From 627798c4928dfc623b6bb011e30761f9c201bbf0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Sat, 24 Jan 2026 00:00:09 +0800 Subject: [PATCH] vault backup: 2026-01-24 00:00:09 --- 编写小组/讲义/正交及二次型（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/正交及二次型（解析版）.md b/编写小组/讲义/正交及二次型（解析版）.md index 5f6a8fb..9c37396 100644 --- a/编写小组/讲义/正交及二次型（解析版）.md +++ b/编写小组/讲义/正交及二次型（解析版）.md @@ -237,7 +237,7 @@ $U = \text{span}\{\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\boldsymbol{\beta}_3 >考虑特征多项式 $|\lambda\boldsymbol E-\boldsymbol A|=\lambda(\lambda-2)^2=0 \implies \lambda_1=0,\lambda_2=\lambda_3=2.$ >$\lambda=0$ 时，$-\boldsymbol A\rightarrow \begin{bmatrix}1&0&1\\0&1&0\\0&0&0\end{bmatrix}$, 取 $\eta_1=(1,0,-1)^T$； >$\lambda=2$ 时，$2\boldsymbol E-\boldsymbol A\rightarrow \begin{bmatrix}1&0&-1\\0&0&0\\0&0&0\end{bmatrix}$, 取 $\eta_2=(1,0,1)^T,\eta_3=(0,1,0)^T$. ->对 $\eta_1,\eta_2,\eta_3$ 进行标准正交化得$$\boldsymbol \varepsilon_1=\dfrac{1}{\sqrt2}(1,0,-1),\boldsymbol \varepsilon_2=\dfrac{1}{\sqrt2}(1,0,1),\boldsymbol \varepsilon_3=(0,1,0)$$ +>对 $\eta_1,\eta_2,\eta_3$ 进行标准正交化得$$\boldsymbol \varepsilon_1=\dfrac{1}{\sqrt2}(1,0,-1),\boldsymbol \varepsilon_2=(0,1,0),\boldsymbol \varepsilon_3=\dfrac{1}{\sqrt2}(1,0,1)$$ >取 $\boldsymbol P_1=\begin{bmatrix}\boldsymbol \varepsilon_3\ \boldsymbol \varepsilon_2\ \boldsymbol \varepsilon_1\end{bmatrix}$, 则有 $\boldsymbol P_1^T\boldsymbol A\boldsymbol P_1=\begin{bmatrix}2&0&0\\0&2&0\\0&0&0\end{bmatrix},$ 再取 $\boldsymbol P_2=\begin{bmatrix}\frac{1}{\sqrt{2}}&0&0\\0&\frac{1}{\sqrt{2}}&0\\0&0&1\end{bmatrix}$, 有$\boldsymbol P_2^T\boldsymbol P_1^T\boldsymbol A\boldsymbol P_1\boldsymbol P_2=(\boldsymbol P_1\boldsymbol P_2)^T\boldsymbol A(\boldsymbol P_1\boldsymbol P_2)=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$, 所以 $\boldsymbol P=\boldsymbol P_1\boldsymbol P_2=\begin{bmatrix}0&\frac{1}{\sqrt{2}}&1\\\frac{1}{\sqrt{2}}&0&0\\0&\frac{1}{\sqrt{2}}&-1\end{bmatrix}$. >（2）由 $\boldsymbol x^T\boldsymbol x=1$ 知 $x_1^2+x_2^2+x_3^2=1$. 由（1）知，存在正交变换 $\boldsymbol x=C\boldsymbol z$，使得 $\displaystyle f\overset{\boldsymbol x=C\boldsymbol z}{=}2(z_1^2+z_2^2)$. 由于正交变换不改变向量长度，有 $z_1^2+z_2^2+z_3^2=1\Rightarrow z_1^2+z_2^2=1-z_3^2$, 带入得 $$f=2(1-z_3^2)\le2, \text{等号当且仅当} z_3=0 \text{时取得.}$$ >由（1），可取正交矩阵 $\displaystyle C=\begin{bmatrix}\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\0 & 1 & 0 \\\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}\end{bmatrix}$，则 $\displaystyle \boldsymbol x=\begin{bmatrix}\frac{z_1}{\sqrt{2}} \\ z_2 \\ \frac{z_1}{\sqrt{2}}\end{bmatrix}$. 当 $x_1=x_2$ 时， $\displaystyle z_1=\sqrt 2z_2=\frac{\sqrt 6}{3}\Rightarrow x_1=x_2=x_3=\frac{1}{\sqrt 3}$. 故 当$\boldsymbol x=\begin{bmatrix}\frac{1}{\sqrt{3}} \\[4pt]\frac{1}{\sqrt{3}} \\[4pt]\frac{1}{\sqrt{3}}\end{bmatrix}$ 时，$f_{\max}=2.$