Appendix_相关例题

1. 已知二次型 $f(x_{1}, x_{2}, x_{3}) = (1-a)x_{1}^{2} + (1-a)x_{2}^{2} +2 x_{3}^{2}+2(1+a)x_{1}x_{2}$ 的秩为 $2$
   1. 求常数 $a$
   2. 求正交变化 $\boldsymbol{X} = \boldsymbol{QY}$，将二次型转化为标准型
   3. 求方程 $f(x_{1},x_{2},x_{3}) = 0$ 的解 $$\begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 1-a & 1+a & 0 \\ 1+a & 1-a & 0 \\ 0 & 0 & 2 \end{smallmatrix} \right), \boldsymbol{X} = \left( \begin{smallmatrix} x_{1} \\ x_{2} \\ x_{3} \end{smallmatrix} \right)\Rightarrow f = \boldsymbol{X}^{\intercal}\boldsymbol{AX}\\ \because & f \text{的秩序为} 2 \\ \therefore & \frac{1-a}{1+a} = \frac{1+a}{1-a} \\ \therefore & a = 0 \\\\ & \det(\lambda\boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda - 1 & -1 & 0 \\ -1 & \lambda-1 & 0 \\ 0 & 0 & \lambda -2 \end{smallmatrix} \right\vert = (\lambda-2)^{2}\lambda \\ \therefore & \lambda_{1} = \lambda_{2} = 2, \lambda_{3} = 0 \\ & 2 \boldsymbol{E} - \boldsymbol{A} = \left( \begin{smallmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{smallmatrix} \right)\\ \therefore & \lambda=2 \text{对应的线性无关特征向量}\boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right), \boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} 0 \\ 0 \\1 \end{smallmatrix} \right) \\ & \boldsymbol{A} \rightarrow \left( \begin{smallmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{smallmatrix} \right) \\ \therefore & \lambda = 0 \text{对应的线性无关特征向量} \boldsymbol{a}_{3} = \left( \begin{smallmatrix} -1 \\ 1 \\ 0 \end{smallmatrix} \right) \\ \therefore & \boldsymbol{\gamma}_{1} = \frac{\sqrt{2}}{2}\boldsymbol{\alpha}_{1}, \boldsymbol{\gamma}_{2} = \boldsymbol{\alpha}_{2}, \boldsymbol{\gamma}_{3} = \frac{\sqrt{2}}{2}\boldsymbol{\gamma}_{3} \\ & \text{令}\boldsymbol{Q} = \left( \begin{matrix} \boldsymbol{\gamma}_{1} & \boldsymbol{\gamma}_{2} & \boldsymbol{\gamma}_{3} \end{matrix} \right) \\ \therefore & \boldsymbol{X}^{\intercal}\boldsymbol{AX} \xlongequal{\boldsymbol{X} = \boldsymbol{QY}} \boldsymbol{Y}^{\intercal}(\boldsymbol{Q}^{\intercal}\boldsymbol{AQ}) \boldsymbol{Y} = 2y_{1}^{2} + 2y_{2}^{2} \\ \\ \because & f(x_{1},x_{2},x_{3}) = 0 \\ \therefore & y_{1} = y_{2} = 0 \\ \because & \boldsymbol{X} = \boldsymbol{QY} \\ \therefore & \boldsymbol{Y} = \boldsymbol{Q}^{\intercal}\boldsymbol{X} \\ \therefore & \begin{cases} y_{1} = \frac{\sqrt{2}}{2}x_{1} + \frac{\sqrt{2}}{2} x_{2} =0 \\ y_{2} = x_{3} = 0 \\ \end{cases} \\ \therefore & \text{解为} k \left( \begin{smallmatrix} -1 \\1 \\ 0 \end{smallmatrix} \right)\\ \end{array}$$
2. 设二次型 $f(x_{1},x_{2},x_{3}) = x_{1}^{2} + x_{2}^{2} - x_{3}^{2} + 4x_{1}x_{3} +2ax_{2}x_{3}$ 经正交变化为标准型为 $f = -3y_{1}^{2}+by_{2}^{2} +3 y_{3}^{2}$
   1. 求常数 $a,b$
   2. 求正交矩阵 $\boldsymbol{Q}$，使得二次型在正交变化 $\boldsymbol{X} = \boldsymbol{QY}$ 下为标准型 $$\begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & 0 & 2 \\ 0 & 1 & a \\ 2 & a & -1 \end{smallmatrix} \right), \boldsymbol{X} = \left( \begin{smallmatrix} x_{1} \\ x_{2} \\ x_{3} \end{smallmatrix} \right) \Rightarrow f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \\ & \det(\lambda \boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda - 1 & 0 & -2 \\ 0 & \lambda -1 & -a \\ -2 & -a & \lambda+1 \end{smallmatrix} \right\vert = (\lambda -1)(\lambda^{2} - a^{2} - 1) - 4(\lambda-1)\\ & = (\lambda-1)(\lambda^{2} -a^{2} - 5) \\ \because & f = -3 y_{1}^{2} + by_{2}^{2} +3y_{3}^{2} \\ \therefore & b = 1, -a^{2} - 5 = -9 \\ \therefore & a = \pm 2 \\\\ & \text{Case1: } a = 2 \\ & \text{Case2: } a = -2 \\ \end{array}$$