AI 中文总结
本文确定振荡矩阵各阶子式首次全正时间的轮廓，提出两角准则，证明尖锐不等式，并给出基于相邻双对角分解的有限刻画与整数幺模实现。
AI 中文摘要
对于 $n\times n$ 振荡矩阵 $A$，设 $e_k(A)$ 为使得 $A^m$ 的所有 $k$ 阶子式均为正的最小正整数 $m$。我们确定了轮廓 $(e_1(A),\ldots,e_n(A))$。对于每个非奇异全非负矩阵，每个复合行中正元素的列指标集在分量序下构成一个区间。端点映射保序且在乘法下复合。这给出了固定阶所有子式正性的两角准则。因此，$e_k(A)$ 是两个远端角子式首次为正时间的较大者。我们证明了尖锐不等式 $e_k(A)\leq\max\{k,n-k\}$（$1\leq k<n$）以及 $|e_{k+1}(A)-e_k(A)|\leq1$（$1\leq k\leq n-2$）。对于 $D=\operatorname{diag}(1,-1,1,-1,\ldots)$，矩阵 $DA^{-1}D$ 是振荡的且满足 $e_k(DA^{-1}D)=e_{n-k}(A)$（$1\leq k<n$）；行列式指数保持为1。相邻双对角分解中正因子的有序位置独立于其值确定轮廓。每个单侧因子序列可由单个置换表示，给出所有轮廓的有限刻画和每个轮廓的整数幺模实现。我们还得到了跨次阶的相容性条件：若 $3\leq k\leq n-3$ 且 $e_k(A)\leq2$，则 $e_j(A)\leq2+\lceil |j-k|/2\rceil$ 对所有 $2\leq j\leq n-2$ 成立。特别地，$e_k(A)\leq2$ 蕴含 $e_{k+2}(A)\leq3$（$3\leq k\leq n-4$），且常数和此范围均为尖锐的。
英文摘要
For an $n\times n$ oscillatory matrix $A$, let $e_k(A)$ be the least positive integer $m$ for which every minor of order $k$ of $A^m$ is positive. We determine the profile $(e_1(A),\ldots,e_n(A))$. For every nonsingular totally nonnegative matrix, the column index sets of positive entries in each compound row form an interval in the componentwise order. The endpoint maps are order-preserving and compose under multiplication. This yields a two-corner criterion for positivity of all minors of a fixed order. Consequently, $e_k(A)$ is the larger of the first positivity times of two remote corner minors. We prove the sharp inequalities $e_k(A)\leq\max{k,n-k}$ for $1\leq k<n$ and $|e_{k+1}(A)-e_k(A)|\leq1$ for $1\leq k\leq n-2$. For $D=\operatorname{diag}(1,-1,1,-1,\ldots)$, the matrix $DA^{-1}D$ is oscillatory and satisfies $e_k(DA^{-1}D)=e_{n-k}(A)$ for $1\leq k<n$; the determinant exponent remains one. The ordered positions of the positive factors in an adjacent bidiagonal factorization determine the profile independently of their values. Each one-sided factor sequence can be represented by a single permutation, giving a finite characterization of all profiles and an integer unimodular realization of each. We also obtain a compatibility condition across minor orders: if $3\leq k\leq n-3$ and $e_k(A)\leq2$, then $e_j(A)\leq2+\lceil |j-k|/2\rceil$ for $2\leq j\leq n-2$. In particular, $e_k(A)\leq2$ implies $e_{k+2}(A)\leq3$ for $3\leq k\leq n-4$, and both the constant and this range are sharp.
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