Monic polynomial with certain properties is the characteristic polynomial? I'm trying to determine whether this statement is true or false:
Assume that $dim V = n$ and let $T \in L(V)$. Let $f(z) \in P_n(F)$ be a
monic polynomial of degree
$n$ such that $f(T) = 0$. Then $f(z)$ is the characteristic polynomial of $T$.
I know that because $f(T) = 0$, the minimal polynomial of T divides it and that the minimal polynomial of T divides the characteristic polynomial of T as well, but I can't seem to put these together in a way that conclusively establishes the truth or falsehood of the above statement.
If anyone could shed some insight on the matter, it would be greatly appreciated. Thanks!
 A: This is not true in general. The minimum polynomial is the smallest degree polynomial that the linear map satisfies, and the characteristic polynomial is defined to be $\det(xI-T)$. If the minimum polynomial is degree $m$, we know that the characteristic polynomial will be degree $n$ so $m\leq n$. If $p(x)$ is the minimum polynomial and $g(x)$ is any other poly, then $T$ will satisfy the polynomial $p(x)g(x)$. So if $m < n$, we can multiply $p(x)$ by any polynomial of degree $(n-m)$ to get a new polynomial of degree $n$ that the linear map satisfies. This will not, in general, be the characteristic polynomial.
The above is all done thinking almost solely about polynomials. Now I'll talk about how we can understand the problem thinking more about linear maps. Working over an algebraically closed field, we know that we can pick a basis with respect to which the matrix of $T$ is in Jordan Normal Form (JNF). The multiplicities of the roots of the characteristic polynomial are the number of times that root appears on the diagonal of the matrix in JNF, but the multiplicities of the roots of the minimum polynomial tell us the size of the largest block corresponding to that root. From looking at the matrix, it is intuitively clear that the matrix will satisfy a given polynomial $f(x)$ iff $p(x)$ divides $f(x)$. Thus we can, as before, multiply $p(x)$ by any polynomial of degree $n-m$ to get one of degree $n$ that the linear map satisfies.
A: Hint Take $T$ to be the $0$ operator. Then $f(T)=0$ for all $f(z)$ divisible by $z$.
