Suppose $M$ is a "hyperbolic" matrix in $GL(n,\mathbb Z)$, i.e., that its characteristic polynomial $p$ is irreducible over $\mathbb Z$ and has no roots of modulus 1.

Is there a closed description of the set of elements of $GL(n,\mathbb Z)$ which commute with $M$?

I have a vague recollection that it is somewhat similar to the Dirichlet theorem on the units of an algebraic field, but it is really vague so a reference would be appreciated.

The case I'm most interested in is when $p$ has only one root of modulus greater than 1. Can $M$ commute with another matrix $M'$ with the same property (and $M, M'$ not being powers of the same matrix in $GL(n,\mathbb Z)$)?


4 Answers 4


This has very little to do with being hyperbolic; the key point is that the characteristic polynomial is irreducible. It is convenient to "close in" on $\mathbb{Z}$ be thinking about easier rings.

Let $M$ be a matrix in $GL(n, \mathbb{C})$ with distinct eigenvalues. Then I claim that the set of $n \times n$ matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{C}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$.

Proof: This statement is clearly invariant under change of basis, so we may assume that $M$ is diagonal, with diagonal entries $\lambda_i$. Then $A M = M A$ if and only if $\lambda_i A_{ij} = \lambda_j A_{ij}$. Since the $\lambda$'s are distinct, this forces $A$ to be diagonal. Since the Vandermonde matrix is invertible, the span of the first $n$ powers of $M$ is precisely the diagonal matrices.

Let $M$ be a matrix in $GL(n, \mathbb{Q})$ whose characteristic polynomial has distinct roots. Then the set of $n \times n$ rational matrices which commute with $M$ is $\mathrm{Span}_{\mathbb{Q}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$.

Proof: Let $A$ be "the set of matrices which commute with $M$" and let $B$ be "the linear span of the first $n$ powers of $M$". Both of these are rational subvector spaces of $\mathrm{Mat}_{n \times n}(\mathbb{Q})$. The previous section shows that $A \otimes \mathbb{C}$ and $B \otimes \mathbb{C}$ are the same subspace of $\mathrm{Mat}_{n \times n}(\mathbb{C})$. A standard lemma is that, if $U$ and $V$ are both $K$-subspaces of a $K$ vector space $W$, and $U \otimes L = V \otimes L$ as subspaces of $W \otimes L$, then $U=W$.

Now, let $p$ be the characteristic polynomial, and assume furthermore that $p$ is irreducible. The $\mathbb{Q}$-span of the powers of $M$ is isomrophic, as a ring, to $\mathbb{Q}[t]/p(t)$. Since $p$ is irreducible, this is some number field, call it $K$. So the set of $\mathbb{Q}$-matrices which commute with $M$ is isomorphic to a number field. Since every element in a field, other than 0, is invertible, we get that the matrices in $GL(n, \mathbb{Q})$ which commute with $M$ are isomorphic to $K^*$.

Let $M$ be a matrix in $GL(n, \mathbb{Z})$ whose characteristic polynomial is irreducible. Let $K$ be the field of $\mathbb{Q}$-matrices which commute with $M$, as discussed above. The set of the matrices whose entries are in $\mathbb{Z}$ forms a lattice $\mathcal{O}$, of rank $n$, in $K$, which is also a subring. Such a subring of a number field is called an order; I don't think there is much to say about this order which is not true of general orders.

Finally, you want to understand those matrices of $\mathcal{O}$ which are in $GL(n, \mathbb{Z})$, meaning that their inverses are also in $\mathcal{O}$. This is the unit group of $\mathcal{O}$. And, indeed, Dirichlet's unit theorem applies to orders: if $K$ has $r$ real places and $s$ complex places, then $\mathcal{O}^*$ is a finite group times $\mathbb{Z}^{r+s-1}$.

Finally, you want to know whether or not you can have $M$ hyperbolic, $N$ hyperbolic commuting with $M$, but $N$ not a power of $M$. The answer is YES. I'll first give a theoretical proof, and then sketch an actual computation. Let the eigenvalues of $M$ be $\lambda_1$, ..., $\lambda_n$, with $|\lambda_1|>1$. Let $N=f(\lambda_i)$, with $f$ a polynomial with rational coefficients. Note that the $\lambda$'s are the Galois orbit of $\lambda_1$. For any Galois automorphism $\sigma$, we have $f(\sigma(\lambda_1)) = \sigma(f(\lambda_1))$. So the eigenvalues of $N$ are the Gaois orbit of $f(\lambda_1)$. In short, your question is equivalent to the following:

Let $\mathcal{O}$ be an order in a number field. Suppose that $\lambda$ and $\mu$ are units such that $|\lambda|$ and $|\mu|>1$, but their Galois conjugates are less than $1$. Can this hapen without $\mu$ not a power of $\lambda$?

It certainly can. I'll give the conceptual proof, then sketch a computation. If you recall the standard proof of Diricihlet's unit theorem, it goes as follows: Map $\mathcal{O}^*$ to $\mathbb{R}^{r+s}$ by $u \mapsto (\log |u|, \log |\sigma_2(u)|, \cdots, \log |\sigma_{s+r}(u)| )$, where the inputs to the logs are the Galois orbit of $u$. Clearly, the image lands in the hyperplane where the coordinates sum to $0$. One proves that the image is a discrete lattice, of rank $r+s-1$, in this hyperplane.

In particular, we are interested in units where $\log |u|>0$ but where all the other coordinates are negative. This is the intersection of our discrete lattice with a full dimensional cone; once $r+s-1>1$, this will be larger than the one dimensional sublattice of the powers of any single unit.

If you want an explicit example, lets take $K=\mathbb{Q}[\cos (2 \pi/7)]$. The Galois action permutes $(\cos (2 \pi/7), \cos (4 \pi/7), \cos (6 \pi/7))$ cyclically. (Note that $\cos (4 \pi/7) = 2 \cos^2 (2 \pi/7) -1$, so it is in the same field, and similarly for $\cos (8 \pi /7 ) = \cos (6 \pi /7 )$.

Set $u=(1-\cos(2 pi/7))/(1-\cos (4 \pi/7))$, $v=(1-\cos(4 pi/7))/(1-\cos (6 \pi/7))$ and $w=(1-\cos(4 pi/7))/(1-\cos (8 \pi/7))$. So the Galois group permutes $(u,v,w)$ cyclically and $u*v*w=1$. You can check that $u$, $v$ and $w$ obey the equation $$t^3 - 6 t^2 + 5t -1=0$$ so $u$, $v$ and $w$ are units. I think they generate the unit group; in any case, the have finite index in it. Also note that $|u|$ and $|v| < 1$, while $|w|>1$. So any matrix with eigenvalues $u$, $v$ and $w$ is hyperbolic.

Numerical experimentation reveals that $|u w^4|>1$, while $|v u^4|$ and $|w v^4|<1$.

So, write down explicit matrices for the action of $w$ and $u w^4$ on the ring of integers of $K$. Then these will be two hyperbolic matrices which commute, but where neither is a power of the other.

  • $\begingroup$ David, thanks! It is clear to me now how these matrices are related to units. As for the second part of my question, I'm afraid I didn't make myself 100% clear... What I meant was whether there exists a hyperbolic $M'$ commuting with $M$ with the <b>same property as $M$</b> - that is, it has only one eigenvalue outside the unit disc. It seems that the basis of the centralizer are matrices from different "clusters" (cluster = $m$ eigenvalues inside the unit disc, $n-m$ outside). Does this mean that the answer to this question is actually no? $\endgroup$ Feb 16, 2011 at 18:47
  • $\begingroup$ Right. Let $M$ have eigenvalues $(u,v,w) \approx (0.307979, 0.643104, 5.04892)$. There will be some other matrix $N$, corresponding to the unit $u w^4$, whose eigenvalues are $(u w^4, v u^4, w v^4) \approx (200.131, 0.0057858, 0.863622)$. So they are both hyperbolic. $\endgroup$ Feb 16, 2011 at 19:03
  • $\begingroup$ Is there a hint what kind of "numerical experimentation" would be needed to find the $f(\lambda_1)$ (i.e. $uw^4$) in the example you gave? $\endgroup$
    – ah--
    Jul 14, 2022 at 6:00
  • 1
    $\begingroup$ All I need is some integers $(a,b,c)$ such that $u^a v^b w^c > 1$ and $v^a w^b u^c$ and $w^a u^b v^c < 1$. I have no memory of why I came up with the specific choice $(1,0,4)$, I assume I tried a few triples until one worked. $\endgroup$ Jul 14, 2022 at 13:35

Suppose $M$ has a simple eigenvalue $\lambda$ with associated eigenvector $v$. Then $M'$ has also eigenvector $v$ associated to an eigenvalue $\rho$. This means that $\lambda$ and $\rho$ are units in the same ring $\mathbb{Z}[\lambda]$. The converse is clear, so a closed description of the $M' \in GL(n,\mathbb{Z})$ commuting with $M$ would be: The multiplicative group of units in the ring $\mathbb{Z}[\lambda]$. If this group is not cyclic then the answer to the second question is yes.


If you do a google search on "centralizer of semisimple element in linear group" you will learn more about this question than you ever wanted. (see, eg, Fleischmann/Janiszczak/Knorr).


$M$ being hyperbolic means that it sits inside a torus, i.e., an algebraic group isomorphic to ${\mathbb G}_m^n$. Then its centralizer in the algebra of $n\times n$ matrices is semisimple, i.e., a product of smaller matrix algebras over skew fields. The $\mathbb Z$-valued points define an order in that algebra and the matrices in $GL_n({\mathbb Z})$ commuting with $M$ form the unit group of that order.

Proofs of these statements can be put together with the material in the books "Associative Algebras" by Pierce and "Maximal Orders" by Reiner.


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