{ "id": "2406.11113", "version": "v1", "published": "2024-06-17T00:22:00.000Z", "updated": "2024-06-17T00:22:00.000Z", "title": "Matrix periods and competition periods of Boolean Toeplitz matrices II", "authors": [ "Gi-Sang Cheon", "Bumtle Kang", "Suh-Ryung Kim", "Homoon Ryu" ], "categories": [ "math.CO" ], "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 $S$ and $T$ of $\\{1,\\ldots,n-1\\}$, an $n\\times n$ Toeplitz matrix $A=T_n\\langle S ; T \\rangle$ is defined to have $1$ as the $(i,j)$-entry if and only if $j-i \\in S$ or $i-j \\in T$. In the previous paper, we have shown that the matrix period and the competition period of Toeplitz matrices $A=T_n\\langle S; T \\rangle$ satisfying the condition ($\\star$) $\\max S+\\min T \\le n$ and $\\min S+\\max T \\le n$ are $d^+/d$ and $1$, respectively, where $d^+= \\gcd (s+t \\mid s \\in S, t \\in T)$ and $d = \\gcd(d, \\min S)$. In this paper, we claim that even if ($\\star$) is relaxed to the existence of elements $s \\in S$ and $t \\in T$ satisfying $s+t \\le n$ and $\\gcd(s,t)=1$, 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 $k, n$ with $2k+1 \\le n$, it is easy to see that $T_n\\langle k, n-k;k+1, n-k-1 \\rangle$ does not satisfies ($\\star$) but satisfies the relaxed condition. Furthermore, we show that the limit of the matrix sequence $\\{A^m(A^T)^m\\}_{m=1}^\\infty$ is $T_n\\langle d^+,2d^+, \\ldots, \\lfloor n/d^+\\rfloor d^+\\rangle$.", "revisions": [ { "version": "v1", "updated": "2024-06-17T00:22:00.000Z" } ], "analyses": { "keywords": [ "toeplitz matrix", "boolean toeplitz matrices", "matrix period", "competition period", "linear algebra appl" ], "note": { "typesetting": "TeX", "pages": 0, "language": "en", "license": "arXiv", "status": "editable" } } }