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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!

SaveMyLife
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  • 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? – Logic_Problem_42 May 04 '18 at 07:44
  • @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. – SaveMyLife May 04 '18 at 07:52
  • What is $A^m$ if $A=diag(\lambda_1,...,\lambda_n)$ and $\lambda_i^m=1$ for all $i$? – Logic_Problem_42 May 04 '18 at 08:07
  • Probably also interesting for You: http://math.mit.edu/classes/18.702/crystrest.pdf – Logic_Problem_42 May 04 '18 at 08:11
  • 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? – ancient mathematician May 04 '18 at 08:47
  • @ancientmathematician thanks, I was really wrong in formulating the hypothesis. This still doesn't give any constructive way to describe $(\lambda_1, ..., \lambda_n)$ – SaveMyLife May 04 '18 at 10:44
  • Yes it does. You are now saying (once you prove your conjecture) that effectively the equivalence classes are described by diagonal block matrices with companion matrices of cyclotomic polynomials on the diagonal. – ancient mathematician May 04 '18 at 11:19
  • @ancientmathematician But how to describe cyclotomic polynomials? – SaveMyLife May 04 '18 at 12:03
  • What's wrong with using $X^n -1 = \prod_{d|n}\Phi_d(X)$ which defines them recursively. I don't think these polynomials have any simpler description. – ancient mathematician May 04 '18 at 12:25
  • @ancientmathematician this is true. So in summary, we can preprocess all irreducible $\Phi_d$ for $d \leq k$ but how $k$ is limited by the size of the matrix? – SaveMyLife May 04 '18 at 13:07
  • There is some sort of answer here https://math.stackexchange.com/questions/101654/lower-bound-for-phin-is-n-5-phi-n-n-for-all-n-1 (said to be Theorem 2.9 in Montgomery and Vaughn's multiplicative number theory.) – ancient mathematician May 04 '18 at 16:19

1 Answers1

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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 $2^2+4+6+8=22$.

  • 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. – SaveMyLife May 04 '18 at 13:20
  • 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. –  May 04 '18 at 18:50
  • 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. – SaveMyLife May 04 '18 at 19:47
  • This kind of description will give an easy way to construct and enumerate all classes of similar matrices. – SaveMyLife May 04 '18 at 19:49