Commuting matrices in GL(n,Z) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:24:46Z http://mathoverflow.net/feeds/question/55620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z Commuting matrices in GL(n,Z) Nikita Sidorov 2011-02-16T14:02:14Z 2011-02-16T17:37:21Z <p>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. </p> <p>Is there a closed description of the set of elements of $GL(n,\mathbb Z)$ which commute with $M$? </p> <p>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. </p> <p>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 $GL(n,\mathbb Z)$)? </p> http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z/55629#55629 Answer by Pedro Martins Rodrigues for Commuting matrices in GL(n,Z) Pedro Martins Rodrigues 2011-02-16T15:40:35Z 2011-02-16T15:40:35Z <p>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.</p> http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z/55635#55635 Answer by Igor Rivin for Commuting matrices in GL(n,Z) Igor Rivin 2011-02-16T16:28:59Z 2011-02-16T16:28:59Z <p>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).</p> http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z/55637#55637 Answer by anton for Commuting matrices in GL(n,Z) anton 2011-02-16T16:32:45Z 2011-02-16T16:32:45Z <p>$M$ 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.</p> <p>Proofs of these statements can be put together with the material in the books "Associative Algebras" by Pierce and "Maximal Orders" by Reiner.</p> http://mathoverflow.net/questions/55620/commuting-matrices-in-gln-z/55646#55646 Answer by David Speyer for Commuting matrices in GL(n,Z) David Speyer 2011-02-16T17:37:21Z 2011-02-16T17:37:21Z <p>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.</p> <hr> <p>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 $\mathrm{Span}_{\mathbb{C}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$. </p> <p>Proof: This statement is clearly invariant under change of basis, so we may assume that $M$ is diagonal, with diagonal entries $\lambda_i$. Then $AM=MA$ if and only if $\lambda_i A_{ij} = \lambda_j A_{ij}$. Since the $\lamdba$'s are distinct, this forces $A$ to be diagonal. Since the <a href="http://en.wikipedia.org/wiki/Vandermonde_matrix" rel="nofollow">Vandermonde matrix</a> is invertible, the span of the first $n$ powers of $M$ is precisely the diagonal matrices.</p> <hr> <p>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 $\mathrm{Span}_{\mathbb{Q}} (\mathrm{Id}, M, M^2, \cdots, M^{n-1})$.</p> <p>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 <code>$\mathrm{Mat}_{n \times n}(\mathbb{Q})$</code>. The previous section shows that $A \otimes \mathbb{C}$ and $B \otimes \mathbb{C}$ are the same subspace of <code>$\mathrm{Mat}_{n \times n}(\mathbb{C})$</code>. 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 $U=W$.</p> <p>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 <code>$K^*$</code>.</p> <hr> <p>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 <a href="http://en.wikipedia.org/wiki/Order_%2528ring_theory%2529" rel="nofollow">order</a>; I don't think there is much to say about this order which is not true of general orders.</p> <p>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 <code>$\mathcal{O}^*$</code> is a finite group times $\mathbb{Z}^{r+s-1}$.</p> <hr> <p>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:</p> <blockquote> <p>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 $\lambda$?</p> </blockquote> <p>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.</p> <p>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.</p> <p>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 $\cos (8 \pi /7 ) = \cos (6 \pi /7 )$. </p> <p>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 $|v| &lt; 1$, while $|w|>1$. So any matrix with eigenvalues $u$, $v$ and $w$ is hyperbolic. </p> <p>Numerical experimentation reveals that $|u w^4|>1$, while $|v u^4|$ and $|w v^4|&lt;1$. </p> <p>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.</p>