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I stuck so much on this question!

I need to describe finite order integer matrices without 1 eigenvalues over $\mathbb C$. I need description in terms of classes of equivalent matrices(such that exist complex matrix C such that $C^{-1}AC = B$)

I came to this results. Let A be the integer matrix of order $m$. First of all A can be diagonalized so is $diag(\lambda_1, ..., \lambda_n)$ in some basis. If $\lambda$ is its eigenvalue, then $\lambda^{m} = 1$.

The minimal polynomial $p$ for A divides $x^m - 1$ and belongs to $\mathbb Z[x]$ because $A \in Mat(\mathbb Z)$. Actually $p(\lambda_i) = 0$. So this should give some limitations on $(\lambda_1, ..., \lambda_n)$. I need to get all possible cases for $(\lambda_1, ..., \lambda_n)$. My hypothesis is "$(\lambda_1, ..., \lambda_n)$ splits into groups where every group is all roots of $\frac{x^k - 1}{x-1}$".

Edit 1. I was really inaccurate with formulating my hypothesis. I really mean that $(\lambda_1, ..., \lambda_n)$ splits into groups where every group is all roots of some irreducible polynomial in decomposition of $\frac{x^k - 1}{x-1}$

Edit 2. I'm sorry if I messed you up, I'll try to give much more readable description of my question. I'm doing a research and found out that for some group all automorpmism are in 1-1 correspondence with the classes of simmilar integer matrices of finite order without 1-eigenvalues. So what I want is to somehow enumerate this classes. I divided the problem in two parts:

1) Prove that the set of classes(that I described above) is finite

2) Give a constructive way to enumerate this classes

Hope now my question is more readable. Any help is appreciated!

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  • $\begingroup$ From $\lambda^m=1$ You know that $\lambda$ is a root of unity. Now assume that each $\lambda_i$ is some root of unity. Is the matrix of finite order? $\endgroup$ – Logic_Problem_42 May 4 '18 at 7:44
  • $\begingroup$ @Logic_Problem_42 Ofc not! This is not what my hypothesis states. All $\lambda$s should be split into groups where each group is a set of all roots of $\frac{x^k -1}{x -1}$. In your case the groups should be 1-element groups, so every lamda is -1. This is the only case for same eigenvalues of matrix with described properties. $\endgroup$ – SaveMyLife May 4 '18 at 7:52
  • $\begingroup$ What is $A^m$ if $A=diag(\lambda_1,...,\lambda_n)$ and $\lambda_i^m=1$ for all $i$? $\endgroup$ – Logic_Problem_42 May 4 '18 at 8:07
  • $\begingroup$ Probably also interesting for You: math.mit.edu/classes/18.702/crystrest.pdf $\endgroup$ – Logic_Problem_42 May 4 '18 at 8:11
  • $\begingroup$ Your hypothesis is a good start, but if $k$ is not prime not all the roots of $\frac{x^k -1}{x -1 }$ have the same properties, so maybe shouldn't all turn up together. Eg, for $k=6$ one is a square root, two are cube roots and two are sixth roots. @Logic_Problem_42 reference shows how a $2\times 2$ matrix can have the pair of $6$-th roots, and not "all the roots of $\frac{x^6 -1}{x -1 }$. Maybe this helps to refine your hypothesis? $\endgroup$ – ancientmathematician May 4 '18 at 8:47
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$A,B$ are not equivalent matrices but similar matrices. It's hard to understand what you're looking for (is it a homework ?); the integer matrices with finite order are well understood (thanks to Minkowski, Taussky and Todd). cf. for example

https://pdfs.semanticscholar.org/d72f/50b01413336e0c2b4f01859b5c39e01d7ad1.pdf

Of course, here we can have $1\in spectrum(A)$; you have to work a little; I hope this will not put your life at stake.

Let $\alpha_n=\{m;$ there is $A\in M_n(\mathbb{Z}) \;s.t.\;A^m=I_n\}$ and $s_n=\max(\alpha_n)$. Some results:

  1. Let $m=p_1^{e_1}\cdots p_t^{e_t}$ be the decomposition of the integer $m$ on the prime factors $p_1<\cdots<p_t$. Then $m\in\alpha_n$ iff

when $p_1^{e_1}=2$: $\sum_i(p_i-1)p_i^{e_i-1}-1\leq n$

otherwise: $\sum_i(p_i-1)p_i^{e_i-1}\leq n$

  1. $\alpha_{2k+1}=\alpha_{2k}$.

  2. $\alpha_2=\alpha_3=\{2,3,4,6\}$, $\alpha_4=\alpha_5=\{2,3,4,5,6,8,10,12\}$.

  3. $s_{22}=2520$. Note that $2520\in\alpha_{22}$ because $2520=2^3.5.7.9$ and $2^2+4+6+8=22$.

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  • $\begingroup$ It's not a homework, it's a mini task in my research. Actually, you got me wrong. We start with fixed $n$ and about to describe all classes of equivalent(similar) matrices $\in M_n(\mathbb Z)$ in terms of sets of eigenvalues. $\endgroup$ – SaveMyLife May 4 '18 at 13:20
  • $\begingroup$ In your opinion: how do we find the $m$ s.t. $A^m=I$ ? What is a fixed $n$ if it is any ? Since you must obtain the polynomials that are annihilated by $A$, read the above reference. You write that "it is a mini task in my research"... I think that your problems will have just begun. $\endgroup$ – loup blanc May 4 '18 at 18:50
  • $\begingroup$ I cannot agree that this is the place my problems just begun. I was describing and automorphism of some group with specific properties and I found a 1-1 correspondence between an automorphism and integer matrix with described properties. So all I want is to get some constructive description of this matrices in terms of eigenvalues. I we come back to your first two questions, we do not find the m. We start with a give n, which is the size of the matrix, and trying to describe the matrices of this fixed size n with given above properties in terms of its eigenvalues. $\endgroup$ – SaveMyLife May 4 '18 at 19:47
  • $\begingroup$ This kind of description will give an easy way to construct and enumerate all classes of similar matrices. $\endgroup$ – SaveMyLife May 4 '18 at 19:49

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