For which values of $a$ and $b$ is the matrix $$ \begin{pmatrix} 0 & a\\ b & 0 \end{pmatrix} $$ diagonalizable over $\mathbb{C}$?
I know that if $a = -1$ and $b = 1$, then the matrix is diagonalizable. However, I am not sure if that is the only solution or how to go about finding solutions in general. Please help.
$$A=\begin{pmatrix}0&a\\b&0\end{pmatrix}\implies p_A(x):=\det(xI-A)=\begin{vmatrix}\;x&\!\!-a\\\!\!-b&\;x\end{vmatrix}=x^2-ab$$
so
$$p_A(x)=0\iff x=\pm\sqrt{ab}\implies$$
$$(1)\;\;\;\;\;\;\;\;\;\;ab\neq0\implies\;\text{there are two different eigenvalues and the matrix is diagonalizable}$$
$$(2)\;\;\;\;\;\;\;\;\;\;ab=0\;,\;a\neq 0\,\,\vee\,\,b\neq0\implies\text{ this is a non-zero nilpotent matrix and thus non-diagonalizable}$$
$$(3)\;\;\;\;\;\;\;\;\;a=b=0\implies\text{ the matrix's already diagonal...}$$