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   
 A: 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.
A: 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.
A: 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. 
