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Let $A \in End (V)$ be a diagonalisable operator on a vector space over $k$ and $B$ an operator which commutes with any $C \in End(V)$ which commutes with $A$.

How to prove that $B=P(A)$ for some polynomial $P(t) \in k[t]$

Any help would be much appreciated.

user1551
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Christophe Chateau
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    Related: http://math.stackexchange.com/questions/568209/matrices-which-commute-with-all-the-matrices-commuting-with-a-given-matrix – user26857 Dec 17 '16 at 20:17

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In the case where $k$ is algebraically closed and $V$ is finite dimensional: let $\lambda$ be an eigenvalue of $A$ and let $V_\lambda$ be the associated eigenspace and let $W_\lambda=\bigoplus_{\mu\neq\lambda}V_\mu$ be the direct sum of all the other eigenspaces of $A$. Since $A$ is diagonalizable, $V=V_\lambda\oplus W_\lambda$. Let $p_\lambda$ be the projection onto $V_\lambda$ along $W_\lambda$. Clearly, $p_\lambda$ and $A$ commute. From your hypothesis, $p_\lambda\circ B=B\circ p_\lambda$ and hence $B(V_\lambda)\subset V_\lambda$. Denote by $B_\lambda$ the endomorphism of $V_\lambda$ obtained by the restriction of $B$ to $V_\lambda$. Let $C_\lambda$ be an endomorphism of $V_\lambda$, and extend it to an endomorphism of $V$ by adding $W_\lambda$ to its kernel. Then $C$ and $A$ commute, hence $B_\lambda$ and $C_\lambda$ commute. It is a standard fact that the only endomorphism that commutes with all endomorphisms is a multiple of the identity, hence $B_\lambda$ is a multiple of the identity of $V_\lambda$, say $B_\lambda=\mu(\lambda)\,\mathrm{id}_{V_\lambda}$. Now let $P$ be any polynomial that extends $\mu$ (such a polynomial exists, e.g., using Lagrange interpolation polynomials—this is where I use the fact that $V$ is finite-dimensional). We clearly have $P(A)=B$.

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gniourf_gniourf
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  • could you detail why:

    $p_\lambda$ and $A$ commute

     – Christophe Chateau Dec 20 '16 at 17:39
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    @ChristopheChateau: let $x\in V$. Decompose $x$ as $x=x_\lambda+x'\lambda$ where $x\lambda\in V_\lambda$ and $x'\lambda\in W\lambda$. Then, on the one hand, $A(p_\lambda x)=Ax_\lambda=\lambda x_\lambda$. On the other hand, notice that $A W_\lambda\subset W_\lambda$, hence $p_\lambda(A x)=p_\lambda(\lambda x_\lambda+A x'\lambda)=p\lambda(\lambda x_\lambda)+p_\lambda(A x'\lambda)=\lambda x\lambda+0_V=\lambda x_\lambda$. Hence $(p_\lambda\circ A)(x)=(A\circ p_\lambda)(x)$ hence $p_\lambda\circ A=A\circ p_\lambda$. – gniourf_gniourf Dec 20 '16 at 17:43
  • could you explain or link me an explanation of why such a polynomial $P$ exists?

    Thanks for the proof it took me some time but I think I got it, it is quite clear.

     – Christophe Chateau Dec 20 '16 at 18:49
  • also, I went too fast at the end, why do we clearly have $P(A)=B$ ? – Christophe Chateau Dec 20 '16 at 18:53
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    @ChristopheChateau: Such a polynomial exists from the Lagrange interpolating polynomials. We have $P(A)=B$ since: if $x\in V$, decompose $x$ on the eigenspaces: $x=x_{\lambda_1}+\cdots+x_{\lambda_n}$ where each $x_{\lambda_i}\in V_{\lambda_i}$. Observe that for all $k\in\mathbb{N}$, $A^k x_{\lambda_i}=\lambda_i^k x_{\lambda_i}$ and from this you'll obtain: $P(A)=P(\lambda_1)x_{\lambda_1}+\cdots+P(\lambda_n)x_{\lambda_n}=\mu(\lambda_1)x_{\lambda_1}+\cdots+\mu(\lambda_n)x_{\lambda_n}=B(x_{\lambda_1})+\cdots+B(x_{\lambda_n})=B(x)$. – gniourf_gniourf Dec 20 '16 at 19:07