English

Matrix periods and competition periods of Boolean Toeplitz matrices II

Combinatorics 2024-10-18 v1

Abstract

This paper is a follow-up to the paper [Matrix periods and competition periods of Boolean Toeplitz matrices, {\it Linear Algebra Appl.} 672:228--250, (2023)]. Given subsets SS and TT of {1,,n1}\{1,\ldots,n-1\}, an n×nn\times n Toeplitz matrix A=TnS;TA=T_n\langle S ; T \rangle is defined to have 11 as the (i,j)(i,j)-entry if and only if jiSj-i \in S or ijTi-j \in T. In the previous paper, we have shown that the matrix period and the competition period of Toeplitz matrices A=TnS;TA=T_n\langle S; T \rangle satisfying the condition (\star) maxS+minTn\max S+\min T \le n and minS+maxTn\min S+\max T \le n are d+/dd^+/d and 11, respectively, where d+=gcd(s+tsS,tT)d^+= \gcd (s+t \mid s \in S, t \in T) and d=gcd(d,minS)d = \gcd(d, \min S). In this paper, we claim that even if (\star) is relaxed to the existence of elements sSs \in S and tTt \in T satisfying s+tns+t \le n and gcd(s,t)=1\gcd(s,t)=1, the same result holds. There are infinitely many Toeplitz matrices that do not satisfy (\star) but the relaxed condition. For example, for any positive integers k,nk, n with 2k+1n2k+1 \le n, it is easy to see that Tnk,nk;k+1,nk1T_n\langle k, n-k;k+1, n-k-1 \rangle does not satisfies (\star) but satisfies the relaxed condition. Furthermore, we show that the limit of the matrix sequence {Am(AT)m}m=1\{A^m(A^T)^m\}_{m=1}^\infty is Tnd+,2d+,,n/d+d+T_n\langle d^+,2d^+, \ldots, \lfloor n/d^+\rfloor d^+\rangle.

Keywords

Cite

@article{arxiv.2406.11113,
  title  = {Matrix periods and competition periods of Boolean Toeplitz matrices II},
  author = {Gi-Sang Cheon and Bumtle Kang and Suh-Ryung Kim and Homoon Ryu},
  journal= {arXiv preprint arXiv:2406.11113},
  year   = {2024}
}