From 0e4780fa9614c5e2c0691ed442709de663605ff7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com>
Date: Thu, 22 Jan 2026 23:39:52 +0800
Subject: [PATCH] vault backup: 2026-01-22 23:39:52

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 素材/整合素材/二次型证明题.md | 10 ----------
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-已知n阶实对称矩阵A为正定矩阵，n阶实矩阵B使得$A-B^{\text{T}}$AB也为正定矩阵，证明B的特征值$\lambda$满足关系式$|\lambda|<1$。
-
-
-**证明**  
-设$\boldsymbol{p}$为B对应于特征值$\lambda$的特征向量，即$B\boldsymbol{p}=\lambda \boldsymbol{p}$，且$\boldsymbol{p}\neq\boldsymbol{0}。由A-B^{\text{T}}$AB为正定矩阵可知
-$\boldsymbol{p}^{\text{T}}(A-B^{\text{T}}AB)\boldsymbol{p}>0.$
-由A为正定矩阵可知$\boldsymbol{p}^{\text{T}}A\boldsymbol{p}>0$。
-再由
-$\boldsymbol{p}^{\text{T}}(A-B^{\text{T}}AB)\boldsymbol{p}= \boldsymbol{p}^{\text{T}}A\boldsymbol{p}-\boldsymbol{p}^{\text{T}}B^{\text{T}}AB\boldsymbol{p}= \boldsymbol{p}^{\text{T}}A\boldsymbol{p}-(B\boldsymbol{p})^{\text{T}}A(B\boldsymbol{p})= \boldsymbol{p}^{\text{T}}A\boldsymbol{p}-\lambda^2 \boldsymbol{p}^{\text{T}}A\boldsymbol{p}$,
-可知$1-\lambda^2>0$，即$|\lambda|<1$。
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