Smith矩阵的无限可除性
Infinite divisibility of Smith matrices
摘要：
给定一个算术函数$f$，用$f(a, b)$和$f[a, b]$分别表示在正整数$a$和$b$的最大公约数$(a, b)$处求值的函数$f$和在最小公倍数$[a, b]$处求值的函数$f$。一个半正定矩阵$A=(a_{ij})$对所有的$i$和$j$都有$a_{ij}\ge 0$，如果对于每个非负实数$r$，其分数Hadamard幂$A^{\circ r}=(a_{ij}^r)$都是半正定的，则称它是无限可分的.设$S=\{x_1, ..., x_n\}$是一组$n$不同的正整数。本文证明了如果$f$是一个乘法函数，使得$(f*μ)(d)\ge 0$对任何$x\in S$都是$d|x$，则$n\times n$矩阵$(f(x_i, x_j))$，$(\frac{1}{f[x_i, x_j]})$和$(\frac{f(x_i, x_j)}{f[x_i, x_j]})$是无限可分的.最后，我们将这些结果扩展到狄利克雷卷积的情况下，产生无限多的例子，无限可分矩阵。我们的结果推广了Bourque，Ligh，Bhatia，Hong，Lee，Lindqvist和Seip的结果.
英文摘要：
Given an arithmetical function $f$, by $f(a, b)$ and $f[a, b]$ we denote the function $f$ evaluated at the greatest common divisor $(a, b)$ of positive integers $a$ and $b$ and evaluated at the least common multiple $[a, b]$ respectively. A positive semi-definite matrix $A=(a_{ij})$ with $a_{ij}\ge 0$ for all $i$ and $j$ is called infinitely divisible if the fractional Hadamard power $A^{\circ r}=(a_{ij}^r)$ is positive semi-definite for every nonnegative real number $r$. Let $S=\{x_1, ..., x_n\}$ be a set of $n$ distinct positive integers. In this paper, we show that if $f$ is a multiplicative function such that $(f*μ)(d)\ge 0$ whenever $d|x$ for any $x\in S$, then the $n\times n$ matrices $(f(x_i, x_j))$, $(\frac{1}{f[x_i, x_j]})$ and $(\frac{f(x_i, x_j)}{f[x_i, x_j]})$ are infinitely divisible. Finally we extend these results to the Dirichlet convolution case which produces infinitely many examples of infinitely divisible matrices. Our results extend the results obtained previously by Bourque, Ligh, Bhatia, Hong, Lee, Lindqvist and Seip.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
