proof about commutative operators and T-cyclic vectors Let $V$ be a finite dimensional vector space over $F$. Let $T:V \to V$ be a linear operator.
Prove that if every linear operator $U$ which commutes with $T$ is a polynomial of $T$, than $T$ has a $T$-cyclic vector.
I don't really know where to start...
can someone please point me in the right direction?

$\newcommand{\complex}{\mathbb{C}}$
$\newcommand{\A}{T}$
$\newcommand{\B}{U}$
$\newcommand{\diag}{\mathrm{diag}}$
When $F = \complex$, I will give a proof that is based on Jordan form theory.
Let $\lambda_1, \lambda_2, \ldots, \lambda_t$ be $\A$'s all distinct eigenvalues, and $\A$'s group of elementary divisors be
\begin{align*}
        & (\lambda - \lambda_1)^{m_{11}}, (\lambda - \lambda_1)^{m_{12}}, \ldots, (\lambda - \lambda_1)^{m_{1k_1}}, \\
        & (\lambda - \lambda_2)^{m_{21}}, (\lambda - \lambda_2)^{m_{22}}, \ldots, (\lambda - \lambda_2)^{m_{2k_2}}, \\
        & \phantom{(\lambda - \lambda_1)^{m_{11}},} \ldots\ldots\ldots\ldots\ldots \\
        & (\lambda - \lambda_t)^{m_{t1}}, (\lambda - \lambda_t)^{m_{t2}}, \ldots, (\lambda - \lambda_t)^{m_{tk_t}},
\end{align*}
where $m_{j1} \geq \cdots \geq m_{jk_j}, j = 1, 2, \ldots, t$. Then there exists a basis $\{\alpha_1, \ldots, \alpha_n\}$ of $V$ under which $\A$'s matrix is $J = \diag(J_{11}, J_{12}, \ldots, J_{1k_1}, \ldots, J_{t1}, J_{t2}, \ldots, J_{tk_t})$,
where $J_{jl}$ is the Jordan block associated with the elementary divisor $(\lambda - \lambda_j)^{m_{jl}}$, $l = 1, 2, \ldots, k_j, j = 1, 2, \ldots, t$.
If $\A$ is not a cyclic operator, then there exists at least one $j \in \{1, 2, \ldots, t\}$ such that $k_j \geq 2$. Otherwise, $\A$'s characteristic polynomial and minimal polynomial are identical, which would imply $\A$ is a cyclic operator. Without loss of generality, we can assume $j = 1$. Suppose $C = \begin{pmatrix} 0_{m_{12} \times {(m_{11} - m_{12}})} & I_{(m_{12})}\end{pmatrix}$ and define
\begin{equation*}
    \tilde{J} = \diag\left(\begin{pmatrix}0 & 0 \\ C & 0 \end{pmatrix},\underbrace{0, \ldots, 0, \ldots, 0, \ldots, 0}_{k_1 + \cdots + k_t - 2 \text{ blocks}}\right).
\end{equation*}
It is easy to verify that $J$ commutes with $\tilde{J}$. Define $\B: V \to V$ by $\B(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)\tilde{J}$, then
\begin{equation*}
    \A\B(\alpha_1, \ldots, \alpha_n) = (\alpha_1, \ldots, \alpha_n)J\tilde{J} = (\alpha_1, \ldots, \alpha_n)\tilde{J}J = \B\A(\alpha_1, \ldots, \alpha_n),
\end{equation*}
i.e., $\B$ commutes with $\A$. On the other hand, for any polynomial $f$,
\begin{equation*}
    f(J) = \diag(f(J_{11}), f(J_{12}), \ldots, f(J_{1k_1}), \ldots, f(J_{t1}), \ldots, f(J_{tk_t}))
\end{equation*}
does not equal to $\tilde{J}$ because the $(2, 1)$ block of $f(J)$ is zero while the $(2, 1)$ block of $\tilde{J}$ is $C \neq 0$. Therefore, $\B$ cannot be expressed as
a polynomial in $\A$, contradiction. This completes the proof.
