例题 4.1. 二次曲面化标准方程

将二次曲面

$$2xy+2xz+2yz-\sqrt{2}x+\sqrt{2}y-1=0$$

化为标准方程, 并指出它是什么曲面.

解

令

$$A=\left(\begin{array}{lll}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$$ $$\mathbf{b}={\left(-\sqrt{2},\sqrt{2},0\right)}^{\mathrm{T}},\mathbf{u}={\left(x,y,z\right)}^{\mathrm{T}},$$

则原方程可以写为

$${\mathbf{u}}^{\mathrm{T}}\mathbf{Au}+{\mathbf{b}}^{\mathrm{T}}\mathbf{u}-1=0.\text{\hspace{1em}(4.12)}$$

求出矩阵 $\mathbf{A}$ 的特征值及对应的标准正交特征向量分别为

• ${\lambda }_1=2$,${\mathbf{q}}_1={\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)}^{\mathrm{T}}$
• ${\lambda }_2={\lambda }_3=-1$, ${\mathbf{q}}_2={\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right)}^{\mathrm{T}},{\mathbf{q}}_3={\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)}^{\mathrm{T}}.$

令

$$\mathbf{Q}=\left({\mathbf{q}}_1,{\mathbf{q}}_2,{\mathbf{q}}_3\right)$$

则有

{\mathbf{Q}}^{\mathrm{T}}\mathbf{A}\mathbf{Q} = \operatorname{diag}\left( {2, - 1, - 1}\right),$$ $${\mathbf{b}}^{\mathrm{T}}\mathbf{Q} = \left( {0, - 2,0}\right).$$ 作正交变换 $\mathbf{u} = \mathbf{Q}\mathbf{v}$ , 其中 $\mathbf{v} = {\left( {x}_{1},{y}_{1},{z}_{1}\right) }^{\mathrm{T}}$ , 则式 (4.12) 变为

{\mathbf{v}}^{\mathrm{T}}{\mathbf{Q}}^{\mathrm{T}}\mathbf{A}\mathbf{Q}\mathbf{v} + {\mathbf{b}}^{\mathrm{T}}\mathbf{Q}\mathbf{v} - 1 = 0

$$即$$

2{x}{1}^{2} - {y}{1}^{2} - {z}{1}^{2} - 2{y}{1} - 1 = 0

$$配方得$$

2{x}{1}^{2} - {\left( {y}{1} + 1\right) }^{2} - {z}_{1}^{2} = 0.

$$作平移变换$$

\left{ \begin{array}{l} {x}{2} = {x}{1} \ {y}{2} = {y}{1} + 1 \ {z}{2} = {z}{1} \end{array}\right.

$$可得$$

2{x}{2}^{2} - {y}{2}^{2} - {z}_{2}^{2} = 0

$$这就是原曲面方程的标准方程,它表示一个圆锥面.$$