# 问题及解答

## 求正交变换 $X=QY$, 将二次型 $f(x_1,x_2,x_3)=2x_1 x_2+2x_1 x_3+2x_2 x_3$ 化为标准型.

Posted by haifeng on 2021-08-28 09:11:53 last update 2021-08-28 09:11:53 | Edit | Answers (1)

求正交变换 $X=QY$, 将二次型 $f(x_1,x_2,x_3)=2x_1 x_2+2x_1 x_3+2x_2 x_3$ 化为标准型.

1

Posted by haifeng on 2021-08-28 09:42:09

$\begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix}$

$\begin{split} |\lambda E-A|&=\begin{vmatrix} \lambda & -1 & -1\\ -1 & \lambda & -1\\ -1 & -1 & \lambda \end{vmatrix}=\begin{vmatrix} \lambda-2 & \lambda-2 & \lambda-2\\ -1 & \lambda & -1\\ -1 & -1 & \lambda \end{vmatrix}\\ &=(\lambda-2)\begin{vmatrix} 1 & 1 & 1\\ -1 & \lambda & -1\\ -1 & -1 & \lambda \end{vmatrix}=(\lambda-2)\begin{vmatrix} 1 & 0 & 0\\ -1 & \lambda+1 & 0\\ -1 & 0 & \lambda+1 \end{vmatrix}\\ &=(\lambda-2)(\lambda+1)^2 \end{split}$

$-E-A=\begin{pmatrix} -1 & -1 & -1\\ -1 & -1 & -1\\ -1 & -1 & -1 \end{pmatrix}\xlongequal{\text{初等变换}} \begin{pmatrix} -1 & -1 & -1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$

$\vec{\alpha}_2=\begin{pmatrix} -1\\ 1\\ 0\end{pmatrix},\quad \vec{\alpha}_3=\begin{pmatrix} -1\\ 0\\ 1\end{pmatrix}.$

$2E-A=\begin{pmatrix} 2 & -1 & -1\\ -1 & 2 & -1\\ -1 & -1 & 2 \end{pmatrix}$

$Q=\begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\ 0 & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\ \end{pmatrix}$

$f=-y_1^2-y_2^2+2y_3^2$

Remark:  来自方守文讲课比赛的笔记.