【ID】1469 【题型】解答题 【类型】考研真题 【来源】2005年全国硕士研究生招生考试试题

(II) 求正交变换 $\boldsymbol{x}=\boldsymbol{Q y}$,把 $f\left(x_{1}, x_{2}, x_{3}\right)$ 化成标准形;
(III) 求方程 $f\left(x_{1}, x_{2}, x_{3}\right)=0$ 的解.

(II) 当 $a=0$ 时, $\boldsymbol{A}=\left[\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]$, $|\lambda \boldsymbol{E}-\boldsymbol{A}|=\left|\begin{array}{ccc}\lambda-1 & -1 & 0 \\ -1 & \lambda-1 & 0 \\ 0 & 0 & \lambda-2\end{array}\right|=(\lambda-2)^{2} \lambda$,

$\boldsymbol{A}$ 的属于 $\lambda_{1}=2$ 的线性无关的特征向量为 $\boldsymbol{\eta}_{1}=(1,1,0)^{\mathrm{T}}, \boldsymbol{\eta}_{2}=(0,0,1)^{\mathrm{T}}$;
$\boldsymbol{A}$ 的属于 $\lambda_{3}=0$ 的线性无关的特征向量为 $\boldsymbol{\eta}_{3}=(-1,1,0)^{\mathrm{T}}$.

$$\boldsymbol{e}_{1}=\frac{1}{\sqrt{2}}(1,1,0)^{\mathrm{T}}, \boldsymbol{e}_{2}=(0,0,1)^{\mathrm{T}}, \boldsymbol{e}_{3}=\frac{1}{\sqrt{2}}(-1,1,0)^{\mathrm{T}},$$

$$f\left(x_{1}, x_{2}, x_{3}\right)=\lambda_{1} y_{1}^{2}+\lambda_{2} y_{2}^{2}+\lambda_{3} y_{3}^{2}=2 y_{1}^{2}+2 y_{2}^{2} \text {. }$$
(III) 解法一 在正交变换 $\boldsymbol{X}=\boldsymbol{Q Y}$ 下, $f\left(x_{1}, x_{2}, x_{3}\right)=0$ 化成 $2 y_{1}^{2}+2 y_{2}^{2}=0$, 解之得 $y_{1}=$ $y_{2}=0$, 从而
$$\boldsymbol{X}=\boldsymbol{Q}\left[\begin{array}{l} 0 \\ 0 \\ y_{3} \end{array}\right]=\left[\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}\right]\left[\begin{array}{l} 0 \\ 0 \\ y_{3} \end{array}\right]=y_{3} \boldsymbol{e}_{3}=k(-1,1,0)^{\mathrm{T}},$$

$$f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}+x_{2}^{2}+2 x_{3}^{2}+2 x_{1} x_{2}=\left(x_{1}+x_{2}\right)^{2}+2 x_{3}^{2}=0,$$