Let $\alpha$ be a linear operator on V, suppose $\alpha$ is bijective and $\alpha^2$ is diagonalisable, and $\mathbb{F}$ is algebraically closed with $\operatorname{char}(\mathbb{F})\neq 2$. Prove that $\alpha$ is diagonalisable.
Some how characteristic field must be used but I am not sure where hence I suspect my approach is wrong:
Let the eigenspace for eigenavalue $\lambda$ of $\alpha$ be $E^{\alpha}_{\lambda}$. Then for each eigenvalue $ \lambda_i$ of $\alpha$ we have $$ E^{\alpha}_{\lambda_i} \subseteq E^{\alpha^2}_{\lambda_i^2}$$
And I need to show the converse, that if v is an eigenvectors of $\alpha^2$ then it is an eigenvector of $\alpha$. I can't prove this direction but once done then the conclusion should be straightforward, and I have no idea how to use the fact that $\alpha$ does not have 0 as an eigenvalue
Since $\alpha^2$ is diagonalizable, its minimal polynomial splits (well, yes, because $\mathbb{F}$ is algebraically closed) and it has no repeated roots. That is, it is of the form $$m(t) = (t-\lambda_1)\cdots (t-\lambda_k)$$ with $\lambda_1,\ldots,\lambda_k$ pairwise distinct.
Now, that means that $\alpha$ satisfies the polynomial $$(t^2-\lambda_1)\cdots(t^2-\lambda_k)$$ and thus, the minimal polynomial of $\alpha$ must divide this polynomial.
Can you take it from here, and do you see why you need all three hypothesis, that $\alpha$ is invertible, that $\mathbb{F}$ is algebraically closed, and that its characteristic is not $2$?