$A^2$ is diagonalizable leads to $A$ diagonalizable?

Question

If $A^2$ is diagonalizable, is it necessary true that $A$ is diagonalizable?

Also, the opposite: If $A$ is diagonalizable, is it necessary true that $A^2$ is diagonalizable?

I'm not sure yet, tried to prove with $A^2 = P^{-1}DP$ , but no success.

Answer 1

The first proposition is false, as for example $A^2 = 0$ for nonzero (nilpotent) $A$.

For example: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.

Of course if $A$ is diagonalizable, then $A^2$ (and indeed any polynomial in $A$) is also diagonalizable: $D = P^{-1} A P$ diagonal implies $D^2 = P^{-1} A^2 P$.

Answer 2

It is actually true iif $A^2$ has only positive eigenvalues.

If $\pi$ is a polynomial that factors completely into distinct factors (hope I'm getting this right in english, please correct otherwise) and for which $\pi(A^2)=0$ (such a polynomial exists because $A^2$ is diagonalizable), ie : $$\pi(X)=\prod_i (X-\lambda_i)$$ with distincts $\lambda_i>0$, and $\pi(A^2)=\prod_i (A^2-\lambda_i I)=0$

Then $P(X)=\pi(X^2)=\prod_i (X^2-\lambda_i)=\prod_i (X-\sqrt{\lambda_i})(X+\sqrt{\lambda_i})$ is also completely factored into distinct factors and verify $P(A)=0$ which imply with kernel lemma that $A$ is also diagonalizable.

The other side is trivial as mentionned by hardmath