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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?

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  • $\begingroup$ swith=commute, $K$-vectoriel=$K$-vector space, polynome=polynomial. $\endgroup$ – Dietrich Burde Mar 1 at 19:40
  • $\begingroup$ Hint: Let $S$ be the set of eigenvalues of $u$. Write $E = \bigoplus_{\lambda \in S} E_\lambda$, where the $E_\lambda$ are the eigenspaces of $\lambda \in S$. Argue that each endomorphism of $E$ that preserves all these eigenspaces $E_\lambda$ must commute with $u$ and thus belongs to $C\left[u\right]$. Thus, in turn, any element of $C\left(C\left[u\right]\right)$ must commute with all such endomorphisms, and hence (check this!) must act as a scalar $k_\lambda$ on each $E_\lambda$. This means (prove this using Lagrange interpolation) that it is a polynomial in $u$. $\endgroup$ – darij grinberg Mar 1 at 20:09
  • $\begingroup$ Where did you encounter this problem? In particular, do the words "Schur's lemma" mean anything to you? $\endgroup$ – Jacob Manaker Mar 1 at 20:28
  • $\begingroup$ @JacobManaker i'v never heard about it before $\endgroup$ – user515918 Mar 1 at 21:09
  • $\begingroup$ @darijgrinberg "Thus, in turn, any element of C(C[U]) must commute with all such endomorphismes " why is it true ? $\endgroup$ – user515918 Mar 1 at 21:11

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