二次型正定

Author : 张一极
19 : 34 - 20201027

一.二次型矩阵化:

设:

\begin{aligned} f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=& a_{11} x_{1}^{2}+a_{12} x_{1} x_{2}+\cdots+a_{1 n} x_{1} x_{n} \\ &+a_{21} x_{2} x_{1}+a_{22} x_{2}^{2}+\cdots+a_{2 n} x_{2} x_{n} \\ &+\cdots \\ &+a_{n 1} x_{n} x_{1}+a_{n 2} x_{n} x_{2}+\cdots+a_{n n} x_{n}^{2} \end{aligned}

$x_1$ ,

$x_2$ ,

$\cdots$

$x_n$ .

\begin{array}{l}=x_{1}\left(a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}\right) \\ +x_{2}\left(a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}\right) \\ +\cdots \\ +x_{n}\left(a_{n 1} x_{1}+a_{n 2} x_{2}+\cdots+a_{n n} x_{n}\right)\end{array}

$[x_1\cdots x_n]$

\left(x_{1}, x_{2}, \cdots, x_{n}\right)\left(\begin{array}{c}a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n} \\ a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n} \\ \vdots \\ a_{n 1} x_{1}+a_{n 2} x_{2}+\cdots+a_{n n} x_{n}\end{array}\right)

后面的因子向量,可以转化为:

\left(\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right)\left(\begin{array}{c}x_{1} \\ x_{2} \\ \vdots \\ x_{n}\end{array}\right)

行列相乘的形式,最终结果:

=\left(x_{1}, x_{2}, \cdots, x_{n}\right)\left(\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right)\left(\begin{array}{c}x_{1} \\ x_{2} \\ \vdots \\ x_{n}\end{array}\right)=x^{T} A x

其中中间的矩阵为实对称矩阵

二.化二次型为标准型:

f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}+4 x_{2}^{2}+x_{3}^{2}-4 x_{1} x_{2}-8 x_{1} x_{3}-4 x_{2} x_{3}

提取二次型矩阵:

A=\left(\begin{array}{ccc}1 & -2 & -4 \\ -2 & 4 & -2 \\ -4 & -2 & 1\end{array}\right)

$A-\lambda E$ :

|A-\lambda E|=\left|\begin{array}{ccc}1-\lambda & -2 & -4 \\ -2 & 4-\lambda & -2 \\ -4 & -2 & 1-\lambda\end{array}\right|=-(\lambda-5)^{2}(\lambda+4)

$\lambda_1=\lambda_2=5,\lambda_3=-4$

代回矩阵:

A-5 E=\left(\begin{array}{rrr}-4 & -2 & -4 \\ -2 & -1 & -2 \\ -4 & -2 & -4\end{array}\right) \rightarrow\left(\begin{array}{ccc}1 & 1 / 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)

求基础解系:

\xi_{1}=\left(\begin{array}{c}-1 / 2 \\ 1 \\ 0\end{array}\right), \xi_{2}=\left(\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right)

两者线性无关,应进行正交化:

施密特正交化:

$\eta_1$ :

\boldsymbol{\eta}_{1}=\xi_{1}

$\eta_2$ =

{\eta}_{2}=\boldsymbol{\xi}_{2}-\frac{\left(\boldsymbol{\xi}_{2}, \boldsymbol{\eta}_{1}\right)}{\left(\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{1}\right)} \boldsymbol{\eta}_{1}

结果 : (单位化后)

\eta_{1}=\frac{1}{\sqrt{5}}\left(\begin{array}{c}-1 \\ 2 \\ 0\end{array}\right), \eta_{2}=\frac{1}{3 \sqrt{5}}\left(\begin{array}{c}-4 \\ -2 \\ 5\end{array}\right)

$\lambda_3=-4$ 的基础解系 :

A+4 E=\left(\begin{array}{ccc}5 & -2 & -4 \\ -2 & 8 & -2 \\ -4 & -2 & 5\end{array}\right) \rightarrow\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right)

解得(单位化:除以模长):

\eta_{3}=\frac{1}{3}(2,1,2)^{T}

最后:

矩阵为三个基础解系正交化或单位化后的结果组合:

\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)=\left(\begin{array}{ccc}-1 / \sqrt{5} & -4 / \sqrt{45} & 2 / 3 \\ 2 / \sqrt{5} & -2 / \sqrt{45} & 1 / 3 \\ 0 & 5 / \sqrt{45} & 2 / 3\end{array}\right)\left(\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right)

可将f化为标准型:

f=5 y_{1}^{2}+5 y_{2}^{2}-4 y_{3}^{2}

三.配方法化二次型为标准型

1.

x_1^2+2x_1x_2+2x_1x_3-x_2^2-2x_2x_3-x_3^3 \\ =x_1^2+x_2^2+x_3^2+2x_1x_2+2x_1x_3+2x_2x_3-2x_2^2-4x_2x_3-2x_3^2 \\=(x_1+x_2+x_3)^2-2(x_2^2+2x_2x_3+x_3^2) \\=(x_1+x_2+x_3)^2-2(x_2+x_3)^2+0\cdot x_3^2

$x_1$ $x_3$ $x_3$ $x_1+x_3+x_3$ 的混合平方

2.凑出前几项的平方,剩下的继续凑平方

得到线性关系

\begin{array}{l}y_{1}=x_{1}+x_{2}+x_{3} \\ y_{2}=x_{2}+x_{3} \\ y_{3}=x_{3}\end{array}

↓

\left\{\begin{array}{l}x_{1}=y_{1}-y_{2} \\ x_{2}=y_{2}-y_{3} \\ x_{3}=y_{3}\end{array}\right.

$x=CY$ ,其中的C:

\left[\begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{array}\right]\cdot[y_1,y_2,y_3]^T

2.如果没有平方项,可创造平方项:

f\left(x_{1}, x_{2}, x_{3}\right)=x_{1} x_{2}+x_{1} x_{3}-x_{2} x_3 \\x_1=y_1+y_2 \\x_2=y_1-y_2\\ x_3=y_3

\begin{array}{l}f(x)=y_{1}^{2}-y_{2}^{2}+y_{1} y_{3}+y_{2} y_{3}-y_{1} y_{3}+y_{2} y_{3}\\=y_{1}^{2}-y_{2}^{2}+2 y_{2} y_{3} \\ =y_{1}^{2}-\left(y_{2}-y_{3}\right)^{2}+y_{3}^{2}\end{array}

设:

z_1=y_1 \\z_2=y_2-y_3 \\z_3=y_3

y_1=z_1 \\y_2= z_2+z_3 \\y_3=z_3

拉回现在的z系列,去表示最开始的x系列,经过的y作为中间变量,可得到对应的CZ

\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & -1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}z_{1} \\ z_{2} \\ z_{3}\end{array}\right]

End
21:44-2020-1028