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:
- 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 $n$.
- 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!
Best Answer
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.
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.