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Let $\mathbb F$ be a finite field. Denote by $M_n(\mathbb F)$ the set of matrices of order $n$ over $\mathbb F$ . For a matrix $A∈M_n(\mathbb F)$ what is the cardinality of $C_{M_n(\mathbb F)} (A)$ , the centralizer of $A$ in $M_n(\mathbb F)$? There are papers about it?

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  • $\begingroup$ Please give a link to MSE question - just for completeness. $\endgroup$
    – Alexander Chervov
    Commented Aug 19, 2012 at 16:00
  • $\begingroup$ math.stackexchange.com/questions/184275/… $\endgroup$
    – zacarias
    Commented Aug 19, 2012 at 16:56
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    $\begingroup$ The question can be asked in the framework of linear algebra (a tag I've added) but is better structured in the framework of Lie algebras of linear algebraic groups in general. In particular, Chevalley's version of "Jordan decomposition" expresses the semisimple and nilpotent (commuting) parts of a given matrix as polynomials in the matrix. But the literature in this area is large, starting with older work by Kolchin, Chevalley, Borel, and many others. And there are more prerequisites for reading most of it. Computations over finite fields are easier but not so helpful. $\endgroup$
    – Jim Humphreys
    Commented Aug 19, 2012 at 22:48

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Let me add some cases in which one has a clear answer:

[R.A. Horn, C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, Cambridge, 1991., Corollary 4.4.18]. Let $F$ be a field and $n$ is a natural number. If $A\in M_n(F)$ is a cyclic matrix, then $C_{M_n(F)}(A)$ is the set of all matrices which are polynomial in $A$ with coefficients in $F$.

Recall that a cyclic matrix in $M_n(F)$ is a matrix whose minimal and characteristic polynomials are the same.

[Lemma 3 of S. Akbari et al. / Linear Algebra and its Applications 390 (2004) 345–355] Let $F$ be a field and $n\geq 2$. If $A$ is a non-scalar matrix in $M_n(F)$ and $C_{M_n(F)}(A)$ has maximum dimension over $F$, then $\dim_F C_{M_n(F)}(A) = n^2 − 2n + 2$ and $A$ is similar to either $aI_1 \oplus bI_{n−1}$ or $aI_n + bE_{12}$, for some $a, b \in F$.

See for the notation the latter mentioned paper.

I suggest you to look for papers on the commuting graphs of rings, you may find some other cases which are treated in the proofs. One paper is quoted above and the another is S. Akbari, P. Raja / Linear Algebra and its Applications 416 (2006) 1038–1047

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    $\begingroup$ Terminology is always a problem in matrix theory, but "cyclic" matrix doesn't suggest much. At least from the viewpoint of Jordan decomposition in linear algebraic groups or their Lie algebras, the current term for such a matrix is "regular semisimple". I recall also an archaic matrix term "nonderogatory" in a similar framework. In any case, Jordan decomposition is a powerful tool to organize centralizers and dimensions even in this classical matrix algebra situation, adapted to Lie algebras over finite fields. $\endgroup$
    – Jim Humphreys
    Commented Aug 19, 2012 at 22:42
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    $\begingroup$ @Jim Humphreys: In this situation "cyclic" is the same as "regular" since any companion matrix is cyclic. In the case of $GL_n$-conjugacy the term cyclic actually suggests more than regular because it comes from the fact that $F^n$ is a cyclic $F[A]$-module. Still I prefer the term "regular" because it highlights the connection to the more general theory. $\endgroup$
    – A Stasinski
    Commented Aug 20, 2012 at 10:12
  • $\begingroup$ @A Stasinski: Yes, it should be just "regular" here (in Steinberg's sense), e.g., for semisimple matrices all eigenvalues are distinct and for nilpotent matrices the Jordan normal form has a single block. A "regular" matrix over an algebraically closed field is one for which the centralizer has smallest possible dimension, here n. $\endgroup$
    – Jim Humphreys
    Commented Aug 20, 2012 at 11:34
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Treat $F^n$ as an $F[t]$-module $M^A$, where $t$ acts by the matrix $A$. Then the centralizer can be thought of as $\mathrm{End}_{F[t]} M^A$. Now, $M^A$ has a primary decomposition

$ M^A = \bigoplus_{p \in \mathrm{Irr}(F[t])} M_p$

where $M_p$ consists of vectors in $M^A$ which are annihilated by some power of $p(A)$. Likewise,

$C_{M_n(F)}(A)= \mathrm{End}_{F[t]} M^A = \bigoplus_p \mathrm{End}_{F[t]} M_p$

So the problem is reduced to the primary case, where the characteristic polynomial of $A$ is a power of some irreducible polynomial $p$.

Now, there exists a unique partition $\lambda=(\lambda_1,\dotsc,\lambda_l)$ such that

$M_p = \bigoplus_{i=1}^l F[t]/(p(t))^{\lambda_i}$.

As a vector space (and even as an $F[t]$-module), the endomorphism algebra of this module is the sum

$\bigoplus_{i,j} \mathrm{Hom}_{F[t]} (F[t]/(p(t))^{\lambda_i},F[t]/(p(t))^{\lambda_j})$.

The $(i,j)$th summand has dimension $(\deg p)\min\{\lambda_i,\lambda_j\}$. Therefore, the endomorphism algebra of this primary part is of dimension

$ (\deg p)\sum_{i,j} \min\{\lambda_i,\lambda_j\}$

To get the centralizer of the original matrix, you would add these numbers over all primary parts. Finally, raising $q$ to this number is the cardinality that you want.

These centralizers are discussed in great detail in Pooja Singla's PhD thesis http://www.hbni.ac.in/phdthesis/allthesis/MATH10200604007_PSingla.pdf and a related paper in J. Algebra 2010 (available on the arXiv at http://arxiv.org/abs/1001.5304v1).

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  • $\begingroup$ Isn't $\textrm{End}_{F[t]} M_p$ and the double direct sum below isomorphic as $F[t]$-module? Is there a reason why you said that those are vector space isomorphic? $\endgroup$
    – Sungjin Kim
    Commented Feb 22, 2013 at 21:19
  • $\begingroup$ @i707104: You are right. $\endgroup$
    – Amritanshu Prasad
    Commented Feb 23, 2013 at 11:04
  • $\begingroup$ @Prasad: Thank you for clarifying. I understand that the OP was asking for $F$-dimension, so "vector space isomorphic" is just enough for answering it, but indeed they are $F[x]$-module isomorphic. $\endgroup$
    – Sungjin Kim
    Commented Feb 23, 2013 at 21:48
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    $\begingroup$ @Prasad: By the way, this idea can be generalized to finding dimension of the solution space to a matrix equation $AX+XB =0$ where $A$ is $n\times n$ matrix, and $B$ is $m\times m$ matrix over $F$, and $X$ is $n\times m$ matrix over $F$. Now, we have to investigate $\textrm{Hom}_{F[t]}(M^{-B}, M^A)$. $\endgroup$
    – Sungjin Kim
    Commented Feb 25, 2013 at 4:27
  • $\begingroup$ @i707104: Nice point; this is the basis of the Kac-Stanley formula for the number of iso classes of representations of a given quiver (see Root systems, representations of quivers, and invariant theory by V. Kac). $\endgroup$
    – Amritanshu Prasad
    Commented Feb 26, 2013 at 11:21
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You asked this on Math Stack Exchange too. One messy case is when $A$ is unipotent. Some cases of tat are dealt with in a famous paper of P. Hall and G. Higman on "Reduction Theorems for Burnside's Problem" (Proceedings of London Mathematical Society, 1956). The centralizer of a semisimple matrix is relatively easy to understand. Therefore, the essence of the question does reside in the structure of the centralizer of a nilpotent matrix.

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    $\begingroup$ As I commented above, all of this becomes more transparent in the framework of algebraic groups and their Lie algebras, though at a higher entry fee. Recently for example the nilpotent cases have been examined much more generally in the new book by Liebeck-Seitz, but in the general linear case the old lecture notes by Springer and Steinberg in Springer LN 131 are still quite useful for getting at the centralizers and conjugacy classes in a relatively down-to-earth way based on semisimple and nilpotent Jordan parts of a matrix (then adapted to finite fields). $\endgroup$
    – Jim Humphreys
    Commented Aug 19, 2012 at 22:54
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I've recently come across this question about the size of the centralizer (I assume that you mean the subgroup of $\text{GL}_n(F)$ which commute with the matrix). It seems to be hard to find. I wonder why it's been omitted from every Algebra book (even Bourbaki!) that I've consulted. There is a derivation of it in Dickson's "Linear Groups with an Exposition of the Galois Field Theory" p. 235 (Dover edition) dating from 1900. A more modern way of seeing this is to let $R = F[x]$, and construct the $R$-module on a vector space $V$ of dimension $n$ (if $A$ is $n \times n$) by having $x v = A v$. Then the centralizer is $\text{Aut}_R(V)$. In MacDonald's book "Symmetric Functions and Hall Polynomials" p. 87 (1979 edition) he derives the formula in terms of the theory of finitely generated modules over a commutative discrete valuation ring. This, incidentally, also gives the formula for the number of automorphisms of a finite abelian group.

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This is just elementary comment, but may be too long for comment, probably you know it, but just for completeness.

I happened to ask almost (but not exactly) the same question some days ago:

Conjugcy classes in GL(F_2) ? GL(F_q)

It sounds a little different - size of conjugacy classes in GL(F_q). But conjugacy class of element "C" is size of GL(F_q)/ size of centralizer of element "C". Because size of any orbit is |G|/|Stabilizer| and here we have action by conjugation and stabilizer is centralizer. So if you know the size of centralizer - you know size of conjugacy class.

The difference is, of course, that you asked about Mat(F_q) while setup above is about non-degenerate matrices GL(F_q).

Let me also write down some elementary facts for completeness.

If you consider element $C$ such that its characteristic polynomial is irreducible, (then it automatically minimal), then size of centralizer in Mat_n(F_q) is q^n (if I understand correctly) and q^n-1 in GL_n(F_q).

One has good way to think about this centralizer: let consiser p(x) - char.pol. of "C". F_q[x]/p(x) is a field F_q^n. To any element of the field "a" in F_q^n one can correspond a matrix M(a) - the matrix of multiplication by "a":F_q^n -> F_q^n.

Map "a->M(a)" is clearly homomorphism of algebras so in particular all matrices M(a) commute among themselves. They provide a centralizer of a matrix M(x).

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