If $A^3 = 0$, how many distinct eigenvalues does $A$ have? The problem is

The matrix $$
A = \begin{bmatrix}
2 & 2 & -2 \\
5 & 1 & -3 \\
1 & 5 & -3
\end{bmatrix}
$$ has the property $A^3 = 0$. How many distinct eigenvalues does $A$ have?

I can, of course, compute all the eigenvalues, but I am wondering what does $A^3 = 0$ tell us here about the number of distinct eigenvalues?
 A: $A^3=0$ is the characteristic equation applied to $A$:
$$p(A)=0$$
The characteristic equation ($p(\lambda)=\lambda^3=0$) has three roots, all roots are zero, so there is a unique (triple) eigenvalue $\lambda =0.$
A: For this particular matrix I don't have anything else to add. So I will look at the general case that you talk about.
Let's suppose that $A$ is a (square) matrix. Then eigenvalues are solutions to the equation $A\vec v = \lambda \vec v$, where $\vec v$ is the corresponding eigenvector. Now it is a relatively simple exercise to show that if $\lambda$ is one such eigenvalue, with corresponding eigenvector $\vec v$, then for each $n\geq 1$, $\lambda ^n$ is an eigenvalue of $A^n$ with eigenvector $\vec v$. (Just multiply each side of the defining equation on the left by $A$ repeatedly.)
Now if $A^3 = 0$, then $\lambda^3$ should be an eigenvalue of $A^3$ and so must be zero. (Note that $A^3\vec v = 0\vec v = \vec 0$ so the only possible eigenvalue is zero.) But then $\lambda^3 = 0$ has only a single solution $\lambda = 0$.
Here are some other questions you might want to think about: what do you think would happen if instead $A^3 = I$ or $A^3 = A$?
