TITLE: If $A^2=2A$, then $A$ is diagonalizable.
QUESTION [11 upvotes]: I think, I should use a double linear transformation but can't find any proper solution. 
Let $\mathbb F$ be a field, $\mathscr M_n (\mathbb F)$, the set of $n\times n$ matrices with elements in $\mathbb F$, and $A\in \mathscr M_n (\mathbb F)$ satisfying the equation
$$
A^2=2A.
$$
Show that $A$ is diagonalizable and find its eigenvalues.
Thank you all!

REPLY [13 votes]: As stated (notably without any restriction on the field*) the result is false. In characteristic $2$, every nilpotent matrix of nilpotency order$~2$ satisfies $A^2=2A$, but such matrices are not diagonalisable. See also this answer to a rather similar question.
*except that it must be denotable using a blackboard-bold font