Eigenvalue of $A$ is root of $p(t)$ I'm working on this proof problem:

I have part (a) done.  I think this is the Cayley Hamilton Theorem, so I got some insight from Wikipedia.  I am a little confused with part (b) though because I feel that the converse of the statement is true, and I guess I can't think of a good counter example?
Any advice?

Hint: Find a $2 \times 2$ matrix $A$ such that $A \neq 0$ but $A^2 = 0$.  
By the way, I don't think that the proof of part a) is only the Cayley-Hamilton Theorem.  You also need to know that the minimal polynomial of $A$ and the characteristic polynomial of $A$ have the same real roots.  
Added: Here is a sketch proof of part a): let $\lambda$ be an eigenvalue of $A$, so there is a nonzero $v \in \mathbb{R}^n$ with $Av = \lambda v$.  Then $v$ is an eigenvector for $p(A)$ with eigenvalue $p(\lambda)$.  On the other hand $p(A) = 0$....
