Prove that every operator G such that GT = TG is a polynomial in T Let be $T:V \rightarrow V$ ($\dim V \lt \infty$) a diagonalizable operator with the algebraic multiplicity of eigenvalues ​​$1$. Then every operator $G$ such that  $GT = TG$ is a polynomial in $T.$
My attempt:
I Know that if $GT = TG$ then the eigenspace of $T$ is G-Invariant. 
In fact, let $v$ an eigenvector of $T$ with eigenvalue $\lambda$, by assumption
$T(G(v)) = G(T(v)) = \lambda G(v)$ $\Rightarrow$ $G(v)$ is eigenvector of $T$ with eigenvalue $\lambda$.
What I do with this?  

$T$ is diagonalizable. So it is easiest if just change basis to one where $T$ is diagonal.
All eigenvalues have multiplicity $1$, so all diagonal entries are distinct. This means that any matrix which commutes with $T$ is also diagonal in this basis (so the subspace of endomorphisms $V\to V$ which commutes with $T$ has the same dimension as $V$).
Also, all diagonal entries of $T$ are distinct, so $I, T, \ldots, T^{n-1}$ are all linearly independent. Therefore the subspace of endomorphisms $V\to V$ consisting of polynomials in $T$ has the same dimension as $V$.
