Proof or counterexample: Let $A$ be a square matrix, then: 
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*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


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*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   

Suppose $A$ is diagonalisable. Then there is a basis of eigenvectors $e_i$ with eigenvalues $\lambda_i$, not necessarily distinct. With respect to this basis, $A^2 e_i = \lambda_i^2 e_i$, so $e_i$ is also a basis of eigenvectors for $A^2$, and hence $A^2$ is diagonal in this basis.
(Or one can use the existence of a unitary or orthogonal matrix $U$ so $UAU^{-1}=D$ is diagonal. Then $UA^2U^{-1} = UAU^{-1}UAU^{-1} = D^2$ is also diagonal.)
Counterexample for the other direction:
$$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$
is not diagonalisable, but $A^2=0$ clearly is.
