Suppose $F(x)$ is a polynomial over $\mathbb{C}$ and that $T$ is a function from a finite dimensional complex vector space to itself. Show that $F(T)$ is invertible if and only if it has no common factors with the minimal polynomial of T.
Here's my attempt:
Definite the minimal polynomial of $T$ as $$M(T) = (T - \lambda_{1}I)^{m_{1}}\dots (T - \lambda_{j}I)^{m_{j}},$$
where $\lambda_{1}, \dots \lambda{j}$ are the distinct eigenvalues of $T$. Since $F(T)$ is a polynomial over $\mathbb{C}$, it factors as $$F(T) = \alpha (T - x_{1}I) \dots (T - x_{k}I),$$
Where $x_{1}, \dots x_{k}$ are not necessarily unique. Suppose that $F(T)$ is invertible. Then $$\det(F(T)) = \alpha ^{n}\det((T - x_{1}I) \dots \det((T - x_{k}I) \neq 0.$$ But this is only true if and only if for each $i$, $\det(T-x_{i}I) \neq 0 $, which implies that $x_{i}$ is not an eigenvalue of $T$. Thus $M(T)$ and $F(T)$ share no common factors.
I believe the other direction is identical, since all the statements I made in the proof are tied together with if and only ifs. I'm not sure if this proof is correct however, since I never made use of any other properties of the minimal polynomial of T, other than it factoring over its eigenvalues. Is there something I'm missing?
Your proof is correct.
A polynomial $F$ having a common factor with the minimal polynomial of $T$ would indeed exactly mean that one of the eigenvalues of $T$ is a root of $F$.
Note that, in place of the minimal polynomial, one could also write the characteristic polynomial here.