Hallo alle zusammen!
Es handelt sich wieder um Matrixmultiplikation, diesmal geht es um die Hintereinanderausführung f•g
Aufgabe: Seien f und g zwei lineare Abbildungen. Bestimme die Abbildungsmatrix von f und, falls möglich die Abbildungsmatrix von f ◦ g.
Könnt ihr mir bitte wieder eine Rückmeldung geben und ggf. meine Fehler korrigieren?
Problem/Ansatz:
\( f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3},\left(\begin{array}{l}x \\ y\end{array}\right) \mapsto\left(\begin{array}{c}-x+2 y \\ 2 y-x \\ 2 x+y\end{array}\right) \)
\( g: \mathbb{R}^{4} \rightarrow \mathbb{R}^{2}\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{1}\end{array}\right) \mapsto\left(\begin{array}{cccc}1 & -1 & 2 & 1 \\ 1 & 0 & 1 & 0\end{array}\right)\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right)= \)
\( \beta_{1}=\left\{\left(\begin{array}{l}1 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 1\end{array}\right)\right\} \)
\( \beta_{2}=\left\{\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)\right\} \)
\( \beta_{3}=\left\{\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 1 \\ 0 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 0 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 0 \\ 0 \\ 1\end{array}\right)\right\} \)
\( f\left(\left[\begin{array}{l}1 \\ 0\end{array}\right]_{B_{1}}\right)=f\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=\left[\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right]_{B_{3}} \)
\( f\left(\left[\begin{array}{l}0 \\ 1\end{array}\right]_{B_{1}}\right)=f\left(\begin{array}{l}0 \\ 1\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)=\left[\begin{array}{l}2 \\ 2 \\ 1\end{array}\right]_{B_{3}} \)
\( F=\left(\begin{array}{cc}-1 & 2 \\ -1 & 2 \\ 2 & 1\end{array}\right) \)
\( G=\left(\begin{array}{cccc}1 & -1 & 2 & 1 \\ 1 & 0 & 1 & 0\end{array}\right) \)
\( f \circ g=F \cdot G=\left(\begin{array}{cc}-1 & 2 \\ -1 & 2 \\ 2 & 1\end{array}\right) \cdot\left(\begin{array}{cccc}1 & -1 & 2 & 1 \\ 1 & 0 & 1 & 0\end{array}\right)=\left(\begin{array}{cccc}1 & 1 & 0 & -1 \\ 1 & 1 & 0 & -1 \\ 3 & -2 & 5 & 2\end{array}\right) \)
d) \( f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},\left(\begin{array}{l}x \\ y \\ \vdots\end{array}\right) \mapsto\left(\begin{array}{c}-x+2 y-3 z \\ 2 y+z\end{array}\right) \)
g : \( \mathbb{R}^{4} \rightarrow \mathbb{R}^{3},\left(\begin{array}{l}x_{1} \\ x_{6} \\ x_{4}\end{array}\right) \mapsto\left(\begin{array}{cccc}2 & 1 & 2 & -1 \\ -1 & 2 & 0 & 1 \\ 2 & 3 & 1 & -1\end{array}\right)\left(\begin{array}{l}x_{1} \\ x_{3} \\ x_{3} \\ x_{4}\end{array}\right)= \)
\( \beta_{1}=\left\{\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)\right\} \)
\( f\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]_{B_{1}}=f\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{c}-1 \\ 0\end{array}\right) \)
\( f\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]_{\beta_{1}}=f\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{l}2 \\ 2\end{array}\right) \)
\( f\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]_{\beta_{1}}=f\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\left(\begin{array}{c}-3 \\ 1\end{array}\right) \)
\( F=\left(\begin{array}{ccc}-1 & 2 & -3 \\ 0 & 2 & 1\end{array}\right) \quad G=\left(\begin{array}{cccc}2 & 1 & 2 & -1 \\ -1 & 2 & 0 & 1 \\ 2 & 3 & 1 & -1\end{array}\right) \)
\( f \cdot g=F \cdot G=\left(\begin{array}{ccc}-1 & 2 & -3 \\ 0 & 2 & 1\end{array}\right)\left(\begin{array}{cccc}2 & 1 & 2 & -1 \\ -1 & 2 & 0 & 1 \\ 2 & 3 & 1 & -1\end{array}\right)=\left(\begin{array}{cccc}-10 & -6 & -5 & 6 \\ 0 & 7 & 1 & 1\end{array}\right) \)
