The matrix semigroup membership problem asks, given square matrices $M,M_1,\ldots,M_k$ of the same dimension, whether $M$ lies in the semigroup generated by $M_1,\ldots,M_k$. It is classical that this problem is undecidable in general but decidable in case $M_1,\ldots,M_k$ commute. In this paper we consider the problem of whether, given $M_1,\ldots,M_k$, the semigroup generated by $M_1,\ldots,M_k$ contains a non-negative matrix. We show that in case $M_1,\ldots,M_k$ commute, this problem is decidable subject to Schanuel's Conjecture. We show also that the problem is undecidable if the commutativity assumption is dropped. A key lemma in our decidability result is a procedure to determine, given a matrix $M$, whether the sequence of matrices $(M^n)_{n\geq 0}$ is ultimately nonnegative. This answers a problem posed by S. Akshay (arXiv:2205.09190). The latter result is in stark contrast to the notorious fact that it is not known how to determine effectively whether for any specific matrix index $(i,j)$ the sequence $(M^n)_{i,j}$ is ultimately nonnegative (which is a formulation of the Ultimate Positivity Problem for linear recurrence sequences).

翻译：矩阵半群成员问题是指：给定相同维度的方阵$M,M_1,\ldots,M_k$，判断$M$是否属于由$M_1,\ldots,M_k$生成的半群。经典结论表明，该问题一般不可判定，但在$M_1,\ldots,M_k$可交换的情形下可判定。本文考虑如下问题：给定$M_1,\ldots,M_k$，判断由它们生成的半群是否包含非负矩阵。我们证明，当$M_1,\ldots,M_k$可交换时，该问题在Schanuel猜想成立的前提下可判定。同时证明，若去除交换性假设，则该问题不可判定。我们可判定性结果中的一个关键引理是：给定矩阵$M$，判断矩阵序列$(M^n)_{n\geq 0}$是否最终非负的算法。这回答了S. Akshay提出的一个问题（arXiv:2205.09190）。该结果与一个众所周知的困难事实形成鲜明对比：对于任意特定矩阵指标$(i,j)$，如何有效判定序列$(M^n)_{i,j}$是否最终非负尚属未知（该问题正是线性递归序列终极正性问题的矩阵形式表述）。