Aufgabe:
4. Gegeben sei die lineare Abbildung \( f: \mathbb{R}^{4} \rightarrow \mathbb{R}^{3} \) mit
$$ \left(\begin{array}{l} {1} \\ {1} \\ {0} \\ {0} \end{array}\right) \stackrel{f}{\rightarrow}\left(\begin{array}{l} {1} \\ {1} \\ {1} \end{array}\right) , \left(\begin{array}{l} {2} \\ {2} \\ {2} \\ {2} \end{array}\right) \stackrel{f}{\rightarrow}\left(\begin{array}{l} {3} \\ {3} \\ {3} \end{array}\right),\left(\begin{array}{c} {3} \\ {1} \\ {-3} \\ {0} \end{array}\right) \stackrel{f}{\rightarrow}\left(\begin{array}{l} {0} \\ {4} \\ {4} \end{array}\right),\left(\begin{array}{c} {0} \\ {-3} \\ {1} \\ {3} \end{array}\right) \stackrel{f}{\rightarrow}\left(\begin{array}{l} {2} \\ {0} \\ {2} \end{array}\right) $$
Bestimmen Sie die darstellende Matrix von \( f \) bezüglich der Basen \( \mathcal{A}=\left\{\left(\begin{array}{l}{1} \\ {0} \\ {0} \\ {0}\end{array}\right),\left(\begin{array}{l}{1} \\ {1} \\ {0} \\ {0}\end{array}\right),\left(\begin{array}{l}{1} \\ {1} \\ {1} \\ {0}\end{array}\right),\left(\begin{array}{l}{1} \\ {1} \\ {1} \\ {1}\end{array}\right)\right\} \) von \( \mathbb{R}^{4} \) und \( \mathcal{B}=\left\{\left(\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right),\left(\begin{array}{l}{0} \\ {1} \\ {1}\end{array}\right), \left(\begin{array}{l}{0} \\ {0} \\ {1}\end{array}\right)\right\} \operatorname{von} \mathbb{R}^{3} \)
Problem:
Wie kommen die auf die Darstellende Matrix...ab "Einsetzen ergibt..."?
Lösung
\( \left(\begin{array}{cccc|c}{1} & {2} & {3} & {0} & {b_{1}} \\ {1} & {2} & {1} & {-3} & {b_{2}} \\ {0} & {2} & {-3} & {1} & {b_{3}} \\ {0} & {2} & {0} & {3} & {b_{4}}\end{array}\right) \quad \underset{\mathbf{IV}-\mathbf{III}}{\mathbf{II}-\mathbf{I}} \)
\( \left(\begin{array}{cccc|c}{1} & {2} & {3} & {0} & {b_{1}} \\ {0} & {0} & {-2} & {-3} & {b_{2}-b_{1}} \\ {0} & {2} & {-3} & {1} & {b_{3}} \\ {0} & {0} & {3} & {2} & {b_{4}-b_{3}}\end{array}\right) \quad \underset{2 \cdot IV}{\text{II++III}}\)
\( \left(\begin{array}{cccc|c}{1} & {2} & {3} & {0} & {b_{1}} \\ {0} & {2} & {-3} & {1} & {b_{3}} \\ {0} & {0} & {-2} & {-3} & {b_{2}-b_{1}} \\ {0} & {0} & {0} & {-5} & {2 b_{4}-2 b_{3}+3 b_{2}-3 b_{1}}\end{array}\right) \)
Für \( b=\left(\begin{array}{l}{1} \\ {0} \\ {0} \\ {0}\end{array}\right) \) ergibt sich \( \lambda_{4}=\frac{3}{5}, \lambda_{3}=-\frac{2}{5} \lambda_{2}=-\frac{9}{10} \) und \( \lambda_{1}=4 . \) Für \( b=\left(\begin{array}{l}{1} \\ {1} \\ {1} \\ {0}\end{array}\right) \)
ergibt sich \( \lambda_{4}=\frac{2}{5}, \lambda_{3}=-\frac{1}{5} \lambda_{2}=-\frac{1}{5} \) und \( \lambda_{1}=4 \)
Setzen wir \( v_{1}=\left(\begin{array}{l}{1} \\ {1} \\ {0} \\ {0}\end{array}\right), v_{2}=\left(\begin{array}{l}{2} \\ {2} \\ {2} \\ {2}\end{array}\right), v_{3}=\left(\begin{array}{c}{3} \\ {1} \\ {-3} \\ {0}\end{array}\right) \) und \( v_{4}=\left(\begin{array}{c}{0} \\ {-3} \\ {1} \\ {3}\end{array}\right), \) so erhalten wir
\( \cdot\left(\begin{array}{l}{1} \\ {0} \\ {0} \\ {0}\end{array}\right)=4 v_{1}-\frac{9}{10} v_{2}-\frac{2}{5} v_{3}+\frac{3}{5} v_{4} ;\left(\begin{array}{l}{1} \\ {1} \\ {0} \\ {0}\end{array}\right)=v_{1} \)
\( \cdot\left(\begin{array}{l}{1} \\ {1} \\ {1} \\ {0}\end{array}\right)=4 v_{1}-\frac{3}{5} v_{2}-\frac{3}{5} v_{3}+\frac{2}{5} v_{4} ; \quad\left(\begin{array}{l}{1} \\ {1} \\ {1}\\{1}\end{array}\right)=\frac{1}{2} v_{2} \)
Wir sehen direkt \( \left(\begin{array}{l}{2} \\ {0} \\ {2}\end{array}\right)=2\left(\left(\begin{array}{l}{1} \\ {1} \\ {1}\end{array}\right)-\left(\begin{array}{l}{0} \\ {1} \\ {1}\end{array}\right)+\left(\begin{array}{l}{0} \\ {0} \\ {1}\end{array}\right)\right) . \) Einsetzen ergibt
$$ M_{\mathcal{A}}^{\mathcal{B}}(f)=\left(\begin{array}{cccc} {5 / 2} & {1} & {3} & {3 / 2} \\ {-14 / 5} & {0} & {-16 / 5} & {0} \\ {6 / 5} & {0} & {4 / 5} & {0} \end{array}\right) $$
