If $A^2$ is diagonalizable then A is also...

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I tried to dis/prove the following:

A is a 2x2 complex matrix

1) If $A^2$ is diag. then $A$ is also.

I mostly tried using $P^{-1} A^2 P = D$ and found out that:

$D=P^{-1} A PP^{-1} A P = \sqrt{D}\sqrt{D}$

I don't think it makes sense though

I also tried using $A={% \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}% }\ $ and to make my life easier I set c=0 to work only on triangular matrices, but I'm not sure I'd obtain a general result.

2) if $A$ is invertible and $A^2$ is diag., then A is also diag.

3) if $A^{-1}$ is diag. then A is also.

$P^{-1} A^{-1} P = D$

$A$ is invertible so $D$ is. I take the $^{-1}$ of both sides to prove.

I need some help, thank you...