\( A:=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\ {a_{21}} & {a_{22}}\end{array}\right) \in \mathbb{R}^{2 \times 2} \)
\( \begin{aligned} \operatorname{det}(A-\lambda E) &=\left|\begin{array}{cc}{a_{11}-\lambda} & {a_{12}} \\ {a_{21}} & {a_{22}-\lambda}\end{array}\right| \\ &=\lambda^{2}-\left(a_{11}+a_{22}\right) \cdot \lambda+a_{11} a_{22}-a_{12} a_{21} \stackrel{!}=0\end{aligned} \)
$$ \begin{array}{l} {\lambda_{1}=\frac{a_{11}+a_{22}}{2}+\sqrt{\left(\frac{a_{11}+a_{22}}{2}\right)^{2}-a_{1} a_{22}+a_{12} a_{21}}} \\ {\lambda_{2}=\frac{a_{11}+a_{22}}{2}-\sqrt{\left(\frac{a_{11}+a_{22}}{2}\right)^{2}-a_{11} a_{22}+a_{12} a_{21}}} \end{array} $$
$$ \begin{aligned} \lambda_{1} \cdot \lambda_{2} &=\left(\frac{a_{11}+a_{22}}{2}\right)^{2}-\left(\left(\frac{a_{11}+a_{22}}{2}\right)^{2}-a_{11} a_{22}+a_{12} a_{21}\right) \\ &=a_{11} a_{22}-a_{12} a_{22} \end{aligned} $$
\( =a_{11} a_{22}-a_{12} a_{21} \)
\( =\operatorname{det}(A) \)
Hey!
Kann mir jemand den zweiten Teil dieses Beweises erklären (ab der dritten binomischen Formel)? Ich verstehe das leider nicht...
