M 编写小组/讲义/正交及二次型（解析版）.md

编写小组/讲义/正交及二次型（解析版）.md

@@ -87,7 +87,7 @@ $U = \text{span}\{\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\boldsymbol{\beta}_3
 >&\boldsymbol u_1=\boldsymbol x_1,\quad\boldsymbol\varepsilon_1=\dfrac{\boldsymbol u_1}{\|\boldsymbol u_1\|}=\dfrac{1}{\sqrt2}(1,0,1,0)^\text T;\\
 >&\boldsymbol u_2=\boldsymbol x_2-\langle\boldsymbol\varepsilon_1,\boldsymbol x_2\rangle\boldsymbol\varepsilon_1,\quad\boldsymbol\varepsilon_2=\dfrac{\boldsymbol u_2}{\|\boldsymbol u_2\|}=\dfrac{1}{\sqrt{10}}(1,-2,-1,2)^\text T;\\
 >&\boldsymbol u_3=\boldsymbol x_3-\langle\boldsymbol\varepsilon_1,\boldsymbol x_3\rangle\boldsymbol\varepsilon_1-\langle\boldsymbol\varepsilon_2,\boldsymbol x_3\rangle\boldsymbol\varepsilon_2,\quad\boldsymbol\varepsilon_3=\dfrac{\boldsymbol u_3}{\|\boldsymbol u_3\|}=\dfrac{1}{\sqrt{35}}(3,4,-3,1)^\text T.
->\end{aligned}$$于是 $U$ 的标准正交基可以为$$\boldsymbol \xi_1=\dfrac{\boldsymbol \alpha_1+\boldsymbol \alpha_2}{\sqrt2},\boldsymbol \xi_2=\dfrac{\boldsymbol \alpha_1-2\boldsymbol \alpha_2-\boldsymbol \alpha_3+2\boldsymbol \alpha_4}{\sqrt{10}},\boldsymbol \xi_3=\dfrac{3\boldsymbol \alpha_1+4\boldsymbol \alpha_2-3\boldsymbol \alpha_3+\boldsymbol \alpha_4}{\sqrt{35}}.$$
+>\end{aligned}$$于是 $U$ 的标准正交基可以为$$\boldsymbol \xi_1=\dfrac{\boldsymbol \alpha_1+\boldsymbol \alpha_3}{\sqrt2},\boldsymbol \xi_2=\dfrac{\boldsymbol \alpha_1-2\boldsymbol \alpha_2-\boldsymbol \alpha_3+2\boldsymbol \alpha_4}{\sqrt{10}},\boldsymbol \xi_3=\dfrac{3\boldsymbol \alpha_1+4\boldsymbol \alpha_2-3\boldsymbol \alpha_3+\boldsymbol \alpha_4}{\sqrt{35}}.$$
 
 >[!example] **例4**
 >已知 $A$ $=$ $[\boldsymbol{\alpha}_1\ \boldsymbol{\alpha}_2\ \boldsymbol{\alpha}_3\ \boldsymbol{\alpha}_4]$ 为正交矩阵，其中
@@ -222,10 +222,10 @@ $U = \text{span}\{\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\boldsymbol{\beta}_3
 >设二次型 $\displaystyle f(x_1,x_2,x_3)=ax_1^2+2x_2^2-2x_3^2+2bx_1x_3,(b>0),$ 其中二次型的矩阵特征值之和为 $1$，之积为 $-12$. 求正交变换将二次型化为标准型. 
 
 >[!note] 解：
->二次型的矩阵为 $\displaystyle \boldsymbol A=\begin{bmatrix}1&0&b\\0&2&0\\b&0&-2\end{bmatrix}$，设其特征值分别为 $\lambda_1,\lambda_2,\lambda_3$, 则有$$\begin{aligned}\lambda_1\lambda_2\lambda_3=|\boldsymbol A|=2(-2a&-b^2)=-12,\lambda_1+\lambda_2+\lambda_3=a+2-2=1\\&\implies a=1,b=2.\end{aligned}$$从而有$\displaystyle|\lambda\boldsymbol E-\boldsymbol A|=(\lambda-2)^2(\lambda-3)=0\implies \lambda_1=\lambda_2=2,\lambda_3=-3.$
+>二次型的矩阵为 $\displaystyle \boldsymbol A=\begin{bmatrix}a&0&b\\0&2&0\\b&0&-2\end{bmatrix}$，设其特征值分别为 $\lambda_1,\lambda_2,\lambda_3$, 则有$$\begin{aligned}\lambda_1\lambda_2\lambda_3=|\boldsymbol A|=2(-2a&-b^2)=-12,\lambda_1+\lambda_2+\lambda_3=a+2-2=1\\&\implies a=1,b=2.\end{aligned}$$从而有$\displaystyle|\lambda\boldsymbol E-\boldsymbol A|=(\lambda-2)^2(\lambda-3)=0\implies \lambda_1=\lambda_2=2,\lambda_3=-3.$
 >考虑 $\lambda=2$，则 $\displaystyle 2\boldsymbol E-\boldsymbol A=\begin{bmatrix}1&0&-2\\0&0&0\\-2&0&4\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-2\\0&0&0\\0&0&0\end{bmatrix}$, 于是特征值 $2$ 的几何重数为 $2$, 对应的线性无关的特征向量可以为 $\displaystyle\boldsymbol\xi_1=(2,0,1)^T,\boldsymbol\xi_2=(0,1,0)^T$.
 >考虑 $\lambda=-3$，则 $\displaystyle -3\boldsymbol E-\boldsymbol A=\begin{bmatrix}-4&0&-2\\0&-5&0\\-2&0&-1\end{bmatrix}\rightarrow\begin{bmatrix}2&0&1\\0&1&0\\0&0&0\end{bmatrix}$，则特征值 $-3$ 几何重数为 $1$, 对应的特征向量为 $\boldsymbol\xi_3=(-1,0,2)^T$.
->对上述三个特征向量正交化得 $\boldsymbol\varepsilon_1=(\frac{2}{\sqrt{5}},0,\frac{1}{\sqrt{5}})^T,\boldsymbol\varepsilon_2=(0,1,0)^T,\boldsymbol\varepsilon_3=(-\frac{1}{\sqrt{5}},0,\frac{2}{\sqrt{5}})^T$，故正交变换为 $\boldsymbol C=\begin{bmatrix}\frac{2}{\sqrt{5}}&0&\frac{1}{\sqrt{5}}\\0&1&0\\-\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\end{bmatrix},$ 标准型为 $\displaystyle f=2y_1^2+2y_2^2-3y_3^2.$
+>对上述三个特征向量正交化得 $\boldsymbol\varepsilon_1=(\frac{2}{\sqrt{5}},0,\frac{1}{\sqrt{5}})^T,\boldsymbol\varepsilon_2=(0,1,0)^T,\boldsymbol\varepsilon_3=(-\frac{1}{\sqrt{5}},0,\frac{2}{\sqrt{5}})^T$，故正交变换为 $\boldsymbol C=\begin{bmatrix}\frac{2}{\sqrt{5}}&0&-\frac{1}{\sqrt{5}}\\0&1&0\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\end{bmatrix},$ 标准型为 $\displaystyle f=2y_1^2+2y_2^2-3y_3^2.$
 
 >[!example] 例题
 >已知二次型 $\displaystyle f(x_1,x_2,x_3)=x_1^2+2x_2^2+ax_3^2+2x_1x_3$ 经可逆线性变换 $\boldsymbol x=\boldsymbol P \boldsymbol y$ 化为 $\displaystyle y_1^2+y_2^2$.