|
|
|
@ -90,6 +90,10 @@
|
|
|
|
|
%% ---- body/quizzes/quiz8.tex; |
|
|
|
|
%% | |
|
|
|
|
%% ---- body/quizzes/quiz9.tex; |
|
|
|
|
%% | |
|
|
|
|
%% ---- body/quizzes/quiz10.tex; |
|
|
|
|
%% | |
|
|
|
|
%% ---- body/quizzes/quiz11.tex; |
|
|
|
|
%% | |
|
|
|
|
%% ---- back/index.tex; |
|
|
|
|
%% | |
|
|
|
@ -344,6 +348,7 @@
|
|
|
|
|
|
|
|
|
|
\newcount\bufferctr |
|
|
|
|
\newcount\bufferreplace |
|
|
|
|
\newcounter{columnanzahl} |
|
|
|
|
|
|
|
|
|
\newlength\rtab |
|
|
|
|
\newlength\gesamtlinkerRand |
|
|
|
@ -657,15 +662,15 @@
|
|
|
|
|
\theoremstyle{nonumberplain} |
|
|
|
|
\theoremseparator{\thmForceSepPt} |
|
|
|
|
\theoremprework{\ra@pretheoremwork} |
|
|
|
|
\@ifundefined{#1@star@basic}{\newtheorem{#1@star@basic}[#4]{#2}}{\renewtheorem{#1@star@basic}[#4]{#2}} |
|
|
|
|
\@ifundefined{#1@star@basic}{\newtheorem{#1@star@basic}[Xdisplaynone]{#2}}{\renewtheorem{#1@star@basic}[Xdisplaynone]{#2}} |
|
|
|
|
%% FOR \BEGIN{THM*}[...] |
|
|
|
|
\theoremstyle{nonumberplain} |
|
|
|
|
\theoremseparator{\thmForceSepPt} |
|
|
|
|
\theoremprework{\ra@pretheoremwork} |
|
|
|
|
\@ifundefined{#1@star@withName}{\newtheorem{#1@star@withName}[#4]{#2}}{\renewtheorem{#1@star@withName}[#4]{#2}} |
|
|
|
|
\@ifundefined{#1@star@withName}{\newtheorem{#1@star@withName}[Xdisplaynone]{#2}}{\renewtheorem{#1@star@withName}[Xdisplaynone]{#2}} |
|
|
|
|
%% GENERATE ENVIRONMENTS: |
|
|
|
|
\umbauenenv{#1}{#3}[#4] |
|
|
|
|
\umbauenenv{#1@star}{#3}[#4] |
|
|
|
|
\umbauenenv{#1@star}{#3}[Xdisplaynone] |
|
|
|
|
%% TRANSFER *-DEFINITION |
|
|
|
|
\rathmtransfer{#1@star}{#1*} |
|
|
|
|
} |
|
|
|
@ -752,6 +757,8 @@
|
|
|
|
|
\ranewthm{fact}{Fakt}{\enndeOnNeutralSign}[X] |
|
|
|
|
\ranewthm{rem}{Bemerkung}{\enndeOnNeutralSign}[X] |
|
|
|
|
\ranewthm{qstn}{Frage}{\enndeOnNeutralSign}[X] |
|
|
|
|
\ranewthm{exer}{Aufgabe}{\enndeOnNeutralSign}[X] |
|
|
|
|
\ranewthm{soln}{Lösung}{\enndeOnNeutralSign}[X] |
|
|
|
|
|
|
|
|
|
\theoremheaderfont{\itshape\bfseries} |
|
|
|
|
\theorembodyfont{\upshape} |
|
|
|
@ -10625,6 +10632,226 @@ für alle linearen Unterräume, $U\subseteq V$.
|
|
|
|
|
$\psi\circ\phi$ surjektiv $\Rightarrow$ \eqcref{it:1:quiz:9}+\eqcref{it:2:quiz:9} gelten. |
|
|
|
|
\end{rem*} |
|
|
|
|
|
|
|
|
|
%% ******************************************************************************** |
|
|
|
|
%% FILE: body/quizzes/quiz10.tex |
|
|
|
|
%% ******************************************************************************** |
|
|
|
|
|
|
|
|
|
\setcounternach{chapter}{10} |
|
|
|
|
\chapter[Woche 10]{Woche 10} |
|
|
|
|
\label{quiz:10} |
|
|
|
|
|
|
|
|
|
Seien |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{cccc} |
|
|
|
|
v_{1} = \begin{vector} 3\\ 2\\\end{vector}, |
|
|
|
|
&v_{2} = \begin{vector} 2\\ 1\\\end{vector}, |
|
|
|
|
&w_{1} = \begin{vector} 2\\ -1\\\end{vector}, |
|
|
|
|
&w_{2} = \begin{vector} 0\\ 5\\\end{vector}. |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
\begin{enumerate}{\bfseries (a)} |
|
|
|
|
%% (a) |
|
|
|
|
\item |
|
|
|
|
\begin{claim*} |
|
|
|
|
$\cal{A}:=(v_{1},\,v_{2})$ |
|
|
|
|
und $\cal{B}:=(w_{1},\,w_{2})$ |
|
|
|
|
sind jeweils Basen von $\reell^{2}$. |
|
|
|
|
\end{claim*} |
|
|
|
|
|
|
|
|
|
\begin{proof} |
|
|
|
|
Da $\dim(\reell^{2})=2$, reicht es aus zu zeigen, |
|
|
|
|
dass $\cal{A}$ und $\cal{B}$ linear unabhängige Systeme sind. |
|
|
|
|
Hierfür reicht es aus \textbf{zu zeigen}, |
|
|
|
|
das $\rank(A)=2$ und $\rank(B)=2$, |
|
|
|
|
wobei |
|
|
|
|
${A:=\left(v_{1}\ v_{2}\right)=\begin{smatrix} |
|
|
|
|
3 &2\\ |
|
|
|
|
2 &1\\ |
|
|
|
|
\end{smatrix}}$ |
|
|
|
|
und |
|
|
|
|
${B:=\left(w_{1}\ w_{2}\right)=\begin{smatrix} |
|
|
|
|
2 &0\\ |
|
|
|
|
-1 &5\\ |
|
|
|
|
\end{smatrix}}$. |
|
|
|
|
Zeilenreduktion liefert uns |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl} |
|
|
|
|
A |
|
|
|
|
&\xrightarrow{ |
|
|
|
|
Z_{2}\mapsfrom 3\cdot Z_{2} - 2\cdot Z_{1} |
|
|
|
|
} |
|
|
|
|
& |
|
|
|
|
\begin{smatrix} |
|
|
|
|
3 &2\\ |
|
|
|
|
0 &-1\\ |
|
|
|
|
\end{smatrix} |
|
|
|
|
\\ |
|
|
|
|
B |
|
|
|
|
&\xrightarrow{ |
|
|
|
|
Z_{2}\mapsfrom 2\cdot Z_{2} + \cdot Z_{1} |
|
|
|
|
} |
|
|
|
|
& |
|
|
|
|
\begin{smatrix} |
|
|
|
|
2 &0\\ |
|
|
|
|
0 &10\\ |
|
|
|
|
\end{smatrix} |
|
|
|
|
\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
Also $\rank(A)=2$ und $\rank(B)=2$, |
|
|
|
|
wie zu zeigen war. |
|
|
|
|
\end{proof} |
|
|
|
|
|
|
|
|
|
%% (b) |
|
|
|
|
\item |
|
|
|
|
Sei $\cal{K}:=(e_{1},\,e_{2})$, die Standardbasis für $\reell^{2}$. |
|
|
|
|
Sei $\phi:\reell^{2}\to\reell^{2}$ |
|
|
|
|
die eindeutige lineare Abbildung, |
|
|
|
|
die $\phi(v_{i})=w_{i}$ für $i\in\{1,2\}$ erfüllt.\footnote{ |
|
|
|
|
Da $\cal{A}$ eine Basis von $\reell^{2}$ ist, |
|
|
|
|
definieren laut \cite[Satz~6.1.13]{sinn2020} diese Bedingungen eine (eindeutige) |
|
|
|
|
lineare Abbildung. |
|
|
|
|
} |
|
|
|
|
\textbf{Zu bestimmen:} die Matrizendarstellung $M:=M_{\cal{K}}^{\cal{K}}(\phi)$. |
|
|
|
|
|
|
|
|
|
\textbf{ANSATZ I}\\ |
|
|
|
|
Wir versuchen, die Standardbasiselement in Bezug auf $\cal{A}$ |
|
|
|
|
umzuschreiben, und berechnen die entsprechenden Outputvektoren: |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccccl} |
|
|
|
|
\mathbf{e}_{1} |
|
|
|
|
&\textoverset{Defn}{=} |
|
|
|
|
&\begin{svector} 1\\ 0\\\end{svector} |
|
|
|
|
&= &2\begin{svector} 2\\ 1\\\end{svector}-\begin{svector} 3\\ 2\\\end{svector} |
|
|
|
|
&= &2v_{2}-v_{1}\\ |
|
|
|
|
\mathbf{e}_{2} |
|
|
|
|
&\textoverset{Defn}{=} |
|
|
|
|
&\begin{svector} 0\\ 1\\\end{svector} |
|
|
|
|
&= &2\begin{svector} 3\\ 2\\\end{svector}-3\begin{svector} 2\\ 1\\\end{svector} |
|
|
|
|
&= &2v_{1}-3v_{2}\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
Also gilt wegen Linearität |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccccccl} |
|
|
|
|
\phi(\mathbf{e}_{1}) |
|
|
|
|
&= &\phi(2v_{2}-v_{1}) |
|
|
|
|
&= &2\phi(v_{2})-\phi(v_{1}) |
|
|
|
|
&= &2w_{2}-w_{1} |
|
|
|
|
&= &\begin{svector} -2\\ 11\\\end{svector}\\ |
|
|
|
|
\phi(\mathbf{e}_{2}) |
|
|
|
|
&= &\phi(2v_{1}-3v_{2}) |
|
|
|
|
&= &2\phi(v_{1})-3\phi(v_{2}) |
|
|
|
|
&= &2w_{1}-3w_{2} |
|
|
|
|
&= &\begin{svector} 4\\ -17\\\end{svector}\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
Da diese Outputvektoren schon in Bezug auf die Standardbasis dargestellt sind, |
|
|
|
|
erhalten wir |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl} |
|
|
|
|
M_{\cal{K}}^{\cal{K}}(\phi) |
|
|
|
|
&= &\boxed{\begin{matrix}{rr} |
|
|
|
|
-2 &4\\ |
|
|
|
|
11 &-17\\ |
|
|
|
|
\end{matrix}}.\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
\textbf{ANSATZ II}\\ |
|
|
|
|
In diesem Ansatz bestimmen wir auf systematische Weise |
|
|
|
|
notwendige Bedingungen dafür, dass eine Matrix, $M$, $\phi$ darstellt. |
|
|
|
|
Per Konstruktion, und da die Vektoren $v_{1},v_{2},w_{1},w_{2}$ bzgl. $\cal{K}$ |
|
|
|
|
dargestellt wurden, muss |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl} |
|
|
|
|
Mv_{i} &= &\phi(v_{i}) &= &w_{i} |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
für alle $i\in\{1,2\}$ gelten. |
|
|
|
|
Mit anderen Worten muss $MA=B$ gelten, |
|
|
|
|
wobei $A,B$ die o.\,s. definierten Matrizen sind. |
|
|
|
|
Also ist eine notwendige Bedingung $M=BA^{-1}$. |
|
|
|
|
Darum ist $BA^{-1}$ \textbf{zu berechnen}. |
|
|
|
|
|
|
|
|
|
Hierfür gibt es mehrere Rechenwege. |
|
|
|
|
Wir arbeiten mit $\left(A^{T}\vert B^{T}\right)$ |
|
|
|
|
und reduzieren, bis in der linken Hälfte die Identitätsmatrix, $\onematrix$, |
|
|
|
|
steht. In der rechten Hälfte steht dann $(A^{T})^{-1}B^{T}$, |
|
|
|
|
also $(BA^{-1})^{T}$. |
|
|
|
|
Das Resultat transponiert liefert uns dann $BA^{-1}$, also $M$.\footnote{ |
|
|
|
|
Wir müssen diesen Umweg gehen, weil das Gaußverfahren uns nur |
|
|
|
|
nach linkst multiplizierte Inverse liefern kann und wir schließendlich |
|
|
|
|
$BA^{-1}$ berechnen wollen, |
|
|
|
|
was eine Rechtsmultiplikation durch das Inverse ist. |
|
|
|
|
} |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcl} |
|
|
|
|
\left(A^{T}\vert B^{T}\right) |
|
|
|
|
= |
|
|
|
|
\begin{matrix}{rr|rr} |
|
|
|
|
3 &2 &2 &-1\\ |
|
|
|
|
2 &1 &0 &5\\ |
|
|
|
|
\end{matrix} |
|
|
|
|
&\xrightarrow{ |
|
|
|
|
Z_{2} \mapsfrom 3\cdot Z_{2}-2\cdot Z_{1} |
|
|
|
|
} |
|
|
|
|
& |
|
|
|
|
\begin{matrix}{rr|rr} |
|
|
|
|
3 &2 &2 &-1\\ |
|
|
|
|
0 &-1 &-4 &17\\ |
|
|
|
|
\end{matrix} |
|
|
|
|
\\ |
|
|
|
|
&\xrightarrow{ |
|
|
|
|
Z_{1} \mapsfrom Z_{1} + 2\cdot Z_{2} |
|
|
|
|
} |
|
|
|
|
& |
|
|
|
|
\begin{matrix}{rr|rr} |
|
|
|
|
3 &0 &-6 &33\\ |
|
|
|
|
0 &-1 &-4 &17\\ |
|
|
|
|
\end{matrix} |
|
|
|
|
\\ |
|
|
|
|
&\xrightarrow{ |
|
|
|
|
\substack{ |
|
|
|
|
Z_{1} \mapsfrom 3^{-1}\cdot Z_{1} |
|
|
|
|
Z_{1} \mapsfrom -1\cdot Z_{2} |
|
|
|
|
} |
|
|
|
|
} |
|
|
|
|
& |
|
|
|
|
\begin{matrix}{rr|rr} |
|
|
|
|
1 &0 &-2 &11\\ |
|
|
|
|
0 &1 &4 &-17\\ |
|
|
|
|
\end{matrix} |
|
|
|
|
\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
Darum gilt notwendigerweise |
|
|
|
|
|
|
|
|
|
\begin{mathe}[mc]{rcccl} |
|
|
|
|
M &= &\begin{smatrix} |
|
|
|
|
-2 &11\\ |
|
|
|
|
4 &-17\\ |
|
|
|
|
\end{smatrix}^{T} |
|
|
|
|
&= &\boxed{\begin{matrix}{rr} |
|
|
|
|
-2 &4\\ |
|
|
|
|
11 &-17\\ |
|
|
|
|
\end{matrix}},\\ |
|
|
|
|
\end{mathe} |
|
|
|
|
|
|
|
|
|
damit $M$ $\phi$ darstellt. |
|
|
|
|
Da es eine eindeutige Darstellungsmatrix für $\phi$ gibt, |
|
|
|
|
gilt somit $M_{\cal{K}}^{\cal{K}}(\phi)=M$. |
|
|
|
|
\end{enumerate} |
|
|
|
|
|
|
|
|
|
%% ******************************************************************************** |
|
|
|
|
%% FILE: body/quizzes/quiz11.tex |
|
|
|
|
%% ******************************************************************************** |
|
|
|
|
|
|
|
|
|
\setcounternach{chapter}{11} |
|
|
|
|
\chapter[Woche 11]{Woche 11} |
|
|
|
|
\label{quiz:11} |
|
|
|
|
|
|
|
|
|
(Siehe Git-Repo $\to$ \textbf{/notes/brerechnungen\_wk12.md}.) |
|
|
|
|
|
|
|
|
|
%% ******************************************************************************** |
|
|
|
|
%% FILE: back/index.tex |
|
|
|
|
%% ******************************************************************************** |
|
|
|
|