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?

*

*$\begin{pmatrix} 
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}$ has a cyclic vector and commutes with $\begin{pmatrix} 
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$ and $\begin{pmatrix} 
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}$ which are polynomials in the operator.

*$\begin{pmatrix} 
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}$ does not have a cyclic vector. It commutes with the same projections, but they are not polynomials in this second operator.
