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There have been many questions in the vein of this one, but I can't find one that answers it specifically.

Suppose $A,B\in M_n(\mathbb C)$ are two matrices such that, for any other matrix $C\in M_n(\mathbb C)$, $$AC=CA\implies BC=CB.$$ Prove that $B=p(A)$ for a polynomial $p\in\mathbb C[t]$.

Nothing is assumed about $A$ being diagonalisable or having distinct eigenvalues or being invertible.

Edit: after further work, this might not actually be true. If it's not, a counterexample would be great.

Rushabh Mehta
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Ian Coley
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    I collected a bunch of equivalent statements (to each other) at http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 Should help. – Will Jagy Sep 18 '13 at 19:23

1 Answers1

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Reference: Lagerstrom, Paco, A Proof of a Theorem on Commutative Matrices, Bulletin of AMS, Volume 51 (1945), 535-536.

This holds for any field $\mathbb{F}$.

Let $M^A$ denote the $\mathbb{F}[x]$-module structure on $\mathbb{F}^n$. We use rational canonical form-invariant factor form. Then we have $$ M^A \simeq \mathbb{F}[x]/P_1 \oplus\cdots \oplus \mathbb{F}[x]/P_r, $$ where $P_i=(p_i)$, $p_i|p_{i+1}$. This gives invariant subspace decomposition, $$ M^A =\bigoplus_{i=1}^r M_i, $$ where $M_i\simeq \mathbb{F}[x]/P_i$.

Let $\pi_i:M^A\longrightarrow M_i$ be the projection, and $\pi_{ij}:M_i\longrightarrow M_j$ be the natural projection for $i>j$. Extend $\pi_{ij}$ linearly to $M^A$ by assigning 0 on all $M_k(k\neq i)$. Then all $\pi_i$ and $\pi_{ij}$ commute with $A$, thus commute with $B$. Therefore, each $M_i$ is $A$-invariant, thus it is also $B$-invariant. Let $e_i\in M_i$ be the element corresponding to $1+P_i\in\mathbb{F}[x]/P_i$. We see that there is $p(x)\in\mathbb{F}[x]$ such that $Be_r=p(A)e_r$. We claim that $Be_i=p(A)e_i$ for all $i<r$, and hence $B=p(A)$. $$ Be_i=B\pi_{ri}e_r=\pi_{ri}Be_r=\pi_{ri}p(A)e_r=p(A)\pi_{ri}e_r=p(A)e_i. $$ This completes the proof.

user26857
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Sungjin Kim
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  • @IanColey PDF at http://www.ams.org/bull/1945-51-08/S0002-9904-1945-08386-4/S0002-9904-1945-08386-4.pdf – Will Jagy Sep 18 '13 at 19:36
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    While I think this is basically correct, I do not agree with calling $\pi_{i,j}$ a natural projection. Indeed to define it one must choose generators of the cyclic modules $M_i$ and $M_j$, and map one generator to the other; the resulting map does depend on the choice made (but by construction it commutes with $A$ for any choice). – Marc van Leeuwen Nov 25 '13 at 08:34
  • Why does $Be_i = p\left(A\right)e_i$ imply that $B = p\left(A\right)$ ? I suspect this isn't hard to prove by establishing some more endomorphisms of $M^A$ and arguing that they therefore commute with $B$; but it isn't really obvious to me. – darij grinberg Sep 21 '16 at 00:43
  • @darij, the original paper is available, I put the link four years ago. Just two pages. – Will Jagy Sep 21 '16 at 00:49
  • In this solution, we have $Be_r = p(A) e_r$. This is just by representing an element $Be_r$ on $M_r$ in terms of its basis. – Sungjin Kim Sep 21 '16 at 00:53