# show that $C(C[u])=K[u]$

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?

• swith=commute, $K$-vectoriel=$K$-vector space, polynome=polynomial. – Dietrich Burde Mar 1 at 19:40
• 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 $u$. – darij grinberg Mar 1 at 20:09
• Where did you encounter this problem? In particular, do the words "Schur's lemma" mean anything to you? – Jacob Manaker Mar 1 at 20:28
• @JacobManaker i'v never heard about it before – user515918 Mar 1 at 21:09
• @darijgrinberg "Thus, in turn, any element of C(C[U]) must commute with all such endomorphismes " why is it true ? – user515918 Mar 1 at 21:11