Is the Converse of the Cayley-Hamilton Theorem true or not? I was thinking about taking the Cayley-Hamilton Theorem in the opposite direction.
Does this hold : If for $F\in$ End$(V)$ $\exists P\in K[t]$ such that $P(F)=0$ then 
$P$ is the characteristic polynomial of $F$ ?
If not could you give a counterexample?
 A: No. If $p_F$ is the characteristic polynomial of $F$ and $q$ is any other polynomial, then
$$P(F) = p_F(F) q(F) = 0 \cdot q(F) = 0.$$
So $P$ is a polynomial with $P(F) = 0$ that is not the characteristic polynomial of $F$ (just take $q$ to be nontrivial).
We can also have something like the following happen: Let
$$F = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Then
$$p_F(t) = (1 - t)^2.$$
But of course if
$$q(t) = 1 - t,$$
we have
$$q(F) = 0.$$
A: For an extreme example in which this fails, consider the zero matrix $\left(0\right) \in M_n(k)$.  Here, the characteristic polynomial $\chi_0(t)=t^k$, corresponding to the fact that $0$ occurs as an eigenvalue of $(0)$ with multiplicity $n$.  Obviously, though, $(0)$ is annihilated by many other endomorphisms, e.g. the identity map.
Your proposition happens to be true if and only if the characteristic polynomial of $A$ equals the minimal polynomial of $A$ (the polynomial that truly satisfies the statement of your proposition), and if you replace "is the characteristic polynomial" with "is a multiple of the characteristic polynomial".
Edit: In the comments below, it's pointed out that $\chi_A$ need not be irreducible for this to work.  Of course, if $\chi_A$ is irreducible, then this is true (and what I meant, originally.)
