master > master: Quizzes 10+11
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%% ---- body/quizzes/quiz8.tex;
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%% ---- body/quizzes/quiz8.tex;
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%% ---- body/quizzes/quiz9.tex;
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%% ---- body/quizzes/quiz9.tex;
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%% |
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%% ---- body/quizzes/quiz10.tex;
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%% ---- body/quizzes/quiz11.tex;
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%% ---- back/index.tex;
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%% ---- back/index.tex;
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\ranewthm{fact}{Fakt}{\enndeOnNeutralSign}[X]
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\ranewthm{fact}{Fakt}{\enndeOnNeutralSign}[X]
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\ranewthm{rem}{Bemerkung}{\enndeOnNeutralSign}[X]
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\ranewthm{rem}{Bemerkung}{\enndeOnNeutralSign}[X]
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\ranewthm{qstn}{Frage}{\enndeOnNeutralSign}[X]
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\ranewthm{qstn}{Frage}{\enndeOnNeutralSign}[X]
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\ranewthm{exer}{Aufgabe}{\enndeOnNeutralSign}[X]
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\ranewthm{soln}{Lösung}{\enndeOnNeutralSign}[X]
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@ -10625,6 +10632,226 @@ für alle linearen Unterräume, $U\subseteq V$.
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$\psi\circ\phi$ surjektiv $\Rightarrow$ \eqcref{it:1:quiz:9}+\eqcref{it:2:quiz:9} gelten.
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$\psi\circ\phi$ surjektiv $\Rightarrow$ \eqcref{it:1:quiz:9}+\eqcref{it:2:quiz:9} gelten.
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\end{rem*}
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\end{rem*}
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%% ********************************************************************************
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%% FILE: body/quizzes/quiz10.tex
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%% ********************************************************************************
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\setcounternach{chapter}{10}
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\chapter[Woche 10]{Woche 10}
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\label{quiz:10}
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Seien
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\begin{mathe}[mc]{cccc}
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v_{1} = \begin{vector} 3\\ 2\\\end{vector},
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&v_{2} = \begin{vector} 2\\ 1\\\end{vector},
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&w_{1} = \begin{vector} 2\\ -1\\\end{vector},
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&w_{2} = \begin{vector} 0\\ 5\\\end{vector}.
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\end{mathe}
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\begin{enumerate}{\bfseries (a)}
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%% (a)
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\item
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\begin{claim*}
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$\cal{A}:=(v_{1},\,v_{2})$
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und $\cal{B}:=(w_{1},\,w_{2})$
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sind jeweils Basen von $\reell^{2}$.
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\end{claim*}
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\begin{proof}
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Da $\dim(\reell^{2})=2$, reicht es aus zu zeigen,
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dass $\cal{A}$ und $\cal{B}$ linear unabhängige Systeme sind.
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Hierfür reicht es aus \textbf{zu zeigen},
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das $\rank(A)=2$ und $\rank(B)=2$,
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wobei
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${A:=\left(v_{1}\ v_{2}\right)=\begin{smatrix}
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3 &2\\
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2 &1\\
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\end{smatrix}}$
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und
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${B:=\left(w_{1}\ w_{2}\right)=\begin{smatrix}
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2 &0\\
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-1 &5\\
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\end{smatrix}}$.
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Zeilenreduktion liefert uns
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\begin{mathe}[mc]{rcl}
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A
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&\xrightarrow{
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Z_{2}\mapsfrom 3\cdot Z_{2} - 2\cdot Z_{1}
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}
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&
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\begin{smatrix}
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3 &2\\
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0 &-1\\
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\end{smatrix}
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\\
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B
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&\xrightarrow{
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Z_{2}\mapsfrom 2\cdot Z_{2} + \cdot Z_{1}
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}
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&
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\begin{smatrix}
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2 &0\\
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0 &10\\
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\end{smatrix}
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\\
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\end{mathe}
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Also $\rank(A)=2$ und $\rank(B)=2$,
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wie zu zeigen war.
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\end{proof}
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%% (b)
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\item
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Sei $\cal{K}:=(e_{1},\,e_{2})$, die Standardbasis für $\reell^{2}$.
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Sei $\phi:\reell^{2}\to\reell^{2}$
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die eindeutige lineare Abbildung,
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die $\phi(v_{i})=w_{i}$ für $i\in\{1,2\}$ erfüllt.\footnote{
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Da $\cal{A}$ eine Basis von $\reell^{2}$ ist,
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definieren laut \cite[Satz~6.1.13]{sinn2020} diese Bedingungen eine (eindeutige)
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lineare Abbildung.
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}
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\textbf{Zu bestimmen:} die Matrizendarstellung $M:=M_{\cal{K}}^{\cal{K}}(\phi)$.
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\textbf{ANSATZ I}\\
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Wir versuchen, die Standardbasiselement in Bezug auf $\cal{A}$
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umzuschreiben, und berechnen die entsprechenden Outputvektoren:
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\begin{mathe}[mc]{rcccccl}
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\mathbf{e}_{1}
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&\textoverset{Defn}{=}
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&\begin{svector} 1\\ 0\\\end{svector}
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&= &2\begin{svector} 2\\ 1\\\end{svector}-\begin{svector} 3\\ 2\\\end{svector}
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&= &2v_{2}-v_{1}\\
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\mathbf{e}_{2}
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&\textoverset{Defn}{=}
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&\begin{svector} 0\\ 1\\\end{svector}
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&= &2\begin{svector} 3\\ 2\\\end{svector}-3\begin{svector} 2\\ 1\\\end{svector}
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&= &2v_{1}-3v_{2}\\
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\end{mathe}
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Also gilt wegen Linearität
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\begin{mathe}[mc]{rcccccccl}
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\phi(\mathbf{e}_{1})
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&= &\phi(2v_{2}-v_{1})
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&= &2\phi(v_{2})-\phi(v_{1})
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&= &2w_{2}-w_{1}
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&= &\begin{svector} -2\\ 11\\\end{svector}\\
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\phi(\mathbf{e}_{2})
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&= &\phi(2v_{1}-3v_{2})
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&= &2\phi(v_{1})-3\phi(v_{2})
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&= &2w_{1}-3w_{2}
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&= &\begin{svector} 4\\ -17\\\end{svector}\\
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\end{mathe}
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Da diese Outputvektoren schon in Bezug auf die Standardbasis dargestellt sind,
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erhalten wir
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\begin{mathe}[mc]{rcl}
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M_{\cal{K}}^{\cal{K}}(\phi)
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&= &\boxed{\begin{matrix}{rr}
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-2 &4\\
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11 &-17\\
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\end{matrix}}.\\
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\end{mathe}
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\textbf{ANSATZ II}\\
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In diesem Ansatz bestimmen wir auf systematische Weise
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notwendige Bedingungen dafür, dass eine Matrix, $M$, $\phi$ darstellt.
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Per Konstruktion, und da die Vektoren $v_{1},v_{2},w_{1},w_{2}$ bzgl. $\cal{K}$
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dargestellt wurden, muss
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\begin{mathe}[mc]{rcccl}
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Mv_{i} &= &\phi(v_{i}) &= &w_{i}
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\end{mathe}
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für alle $i\in\{1,2\}$ gelten.
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Mit anderen Worten muss $MA=B$ gelten,
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wobei $A,B$ die o.\,s. definierten Matrizen sind.
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Also ist eine notwendige Bedingung $M=BA^{-1}$.
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Darum ist $BA^{-1}$ \textbf{zu berechnen}.
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Hierfür gibt es mehrere Rechenwege.
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Wir arbeiten mit $\left(A^{T}\vert B^{T}\right)$
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und reduzieren, bis in der linken Hälfte die Identitätsmatrix, $\onematrix$,
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steht. In der rechten Hälfte steht dann $(A^{T})^{-1}B^{T}$,
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also $(BA^{-1})^{T}$.
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Das Resultat transponiert liefert uns dann $BA^{-1}$, also $M$.\footnote{
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Wir müssen diesen Umweg gehen, weil das Gaußverfahren uns nur
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nach linkst multiplizierte Inverse liefern kann und wir schließendlich
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$BA^{-1}$ berechnen wollen,
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was eine Rechtsmultiplikation durch das Inverse ist.
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}
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\begin{mathe}[mc]{rcl}
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\left(A^{T}\vert B^{T}\right)
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=
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\begin{matrix}{rr|rr}
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3 &2 &2 &-1\\
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2 &1 &0 &5\\
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\end{matrix}
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&\xrightarrow{
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Z_{2} \mapsfrom 3\cdot Z_{2}-2\cdot Z_{1}
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}
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&
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\begin{matrix}{rr|rr}
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3 &2 &2 &-1\\
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0 &-1 &-4 &17\\
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\end{matrix}
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\\
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&\xrightarrow{
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Z_{1} \mapsfrom Z_{1} + 2\cdot Z_{2}
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}
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&
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\begin{matrix}{rr|rr}
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3 &0 &-6 &33\\
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0 &-1 &-4 &17\\
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\end{matrix}
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\\
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&\xrightarrow{
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\substack{
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Z_{1} \mapsfrom 3^{-1}\cdot Z_{1}
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Z_{1} \mapsfrom -1\cdot Z_{2}
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}
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}
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&
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\begin{matrix}{rr|rr}
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1 &0 &-2 &11\\
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0 &1 &4 &-17\\
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\end{matrix}
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\\
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\end{mathe}
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Darum gilt notwendigerweise
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\begin{mathe}[mc]{rcccl}
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M &= &\begin{smatrix}
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-2 &11\\
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4 &-17\\
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\end{smatrix}^{T}
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&= &\boxed{\begin{matrix}{rr}
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-2 &4\\
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11 &-17\\
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\end{matrix}},\\
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\end{mathe}
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damit $M$ $\phi$ darstellt.
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Da es eine eindeutige Darstellungsmatrix für $\phi$ gibt,
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gilt somit $M_{\cal{K}}^{\cal{K}}(\phi)=M$.
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\end{enumerate}
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%% ********************************************************************************
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%% FILE: body/quizzes/quiz11.tex
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%% ********************************************************************************
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\setcounternach{chapter}{11}
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\chapter[Woche 11]{Woche 11}
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\label{quiz:11}
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(Siehe Git-Repo $\to$ \textbf{/notes/brerechnungen\_wk12.md}.)
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: back/index.tex
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%% FILE: back/index.tex
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%% ********************************************************************************
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%% ********************************************************************************
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