master > master: Woche 9 (ÜB + quiz)
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%% |
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%% — body/uebung/ueb8.tex;
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%% — body/uebung/ueb9.tex;
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%% — body/ska/ska4.tex;
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%% — body/ska/ska5.tex;
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@ -82,6 +84,8 @@
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%% — body/quizzes/quiz7.tex;
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%% — body/quizzes/quiz8.tex;
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%% — body/quizzes/quiz9.tex;
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%% — back/index.tex;
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@ -2708,10 +2712,11 @@ und daraus die Parameter abzulesen.
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\begin{mathe}[mc]{c}
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\begin{smatrix}
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1&-2&4&0\\
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0&11&-15&1\\
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0&0&-7&1\\
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\end{smatrix}\\
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1&-2&4&0\\
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0&11&-15&1\\
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0&0&-7&1\\
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\end{smatrix}
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\\
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\end{mathe}
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Wende die Zeilentransformation
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@ -2720,10 +2725,11 @@ und daraus die Parameter abzulesen.
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\begin{mathe}[mc]{c}
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\begin{smatrix}
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1&-2&4&0\\
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0&11&-8&0\\
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0&0&-7&1\\
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\end{smatrix}\\
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1&-2&4&0\\
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0&11&-8&0\\
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0&0&-7&1\\
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\end{smatrix}
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\\
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\end{mathe}
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Aus der Zeilenstufenform erschließt sich, dass $t_{4}$ frei ist.
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@ -5865,6 +5871,489 @@ Seien $n\in\ntrlpos$ und $\mathbf{v}_{i}\in V$ für $i\in\{1,2,\ldots,n\}$.
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eine Basis für $W:=\kmplx^{2}$, wenn dies als $\reell$-Vektorraum betrachtet wird.
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Insbesondere gilt $\dim(W)=4$.
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%% ********************************************************************************
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%% FILE: body/uebung/ueb9.tex
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%% ********************************************************************************
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\setcounternach{chapter}{9}
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\chapter[Woche 9]{Woche 9}
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\label{ueb:9}
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%% AUFGABE 9-1
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\let\altsectionname\sectionname
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\def\sectionname{Aufgabe}
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\section[Aufgabe 1]{}
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\label{ueb:9:ex:1}
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\let\sectionname\altsectionname
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Seien
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\begin{mathe}[mc]{cqcqcqcqc}
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u_{1} := \begin{svector}1\\2\\-1\\1\\\end{svector},
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&u_{2} := \begin{svector}-1\\-2\\1\\2\\\end{svector},
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&v_{1} := \begin{svector}1\\2\\-1\\-2\\\end{svector},
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&v_{2} := \begin{svector}-1\\3\\0\\-2\\\end{svector},
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&v_{3} := \begin{svector}2\\-1\\-1\\1\\\end{svector}.\\
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\end{mathe}
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Vektoren in $\reell^{4}$ und setze
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\begin{mathe}[mc]{cqc}
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U:=\vectorspacespan\{u_{1},u_{2}\}
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&V:=\vectorspacespan\{v_{1},v_{2},v_{3}\}.\\
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\end{mathe}
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\begin{schattierteboxdunn}
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\begin{claim}
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\makelabel{claim:1:ueb:9:ex:1}
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$U\subseteq V$.
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\end{claim}
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\end{schattierteboxdunn}
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\begin{einzug}[\rtab][\rtab]
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\begin{proof}
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Es reicht aus zu zeigen, dass $u_{1},u_{2}\in\vectorspacespan\{v_{1},v_{2},v_{3}\}$.
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Zu diesem Zwecke reicht es aus
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das homogene LGS $A\mathbf{x}=\zerovector$
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in Zeilenstufenform zu bringen,
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wobei
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\begin{mathe}[mc]{rcccl}
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A &= &\left(
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v_{1}\:v_{2}\:v_{3}\:u_{1}\:u_{2}
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\right)
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&= &\begin{smatrix}
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1&-1&2&1&-1\\
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2&3&-1&2&-2\\
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-1&0&-1&-1&1\\
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-2&-2&1&1&2\\
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\end{smatrix}\\
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\end{mathe}
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und \textbf{zu zeigen}, dass $x_{4},x_{5}$ darin freie Unbekannte sind.
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\begin{algorithm}[\rtab][\rtab]
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Zeilenoperationen
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${Z_{2}\leftsquigarrow Z_{2} - 2\cdot Z_{1}}$;
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${Z_{3}\leftsquigarrow Z_{3} + Z_{1}}$
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und
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${Z_{4}\leftsquigarrow Z_{4} + 2\cdot Z_{1}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &-1 &2 &1 &-1\\
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0 &5 &-5 &0 &0\\
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0 &-1 &1 &0 &0\\
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0 &-4 &5 &3 &0\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{3}\leftsquigarrow Z_{3} + \frac{1}{5}\cdot Z_{2}}$
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und
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${Z_{4}\leftsquigarrow Z_{4} + \frac{4}{5}\cdot Z_{2}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &-1 &2 &1 &-1\\
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0 &5 &-5 &0 &0\\
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0 &0 &0 &0 &0\\
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0 &0 &1 &3 &0\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{3}\leftrightsquigarrow Z_{4}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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\boxed{1} &-1 &2 &1 &-1\\
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0 &\boxed{5} &-5 &0 &0\\
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0 &0 &\boxed{1} &3 &0\\
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0 &0 &0 &0 &0\\
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\end{matrix}.\\
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\end{mathe}
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\end{algorithm}
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Darum sind $x_{4},x_{5}$ im homogenen LGS frei.
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Folglich gelten $u_{1},u_{2}\in\vectorspacespan\{v_{1},v_{2},v_{3}\}=V$
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und damit gilt $U=\vectorspacespan\{u_{1},u_{2}\}\subseteq V$.
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\end{proof}
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\end{einzug}
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\begin{rem}
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\makelabel{rem:1:ueb:9:ex:1}
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Im Beweis von \Cref{claim:1:ueb:9:ex:1} wurde aus der Feststellung,
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dass $x_{4},x_{5}$ im LGS $A\mathbf{x}=\zerovector$ frei sind,
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schlussfolgert,
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dass $u_{1},u_{2}\in\vectorspacespan\{v_{1},v_{2},v_{3}\}$.
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Diese Schlussfolgerung lässt sich rechtfertigen:
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\begin{kompaktitem}[\rtab][\rtab]
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\item Sei $u\in U=\vectorspacespan\{u_{1},u_{2}\}$.
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Dann existieren $c_{1},c_{2}\in\reell$,
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so dass \fbox{$u=c_{1}u_{1}+c_{2}u_{2}$}.
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\item
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Setze nun im homogenen LGS $x_{4}:=-c_{1}$ und $x_{5}:=-c_{2}$
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und bestimme $x_{1},x_{2},x_{3}\in\reell$, gemäß der Zeilenstufenform.
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Dies liefert uns eine Lösung zu $A\mathbf{x}=\zerovector$.
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Wegen der Konstruktion von $A$ heißt dies
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\begin{mathe}[mc]{rcl}
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\eqtag[eq:1:ueb:9:ex:1]
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x_{1}v_{1}+x_{2}v_{2}+x_{3}v_{3}+x_{4}u_{1}+x_{5}u_{2} &= &\zerovector\\
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\end{mathe}
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\item
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Folglich gilt
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\begin{mathe}[mc]{rcl}
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u &= &c_{1}u_{1}+c_{2}u_{2}\\
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&= &-x_{4}u_{1} + -x_{5}u_{2}
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\quad\text{(per Konstruktion)}\\
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&\eqcrefoverset{eq:1:ueb:9:ex:1}{=}
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&x_{1}v_{1}+x_{2}v_{2}+x_{3}v_{3}
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\in \vectorspacespan\{v_{1},v_{2},v_{3}\}=V.\\
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\end{mathe}
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Da $u\in U$ beliebig gewählt wurde,
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gilt $U\subseteq V$.
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\end{kompaktitem}
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Beachte hier, dass die Freiheit von $x_{4},x_{5}$ im LGS eine kritische Rolle spielt.
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\end{rem}
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\begin{schattierteboxdunn}
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\begin{claim}
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\makelabel{claim:2:ueb:9:ex:1}
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$\{v_{2}+U\}$ ist eine Basis für $V/U$.
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Insbesondere gilt $\dim(V/U)=1$.
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\end{claim}
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\end{schattierteboxdunn}
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\begin{einzug}[\rtab][\rtab]
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\begin{proof}
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Unser Ziel ist es, zu bestimmen,
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in wiefern sich das\footnote{evtl. nicht linear unabhängiges} System $\{u_{1},u_{2}\}$
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durch die Vektoren
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$v_{1},v_{2},v_{3}$
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erweitern lässt,
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und dabei die linear abhängigen Vektoren zu entfernen
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und die linear unabhängigen zu behalten
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(vgl. Ansatz im Beweis von \cite[Satz~5.5.3]{sinn2020}).
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Zu diesem Zwecke untersuchen wir das homogene LGS,
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$B\mathbf{x}=\zeromatrix$,
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wobei
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\begin{mathe}[mc]{rcccl}
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B &= &\left(
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u_{1}\:u_{2}\:v_{1}\:v_{2}\:v_{3}
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\right)
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&= &\begin{smatrix}
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1&-1&1&-1&2\\
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2&-2&2&3&-1\\
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-1&1&-1&0&-1\\
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1&2&-2&-2&1\\
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\end{smatrix}.\\
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\end{mathe}
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Es reicht aus hier \textbf{zu zeigen},
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dass $x_{3},x_{5}$ frei sind und $x_{4}$ nicht frei ist.
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\begin{algorithm}[\rtab][\rtab]
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Zeilenoperationen
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${Z_{2}\leftsquigarrow Z_{2} - 2\cdot Z_{1}}$;
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${Z_{3}\leftsquigarrow Z_{3} + Z_{1}}$
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und
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${Z_{4}\leftsquigarrow Z_{4} - Z_{1}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &-1 &1 &-1 &2\\
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0 &0 &0 &5 &-5\\
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0 &0 &0 &-1 &1\\
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0 &3 &-3 &-1 &-1\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{2}\leftrightsquigarrow Z_{4}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &-1 &1 &-1 &2\\
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0 &3 &-3 &-1 &-1\\
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0 &0 &0 &-1 &1\\
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0 &0 &0 &5 &-5\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{4}\leftsquigarrow Z_{4} + 5\cdot Z_{3}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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\boxed{1} &-1 &1 &-1 &2\\
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0 &\boxed{3} &-3 &-1 &-1\\
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0 &0 &0 &\boxed{-1} &1\\
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0 &0 &0 &0 &0\\
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\end{matrix}.\\
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\end{mathe}
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\end{algorithm}
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Darum sind $x_{3},x_{5}$ frei und $x_{1},x_{2},x_{4}$ nicht.
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Also ist $\{v_{2}+U\}$ eine (einelementige) Basis für $V/U$.
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\end{proof}
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\end{einzug}
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\begin{rem*}
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Wie in \Cref{rem:1:ueb:9:ex:1} erklärt wurde,
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sind die Spalten/Vektoren entsprechend den freien Variablen,
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also $v_{1},v_{3}$,
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linear abhängig von den Spalten/Vektoren entsprechend den nicht-freien Variablen,
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also $u_{1},u_{2},u_{3}$.
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Folglich gelten
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\begin{mathe}[mc]{rcl}
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v_{1}+U,v_{2}+U &\in &\vectorspacespan\{v_{3}+U\}\\
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\end{mathe}
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Und da $x_{4}$ nicht frei ist, hängt $v_{3}$ von $u_{1},u_{2}$ nicht ab.
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Also gilt $v_{3}+U\neq\zerovector_{V/U}$.
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\end{rem*}
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%% AUFGABE 9-2
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\clearpage
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\let\altsectionname\sectionname
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\def\sectionname{Aufgabe}
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\section[Aufgabe 2]{}
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\label{ueb:9:ex:2}
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\let\sectionname\altsectionname
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Seien
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\begin{mathe}[mc]{cqcqcqcqc}
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\mathbf{v}_{1} := \begin{svector}1\\-2\\3\\1\\\end{svector},
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&\mathbf{v}_{2} := \begin{svector}2\\-5\\7\\0\\\end{svector},
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&\mathbf{v}_{3} := \begin{svector}-2\\6\\-9\\-3\\\end{svector}.\\
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\end{mathe}
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Vektoren in $\reell^{4}$ und sei $\phi:\reell^{3}\to\reell^{4}$
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die eindeutig definierte lineare Abbildung mit $\phi(\mathbf{e}_{i})=\mathbf{v}_{i}$
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für $i\in\{1,2,3\}$,
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wobei $\{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$
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die kanonische Basis für $\reell^{3}$ ist.
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\begin{enumerate}{\bfseries (a)}
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%% AUFGABE 9-2(a)
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\item
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Wegen Linearität gilt für alle $x_{1},x_{2},x_{3}\in\reell$
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\begin{mathe}[mc]{rcl}
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\phi(x_{1},x_{2},x_{3})
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&= &\phi(x_{1}\mathbf{e}_{1}+x_{2}\mathbf{e}_{2}+x_{3}\mathbf{e}_{3})\\
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&= &x_{1}\phi(\mathbf{e}_{1})+x_{2}\phi(\mathbf{e}_{2})+x_{3}\phi(\mathbf{e}_{3})\\
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&= &x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+x_{3}\mathbf{v}_{3}\\
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&= &A\mathbf{x}\\
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\end{mathe}
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wobei
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\begin{mathe}[mc]{rcl}
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A &:= &\begin{smatrix}
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1&2&-2\\
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-2&-5&6\\
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3&7&-9\\
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1&0&-3\\
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\end{smatrix}
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\end{mathe}
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Insbesondere gilt \fbox{$\phi=\phi_{A}$}.
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Im nächsten Aufgabenteil nutzen wir diese Darstellung aus.
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%% AUFGABE 9-2(b)
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\item
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Um zu bestimmen, ob $\phi=\phi_{A}$ injektiv, surjektiv, bijektiv ist,
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berechnen wir die Zeilenstufenform von $A$.
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Es gilt
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\begin{algorithm}[\rtab][\rtab]
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Zeilenoperationen
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${Z_{2}\leftsquigarrow Z_{2} + 2\cdot Z_{1}}$;
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${Z_{2}\leftsquigarrow Z_{2} - 3\cdot Z_{1}}$
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und
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${Z_{3}\leftsquigarrow Z_{3} - Z_{1}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &2 &-2\\
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0 &-1 &2\\
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0 &1 &-3\\
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0 &-2 &-1\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{3}\leftsquigarrow Z_{3} + Z_{2}}$
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und
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${Z_{4}\leftsquigarrow Z_{4} - 2\cdot Z_{2}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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1 &2 &-2\\
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0 &-1 &2\\
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0 &0 &-1\\
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0 &0 &-5\\
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\end{matrix}.\\
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\end{mathe}
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Zeilenoperation
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${Z_{4}\leftsquigarrow Z_{4} - 5\cdot Z_{3}}$
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anwenden:
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\begin{mathe}[mc]{rcl}
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\begin{matrix}{ccccc}
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\boxed{1} &2 &-2\\
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0 &\boxed{-1} &2\\
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0 &0 &\boxed{-1}\\
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0 &0 &0\\
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\end{matrix}.\\
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\end{mathe}
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$\Rightarrow$ $\rank(A)=\text{\upshape Zeilenrang}(A)=3$
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(siehe \cite[Satz~6.3.11]{sinn2020}).
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\end{algorithm}
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Also gilt $\phi=\phi_{A}$,
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wobei $A$ eine $m\times n$-Matrix ist,
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wobei $m=4$, $n=3$,
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und $\rank(A)=3$.
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Laut \cite[Korollar~6.3.15]{sinn2020} erhalten wir also
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\begin{kompaktitem}
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\item
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$\phi$ \fbox{ist injektiv},
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weil $\rank(A)=3\geq 3=n$.
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\item
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$\phi$ ist \fbox{nicht surjektiv},
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weil $\rank(A)=3\ngeq 4=m$.
|
||||
\item
|
||||
$\phi$ ist \fbox{nicht bijektiv},
|
||||
weil $m\neq n$.
|
||||
\end{kompaktitem}
|
||||
\end{enumerate}
|
||||
|
||||
%% AUFGABE 9-3
|
||||
\clearpage
|
||||
\let\altsectionname\sectionname
|
||||
\def\sectionname{Aufgabe}
|
||||
\section[Aufgabe 3]{}
|
||||
\label{ueb:9:ex:3}
|
||||
\let\sectionname\altsectionname
|
||||
|
||||
Seien $U,V,W$ Vektorräume über einem Körper $K$.
|
||||
Seien ${\phi:U\to V}$ und ${\psi:V\to W}$ lineare Abbildungen.
|
||||
|
||||
\begin{schattierteboxdunn}
|
||||
\begin{claim*}
|
||||
${\psi\circ \phi:U\to W}$
|
||||
ist injektiv $\Leftrightarrow$
|
||||
$\phi$ injektiv und $\ker(\psi)\cap\range(\phi)=\{0\}$.
|
||||
\end{claim*}
|
||||
\end{schattierteboxdunn}
|
||||
|
||||
\begin{proof}
|
||||
\hinRichtung
|
||||
Angenommen, $\psi\circ\phi$ sei injektiv.
|
||||
\textbf{Zu zeigen:} (i)~$\phi$ injektiv und (ii)~$\ker(\psi)\cap\range(\phi)=\{0\}$.\\
|
||||
|
||||
\begin{enumerate}{\bfseries (i)}[\rtab][\rtab]
|
||||
\item
|
||||
Seien $x,x'\in U$ beliebig. Dann gilt
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
\phi(x)=\phi(x')
|
||||
&\Longrightarrow
|
||||
&\psi(\phi(x))=\psi(\phi(x'))\\
|
||||
&\Longrightarrow
|
||||
&(\psi\circ \phi)(x)=(\psi\circ \phi)(x')\\
|
||||
&\textoverset{$\psi\circ\phi$ inj.}{\Longrightarrow}
|
||||
&x=x'.\\
|
||||
\end{mathe}
|
||||
|
||||
Folglich ist $\phi$ injektiv.
|
||||
\item
|
||||
Sei $y\in V$ beliebig. Es gilt
|
||||
|
||||
\begin{longmathe}[mc]{RCL}
|
||||
y\in\ker(\psi)\cap\range(\phi)
|
||||
&\Longleftrightarrow
|
||||
&y\in\range(\phi)\,\text{und}\,y\in\ker(\psi)\\
|
||||
&\Longleftrightarrow
|
||||
&(\exists{x\in U:~}y=\phi(x))\,\text{und}\,y\in\ker(\psi)\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,y\in\ker(\psi))\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,\psi(y)=0)\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,\psi(\phi(x))=0)\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,(\psi\circ\phi)(x)=0)\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,x\in\ker(\psi\circ\phi))\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,x\in\{0\})\\
|
||||
&&\text{(wegen Injektivität von $\psi\circ\phi$ + \cite[Lemma~6.1.4(1)]{sinn2020})}\\
|
||||
&\Longleftrightarrow
|
||||
&\exists{x\in U:~}(y=\phi(x)\,\text{und}\,x=0)\\
|
||||
&\Longleftrightarrow
|
||||
&y=\phi(0)=0
|
||||
\Longleftrightarrow
|
||||
y\in\{0\}.\\
|
||||
\end{longmathe}
|
||||
|
||||
Darum gilt $\ker(\psi)\cap\range(\phi)=\{0\}$.
|
||||
\end{enumerate}
|
||||
|
||||
\herRichtung
|
||||
Angenommen, (i)~$\phi$ sei injektiv und (ii)~$\ker(\psi)\cap\range(\phi)=\{0\}$.
|
||||
\textbf{Zu zeigen:} $\psi\circ\phi$ injektiv.\\
|
||||
Hierfür wenden wir \cite[Lemma~6.1.4(1)]{sinn2020} an.
|
||||
Sei $x\in U$ beliebig. Es gilt
|
||||
|
||||
\begin{longmathe}[mc]{RCL}
|
||||
x\in\ker(\psi\circ\phi)
|
||||
&\Longleftrightarrow
|
||||
&\psi(\phi(x))=(\psi\circ\phi)(x)=0\\
|
||||
&\Longleftrightarrow
|
||||
&\phi(x)\in\ker(\psi)\\
|
||||
&\Longleftrightarrow
|
||||
&\phi(x)\in\ker(\psi)\cap\range(\phi)
|
||||
\quad
|
||||
\text{($\phi(x)$ ist immer in $\range(\phi)$)}\\
|
||||
&\textoverset{(ii)}{\Longleftrightarrow}
|
||||
&\phi(x)\in\{0\}\\
|
||||
&\Longleftrightarrow
|
||||
&\phi(x)=0\\
|
||||
&\Longleftrightarrow
|
||||
&x\in\ker(\phi)\\
|
||||
&\Longleftrightarrow
|
||||
&x\in\{0\}
|
||||
\quad\text{(wegen (i) + \cite[Lemma~6.1.4(1)]{sinn2020})}\\\
|
||||
\end{longmathe}
|
||||
|
||||
Darum gilt $\ker(\psi\circ\phi)=\{0\}$
|
||||
und laut \cite[Lemma~6.1.4(1)]{sinn2020}
|
||||
ist dies zur Injektivität von $\psi\circ\phi$ äquivalent.
|
||||
\end{proof}
|
||||
|
||||
\setcounternach{part}{2}
|
||||
\part{Selbstkontrollenaufgaben}
|
||||
|
||||
@ -7034,22 +7523,22 @@ Für jeden Fall berechnen wir $\ggT(a,b)$ mittels des Euklidischen Algorithmus
|
||||
$a$ &$b$ &Restberechnung (symbolisch) &Restberechnung (Werte)\\
|
||||
\hline
|
||||
\endhead
|
||||
$1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\
|
||||
\hline
|
||||
$13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\
|
||||
\hline
|
||||
$210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\
|
||||
\hline
|
||||
$1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\
|
||||
&&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\
|
||||
&&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\
|
||||
$1529$ &$170$ &$a = b\cdot q_{1} + r_{1}$ &$1529 = 170\cdot 8 + 169$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$170 = 169\cdot 1 + \boxed{\mathbf{1}}$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$169 = 1\cdot 169 + 0$\\
|
||||
\hline
|
||||
$13758$ &$21$ &$a = b\cdot q_{1} + r_{1}$ &$13758 = 21\cdot 655 + \boxed{\mathbf{3}}$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 3\cdot 7 + 0$\\
|
||||
\hline
|
||||
$210$ &$45$ &$a = b\cdot q_{1} + r_{1}$ &$210 = 45\cdot 4 + 30$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$45 = 30\cdot 1 + \boxed{\mathbf{15}}$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$30 = 15\cdot 2 + 0$\\
|
||||
\hline
|
||||
$1209$ &$102$ &$a = b\cdot q_{1} + r_{1}$ &$1209 = 102\cdot 11 + 87$\\
|
||||
&&$b = r_{1}\cdot q_{2} + r_{2}$ &$102 = 87\cdot 1 + 15$\\
|
||||
&&$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$87 = 15\cdot 5 + 12$\\
|
||||
&&$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$15 = 12\cdot 1 + \boxed{\mathbf{3}}$\\
|
||||
&&$r_{3} = r_{4}\cdot q_{5} + r_{5}$ &$12 = 3\cdot 4 + 0$\\
|
||||
\hline
|
||||
\hline
|
||||
\end{longtable}
|
||||
@ -7070,18 +7559,18 @@ Wir verwenden die Berechnungen aus der Tabelle in SKA \ref{ska:5:ex:6}.
|
||||
$a$ &$b$ &Rest (symbolisch) &Rest (Werte)\\
|
||||
\hline
|
||||
\endhead
|
||||
$1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\
|
||||
\hline
|
||||
$13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\
|
||||
\hline
|
||||
$210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\
|
||||
\hline
|
||||
$1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\
|
||||
&&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\
|
||||
&&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\
|
||||
$1529$ &$170$ &$r_{1} = a - 8\cdot b$ &$169 = 1\cdot a + -8\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{1 = \mathbf{-1}\cdot a + \mathbf{9}\cdot b}$\\
|
||||
\hline
|
||||
$13758$ &$21$ &$r_{1} = a - 655\cdot b$ &$\boxed{3 = \mathbf{1}\cdot a + \mathbf{-655}\cdot b}$\\
|
||||
\hline
|
||||
$210$ &$45$ &$r_{1} = a - 4\cdot b$ &$30 = 1\cdot a + -4\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$\boxed{15 = \mathbf{-1}\cdot a + \mathbf{5}\cdot b}$\\
|
||||
\hline
|
||||
$1209$ &$102$ &$r_{1} = a - 11\cdot b$ &$87 = 1\cdot a + -11\cdot b$\\
|
||||
&&$r_{2} = b - 1\cdot r_{1}$ &$15 = -1\cdot a + 12\cdot b$\\
|
||||
&&$r_{3} = r_{1} - 5\cdot r_{2}$ &$12 = 6\cdot a + -71\cdot b$\\
|
||||
&&$r_{4} = r_{2} - 1\cdot r_{3}$ &$\boxed{3 = \mathbf{-7}\cdot a + \mathbf{83}\cdot b}$\\
|
||||
\hline
|
||||
\hline
|
||||
\end{longtable}
|
||||
@ -7748,10 +8237,10 @@ Für $x=[2]$ und $y=[3]$ gilt $x,y\neq [0]$ und aber $xy=[2\cdot 3]=[6]=[0]$.
|
||||
Restberechnung (symbolisch) &Restberechnung (Werte)\\
|
||||
\hline
|
||||
\endhead
|
||||
$a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\
|
||||
$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\
|
||||
$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\
|
||||
$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\
|
||||
$a = b\cdot q_{1} + r_{1}$ &$103 = 21\cdot 4 + 19$\\
|
||||
$b = r_{1}\cdot q_{2} + r_{2}$ &$21 = 19\cdot 1 + 2$\\
|
||||
$r_{1} = r_{2}\cdot q_{3} + r_{3}$ &$19 = 2\cdot 9 + \boxed{\mathbf{1}}$\\
|
||||
$r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\
|
||||
\hline
|
||||
\hline
|
||||
\end{longtable}
|
||||
@ -7765,9 +8254,9 @@ $r_{2} = r_{3}\cdot q_{4} + r_{4}$ &$2 = 1\cdot 2 + 0$\\
|
||||
Rest (symbolisch) &Rest (Werte)\\
|
||||
\hline
|
||||
\endhead
|
||||
$r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\
|
||||
$r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\
|
||||
$r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\
|
||||
$r_{1} = a - 4\cdot b$ &$19 = 1\cdot a + -4\cdot b$\\
|
||||
$r_{2} = b - 1\cdot r_{1}$ &$2 = -1\cdot a + 5\cdot b$\\
|
||||
$r_{3} = r_{1} - 9\cdot r_{2}$ &$\boxed{1 = \mathbf{10}\cdot a + \mathbf{-49}\cdot b}$\\
|
||||
\hline
|
||||
\hline
|
||||
\end{longtable}
|
||||
@ -8532,6 +9021,116 @@ für alle Teilmengen, $U\subseteq V$, und von
|
||||
für alle linearen Unterräume, $U\subseteq V$.
|
||||
\end{rem*}
|
||||
|
||||
%% ********************************************************************************
|
||||
%% FILE: body/quizzes/quiz9.tex
|
||||
%% ********************************************************************************
|
||||
|
||||
\setcounternach{chapter}{9}
|
||||
\chapter[Woche 9]{Woche 9}
|
||||
\label{quiz:9}
|
||||
|
||||
\begin{claim*}
|
||||
Seien $U,V,W$ Vektorräume über einem Körper, $K$.
|
||||
Seien
|
||||
${\phi:U\to V}$
|
||||
und
|
||||
${\psi:V\to W}$
|
||||
linear.
|
||||
Falls
|
||||
|
||||
\begin{kompaktenum}{\bfseries (i)}[\rtab][\rtab]
|
||||
\item\label{it:1:quiz:9}
|
||||
$\psi$ surjektiv ist; und
|
||||
\item\label{it:2:quiz:9}
|
||||
$\ker(\psi)+\range(\phi)=V$,
|
||||
\end{kompaktenum}
|
||||
|
||||
dann ist ${\psi\circ\phi:U\to W}$ surjektiv.
|
||||
\end{claim*}
|
||||
|
||||
\begin{einzug}[\rtab][\rtab]
|
||||
\begin{proof}
|
||||
Es reicht aus, für alle $z\in W$
|
||||
\textbf{zu zeigen}, dass ein $x\in U$ existiert mit
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
\eqtag[eq:0:quiz:9]{$\ast$}
|
||||
(\psi\circ\phi)(x) &= &z.\\
|
||||
\end{mathe}
|
||||
|
||||
Sei also $z\in W$ beliebig.
|
||||
|
||||
\begin{einzug}[\rtab]
|
||||
Wegen \eqcref{it:1:quiz:9} existiert ein $y\in V$,
|
||||
so dass
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
\eqtag[eq:1:quiz:9]
|
||||
\phi(y) &= &z.\\
|
||||
\end{mathe}
|
||||
|
||||
Da $y\in V$ und laut \eqcref{it:2:quiz:9} $V=\ker(\psi)+\range(\phi)$,
|
||||
es existieren $y_{0}\in\ker(\psi)$ und $y_{1}\in\range(\phi)$,
|
||||
so dass
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
\eqtag[eq:2:quiz:9]
|
||||
y &= &y_{0}+y_{1}.\\
|
||||
\end{mathe}
|
||||
|
||||
Da $y_{1}\in\range(\phi)$, existiert nun ein \fbox{$x\in U$},
|
||||
so dass $\phi(x)=y_{1}$. Wir berechnen nun
|
||||
|
||||
\begin{mathe}[mc]{rcl}
|
||||
(\psi\circ\phi)(x)
|
||||
&= &\psi(\phi(x))\\
|
||||
&= &\psi(y_{1})\\
|
||||
&\eqcrefoverset{eq:2:quiz:9}{=}
|
||||
&\psi(y-y_{0})\\
|
||||
&= &\psi(y)-\psi(y_{0})\\
|
||||
&= &\psi(y)-0,
|
||||
\quad\text{da $y_{0}\in\ker(\psi)$}\\
|
||||
&\eqcrefoverset{eq:1:quiz:9}{=}
|
||||
&z.\\
|
||||
\end{mathe}
|
||||
|
||||
Damit haben wir \eqcref{eq:0:quiz:9} gezeigt.
|
||||
\end{einzug}
|
||||
|
||||
Also ist $\psi\circ\phi$ surjektiv.
|
||||
\end{proof}
|
||||
\end{einzug}
|
||||
|
||||
\begin{rem*}
|
||||
Wir können in der Tat zeigen, die umgekehrte Richtung auch gilt:
|
||||
Angenommen, $\psi\circ\phi$ sei surjektiv.
|
||||
Dann gilt
|
||||
$W\supseteq\psi(V)\supseteq\psi(\phi(U))=(\psi\circ\phi)(U)=W$,
|
||||
und somit $\psi(V)=W$,
|
||||
sodass \eqcref{it:1:quiz:9} gilt.
|
||||
Und für alle $y\in V$, wegen Surjektivität von $\psi\circ\phi$,
|
||||
existiert ein $x\in U$, so dass $\psi(y)=(\psi\circ\phi)(x)$.
|
||||
Daraus folgt
|
||||
|
||||
\begin{mathe}[mc]{rcccl}
|
||||
\psi(y-\phi(x))
|
||||
&= &\psi(y)-\psi(\phi(x))
|
||||
&= &0,\\
|
||||
\end{mathe}
|
||||
|
||||
sodass \fbox{$y-\phi(x)\in\ker(\psi)$} gilt.
|
||||
Darum
|
||||
|
||||
\begin{mathe}[mc]{rcccl}
|
||||
y &= &\underbrace{y-\phi(x)}_{\in\ker(\psi)}
|
||||
+\underbrace{\phi(x)}_{\in\range(\phi)}
|
||||
&\in &\ker(\psi)+\range(\phi).\\
|
||||
\end{mathe}
|
||||
|
||||
Also gilt die $\supseteq$-Inklusion in \eqcref{it:2:quiz:9}.
|
||||
Und offensichtlich gilt die $\subseteq$-Inklusion in \eqcref{it:1:quiz:9}.
|
||||
\end{rem*}
|
||||
|
||||
%% ********************************************************************************
|
||||
%% FILE: back/index.tex
|
||||
%% ********************************************************************************
|
||||
|
||||
Loading…
Reference in New Issue
Block a user