TITLE: Proof or counterexample: Let $A$ be a square matrix, then:
QUESTION [5 upvotes]: If $A$ is diagonalizable, then so is $A^2$

I answered yes. I argued that since $A$ is diagonalizable there exists an eigenbasis, and since $A^2$ has the same eigenvectors than $A$, and its eigenvalues are those of $A$ squared, there is also an eigenbasis for $A^2$, so it is diagonalizable

If $A^2$ is diagonalizable, then so is $A$

I am pretty sure the answer is no, but I can't think of a counterexample. 
Thank you in advance

REPLY [3 votes]: If $A$ is diagonalizable, then $A = P^{-1}DP$ for $D$ diagonal, so $A^2 = P^{-1}D^2P$, and $D^2$ is certainly still diagonal.
The second assertion is not true. Consider $A = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$. Then $A^2 = 0$, which is diagonalizable. On the other hand, the minimal polynomial of $A$ is $x^2$, so $A$ is already in Jordan canonical form, but $A$ is not diagonal, hence is not diagonalizable.