For a linear map $f: V \to V$ if $f^2$ is diagonalizable and $\ker f = \ker f^2$ then is $f$ diagonalizable?

Question

Here $V$ is a finite dimensional vector space, of dimension $n$, over an algebraically closed field $F$. My original approach was to use a minimal polynomial argument by showing that $\pi_f$ (which I will let denote the minimal polynomial of a linear map $f$) divides a polynomial of simple roots. We know that $f^2$ being diagonalizable implies that $\pi_{f^2} = \prod (X - \lambda_i)$ for distinct eigenvalues $\lambda_i$ of $f^2$ so that implies that there exists some $P(X) = \prod (X^2 - \lambda_i)$ such that $P(f) = 0$. Since $P$ annihilates $f$ and we can rewrite $P(X) = \prod (X - \sqrt{\lambda_i})(X + \sqrt{\lambda_i})$ we have that $\pi_f \mid P \Longrightarrow$ $\pi_f$ has simple roots. This prior argument doesn't necessarily hold because $P(X)$ could have a factor of $X^2$ and so we can't conclude that $\pi_f$ will have simple roots.

I don't see any easy way to incorporate the fact that $\ker f = \ker f^2$ into the above argument when $n-2 \geq \dim(\ker f) \geq 2$ (for dimension 1 it divides the characteristic polynomial which will have $X$ as a factor only once, and dimension being $n$ implies they are both the $0$ map). Other approaches I have considered are:

1. Attempting to relate the dimensions of the Eigenspaces of $f^2$ and $f$ since we know that the sum of the dimensions of the Eigenspaces of $f^2$ give us $n$.
2. Attempting to show for each distinct eigenvalue $\lambda_i$ of $f^2$, that $f_{\upharpoonright E(\sqrt{\lambda_i})} - \sqrt{\lambda_i} I \equiv 0$. (without loss of generality just let $\sqrt{\lambda_i}$ be eigenvalue of $f$ when $\lambda_i$ is eigenvalue of $f^2$.)

I would very much appreciate some hints on how to approach this problem, or how to go off of approaches I've already considered. Thank you!

Answer 1

The argument in your first paragraph works fine: let $p(x) = \prod_i (x - \lambda_i)$ be the minimal polynomial of $f^2$. Certainly $p(f^2) = 0$, so setting $q(x) := \prod_i(x^2 - \lambda_i)$ we have $q(f) = 0$, so that the minimal polynomial of $f$ divides $q(x)$. As you say, then all of the roots of the minimal polynomial of $f$ are thus distinct except for possibly the root $0$ (which could have multiplicity 2).

But if the root $0$ has multiplicity $2$ in the minimal polynomial of $f$ then $\dim \ker f < \dim \ker f^2$. This contradicts the hypothesis that $\ker f = \ker f^2$. If you like, this follows from the fact that the multiplicity of a root $\lambda$ of the minimal polynomial of $f$ is the smallest $n$ such $\dim \ker (f - \lambda)^n = \dim \ker (f - \lambda)^N$ for all $N \geq n$.

Answer 2

The answer is no.

Counter-example: let $\mathbb F$ be the algebraic closure of $\mathbb F_2$, $n=2$ and $f$ have matrix representation

$A= \displaystyle \left[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right]$

Then $f$ is an involution but not diagonalizable.

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Now when $\text{char } \mathbb F \neq 2$ the the claim is true since the eigenvalue 0 is semi-simple and when we restrict to the subspace on which $f$ is invertible, and reference e.g. my triangularization argument here: $A \in M_n(\mathbb{C})$ invertible and $A^2$ is diagonalizable. Prove $A$ is diagonalizable we see $f$ must be diagonalizable on this subspace.