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

*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   

The first one is true. One has $D = S^{-1}AS$ for a diagonal matrix $D$ and a base transformation $S$. So clearly it is $D^2 = S^{-1}ASS^{-1}AS = S^{-1}A^2S $ and therefore $A^2$ is diagonalizable.
The answer to the second claim is no. Choose for example
$$
A = \begin{pmatrix}
0 & 1 \\
0 & 0 
\end{pmatrix} $$
Clearly $A^2$ is diagonalizable but $A$ is not. 
