begin_inset Formula $V$ +\end_inset + + er forskellig fra +\begin_inset Formula $\{\boldsymbol{0}\}$ +\end_inset + + og kan udspændes af +\begin_inset Formula $n$ +\end_inset + +elementer, men ikke af færre end +\begin_inset Formula $n$ +\end_inset + + elementer, så siger vi, at dimensionen af +\begin_inset Formula $V$ +\end_inset + +er lig +\begin_inset Formula $n$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Hivs +\begin_inset Formula $V$ +\end_inset + + ikke kan udspændes af en endelig mængde, så siges +\begin_inset Formula $V$ +\end_inset + +at have uendelig dimension. +\end_layout + +\begin_layout Standard +Dimensionen af +\begin_inset Formula $V$ +\end_inset + +betegnes med +\begin_inset Formula $\dim(V)$ +\end_inset + +. + Uendelig dimension beskrives ved +\begin_inset Formula $\dim(V)=\infty$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Formel 5.25 (Matrixprodukter og linearkombinationer) +\end_layout + +\begin_layout Standard +\begin_inset Quotes ald +\end_inset + +Ekstrem vigtig sammenhæng +\begin_inset Quotes ard +\end_inset + + mellem linearkombinationer og matrixproduktet. +\end_layout + +\begin_layout Standard +\begin_inset Formula $A=(a_{ij})\in{\rm Mat}_{m,n}(\mathbb{F})$ +\end_inset + +. + Da vil +\begin_inset Formula +\begin{align*} +A\cdot\begin{pmatrix}\alpha_{1}\\ +\alpha_{2}\\ +\vdots\\ +\alpha_{n} +\end{pmatrix} & =\begin{pmatrix}a_{11}\alpha_{1}+a_{12}\alpha_{2}+\cdots+a_{1n}\alpha_{n}\\ +a_{21}\alpha_{1}+a_{22}\alpha_{2}+\cdots+a_{2n}\alpha_{n}\\ +\vdots\\ +a_{m1}\alpha_{1}+a_{m2}\alpha_{2}+\cdots+a_{mn}\alpha_{n} +\end{pmatrix}\\ + & =\alpha_{1}\begin{pmatrix}a_{11}\\ +a_{21}\\ +\vdots\\ +a_{m1} +\end{pmatrix}+\alpha_{2}\begin{pmatrix}a_{12}\\ +a_{22}\\ +\vdots\\ +a_{m2} +\end{pmatrix}+\cdots+\alpha_{n}\begin{pmatrix}a_{1n}\\ +a_{2n}\\ +\vdots\\ +a_{mn} +\end{pmatrix}\\ + & =\alpha_{1}\cdot\boldsymbol{a}_{1}+\alpha_{2}\cdot\boldsymbol{a}_{2}+\cdots+\alpha_{n}\cdot\boldsymbol{a}_{n} +\end{align*} + +\end_inset + +Dette giver også at +\begin_inset Formula +\[ +{\rm Span}(\boldsymbol{a_{1},a_{2},\dots,a_{n}})=\{A\cdot\boldsymbol{v}\mid\boldsymbol{v}\in\mathbb{F}^{n}\}. +\] + +\end_inset + + +\end_layout + +\begin_layout Subsection +Proposition 5.17 (Vektorrum) +\end_layout + +\begin_layout Standard +Dimensionen af +\begin_inset Formula $\mathbb{F}^{n}$ +\end_inset + + er +\begin_inset Formula $n$ +\end_inset + + +\end_layout + +\begin_layout Paragraph +Bevis +\end_layout + +\begin_layout Standard +Det er allerede bemærket at +\begin_inset Formula $\mathbb{F}^{n}$ +\end_inset + + kan udspændes af +\begin_inset Formula $n$ +\end_inset + +elementer +\begin_inset Formula $(\boldsymbol{e}_{1},\boldsymbol{e}_{2},\dots,\boldsymbol{e}_{n})$ +\end_inset + +. + Det ønskes derfor kun vist at +\begin_inset Formula $\mathbb{F}^{n}$ +\end_inset + + ikke kan udspændes af +\emph on +færre end +\begin_inset Formula $n$ +\end_inset + + elementer. +\end_layout + +\begin_layout Standard +Det antages at +\begin_inset Formula $\mathbb{F}^{n}$ +\end_inset + + udspændes af +\begin_inset Formula $m$ +\end_inset + + elementer +\begin_inset Formula $\boldsymbol{a_{1},a_{2},\dots,a_{m}}\in\mathbb{F}^{n}$ +\end_inset + +, dvs. +\begin_inset Formula +\[ +{\rm Span}(\boldsymbol{a_{1},a_{2},\dots,a_{m}})=\mathbb{F}^{n} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +Lad nu +\begin_inset Formula $A\in{\rm Mat}_{n,m}(\mathbb{F})$ +\end_inset + + beskrive matricen, hvis +\begin_inset Formula $i$ +\end_inset + +'te søjle er givet ved +\begin_inset Formula $\boldsymbol{a}_{i}$ +\end_inset + +. + Da vil ligningssystemet +\begin_inset Formula $A\cdot\boldsymbol{x}=\boldsymbol{b}$ +\end_inset + + have en løsning for ethvert +\begin_inset Formula $\boldsymbol{b}\in\mathbb{F}^{n}$ +\end_inset + +, grundet Lemma 5.16 samholdt med ovenstående. +\end_layout + +\begin_layout Standard +Vi lader nu +\begin_inset Formula $\boldsymbol{e}_{i}\in\mathbb{F}^{n}$ +\end_inset + + betegne den +\begin_inset Formula $i$ +\end_inset + +'te søjle i +\begin_inset Formula $I_{n}$ +\end_inset + + og +\begin_inset Formula $\boldsymbol{b}_{i}\in\mathbb{F}^{m}$ +\end_inset + + for +\begin_inset Formula $i=[1,n]$ +\end_inset + + være en løsning til ligningssystemet +\begin_inset Formula $A\cdot\boldsymbol{x}=\boldsymbol{e}_{i}$ +\end_inset + +. + Lad endvidere +\begin_inset Formula $B\in{\rm Mat}_{m,n}(\mathbb{F})$ +\end_inset + + betegne matricen hvis +\begin_inset Formula $i$ +\end_inset + +'te søjle er +\begin_inset Formula $\boldsymbol{b}_{i}$ +\end_inset + + for +\begin_inset Formula $i=[1,n]$ +\end_inset + +. + Da gælder der at +\begin_inset Formula +\begin{align*} +A\cdot B & =\left(A\cdot\boldsymbol{b}_{1}\mid A\cdot\boldsymbol{b}_{2}\mid\cdots\mid A\cdot\boldsymbol{b}_{n}\right)\\ + & =\left(\boldsymbol{e}_{1}\mid\boldsymbol{e}_{2}\mid\cdots\mid\boldsymbol{e}_{n}\right)\\ + & =I_{n}. +\end{align*} + +\end_inset + + +\end_layout + +\begin_layout Standard +Det homogene ligningssystem +\begin_inset Formula $B\cdot\boldsymbol{x}=\boldsymbol{0}$ +\end_inset + + har kun har nulvektoren som løsning, da +\begin_inset Formula $B$ +\end_inset + + er invertibel samholdt med Lemma 4.4 og 4.6. + Proposition 1.14 implicerer da at +\begin_inset Formula $m\geq n$ +\end_inset + +, hvilket fuldender beviset. +\end_layout + +\begin_layout Subsubsection +Hjælpesætning - Lemma 5.16 +\end_layout + +\begin_layout Standard +Et lineært ligningssystem +\begin_inset Formula $A\cdot\boldsymbol{x}=\boldsymbol{b}$ +\end_inset + + har en løsning hvis og kun hvis +\begin_inset Formula $\boldsymbol{b}\in R(A)$ +\end_inset + + +\end_layout + +\begin_layout Paragraph +Bevis +\end_layout + +\begin_layout Standard +Det at +\begin_inset Formula $\boldsymbol{b}$ +\end_inset + + er i søjlerummet for +\begin_inset Formula $A$ +\end_inset + + betyder at +\begin_inset Formula $\boldsymbol{b}$ +\end_inset + + er på formen +\begin_inset Formula $A\cdot\boldsymbol{v}$ +\end_inset + + for et passende +\begin_inset Formula $\boldsymbol{v}\in\mathbb{F}^{n}$ +\end_inset + +. + Dette er netop betingelsen for at +\begin_inset Formula $A\cdot\boldsymbol{x}=\boldsymbol{b}$ +\end_inset + + har en løsning. +\end_layout + +\begin_layout Subsection +Lemma 5.12 (Hvis tid?) +\end_layout + +\begin_layout Standard +Mængden +\begin_inset Formula ${\rm Span}(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n})$ +\end_inset + + udgør et underrum i +\begin_inset Formula $V$ +\end_inset + + indeholdende alle elementerne +\begin_inset Formula $\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}$ +\end_inset + +. + Ethvert underrum af +\begin_inset Formula $V$ +\end_inset + + indeholdende +\begin_inset Formula $\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n}$ +\end_inset + + vil indeholde +\begin_inset Formula ${\rm Span}(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n})$ +\end_inset + + som delmængde. +\end_layout + +\begin_layout Paragraph +Meget uformel begrundelse +\end_layout + +\begin_layout Standard +Dette er på grund af spannets og underrummets definition. + Underrum er stabile overfor netop addition og skalarmultiplikation, hvilket + er det linearkombinationer benytter sig af. +\end_layout + +\begin_layout Subsection +Noter +\end_layout + +\begin_layout Standard +Skriv måske mere til Lemma 5.12 +\end_layout + +\begin_layout Standard +\begin_inset Newpage newpage +\end_inset + + +\end_layout + +\begin_layout Section +Basis for vektorrum +\end_layout + +\begin_layout Subsection +Definition 7.1 ( +\emph on +Udspænde +\emph default +, +\emph on +lineær +\emph default + +\emph on +uafhængighed +\emph default + og +\emph on +basis +\emph default +) +\end_layout + +\begin_layout Standard +For en samling af elementer +\begin_inset Formula $\mathcal{V}=(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n})$ +\end_inset + + i et +\begin_inset Formula $\mathbb{F}$ +\end_inset + +-vektorrum +\begin_inset Formula $V$ +\end_inset + +defineres: +\end_layout + +\begin_layout Itemize +Udspænding af +\begin_inset Formula $V$ +\end_inset + +: +\begin_inset Formula ${\rm Span}(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n})=V$ +\end_inset + + +\end_layout + +\begin_layout Itemize +Lineær uafhængighed: +\begin_inset Formula $\begin{pmatrix}\alpha_{1}\\ +\alpha_{2}\\ +\vdots\\ +\alpha_{n} +\end{pmatrix}\cdot\begin{pmatrix}\vdots & \vdots & & \vdots\\ +\boldsymbol{v}_{1} & \boldsymbol{v}_{2} & \cdots & \boldsymbol{v}_{n}\\ +\vdots & \vdots & & \vdots +\end{pmatrix}=\boldsymbol{0}\iff\alpha_{i}=0\:{\rm for}\:i=1,2,\dots,n$ +\end_inset + + +\end_layout + +\begin_layout Itemize +Basis for +\begin_inset Formula $V$ +\end_inset + + såfremt +\begin_inset Formula $\mathcal{V}$ +\end_inset + + udspænder +\begin_inset Formula $V$ +\end_inset + + samt er lineært uafhængig. +\end_layout + +\begin_layout Subsection +Lemma 7.2 (Relation af samlinger til afbildningsbegreber) +\end_layout + +\begin_layout Standard +For en samling af elementer +\begin_inset Formula $\mathcal{V}=(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{n})$ +\end_inset + + i et +\begin_inset Formula $\mathbb{F}$ +\end_inset + +-vektorrum +\begin_inset Formula $V$ +\end_inset + +gælder det at +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\mathcal{V}$ +\end_inset + +udspænder +\begin_inset Formula $V$ +\end_inset + + hvis og kun hvis +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er surjektiv +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\mathcal{V}$ +\end_inset + + er lineært uafhængig hvis og kun hvis +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er injektiv +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\mathcal{V}$ +\end_inset + + er en basis for +\begin_inset Formula $V$ +\end_inset + + hvis og kun hvis +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er en isomorfi +\end_layout + +\begin_layout Standard +Dette giver endvidere at +\begin_inset Formula $\mathcal{V}$ +\end_inset + + er en basis for +\begin_inset Formula $V$ +\end_inset + + netop når ethvert element i +\begin_inset Formula $V$ +\end_inset + + +\emph on +på entydig vis +\emph default + er en linearkombination af vektorerne i +\begin_inset Formula $\mathcal{V}$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph +Bevis +\end_layout + +\begin_layout Standard +(1): Billedet af +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er lig +\begin_inset Formula ${\rm Span}(\mathcal{V})$ +\end_inset + +. + +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er surjektiv hvis og kun hvis +\begin_inset Formula ${\rm Span}(\mathcal{V})=V$ +\end_inset + +. +\end_layout + +\begin_layout Standard +(2): +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er injektiv hvis og kun hvis at kernen for +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + er lig +\begin_inset Formula $\{\boldsymbol{0}\}$ +\end_inset + + jf. + Sætning 6.14. + Vektoren +\begin_inset Formula $(\alpha_{1},\alpha_{2},\dots,\alpha_{n})^{T}$ +\end_inset + + er et element i kernen for +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + hvis og kun hvis identiteten +\begin_inset Formula $\alpha_{1}\cdot\boldsymbol{v}_{1}+\alpha_{2}\cdot\boldsymbol{v}_{2}+\cdots+\alpha_{n}\cdot\boldsymbol{v}_{n}=0$ +\end_inset + + er opfyldt. +\end_layout + +\begin_layout Standard +For at +\begin_inset Formula $L_{\mathcal{V}}$ +\end_inset + + kan være injektiv må identiteten kun være opfyldt hvis alle +\begin_inset Formula $\alpha_{i}=0$ +\end_inset + +, hvilket netop er definitionen på, at +\begin_inset Formula $\mathcal{V}$ +\end_inset + + er lineært uafhængig. +\end_layout + +\begin_layout Standard +(3): Følger af (1) og (2), da isomorfi kræver samtidig surjektivitet og + injektivitet ligesom basis kræver samtidig udspænding (surjektivitet) og + lineær uafhængighed (injektivitet). +\end_layout + +\begin_layout Subsection +Sætning 7.12 ( +\emph on +Udtynding +\emph default + og +\emph on +udvidelse +\emph default +) +\end_layout + +\begin_layout Standard +\begin_inset Formula $V$ +\end_inset + +er et vektorrum af endelig dimension +\begin_inset Formula $n>0$ +\end_inset + + og +\begin_inset Formula $\mathcal{W}=(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\dots,\boldsymbol{v}_{m})$ +\end_inset + + er en samling af +\begin_inset Formula $m$ +\end_inset + + elementer i vektorrummet +\begin_inset Formula $V$ +\end_inset + +. + Da er to ting mulige: +\end_layout + +\begin_layout Enumerate +Hvis +\begin_inset Formula $\mathcal{W}$ +\end_inset + + udspænder +\begin_inset Formula $V$ +\end_inset + +, så er +\begin_inset Formula $n\leq m$ +\end_inset + + og +\begin_inset Formula $\mathcal{W}$ +\end_inset + + kan +\emph on +udtyndes +\emph default + til en basis for +\begin_inset Formula $V$ +\end_inset + +. + Det vil sige at nogle af vektorerne i +\begin_inset Formula $\mathcal{W}$ +\end_inset + + kan +\emph on +fjernes +\emph default + for at gøre +\begin_inset Formula $\mathcal{W}$ +\end_inset + + til en basis for +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Enumerate +Hvis +\begin_inset Formula $\mathcal{W}$ +\end_inset + + er lineært uafhængig, så er +\begin_inset Formula $m\leq n$ +\end_inset + + og +\begin_inset Formula $\mathcal{W}$ +\end_inset + + kan +\emph on +udvides +\emph default + til en basis for +\begin_inset Formula $V$ +\end_inset + +. + Det vil sige at der kan +\emph on +tilføjes +\emph default + vektorerne fra +\begin_inset Formula $V$ +\end_inset + + til +\begin_inset Formula $\mathcal{W}$ +\end_inset + + for at gøre +\begin_inset Formula $\mathcal{W}$ +\end_inset + + til en basis for +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Paragraph +Bevis +\end_layout + +\begin_layout Standard +Beviset foregår i to dele. + Først vises udtynding og derefter udvidelse. +\end_layout + +\begin_layout Subparagraph +(1) +\end_layout + +\begin_layout Standard +Der argumenteres genmem induktion i +\begin_inset Formula $m>0$ +\end_inset + +. + Hvis +\begin_inset Formula $m=1$ +\end_inset + +, så vil +\begin_inset Formula $\mathcal{W}=(\boldsymbol{v}_{1})$ +\end_inset + + pr. + antagelse udspænde +\begin_inset Formula $V,$ +\end_inset + + og da +\begin_inset Formula $V$ +\end_inset + + ikke er nulrummet (da +\begin_inset Formula $n>0$ +\end_inset + +) vil +\begin_inset Formula $\mathcal{W}$ +\end_inset + + være lineært uafhængig (jf. + Eks 7.6(A)). + Derfor vil +\begin_inset Formula $\mathcal{W}$ +\end_inset + + i dette tilfælde være en basis for +\begin_inset Formula $V$ +\end_inset + +. + Proposition +\begin_inset Formula $7.8$ +\end_inset + + giver at +\begin_inset Formula $n$ +\end_inset + + derved er lig +\begin_inset Formula $1$ +\end_inset + +. + Derfor kan det anvendes at +\begin_inset Formula $\mathcal{V}=\mathcal{W}$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Det antages nu at +\begin_inset Formula $m>1$ +\end_inset + + og at udsagnet er vist i tilfældet +\begin_inset Formula $m-1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Hvis +\begin_inset Formula $\mathcal{W}$ +\end_inset + + er lineært +\emph on +uafhængig +\emph default +, så er +\begin_inset Formula $W$ +\end_inset + + en basis og dermed er +\begin_inset Formula $n=m$ +\end_inset + + ifølge Proposition +\begin_inset Formula $7.8$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Hvis +\begin_inset Formula $\mathcal{W}$ +\end_inset + + er lineært +\emph on +afhængig +\emph default +, så eksisterer der jf. + Lemma 7.7(1) et +\begin_inset Formula $i$ +\end_inset + +, +\begin_inset Formula $1\leq i\leq m$ +\end_inset + +, så kan en ny samling +\begin_inset Formula $\mathcal{W}^{\prime}$ +\end_inset + + skabes ud fra +\begin_inset Formula $\mathcal{W}$ +\end_inset + + ved fjernelse af vektoren +\begin_inset Formula $\boldsymbol{v}_{i}$ +\end_inset + + ift. + +\begin_inset Formula $\mathcal{W}$ +\end_inset + + således at +\begin_inset Formula $\mathcal{W}^{\prime}$ +\end_inset + + udspænder +\begin_inset Formula $V$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Pr. + induktion kan det nye +\begin_inset Formula $\mathcal{W}^{\prime}$ +\end_inset + + udtyndes til en basis for +\begin_inset Formula $V$ +\end_inset + +, og denne basis vil også være en udtynding af +\begin_inset Formula $\mathcal{W}$ +\end_inset + +. +\end_layout + +\begin_layout Subparagraph +(2) +\end_layout + +\begin_layout Standard +At +\begin_inset Formula $m\leq n$ +\end_inset + + følger af Lemma 7.10. +\end_layout + +\begin_layout Standard +Der argumenteres ved induktion i tallet +\begin_inset Formula $n-m\geq0$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Hvis +\begin_inset Formula $n-m=0$ +\end_inset + +, så er +\begin_inset Formula $\mathcal{W}$ +\end_inset + + en basis for +\begin_inset Formula $V$ +\end_inset + + jf. + Proposition 7.11. +\end_layout + +\begin_layout Standard +Det antages nu at +\begin_inset Formula $n-m>0$ +\end_inset + + og at udsagnet er vist i tilfældet +\begin_inset Formula $(n-m)-1$ +\end_inset + +. +\end_layout + +\begin_layout Standard +Da +\begin_inset Formula $m\neq n$ +\end_inset + + er +\begin_inset Formula $\mathcal{W}$ +\end_inset + + ikke en basis for +\begin_inset Formula $V$ +\end_inset + + jf. + Proposition +\begin_inset Formula $7.8$ +\end_inset + + (størrelse af basis skal være lig dimension af rum). + +\begin_inset Formula $\mathcal{W}$ +\end_inset + + kan da ikke udspænde +\begin_inset Formula $V$ +\end_inset + +. + Det må derfor være muligt at vælge et element +\begin_inset Formula $\boldsymbol{v}^{\prime}$ +\end_inset + + i +\begin_inset Formula $V$ +\end_inset + + som ikke er indeholdet i +\begin_inset Formula ${\rm Span}(\mathcal{W})$ +\end_inset + +. + Pr. + Lemma 7.7(2) vil +\begin_inset Formula $\mathcal{W}^{\prime}=\mathcal{W}+\boldsymbol{v}^{\prime}$ +\end_inset + + være lineært uafhængig. +\end_layout + +\begin_layout Standard +Pr. + induktion så kan +\begin_inset Formula $\mathcal{W}^{\prime}$ +\end_inset + + nu udvides til en basis for +\begin_inset Formula $V$ +\end_inset + +. + En sådan udvidelse vil samtidig være en udvidelse af det oprindelige +\begin_inset Formula $\mathcal{W}$ +\end_inset + + +\end_layout + +\begin_layout Subsection +Noter +\end_layout + +\begin_layout Standard +Overvej at droppe Lemma 7.2 fra dispositionen. +\end_layout + \end_body \end_document