Die zugehörigen Aufgaben stehen immer im ersten Satz.
Hab ich das richtig gemacht?
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\( A=\left(\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right), \quad \ell_{A}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, x \mapsto A x \)
\( \begin{array}{l} \text { a) } \varphi_{A} \circ \ell_{A}=i d \mathbb{R}^{2} \quad\left(i d \mathbb{R}^{2}: \mathbb{R}^{2} \rightarrow \mathbb{R}_{1}^{2} x \mapsto x\right) \\ \varphi_{A}(x)=A x=\left(\begin{array}{cc} \hat{0} & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{c} x_{1}+x_{2} \\ -x_{2} \end{array}\right) \\ \Rightarrow \ell_{A} \circ \varphi_{A}=\varphi_{A}\left(\varphi_{A}(\vec{x})\right)=\varphi_{A}(A x)=A(A x) \\ =\left(\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} x_{1}+x_{2} \\ -x_{2} \end{array}\right)=\left(\begin{array}{c} x_{1}+x_{2}-x_{2} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=x=i d \mathbb{R}^{2} \end{array} \)
b) Ledes \( x \in \mathbb{R}^{2} \), hat eine eindeutige Darstellung
\( x=u+v \text { mit } \hat{e}_{A}(u)=u \quad \&{ }^{2} l_{A}(v)=-v \)
Es gilt:
\( \begin{array}{l} \ell_{A}(u)=u=\left(\begin{array}{l} u_{1} \\ u_{2} \end{array}\right)=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} u_{1} \\ u_{2} \end{array}\right)=A u \\ \ell_{A}(v)=-v=\left(\begin{array}{l} -v_{1} \\ -v_{2} \end{array}\right)=\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} v_{1} \\ v_{2} \end{array}\right)=A v \\ l_{A}(x)=l_{A}(u+v)=A(u+v) \\ =\left(\begin{array}{ll} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} u_{1}+v_{1} \\ u_{2}+v_{2} \end{array}\right)=\left(\begin{array}{cc} u_{1}+u_{2} & v_{1}+v_{2} \\ -u_{2} & -v_{2} \end{array}\right) \\ =\left(\begin{array}{ll} u_{1}+u_{2} \\ -u_{2} \end{array}\right)+\left(\begin{array}{l} v_{1}+v_{2} \\ -v_{2} \end{array}\right)=\left(\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} u_{1} \\ u_{2} \end{array}\right)+\left(\begin{array}{ll} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} v_{1} \\ v_{2} \end{array}\right) \\ =\left(\begin{array}{ll} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} u_{1}+v_{1} \\ u_{2}+v_{2} \end{array}\right) \end{array} \)
\( \begin{array}{l} l_{A}(u)+l_{A}(v)=A u+A v=\left(\begin{array}{cc} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} u_{1} \\ u_{2} \end{array}\right)+\left(\begin{array}{ll} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} v_{1} \\ v_{2} \end{array}\right) \\ =\left(\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} u_{1}+v_{1} \\ u_{2}+v_{2} \end{array}\right) \end{array} \)
Somit gilt \( l_{A}(u)+l_{A}(v)=l_{A}(u+\gamma)=l_{A}(x) \) Also hat jedes \( x \) eire Darstelleng.
