If A is idempotent and $A^5=3A^3-2A$ Stuck at this question for a while.
Question : If A is idempotent and $A^5=3A^3-2A$ , Prove $A=0$
My try :
By $A^2-A=0$ , we know that the minimal polynomial devides $\Rightarrow x(x-1)$
So the options for the minimal polynomial is only : $x , x-1 , x(x-1)$
I tried assuming by contradiction that the minimal polynomial is $x$ and then we get $A=0$ and we are finished.
if $x-1$ is the minimal then $A-I=0 \Rightarrow A=I$ , but we have an eigenvalue that is $0$ so $|A|=0$ and the matrix can't be $I$.
but I'm stuck with $x(x-1)$.
any hints? Thank you
 A: The claim is false. Just take any nonzero idempotent matrix.
Indeed, $A^2=A$ implies $A^n=A$ for all $n\ge 1$. Therefore $A^5-3A^3+2A=A-3A+2A=0$ and so $A^5=3A^3-2A$ gives no new information.
A: I agree with the other answer that the claim is false, and won't repeat that argument.
However, assuming that there is a typo in the problem, and it should've been e.g. $A^5=3A^3+2A^2$... Here is a procedure that helps in problems like this one:
We know that $p(A)=0$ and $q(A)=0$, where $p(X)=X^5-3X^3-2X^2$ and $q(X)=X^2-X$. Let's do Euclidean division of $p$ by $q$ (with a remainder):
$$X^5-3X^3-2X^2 = (X^2-X)(X^3+X^2-2X-4)-4X$$
So $p(X)=q(X)s(X)+r(X)$ where $s(X)=X^3+X^2-2X-4$ and $r(x)=-4X$.
This immediately implies that $r(A)=p(A)-q(A)s(A)=0-0\cdot s(A)=0$, i.e. $-4A=0$. I hope you are working in the field $\mathbb R$ (or another field of characteristic $\ne 2$), in which $-4\ne 0$ and so $A=0$.
(Note in a field of characteristic $2$ the claim is just as false as with the original problem, as in characteristic $2$ we have not really fixed the typo - the "fixed" problem is the same as the original one. This makes me think that, perhaps, there is a different typo in the original problem.)
