矩阵的最终非负性属于P
Eventual Nonnegativity of a Matrix Is in P
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中文总结 AI 辅助
该论文证明，判断有理矩阵的足够大幂是否逐项非负可在多项式时间内完成，通过检查单一等差序列并利用特征值分组与伽罗瓦等变性实现高效判定。
中文摘要 AI 辅助
给定一个有理矩阵$A$，是否每个足够大的幂$A^n$逐项非负？我们证明该问题在确定性多项式时间内可判定。这与判定单个指定项序列$(A^n)_{ij}$的最终非负性形成对比，后者等价于线性递推序列的终极正性问题，其可判定性仍是开放的。先前的可判定性程序（D'Costa, Ouaknine和Worrell，STACS 2024）将矩阵幂按模一个扭转指数$D$的剩余类拆分，$D$的值可能随输入规模呈指数增长。我们证明只需检查单个等差序列$A^{Dk+1}$：满足$A^n \geq 0$的指数$n$对加法封闭，且该序列中两个连续指数互素，因此它们生成所有足够大的指数。沿该序列，特征值按其共同的$D$次幂分组，每组所需的系数无需构造$A^D$、任何$\lambda^D$或分裂域即可计算：每个求和项在其自身的根域$\mathbb{Q}(\lambda)$中求值，伽罗瓦等变性使每个类和的次数和高度保持多项式有界，从而实现可验证的零和符号测试。同一算法在拒绝一种情形后判定最终正性，其变体判定矩阵是否具有任何非负幂。
英文摘要
Given a rational matrix $A$, is every sufficiently large power $A^n$ entrywise nonnegative? We prove that this problem is decidable in deterministic polynomial time. This is in contrast with deciding eventual nonnegativity of a single prescribed entry sequence $(A^n)_{ij}$, which amounts to the Ultimate Positivity Problem for linear recurrence sequences, whose decidability is open. The previous decidability procedure (D'Costa, Ouaknine and Worrell, STACS 2024) splits the matrix powers into residue classes modulo a torsion exponent $D$ whose value can be exponential in the input size. We show that it is sufficient to check a single progression $A^{Dk+1}$: the exponents $n$ with $A^n \geq 0$ are closed under addition, and two consecutive exponents of the progression are coprime, so they generate every sufficiently large exponent. Along the progression, eigenvalues are grouped by their common $D$th power, and the required coefficients for each group are computed without forming $A^D$, any $λ^D$, or a splitting field: each summand is evaluated in its own root field $\mathbb{Q}(λ)$, and Galois equivariance keeps the degree and height of every class sum polynomially bounded, enabling certified zero and sign tests. The same algorithm with one case rejected decides eventual positivity, and a variant decides whether the matrix has any nonnegative power at all.
