If $A^3=A^2$ then $A^2$ is diagonalizable. 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.
 A: Notice that $A^4 = A^3 = A^2$, so the matrix $B = A^2$ satisfies $(B-I)B = B^2 - B = 0$. By Jordan Normal Form, since it satisfies a polynomial with no repeated roots, we must have $B$ is diagonalizable.
To proceed without heavy machinery:
Any matrix satisfying $B^2 = B$ is called an idempotent matrix. One can show
-$B$ must only have eigenvalues $0$ and $1$
-The eigenspace for $0$ corresponds to the null space by definition
-The eigenspace for $1$ corresponds to the column space by idempotency
Then by rank nullity, the dimension of the null space and the column space add to the dimension of your vector space, so your matrix is diagonalizable.
A: 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. 
A: Using the Jordan form, we see we have Jordan blocks of size at most $2$, hence by decomposing you know any $0$-eigenvalue blocks of size $2$ square to all $0$ blocks, and any $1$ blocks have size at most $1$, and so remain unchanged.
