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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 ?

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marked as duplicate by Marc van Leeuwen Jan 26 at 10:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is wrong with calculating? –  Sergio Parreiras Nov 15 '13 at 14:45
@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
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
@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
@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 8:03

2 Answers 2

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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The result is true. You can find a proof, in french, here

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Firstly, I am not interested by your bounty. Secondly, you ask a question and I give a reference which is a good one. Sorry, but I have not read all the previous comments. –  loup blanc Nov 20 '13 at 23:15
Could the downvoters explain why they downvoted?? –  Etienne Nov 24 '13 at 10:28
The link seems to be broken, I think that's why. (I did not downvote this) –  i707107 Nov 25 '13 at 7:19
It's not broken; it only has a broken security certificate. If you remove "s" from "https" or just accept a certificate exception, it will work. However, the PDF is in French, which I suspect is not very popular as a link-only answer on an English speaking Q&A site. (I too didn't downvote, so this is just my assumption on why the downvotes) –  Vedran Šego Nov 25 '13 at 14:38

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