$$\begin{matrix} x_1' &= &a_{11}x_1 + a_{12}x_2 \cr x_2' &= &a_{21}x_1 + a_{22}x_2 \cr \cr \vec{x}' &= &\begin{pmatrix}a_{11} &a_{12}\cr a_{21} &a_{22}\cr\end{pmatrix} \vec{x}\cr\end{matrix}$$ a. Bestimmung der Eigenwerte $$\begin{matrix} \lambda_{1,2} &= &\frac{a_{11} + a_{22}}{ 2} \pm \sqrt{\left(\frac{a_{11} + a_{22}}{ 2}\right)^2 - \left(a_{11} a_{22} - a_{12} a_{21}\right)}\cr \cr \lambda_{1,2} &= &\frac{a_{11} + a_{22}}{ 2} \pm \sqrt{ \frac{a_{11}^2+2a_{11}a_{22} + a_{22}^2}{ 4} - \frac{4a_{11}a_{22}}{4} + a_{12} a_{21}}\cr \cr \lambda_{1,2} &= &\frac{a_{11} + a_{22}}{ 2} \pm \sqrt{\left(\frac{a_{11} - a_{22}}{ 2}\right)^2 + a_{12} a_{21}}\cr \end{matrix}$$ b. Bestimmung der Eigenvektoren $$\begin{pmatrix}a_{11}-\lambda_{1,2} &a_{12}\cr a_{21} & a_{22}-\lambda_{1,2}\cr\end{pmatrix} \begin{pmatrix}x_1\cr x_2\cr\end{pmatrix} = \begin{pmatrix}0\cr 0\cr\end{pmatrix}$$ Für die beiden Eigenvektoren ergibt sich also: $$\begin{matrix} (a_{11}-\lambda_{1,2})x_1 &= &-a_{12}x_2 \cr \cr \vec{u_{1,2}} &= &C\begin{pmatrix}a_{12} \cr \lambda_{1,2} -a_{11}\cr\end{pmatrix}\cr\end{matrix} $$ c. Allgemeine Lösung bei positiver Diskriminante $$\vec{x} = C_1\begin{pmatrix}a_{12} \cr \lambda_1 -a_{11}\cr\end{pmatrix} e^{\lambda_1 t} + C_2\begin{pmatrix}a_{12} \cr \lambda_2 -a_{11}\cr\end{pmatrix} e^{\lambda_2 t}$$