用深度学习发现矩阵单调性的简单判别规则，准确率达95%。
Explaining Deep Network Classification of Matrices: A Case Study on Monotonicity
- 结合神经网络与可解释AI，从矩阵特征中提炼出判别规则。
- 仅用特征多项式前两项系数绝对值，分类准确率达95%。
- 发现矩阵迹的倒数大于5.7是单调矩阵的强判据，适合数学与算法研究者。
本文提出一种基于深度学习的矩阵分类方法，用于发现抽象代数性质下的实用判别准则。针对单调矩阵（其逆矩阵元素非负）这一难题，我们构建了18,000个随机生成的7×7矩阵数据集，通过深度神经网络对矩阵进行单调性分类，并利用集成梯度等可解释性技术识别关键特征。结果表明，仅凭特征多项式的前两个系数绝对值 $|c_0|$ 与 $|c_1|$ 即可实现95%的分类准确率。进一步分析显示，单调矩阵满足 $|c_0/c_1| \leq 0.18$ 概率超过99.98%，而由于 $|c_0/c_1| = 1/\mathrm{tr}(A^{-1})$，该条件等价于 $\mathrm{tr}(A^{-1}) \geq 5.7$，首次系统揭示了单调矩阵的可计算判据。
原文摘要 · Abstract (English)
This work demonstrates a methodology for using deep learning to discover simple, practical criteria for classifying matrices based on abstract algebraic properties. By combining a high-performance neural network with explainable AI (XAI) techniques, we can distill a model's learned strategy into human-interpretable rules. We apply this approach to the challenging case of monotone matrices, defined by the condition that their inverses are entrywise nonnegative. Despite their simple definition, an easy characterization in terms of the matrix elements or the derived parameters is not known. Here, we present, to the best of our knowledge, the first systematic machine-learning approach for deriving a practical criterion that distinguishes monotone from non-monotone matrices. After establishing a labelled dataset by randomly generated monotone and non-monotone matrices uniformly on $(-1,1)$, we employ deep neural network algorithms for classifying the matrices as monotone or non-monotone, using both their entries and a comprehensive set of matrix features. By saliency methods, such as integrated gradients, we identify among all features, two matrix parameters which alone provide sufficient information for the matrix classification, with $95\%$ accuracy, namely the absolute values of the two lowest-order coefficients, $c_0$ and $c_1$ of the matrix's characteristic polynomial. A data-driven study of 18,000 random $7\times7$ matrices shows that the monotone class obeys $\lvert c_{0}/c_{1}\rvert\le0.18$ with probability $>99.98\%$; because $\lvert c_{0}/c_{1}\rvert = 1/\mathrm{tr}(A^{-1})$ for monotone $A$, this is equivalent to the simple bound $\mathrm{tr}(A^{-1})\ge5.7$.
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