Hallo,
Die Eigenwerte stimmen.
Du setzt für λ die Eigenwerte ein:
Eigenvektoren:
\( \left|\begin{array}{cc}{1-λ} & {2} \\ {-2} & {1-λ}\end{array}\right|=0 \)
\( E V_{1}=1+2 i \)
\( \left(\begin{array}{cc}{-2 i} & {2} \\ {-2} & {-2 i}\end{array}\right)\left(\begin{array}{c}{y_{1}} \\ {y_2}\end{array}\right)=\left(\begin{array}{l}{0} \\ {0}\end{array}\right) \)
A) \( -2 i y_{1}+2 y_{2}=0 \)
B) \( -2 y_{1}-2 i y_{2}=0 \quad \Rightarrow \) redundant
A) \( \begin{aligned}-2i y_{1}+2 y_{2} &=0 \\-2 i y_{1} &=-2 y_2 \end{aligned} \)
\( \begin{array}{ll}y_{1} & =-i y_{2} \\ y_{2} & =a\end{array} \)
\( E V_{1}=\left(\begin{array}{r}-i a \\ a\end{array}\right)=a\left(\begin{array}{r}-i \\ 1\end{array}\right)=\left(\begin{array}{r}-i \\ 1\end{array}\right) \)
\( E V_{2}=1-2 i \)
A) \( \left(\begin{array}{cc}2 i & 2 \\ -2 & 2 i\end{array}\right)\left(\begin{array}{l}y_{1} \\ y_{2}\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right) \)
\( E V_{2}=\left(\begin{array}{l}i \\ 1\end{array}\right) \)
