If $A^2$ is diagonalizable, must $A$ be such as well? Given a diagonalizable matrix $A^2$, must the matrix $A$ be diagonalizable as well?
I can prove that this is true for when $A\in M_{n\times n} (\mathbb{C})$ by using the theorem that the Minimal polynomial for $A^2$ is expressed as a multiplication of linear attributes, and we can simply take $\pm \sqrt{\lambda_i}$ and show that $A$'s minimal polynomial is also a multiplication of linear attributes, thus making $A$ diagonalizable as well. 
The problem is that I do now know whether or not this statement is correct for $A\in M_{n\times n} (\mathbb{R})$, I know my proof won't work for when $\lambda_i < 0$, but perhaps there is another proof for this? Or a counterexample?

$$
A = \begin{pmatrix}
0&1\\
0&0
\end{pmatrix}
$$
is not diagonalizable (it has 1 eigenvalue and is not diagonal) but $A^2 = 0$ is.
