Matrix $A$ such that $A^{3} - A^{2} = 0$ and its eigenvalues We have the matrices $A \in M_{n \times n}$ and $X \in M_{n \times 1}$ such that $A \cdot X = \lambda \cdot X$ for any scalar $\lambda$. Let $q(x) = \alpha_{n} \cdot x^{r} + \alpha_{n-1} \cdot x^{r-1} + \cdots + \alpha_{1} \cdot x + \alpha$ be a polynomial equation and $q(A) = \alpha_{n} \cdot A^{r} + \alpha_{n-1} \cdot A^{r-1} + \cdots + \alpha_{1} \cdot A + \alpha \cdot I_{n}$.
In the book I'm studying there is a property of eigenvalues that goes like this: If $\lambda$ is an eigenvalue of $A$ and $q(A) = 0$ then $q(\lambda) = 0$.
If we know that $A$ is a matrix such that $A^{3} - A^{2} = 0$ and we have a polynomial equation $q(x) = x^{3} - x^{2}$ then we know that $q(A) = 0$. Shouldn't this tell us that the only possible eigenvalues of $A$ for any eigenvector are either $0 \lor 1$?
Because the only solutions for $q(\lambda) = 0$ are $\lambda = 0 \lor \lambda = 1$.
 A: YES! The only possible eigenvalues for $A$ are $0$ and $1$ because $q(A)=0$ where $q(\lambda)=\lambda^2(\lambda-1)$. As you noted, that does not mean that $\lambda=0$ is an eigenvalue, and it does not mean that $\lambda=1$ is an eigenvalue.
One of the two values $0$ or $1$ is an eigenvalue. To see why, suppose $1$ is not an eigenvalue, then $A-I$ is invertible. Therefore,
$$
                   (A-I)A^2=0 \implies A^2=0,
$$
and that means that either $Ax=0$ or $A(Ax)=0$ for $x\ne 0$, which means $0$ an eigenvalue for $A$.
On the other hand, suppose that $0$ is not an eigenvalue. Then $A^2$ is invertible, which gives $A-I=0$, and that forces $1$ to be an eigenvalue of $A$.
A: if the minimal polynomial is $x^2(x-1)$ the Jordan form can be of this type:
$$
\left(
\begin{array}{ccc|cc|cc|ccc}
1&0&0&0&0&0&0&0&0&0 \\
0&1&0&0&0&0&0&0&0&0 \\
0&0&1&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&1&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&0&0&1&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&0&0&0&0&0&0 \\ 
0&0&0&0&0&0&0&0&0&0 \\ 
0&0&0&0&0&0&0&0&0&0 \\ 
\end{array}
\right)
$$
The degree of $k$ each factor $(x-\lambda)^k$ in the minimal polynomial is the size of the largest Jordan block with that eigenvalue. Here, we are allowed one or more 2 by 2 Jordan blocks of Dietrich's type (in comment above).
