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Let $A$ be an $n \times n$ matrix with entries from an arbitrary field $F$ and let $C(A)$ denote the set of all matrices which commute with $A$.

Is it true that $C(C(A))= \{ \alpha_1 + \alpha_2 A + \cdots + \alpha_{n-1}A^{n-1}\mid\alpha_i \in F \}$ ?

This problem is a generalized version of Problems 6.3.13, 6.3.14 of Herstein's Topics in Algebra which I am unable to do. Herstein asks to prove that this holds for the cases $n=2,3$.

Of course, one way to go about it is by brute force calculation (which I have not tried), but I guess there is a more conceptual way to do this which I am unable to find.

Clearly all the matrices of the type in R.H.S. are in $C(C(A))$ so the part I am unable to do is that these are the only matrices.

I can see that $C(A)$ and $C(C(A))$ are themselves vector spaces but what to do further ?

user90041
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    What is wrong with calculating? – Sergio Parreiras Nov 15 '13 at 14:45
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    @SergioParreiras I just wish to know if there is an alternate more conceptual/insightful way of doing it. – user90041 Nov 15 '13 at 14:48
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    You didn't define what $C(X)$ should be for a set $X$, but I'm guessing from context you just want it to be the elements that commute with each element of $X$. Is that right? – rschwieb Nov 15 '13 at 15:03
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    @user90041 Sometimes we find or are able to see the clever way only after we bother to get paper and pen and starting working on the obvious route. :-) – Sergio Parreiras Nov 15 '13 at 15:14
  • Since the question is basis independent, you might as well assume that $A$ is in Jordan canonical form. – Jim Belk Nov 15 '13 at 19:24
  • @rschwieb Sorry, I did not define that. You are right in your interpretation. And thanks for pointing it out. – user90041 Nov 15 '13 at 19:37
  • Btw, the answer to your question is positive and can be find in Horn and Johnson, Topics in Matrix Theory, but their proof is far from "conceptual", that's why I think the non-commutative algebraists could show us a proof using the Double Centralizer Theorem. –  Nov 16 '13 at 11:00
  • @user Thanks a lot for the reference. I looked at the book and the contents look interesting. And I found that this is proved in Theorem 4.4.19 of the book. It will take me some time to read the proof though. – user90041 Nov 16 '13 at 14:11
  • @SergioParreiras I take your advice and could actually do it by calculation in case $n=2$, shall try for higher cases too. Thanks a lot :-) – user90041 Nov 16 '13 at 14:16
  • Yes, this is true as you already know by now. If you are interested in infinite-dimensional generalizations, see this thread. – Julien Nov 17 '13 at 16:00
  • I posted an answer to this before. http://math.stackexchange.com/questions/497806/matrices-that-commute-with-two-distinct-matrices/497818#497818 – Sungjin Kim Nov 25 '13 at 07:18
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    @JimBelk: One cannot assume Jordan canonical form because the field $F$ is not assumed to be algebraically closed. I think one might nonetheless reduce to the case, because the centraliser $C(A)$ is a vector space, so the centraliser of the algebraic closeur $\overline F$ of$~F$ is just $C(A)\otimes_F\overline F$ and the result for $\overline F$ then "inherits" down to$~F$. – Marc van Leeuwen Nov 25 '13 at 08:03

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As user i7071707 pointed out in a comment, this question has already been asked, and has received a correct answer there, based on this paper by Paco Lagerstrom (1945). I'm just posting this answer because questions with an open bounty cannot be flagged as duplicates.

Here is a quick sketch of the cited proof. Viewing $F^n$ as an $F[X]$-module through the action of$~A$ (so $Xv=A\cdot v$) it decomposes by the structure theorem as a finite direct sum of modules $M_i$, each cyclic so there exists $v_i\in M_i$ generating it, and such that the minimal polynomials $P_i$ of$~v_i$ (the generator of the ideal of $F[X]$ acting as$~0$ on$~v_i$, and therefore on$~M_i$) are divisible by any $P_j$ with $j>i$ (this is opposite to the usual order; it makes $P_1$ the (global) minimal polynomial of$~A$).

Elements of $C(A)$ are precisely the endomorphsims of this $F[X]$-module; I shall call them $A$-endomorphisms, and call elements of $C(C(A))$ central $A$-endomorphisms. The decomposition into $F^n$ as a direct sum of submodules $M_i$ is not canonical, so the $M_i$ need not be stable under all $A$-endomorphisms (indeed $M_1$ is not stable under the $A$-endomorphism $f$ below). However, since the projections onto the $M_i$ according to the direct sum decomposition are particular $A$-endomorphisms, any central $A$-endomorphism$~\zeta$ must stabilise the eigenspaces of these projections, and therefore each$~M_i$. Then $\zeta$ is completely determined by the images $\zeta(v_i)\in M_i$.

But $\zeta$ is in fact completely determined by the single image $\zeta(v_1)\in M_1$. To see this fix $k>1$. There is for each$~i$ a unique $F[X]$-module map $F[X]\to M_i$ sending $X\mapsto v_i$, with kernel $(P_i)$, and since $(P_k)$ contains$~P_1$, the map $F[X]\to M_k$ passes to the quotient to give a $F[X]$-module map $f:M_1\to M_k$, which by construction satisfies $f(v_1)=v_k$. Extend $f$ to an $A$-endomorphism by setting it zero on $\bigoplus_{i>1}M_i$. Then $\zeta(v_k)=\zeta(f(v_1))=f(\zeta(v_1))$ because $\zeta$ is a central $A$-endomorphism, and this is determined by $\zeta(v_1)$ as promised.

Finally, since $M_1$ is cyclic there exists $P\in F[X]$ with $\zeta(v_1)=P[A]\cdot v_1$. By the above it follows that $\zeta(v_i)=P[A]\cdot v_i$ for all$~i$, and therefore $\zeta=P[A]$.

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