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\documentclass[uebung]{../../../lecture}
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\author{Christian Merten}
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\title{Lineare Algebra 1: Übungsblatt Nr. 9}
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\usepackage[]{gauss}
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\begin{document}
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\punkte
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Es seien $f\colon V \to W$ und $g: W \to V$ lineare Abbildungen
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zwischen $V$ und $W$ Vektorräumen.
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Beh.: Es existiert genau dann ein $v \in V \setminus \{0\} $ mit $(g \circ f)(v) = v$, wenn
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es ein $w \in W \setminus \{0\} $ gibt mit $(f \circ g)(w) = w$.
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\begin{proof}
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,,$\implies$'' Es sei $w \in W$ mit $(f \circ g)(w) = w$. Dann
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definiere $v := g(w)$. Wegen $f(g(w)) = f(v) = w$ folgt $g(f(v)) = g(w) = v$.
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,,$\impliedby$'' folgt analog.
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\end{proof}
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\item Es sei $A \in M_{n,m}(K)$ und $B \in M_{m,n}(K)$.
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Beh.: $E_n - AB$ invertierbar $\iff$ $E_m - BA$ invertierbar.
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\begin{proof}
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,,$\implies$'' Es seien $a\colon K^{m} \to K^{n}$ und $b\colon K^{n} \to K^{m}$ die
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zu $A$ und $B$ gehörigen Abbildungen.
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Da $E_{n} - AB$ invertierbar, folgt $id_{K^{n}} - a \circ b$ ist Automorphismus.
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Also ist zu zeigen, dass
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der Endomorphismus $id_{K^{m}} - b \circ a$ bijektiv ist.
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Da $id_{K^{n}} - a \circ b$ bijektiv, insbesondere injektiv ist, folgt
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\begin{align*}
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&\text{ker}(id_{K^{n}} - a \circ b) = \{0\} \\
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\implies & id_{K^{n}}(v) - a(b(v)) \neq 0 \quad \forall v \in V \setminus \{0\} \\
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\implies &v \neq a(b(v)) \quad \forall v \in V \setminus \{0\}
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\intertext{Sei nun $w \in K^{m}$ mit $id_{K^{m}} - b(a(w)) = 0$.}
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\implies & w = b(a(w)) \\
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\stackrel{\text{1a)}}{\implies} &w = 0
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.\end{align*}
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Damit ist $id_{K^{m}} - b \circ a$ ein injektiver Endomorphismus, also
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auch bijektiv, also Automorphismus.\\
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$\implies E_m - BA$ invertierbar.
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,,$\impliedby$'' folgt analog.
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe} Es sei $K$ Körper und $A \in M_{n,m}(K)$ und $B \in M_{m,n}(K)$ mit $ABA = A$.
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\begin{enumerate}[(a)]
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\item Beh.: $\text{ker } A = \{x - BAx \mid x \in K^{m}\} $
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\begin{proof}
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Zz.: $\text{ker } A \subset \{x - BAx \mid x \in K^{m}\} $
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Sei $x \in \text{ker } A$, d.h. $Ax = 0$, damit:
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\[
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x - BAx = x - B\cdot 0 = x
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.\]
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Zz.: $\{x - BAx \mid x \in K^{m}\} \subset \text{ker } A$
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Sei $r \in K^{m}$, dann $x := r - BAr$. Damit folgt:
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\begin{align*}
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Ax = Ar - ABAr \stackrel{ABA=A}{=} Ar - Ar = 0
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.\end{align*}
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\end{proof}
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\item Beh.: $Ax = b$ hat eine Lösung $\iff ABb = b$
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\begin{proof}
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\begin{align*}
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& Ax = b \text{ hat eine Lösung} \\
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\iff & b \in \text{Bild}(A) \\
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\iff & \exists x \in K^{m}\colon Ax = ABAx = AB(Ax) = b \\
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\iff & ABb = b
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.\end{align*}
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\end{proof}
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Beh.: $L := \{x \in K^{m} \mid Ax = b\} = \{Bb + x' - BAx' \mid x' \in K^{m}\} $
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\begin{proof}
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\begin{enumerate}[(i)]
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\item Zz.: $L \subset \{Bb + x - BAx \mid x \in K^{m}\} $,
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Sei $x \in L$ beliebig, d.h. $Ax = b$. Nun g.z.z
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$\exists r \in K^{m}\colon x = Bb + r - BAr$. Wähle $k := x - Bb \in K^{m}$. Damit:
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\begin{align*}
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&Ak = Ax - ABb \stackrel{ABb = b}{=} b - b = 0 \\
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\implies &k \in \text{ker}(A)\\
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\stackrel{(a)}{\implies} & \exists r \in K^{m}\colon k = r - BAr. \text{ Fixiere }r \\
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\implies & Bb + r - BAr = Bb + k = Bb + x - Bb = x
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.\end{align*}
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\item Zz.: $\{Bb + x - BAx \mid x \in K^{m}\} \subset L$.
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Sei $r \in K^{m}$ beliebig, dann definiere $x := Bb + r - BAr \in K^{m}$.
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Nun g.z.z. $Ax = b$.
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\begin{align*}
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Ax = ABb + Ar - ABAr \stackrel{ABb = b}{=} b + Ar - ABAr
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\stackrel{ABA = A}{=} b
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.\end{align*}
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\end{enumerate}
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{align*}
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&\begin{gmatrix}[p]
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1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0
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\rowops
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\add[-1]{0}{2}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0
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\rowops
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\add[-1]{1}{0}
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\end{gmatrix}
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\to
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\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}
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\intertext{$\implies$ Rang 2}
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&\begin{gmatrix}[p]
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1 & 1 & 0 \\ 2 & 1 & 1 \\ 1 & 1 & 1
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\rowops
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\add[-2]{0}{1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 1 & 0 \\ 0 & -1 & 0 \\ 1 & 1 & 1
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\rowops
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\add[-1]{0}{2}
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\mult{1}{\scriptstyle\cdot-1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
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\rowops
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\add[-1]{1}{0}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1
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\end{gmatrix}
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\intertext{$\implies$ Rang 3}
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&\begin{gmatrix}[p]
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1 & 0 & 2 \\ 1 & 1 & 4
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\rowops
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\add[-1]{0}{1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 0 & 2 \\ 0 & 1 & 2
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\rowops
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\end{gmatrix}
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\intertext{$\implies$ Rang 2}
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&\begin{gmatrix}[p]
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1 & 2 & 1 \\ 2 & 4 & 2
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\rowops
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\add[-2]{0}{1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 2 & 1 \\ 0 & 0 & 0
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\end{gmatrix}
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\intertext{$\implies$ Rang 1}
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\intertext{Für $a = 1$ folgt direkt:}
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&\begin{gmatrix}[p]
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a & 1 & a \\ 1 & a & 1 \\ a & 1 & a
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\end{gmatrix}
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=
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\begin{gmatrix}[p] 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1
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\rowops
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\add[-1]{0}{1}
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\add[-1]{0}{2}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0
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\end{gmatrix}
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\intertext{$\implies$ Rang 1 \\Für $a = -1$ folgt}
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&\begin{gmatrix}[p]
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a & 1 & a \\ 1 & a & 1 \\ a & 1 & a
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\end{gmatrix}
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=
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\begin{gmatrix}[p] -1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & -1
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\rowops
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\add{0}{1}
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\add[-1]{0}{2}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0
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\end{gmatrix}
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\intertext{$\implies$ Rang 0\\ Für $a \neq 1 \land a \neq -1 \implies 1 - a^2 \neq 0$, damit:}
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&\begin{gmatrix}[p]
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a & 1 & a \\ 1 & a & 1 \\ a & 1 & a
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\rowops
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\add[-1]{0}{2}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] a & 1 & a \\ 1 & a & 1 \\ 0 & 0 & 0
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\rowops
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\swap{0}{1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & a & 1 \\ a & 1 & a \\ 0 & 0 & 0
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\rowops
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\add[-a]{0}{1}
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\end{gmatrix}
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\to
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\begin{gmatrix}[p] 1 & a & 1 \\ 0 & 1-a^2 & 0 \\ 0 & 0 & 0
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\rowops
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\mult{1}{\scriptstyle\cdot \frac{1}{1-a^2}}
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\end{gmatrix}\\
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\to
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&\begin{gmatrix}[p] 1 & a & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0
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\rowops
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\add[-a]{1}{0}
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\end{gmatrix}
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\to
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\begin{pmatrix}
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1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0
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\end{pmatrix}
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.\end{align*}
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$\implies$ Rang 2
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Beh.: $\underline{v} = \left( (1,2)^{t}, (0, -1)^{t} \right) $ ist Basis von $\Q^{2}$.
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\begin{proof}
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Zu zeigen.: $\underline{v}$ ist linear unabhängig
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Seien $a, b \in \Q$ mit
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\begin{align*}
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&a \cdot \binom{1}{2} + b \binom{0}{-1} = 0 \\
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\implies & a = 0 \land 2a -b = 0 \implies b = 0
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.\end{align*}
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$\implies$ $\underline{v}$ ist linear unabhängig wegen $\text{dim } \Q^{2} = 2$ eine Basis
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von $\Q^{2}$.
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\end{proof}
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Beh.: $\underline{w} = \left( (1,1)^{t}, (3,2)^{t} \right) $ is Basis von $\Q^{2}$
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\begin{proof}
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Zu zeigen.: $\underline{v}$ ist linear unabhängig
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Seien $a, b \in \Q$ mit
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\begin{align*}
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&a \cdot \binom{1}{1} + b \binom{3}{2} = 0 \\
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\implies & a + 3b = 0 \land a +2b = 0
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\implies b = a = 0
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.\end{align*}
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$\implies$ $\underline{v}$ ist linear unabhängig wegen $\text{dim } \Q^{2} = 2$ eine Basis
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von $\Q^{2}$.
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\end{proof}
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Beh.:
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\[
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T = M_{\underline{e}}^{\underline{v}}(\text{id}_V) =
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\begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}
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.\]
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\begin{proof}
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Zu Überprüfen für die zwei Basisvektoren aus $\underline{v}$.
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\begin{enumerate}[(i)]
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\item $v_1 = (1,2)^{t}$. $\phi(v_1) = (1,0)^{t}$.
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\begin{align*}
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\begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}
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\cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix}
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= \begin{pmatrix} 1 \\ 2 \end{pmatrix}
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.\end{align*}
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\item $v_2 = (0,-1)^{t}$. $\phi(v_2) = (0,1)^{t}$.
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\begin{align*}
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\begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix}
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\cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix}
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= \begin{pmatrix} 0 \\ -1 \end{pmatrix}
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.\end{align*}
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\end{enumerate}
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\end{proof}
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|
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|
Beh.:
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\[
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T = M_{\underline{e}}^{\underline{w}}(\text{id}_V) =
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\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}
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.\]
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\begin{proof}
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|
Zu Überprüfen für die zwei Basisvektoren aus $\underline{w}$.
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\begin{enumerate}[(i)]
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|
\item $w_1 = (1,1)^{t}$. $\phi(w_1) = (1,0)^{t}$.
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|
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\begin{align*}
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\begin{pmatrix} 1 & 3 \\ 1 & 2 \end{pmatrix}
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\cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix}
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= \begin{pmatrix} 1 \\ 1 \end{pmatrix}
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.\end{align*}
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\item $w_2 = (3,2)^{t}$. $\phi(w_2) = (0,1)^{t}$.
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|
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\begin{align*}
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\begin{pmatrix} 1 & 3 \\ 1 & 2 \end{pmatrix}
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\cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix}
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= \begin{pmatrix} 3 \\ 2 \end{pmatrix}
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.\end{align*}
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\end{enumerate}
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\end{proof}
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|
\item
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\begin{align*}
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&\begin{gmatrix}[p]
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1 & 0 \\ 2 & -1
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\end{gmatrix}
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|
\begin{gmatrix}[p]
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|
1 & 0 \\ 0 & 1
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\rowops
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\add[-2]{0}{1}
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\mult{1}{\scriptstyle\cdot -1}
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\end{gmatrix}
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|
\to
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|
\begin{gmatrix}[p]
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|
1 & 0 \\ 0 & 1
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|
\end{gmatrix}
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|
\begin{gmatrix}[p]
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|
1 & 0 \\ 2 & -1
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|
\end{gmatrix}
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|
\intertext{$\implies T = \left(M_{\underline{e}}^{\underline{v}}\right)^{-1} =
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\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix}$ }
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|
&\begin{gmatrix}[p]
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|
1 & 3 \\ 1 & 2
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|
\end{gmatrix}
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||||||
|
\begin{gmatrix}[p]
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||||||
|
1 & 0 \\ 0 & 1
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||||||
|
\rowops
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||||||
|
\add[-1]{0}{1}
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||||||
|
\mult{1}{\scriptstyle\cdot -1}
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||||||
|
\end{gmatrix}
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|
\to
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|
\begin{gmatrix}[p]
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||||||
|
1 & 3 \\ 0 & 1
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||||||
|
\end{gmatrix}
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||||||
|
\begin{gmatrix}[p]
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||||||
|
1 & 0 \\ 1 & -1
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||||||
|
\rowops
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||||||
|
\add[-3]{1}{0}
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||||||
|
\end{gmatrix}
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||||||
|
\to
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||||||
|
\begin{gmatrix}[p]
|
||||||
|
1 & 0 \\ 0 & 1
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||||||
|
\end{gmatrix}
|
||||||
|
\begin{gmatrix}[p]
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||||||
|
-2 & 3 \\ 1 & -1
|
||||||
|
\end{gmatrix}
|
||||||
|
.\end{align*}
|
||||||
|
$\implies S = \left(M_{\underline{e}}^{\underline{w}}\right)^{-1} =
|
||||||
|
\begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix}$
|
||||||
|
\item $M_{\underline{e}}^{\underline{e}}(f) = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} $
|
||||||
|
durch ablesen, die restlichen Matrizen ergeben sich durch Multiplikation:
|
||||||
|
\begin{align*}
|
||||||
|
&M_{\underline{v}}^{\underline{v}}(f)
|
||||||
|
= M_{\underline{v}}^{\underline{e}}(\text{id}_V) \cdot M_{\underline{e}}^{\underline{e}}(f)
|
||||||
|
\cdot M_{\underline{e}}^{\underline{v}}(\text{id}_V)
|
||||||
|
= \begin{pmatrix} 5 & -2 \\ -3 & 1 \end{pmatrix} \\
|
||||||
|
&M_{\underline{w}}^{\underline{w}}(f)
|
||||||
|
= M_{\underline{w}}^{\underline{e}}(\text{id}_V) \cdot M_{\underline{e}}^{\underline{e}}(f)
|
||||||
|
\cdot M_{\underline{e}}^{\underline{w}}(\text{id}_V)
|
||||||
|
= \begin{pmatrix} -12 & -29 \\ 5 & 12 \end{pmatrix} \\
|
||||||
|
&M_{\underline{v}}^{\underline{w}}(\text{id}_{V})
|
||||||
|
= M_{\underline{v}}^{\underline{e}}(\text{id}_V)
|
||||||
|
\cdot M_{\underline{e}}^{\underline{w}}(\text{id}_V)
|
||||||
|
= \begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix}
|
||||||
|
.\end{align*}
|
||||||
|
\item
|
||||||
|
\begin{align*}
|
||||||
|
AC - CB = M_{\underline{v}}^{\underline{v}}(f)
|
||||||
|
\cdot M_{\underline{v}}^{\underline{w}}(\text{id}_V)
|
||||||
|
- M_{\underline{v}}^{\underline{w}}(\text{id}_V)
|
||||||
|
\cdot M_{\underline{w}}^{\underline{w}}(f)
|
||||||
|
= M_{\underline{v}}^{\underline{w}}(f)
|
||||||
|
- M_{\underline{v}}^{\underline{w}}(f)
|
||||||
|
= 0
|
||||||
|
.\end{align*}
|
||||||
|
\end{enumerate}
|
||||||
|
\end{aufgabe}
|
||||||
|
|
||||||
|
\end{document}
|
||||||
Reference in New Issue
Block a user