$A^3 = A^2$ How can $A$'s minimal polynomial look like?

Question

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

1. $A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.
2. $A = E$, 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.

Answer 1

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.