Prove: $A^2\in M_n(\mathbb F)$ is diagonalizable $\implies A$ is diagonalizable (A is invertible) I do not understand intuitively why this is true but I have a feeling this should be proved using the Cayley-Hamilton theorem.
I know that 

A matrix or linear map is diagonalizable over the field $\mathbb F$ if and only if its minimal polynomial is a product of distinct linear factors over $\mathbb F$.

And if I understand it correctly it means that the algebraic multiplicity of every linear factor in the minimal polynomial is $1$.
But I'm not even sure that's the right direction. 
Any hints?
 A: It is not actually true. $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ is not diagonalizable, but $A^2$ is the zero matrix, which is even diagonal itself.
It does* hold when $A$ is invertible, and when the geometric multiplicity of $0$ as an eigenvalue equals the algebraic multiplicity. This can be seen, for example, by switching to a basis where $A$ is on Jordan normal form (working over an algebraic closure of $\mathbb F$ if necessary) and considering what $A^2$ looks like in that basis.
* Except in characteristic 2, where they use different sockets and everything works funny.
A: Well, In fact it is false. Hint: Prove that if a matrix $A$ has minimal polynomial $x^2$, then $A^2$ has minimal polynomial $x$. From this, you should be able to derive a counterexample.
A: It is true if the matrix $ A $ is invertible, the field is algebraically closed, and it is of characteristic $ \neq 2 $. To see this, let $ f(X) $ be the minimal polynomial of $ A^2 $, then $ f(X^2) $ is a polynomial which has $ A $ as a root. Its formal derivative is $ 2X f'(X^2) $ (this vanishes if the field is of characteristic 2!). Let $ \alpha \neq 0 $ be a root of $ f(X^2) $. Since $ f(X) $ and $ f'(X) $ share no roots by the assumption that $ A^2 $ is diagonalizable, $ f'(\alpha^2) \neq 0 $, and therefore $ 2\alpha f'(\alpha^2) \neq 0 $. This shows that the polynomial $ f(X^2) $ has distinct roots with the possible exception of $\alpha = 0$. Since the minimal polynomial $ g(X) $ of $ A $ must divide $ f(X^2) $, and $ g(0) \neq 0 $ by the assumption that $ A $ is invertible, this gives us that $ g(X) $ splits into distinct linear factors over the algebraically closed field $ \mathbb F $.
Other answers give counterexamples for the cases when one (or both) of the assumptions doesn't hold.
A: As $\mathbb F$ may not be algebraically closed, $A$ can be non-diagonalizable simply because its characteristic polynomial does not split in $\mathbb F$, e.g. when $A=\pmatrix{0&-1\\ 1&0}$ and $\mathbb F=\mathbb R$.
