# 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?

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.

• It is not true in characteristic $2$; that's quite a complication. Jul 3 '16 at 9:58
• @darij: Better now? Jul 3 '16 at 10:03
• I actually forgot to mention that A is invertible but the second part of the question was whether this statement is true for a singular matrix, so thanks Jul 3 '16 at 10:04
• @PanthersFan92: You still need to take note of the characteristic 2 exception, where, for example, $({}^1_0\;{}^1_1)^2=I$. Jul 3 '16 at 10:06

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.

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.

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 $\mathbb F=\mathbb R$.