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?
 A: Here's a different option.
Instead of saying to yourself "Those were tedious computations", you could instead say "What have I really proved?"

Lemma: The characteristic polyomial of a $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is
$$\lambda^2 - \text{tr}(M) \lambda + \text{det}(M)
$$
where $\text{tr}(M)$ and $\text{det}(M)$ denote the trace and the determinant:
$$\text{trace}(M) = a + d, \qquad \text{det}(M) = ad - bc
$$

Proof: (tedious but suddenly, perhaps, more interesting computations)

Theorem: If a $2 \times 2$ matrix $M$ satisfies the inequality
$$\text{trace}^2(M) - 4 \, \text{det}(M) > 0
$$
then it is diagonalizable.

Proof: $\text{trace}^2(M) - 4 \, \text{det}(M)$ is the discriminant of the characteristic polynomial, therefore if it is positive then there are two distinct characteristic roots. QED

Corollary: If $M = \pmatrix{a & b \\ b & d}$ and $b \ne 0$ then $M$ is diagonalizable.

Proof: $\text{trace}^2(M) - 4 \, \text{det}(M) = (a+d)^2 - 4 (ad-bb^2) =$ (still one last tedious, but perhaps more enlightening, computation) $= (a-d)^2 + 4b^2 > 0$.
And now you can start to ask yourself: What happens with $3 \times 3$ matrices?.... or $4 \times 4$? ............
A: *

*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).
