AI 中文总结
该研究证明四阶实矩阵的帕累托特征值最大数目为23，通过分情况论证排除25的可能，结合扰动与符号计算完成结论验证。
AI 中文摘要
对于给定的实矩阵$A\in\mathbb{R}^{n\times n}$，$A$的帕累托特征值是满足以下条件的实数$\lambda\in\mathbb{R}$：存在非零向量$x\in\mathbb{R}^n\setminus\{0\}$，使得$0\leq x\perp Ax-\lambda x\geq0$。本文证明所有四阶实矩阵$A\in\mathbb{R}^{4\times4}$至多有23个不同的帕累托特征值。首先，针对满足两个条件的四阶矩阵（每个主子矩阵的实特征值均为单根，且任意两个不同主子矩阵无公共实特征值），通过不动点论证可知其帕累托特征值数目为奇数。此前已知四阶矩阵的帕累托容量介于23到26之间，因此仅需排除25这一可能值。本文利用三阶支撑轮廓及主子矩阵特征向量相关恒等式排除该情况，再通过扰动与不动点指数论证将该上界推广至所有四阶实矩阵，且精确符号计算验证了某显式四阶矩阵恰好有23个不同的正则帕累托特征值。
英文摘要
For a given real matrix $A\in\R^{n\times n}$, a Pareto eigenvalue of $A$ is a real number $λ\in\R$ for which there exists a nonzero vector $x\in\R^n\setminus\{0\}$ such that $$ 0\leq x\perp Ax-λx\geq0. $$ We prove that every matrix $A\in\R^{4\times4}$ has at most $23$ distinct Pareto eigenvalues. We first prove the result for matrices of order $4$ satisfying two conditions: every real eigenvalue of a principal submatrix is simple, and two different principal submatrices have no real eigenvalue in common. For these matrices, a fixed point argument shows that the number of Pareto eigenvalues is odd. Previous known results show that the Pareto capacity of order $4$ is between $23$ and $26$. Thus only $25$ remains to exclude. We exclude this case by using the support profiles in order $3$ and identities involving eigenvectors of principal submatrices. A perturbation and fixed point index argument then extends the bound to all real matrices of order $4$. An exact symbolic computation certifies that an explicit matrix of order $4$ has exactly $23$ distinct regular Pareto eigenvalues.
