$A^3 = A^2$ How can $A$'s minimal polynomial look like? 
Let $K$ be a field and $A \in K^{n \times n}$ a matrix with $A^3 = A^2$.
  How can $A$'s minimal polynomial $\mu_A$ look like?

The only possibilities I could think of are


*

*$A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.

*$A = E$, where $E$ is the $n \times n$ identity matrix.  Then the characteristic polynomial is $P_A(t) = (1 -t)^n$.


I am surely missing some possibilities.
How can I draw a conclusion about the minimal polynomial from characteristic polynomial, knowing that the exponents in the first are the sizes of the largest Jordan blocks.

Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also


*

*$x(x-1)$ for example in $$\begin{bmatrix}
    1       & 0 \\
    0       & 0 
\end{bmatrix}$$

*$x^2$ for example in $$\begin{bmatrix}
    0       & 1 \\
    0       & 0 
\end{bmatrix}$$

*$x^2(x-1)$ for example in $$\begin{bmatrix}
    1       & 0 & 0 \\
    0       & 0 & 1 \\
    0       & 0 & 0
\end{bmatrix}$$


In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
