master > master: Quizzes 10+11
This commit is contained in:
parent
e320f430a5
commit
97b3a8738c
Binary file not shown.
@ -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
|
||||
%% ********************************************************************************
|
||||
|
||||
Loading…
Reference in New Issue
Block a user