<s>[INST] How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with $b\neq 0$ is diagonalizable? 
How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with
$b\neq 0$ is diagonalizable?

Sei $b\neq 0$, dann gilt:$$\det(A-\lambda E)=\det \begin{pmatrix} a-\lambda & b \\ b & d-\lambda \end{pmatrix}=(a-\lambda)(d-\lambda)-b^2=\lambda ^2-\lambda (a+d)+ad-b^2 $$
$$\lambda_{1,2}=\frac{a+d}{2}\pm\sqrt{\frac{(a+d)^2}{4}-ad+b^2}=\frac{a+d}{2}\pm \sqrt{\frac{1}{4}(d-a)^2+b^2}$$
Since  $\frac{1}{4}(d-a)^2\geq 0$ for all $d,a\in \mathbb{R}$ and $b\neq 0$ (and thus $b^2>0$), we have $\sqrt{\frac{1}{4}(d-a)^2+b^2}>0$. Hence, we have to different Eigenvalues.
Do I have to brute force this problem? Or can I show it without going through all these tedious computations?
 [/INST] *

*The characteristic equation is $(a-\lambda)(d-\lambda)-b^2=\lambda^2-(a+d)\lambda+ad-b^2$;


*The roots a distinct because the discriminant $(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2$ is certainly non-zero.


*$v=(b,d-\lambda)$ solves the system $Av=\lambda v$, for both $\lambda$s (check by direct subtitution).
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