vault backup: 2026-01-23 21:12:45

编写小组/讲义/正交及二次型（解析版）.md

@@ -117,7 +117,7 @@ $$\|\boldsymbol{\gamma}_3\|=\sqrt{\dfrac{3}{2}},\boldsymbol{\varepsilon}_3=\dfra
 >$$\boldsymbol{\xi}_1=(2,1,4,0)^T,\quad
 \boldsymbol{\xi}_2=(0,1,-1,1)^T$$
 >用施密特正交化法得 $$\boldsymbol u_1=\boldsymbol \xi_1,\boldsymbol\varepsilon_1=\dfrac{\boldsymbol u_1}{\|\boldsymbol u_1\|}=\dfrac{1}{\sqrt{21}}(2,1,4,0)^\text{T},$$
->$$\boldsymbol u_2=\boldsymbol \xi_2-\langle\boldsymbol \xi_2,\boldsymbol \varepsilon_1\rangle\boldsymbol \varepsilon_1=\dfrac{1}{7}(2,8,-3,7)^\text{T},\boldsymbol \varepsilon_2=\dfrac{\boldsymbol u_2}{\|\boldsymbol u_2\|}=\dfrac{1}{4\sqrt 5}(2,0,-3,7)^\text T$$
+>$$\boldsymbol u_2=\boldsymbol \xi_2-\langle\boldsymbol \xi_2,\boldsymbol \varepsilon_1\rangle\boldsymbol \varepsilon_1=\dfrac{1}{7}(2,0,-3,7)^\text{T},\boldsymbol \varepsilon_2=\dfrac{\boldsymbol u_2}{\|\boldsymbol u_2\|}=\dfrac{1}{4\sqrt 5}(2,0,-3,7)^\text T$$
 >满足条件的一组标准正交向量为：$$\boldsymbol{\varepsilon}_1 = \frac{1}{\sqrt{21}}\begin{bmatrix}2\\1\\4\\0\end{bmatrix},\quad\boldsymbol{\varepsilon}_2 = \frac{1}{4\sqrt{5}}\begin{bmatrix}2\\0\\-3\\7\end{bmatrix}.$$
 
 # Section 2  实对称矩阵的正交变换与二次型