Let $V$ be a finite dimensional vector space, $T$ a linear operator on $V$, $C(T)=\{S: ST=TS\}$, and $\mathbb{F}(T)=\{f(T):f\in\mathbb{F}[x]\}$. Is it true that if $C(T)=\mathbb{F}(T)$, then the minimal polynomial and characteristic polynomial of $T$ are equal?
I think the answer is yes, but i'm not sure how to prove it. It suffices to show that for some $\alpha\in V$, $V=Z(\alpha;T)=\{g(T)\alpha:g\in\mathbb{F}[x]\}$. Since $C(T)=\mathbb{F}(T)$, $Z(\alpha;T)=\{S\alpha:ST=TS\}$. Thus, for any $v\in V$, we just need to find an operator $S$ such that $S\alpha=v$ and $S$ commutes with $T$.
Let $\{e_i\}_i$ be a basis for $V$, and let $\alpha\in V$ be such that $\alpha=\sum a_ie_i$, where each $a^i\not=0$. Then for any $v=\sum v_ie_i\in V$, define $S(e_i)=\frac{v_i}{a_i}e_i$, so $S(\alpha)=v$. This also says that $S$ is diagonalizable (since the basis $\{e_i\})$ consists of eigenvectors of $S$). Now if $T$ is diagonalizable, we can conclude $ST=TS$ since we can just take $\{e_i\}$ to be the common basis of eigenvectors so $S$ and $T$ are simultaneously diagonalizable. But I don't think we can infer $ST=TS$ in general.
Can someone help? Thank you.
