$A^2$ is diagonalizable and so is $A$

Question

If $A$ is not diagonalizable, then $A^2$ is not diagonalizable or $A$ is singular.

I don't know if I'm right but $A^2=A$. So if $A$ is not diagonalizable then $A^2$ is not diagonalizable. Also If $A$ is diagonalizable then there exist a nonsingular matrix such that $P^{-1}AP = D$ where $D$ is a diagonal matrix. I don't know how $A$ is singular.

Can someone correct and enlighten me with the proof. Thanks!

Answer 1

This is false over the reals. Let $A$ be the $90$-degree rotation in the plane. Then $A^2=-I$ is diagonalizable but $A$ is not diagonalizable.

Answer 2

Over $\mathbb C$

$\det A \neq 0 \iff \det A^2 \neq 0 \iff 0 \notin Sp(A^2)$

Suppose now that $A$ is invertible and $A^2$ is diagonalizable.

$A^2$ is diagonalizable means there exists a polynomial $P$ with distinct simple roots that vanishes $A^2$.

Take for $P$ the minimal polynomial of $A^2$, so $0$ is not one of its roots.

Say $P = \prod_j (X-\lambda_j) $

$\prod_j (A^2-\lambda_jI) =0$

Hence $\prod_j (X^2-\lambda_j)$ vanishes $A$.

Since for all $j$, $\lambda_j \neq 0$, we get a polynomial with distinct simple complex roots that vanishes $A$ i.e. $A$ is diagonalizable.