$A^2$ and $A^3+A$ are diagonalizable, prove $A$ is diagonalizable In a finite dimensional vector space, suppose $A^2$ and $A^3+A$ are diagonalizable, how to prove $A$ is diagonalizable?

What I can know is the minimal polynomial of $A^2$ and $A^3+A$, but I don't know how to proceed
 A: Hint: Suppose $A$ is in Jordan form. Since $A^2$ is diagonalizable, any Jordan blocks associated with a non-zero eigenvector have size $1$. Since $A^3+A$ is diagonalizable, any eigenvalues associated with $0$ are of size $1$. 
The conclusion follows.

For instance, consider for any $\lambda\neq 0$
$$
A=\pmatrix{\lambda & 1\\ 0&\lambda} \implies
A^2= \pmatrix{\lambda^2 & 2\lambda\\ 0&\lambda^2}
$$
Note that $A$ has only $1$ eigenvector (more accurately, a $1$-dimensional eigenspace) associated with $\lambda^2$ and no other eigenvalues. So, $A^2$ is not diagonalizable, and no matrix with $A$ as a diagonal subblock is diagonalizable.
A: The example $A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ over $\mathbb{R}$ shows that the assertion is false without some additional assumptions. After all,
$$ A^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix} $$
is diagonalizable, as is
$$ A^3+A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}+\begin{bmatrix}0&-1\\1&0\end{bmatrix}=0$$
but $A$ is not diagonalizable. 
If the field is algebraically closed, then Omnomnomnom's answer works.
