# Commuting matrices in GL(n,Z)

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 $GL(n,\mathbb Z)$)?

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.

• 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? Feb 16, 2011 at 18:47
• 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. Feb 16, 2011 at 19:03
• 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?
– ah--
Jul 14, 2022 at 6:00
• 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. 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.

$M$ 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.