设A为实对称矩阵，且IAI＜0，试证 存在非零n维列向量X，使得X的转置AX<0.

Question

如题所述

Answer 1

证明: 由A为实对称矩阵,
则存在正交矩阵P满足 P'AP=diag(a1,a2,...,an). [P'=P^-1]
其中a1,a2,...,an是A的特征值.
又因为 |A|=a1a2...an<0
所以a1,a2,...,an中必有负数.
不妨设 a1<0. [注:可调整P的列向量的顺序实现]
令X=P(1,0,0,...,0)'
则 X'AX=[P(1,0,0,...,0)']A[P(1,0,0,...,0)']
= (1,0,0,...,0)P'AP(1,0,0,...,0)'
= (1,0,0,...,0)diag(a1,a2,...,an)(1,0,0,...,0)'
= a1 < 0.

Answer 2

太繁了