TITLE: If $A^3=A^2$ then $A^2$ is diagonalizable.
QUESTION [5 upvotes]: Let $A\in \mathbb{k}^{n\times n} $.
Prove that if $A^3=A^2$ then $A^2$ is diagonalizable.
Could you give me any hints on how to prove it?. I can't use the minimal polynomial, since we haven't seen it in class.
Thanks.

REPLY [2 votes]: Let $A^2=B$, then $B^2 =B$. (In other words, $B^2$ is a projection.) Now, if $y=B(x)\in\mathop{\mathrm{im}} B$, then
$$y = B(x) = B(B(x)) = B(y).$$
So $y$ is an eigenvector corresponding to the eigenvalue $1$ of $B$. Also, if $y\in \ker B$, then $y$ is an eigenvalue corersponding to value $0$. 
On the other hand, let $x\in k^{n\times n}$, we have
$$B(x) = B(B(x)).\tag{1}$$
so $x- B(x)\in \ker(B)$. Thus, 
$$k^{n\times n} = \mathop{\mathrm{im}} B+\ker B.$$
Lastly, if $x \in \ker B \cap \mathop{\mathrm{im}} B$, then
$$x=B(x) = B(B(x)) = 0.$$
So
$$k^{n\times n} = \mathop{\mathrm{im}} B \oplus \ker B.$$
Therefore, $B$ is diagonalizable.