例题 1.3. 二次型化成标准形

把下面的二次型化成标准形

$$f\left(x_1,x_2,x_3\right)=x_1{x}_2+2x_1{x}_3+4x_2{x}_3.$$

解

作非退化的线性替换

$$\{\begin{array}{l}{x}_1=y_1+y_2\\ x_2=y_1-y_2\\ x_3=y_3\end{array}$$

代入原二次型得

$$\begin{array}{rl}f& =y_1^2-y_2^2+2\left(y_1+y_2\right)y_3+4\left(y_1-y_2\right)y_3\\ & =y_1^2-y_2^2+6y_1{y}_3-2y_2{y}_3\\ & =y_1^2+6y_1{y}_3+9y_3^2-y_2^2-2y_2{y}_3-9y_3^2\\ & ={\left(y_1+3y_3\right)}^2-\left(y_2^2+2y_2{y}_3+y_3^2\right)-8y_3^2.\end{array}$$

令

$$\{\begin{array}{l}{z}_1=y_1+3y_3\\ z_2=y_2+y_3\\ z_3=y_3\end{array}$$

则

$$f=z_1^2-z_2^2-8z_3^2$$

由 $x_1,x_2,x_3$ 到 $z_1,z_2,z_3$ 的线性替换为

$$\{\begin{array}{l}{x}_1=z_1+z_2-4z_3\\ x_2=z_1-z_2-2z_3\\ x_3=z_3\end{array}$$

记原二次型的矩阵为 $\mathbf{A}$, 化成标准形后的矩阵为 $\mathbf{B}$, 线性替换的矩阵为 $\mathbf{C}$, 则

$$\mathbf{A}=\left(\begin{array}{rrr}0& \frac{1}{2}& 1\\ \frac{1}{2}& 0& 2\\ 1& 2& 0\end{array}\right),\mathbf{B}=\left(\begin{array}{rrr}1& & \\ & -1& \\ & & -8\end{array}\right),\mathbf{C}=\left(\begin{array}{rrr}1& 1& -4\\ 1& -1& -2\\ 0& 0& 1\end{array}\right).$$

经计算, 确有

$$C^{\mathrm{T}}AC=B\text{.}$$