Let $E$ be a $K$-vector space of dimension $n$, and $u$ be a diagonalizable linear map from $E$ to $E$.
Let $C[u]$ be the set of linear maps from $E$ to $E$ which commute with $u$. In other words,
$C[u] = \left\{ v \text{ linear map from $E$ to $E$ such that $ v \circ u=u \circ v$} \right\}$.
More generally, for any subset $X$ of $L(E) = \left(\text{vector space of linear maps from $E$ to itself}\right)$, we let $C(X)$ be the set of all elements $\in L(E)$ which commute with all elements of $X$.
How can we show that $C(C[u])=K[u]$?
Here, $K[u]$ is the set of all $P(u) \in L(E)$ such that $P$ is a polynomial.
I see that any element of $K[u]$ can commute with all the linear maps which can commute with $u$
but showing that any of them is polynomial of $u$ seems an interesting result.
I can write $E=E_{\lambda_1} \oplus... \oplus E_{\lambda_k}$ such as $sp(u)=$ { $\lambda_{1}$, ... , $\lambda_{k}$} (spectra)
How can we continue?