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用正交变换化二次型成标准形 , 具有保持几何形状不变的优点

如果不限于用正交变换

那么还可以有多种方法(对应有多个可逆的线性变换)把二次型化成标准形

这里只介绍拉格朗日配方法 , 下面举例来说明这种方法

例15 用配方法化二次型f=x12x_{1}^{2}+2x22x_{2}^{2}+5x32x_{3}^{2}+2x1x2+2x1x3+6x2x3成标准形

并求所用的变换矩阵

1. 写出二次型的矩阵

二次型f=x12+2x22+5x32+2x1x2+2x1x3+6x2x3f = x_{1}^{2} + 2x_{2}^{2} + 5x_{3}^{2} + 2x_{1}x_{2} + 2x_{1}x_{3} + 6x_{2}x_{3}

二次型的一般形式 f=∑i,jaijxixjf = \sum_{i,j}^{}a_{ij}x_{i}x_{j} 且 aij=ajia_{ij} = a_{ji}。

所以对称矩阵 AA 为：A=(111123135)A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{pmatrix}

2. 配方法（按 x1,x2,x3x_{1},x_{2},x_{3} 顺序配方）

f=x12+2x22+5x32+2x1x2+2x1x3+6x2x3f = x_{1}^{2} + 2x_{2}^{2} + 5x_{3}^{2} + 2x_{1}x_{2} + 2x_{1}x_{3} + 6x_{2}x_{3}

首先把所有含 x1x_{1} 的项放在一起：

f=x12+2x1x2+2x1x3+[2x22+5x32+6x2x3]f = x_{1}^{2} + 2x_{1}x_{2} + 2x_{1}x_{3} + \left\lbrack 2x_{2}^{2} + 5x_{3}^{2} + 6x_{2}x_{3} \right\rbrack

配 x1x_{1} 平方：x12+2x1(x2+x3)x_{1}^{2} + 2x_{1}(x_{2} + x_{3})

配成：[x1+(x2+x3)]2−(x2+x3)2\lbrack x_{1} + (x_{2} + x_{3})\rbrack^{2} - (x_{2} + x_{3})^{2}

因此：f=[x1+x2+x3]2−x22−2x2x3−x32+2x22+5x32+6x2x3f = \lbrack x_{1} + x_{2} + x_{3}\rbrack^{2} - x_{2}^{2} - 2x_{2}x_{3} - x_{3}^{2} + 2x_{2}^{2} + 5x_{3}^{2} + 6x_{2}x_{3}

=(x1+x2+x3)2+x22+4x32+4x2x3= (x_{1} + x_{2} + x_{3})^{2} + x_{2}^{2} + 4x_{3}^{2} + 4x_{2}x_{3}

再对后面的 x2,x3x_{2},x_{3} 部分配方：x22+4x2x3+4x32x_{2}^{2} + 4x_{2}x_{3} + 4x_{3}^{2}

注意这是 x22+4x2x3+4x32x_{2}^{2} + 4x_{2}x_{3} + 4x_{3}^{2}，刚好可以配方：x22+4x2x3+4x32=(x2+2x3)2x_{2}^{2} + 4x_{2}x_{3} + 4x_{3}^{2} = (x_{2} + 2x_{3})^{2}

验证：(x2+2x3)2=x22+4x2x3+4x32(x_{2} + 2x_{3})^{2} = x_{2}^{2} + 4x_{2}x_{3} + 4x_{3}^{2}，完全正确。

因此：f=(x1+x2+x3)2+(x2+2x3)2f = (x_{1} + x_{2} + x_{3})^{2} + (x_{2} + 2x_{3})^{2}

3. 标准形

令{y1=x1+x2+x3y2=x2+2x3y3=x3(自由变量，为了可逆变换，可令y3=x3)\left\{ \begin{matrix} y_{1} = x_{1} + x_{2} + x_{3} \\ y_{2} = x_{2} + 2x_{3} \\ y_{3} = x_{3}\quad(\text{自由变量，为了可逆变换，可令}y_{3} = x_{3}) \end{matrix} \right.\

这里我们需要第三个变量 y3y_{3} 使变换可逆。

从 y2=x2+2x3y_{2} = x_{2} + 2x_{3}，y3=x3y_{3} = x_{3} 得：x3=y3,x2=y2−2y3x_{3} = y_{3},\quad x_{2} = y_{2} - 2y_{3}

再代到 y1=x1+x2+x3y_{1} = x_{1} + x_{2} + x_{3}：

y1=x1+(y2−2y3)+y3=x1+y2−y3y_{1} = x_{1} + (y_{2} - 2y_{3}) + y_{3} = x_{1} + y_{2} - y_{3}

x1=y1−y2+y3x_{1} = y_{1} - y_{2} + y_{3}

于是：(x1x2x3)=(1−1101−2001)(y1y2y3)\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}

4. 此时的二次型

代入 ff 得：f=y12+y22+0⋅y32f = y_{1}^{2} + y_{2}^{2} + 0 \cdot y_{3}^{2}

所以标准形是：f=z12+z22f = z_{1}^{2} + z_{2}^{2}

第三个平方项系数为 0，即该二次型秩为 2，有一个零特征值。

但原题要求用配方法化标准形，一般标准形会写成 d1y12+d2y22+d3y32d_{1}y_{1}^{2} + d_{2}y_{2}^{2} + d_{3}y_{3}^{2} 的形式，

并求所用变换矩阵。这里第三个变量 y3y_{3} 在 ff 中没有出现，所以 d3=0d_{3} = 0。

也可取正交变换对角化，但配方法得出的可逆线性变换不一定是正交变换。

5. 写出所用变换矩阵（配方法的 C）

我们设 X=CYX = CY，其中 Y=(y1,y2,y3)TY = (y_{1},y_{2},y_{3})^{T}。

从上面我们已经得到：

x1=y1−y2+y3x_{1} = y_{1} - y_{2} + y_{3}

x2=y2−2y3x_{2} = y_{2} - 2y_{3}

x3=y3x_{3} = y_{3}

写成矩阵：(x1x2x3)=(1−1101−2001)(y1y2y3)\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}

所以 C=(1−1101−2001)C = \begin{pmatrix} 1 & - 1 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{pmatrix}，且 f=y12+y22f = y_{1}^{2} + y_{2}^{2}。

例16 用配方法化二次型f=2x1x2+2x1x3-6x2x3成规范形 , 并求所用的变换矩阵

解：

一、问题分析

已知二次型f=2x1x2+2x1x3−6x2x3f = 2x_{1}x_{2} + 2x_{1}x_{3} - 6x_{2}x_{3}

不含平方项（即没有x12,x22,x32x_{1}^{2},x_{2}^{2},x_{3}^{2}），因此不能直接配方。

我们采用配方法，先通过可逆线性变换引入平方项，再逐步配方化为规范形，

并求出所用的变换矩阵。

二、引入新变量构造平方项

令{x1=y1+y2x2=y1−y2x3=y3\left\{ \begin{matrix} x_{1} = y_{1} + y_{2} \\ x_{2} = y_{1} - y_{2} \\ x_{3} = y_{3} \end{matrix} \right.\ ，该变换可逆，目的是将x1x2x_{1}x_{2}转化为平方差形式。

计算各项：

x1x2=(y1+y2)(y1−y2)=y12−y22x_{1}x_{2} = (y_{1} + y_{2})(y_{1} - y_{2}) = y_{1}^{2} - y_{2}^{2}

x1x3=(y1+y2)y3=y1y3+y2y3x_{1}x_{3} = (y_{1} + y_{2})y_{3} = y_{1}y_{3} + y_{2}y_{3}

x2x3=(y1−y2)y3=y1y3−y2y3x_{2}x_{3} = (y_{1} - y_{2})y_{3} = y_{1}y_{3} - y_{2}y_{3}

代入原式：

f=2(y12−y22)+2(y1y3+y2y3)−6(y1y3−y2y3)f = 2(y_{1}^{2} - y_{2}^{2}) + 2(y_{1}y_{3} + y_{2}y_{3}) - 6(y_{1}y_{3} - y_{2}y_{3})

=2y12−2y22+2y1y3+2y2y3−6y1y3+6y2y3= 2y_{1}^{2} - 2y_{2}^{2} + 2y_{1}y_{3} + 2y_{2}y_{3} - 6y_{1}y_{3} + 6y_{2}y_{3}

=2y12−2y22−4y1y3+8y2y3= 2y_{1}^{2} - 2y_{2}^{2} - 4y_{1}y_{3} + 8y_{2}y_{3}

三、逐步配方

1.对含y1y_{1}的项配方

2y12−4y1y3=2(y12−2y1y3)=2[(y1−y3)2−y32]=2(y1−y3)2−2y322y_{1}^{2} - 4y_{1}y_{3} = 2(y_{1}^{2} - 2y_{1}y_{3}) = 2\left\lbrack (y_{1} - y_{3})^{2} - y_{3}^{2} \right\rbrack = 2(y_{1} - y_{3})^{2} - 2y_{3}^{2}

2.对含y2y_{2}的项配方

−2y22+8y2y3=−2(y22−4y2y3)=−2[(y2−2y3)2−4y32]- 2y_{2}^{2} + 8y_{2}y_{3} = - 2(y_{2}^{2} - 4y_{2}y_{3}) = - 2\left\lbrack (y_{2} - 2y_{3})^{2} - 4y_{3}^{2} \right\rbrack

=−2(y2−2y3)2+8y32= - 2(y_{2} - 2y_{3})^{2} + 8y_{3}^{2}

3.合并结果

f=[2(y1−y3)2−2y32]+[−2(y2−2y3)2+8y32]f = \left\lbrack 2(y_{1} - y_{3})^{2} - 2y_{3}^{2} \right\rbrack + \left\lbrack - 2(y_{2} - 2y_{3})^{2} + 8y_{3}^{2} \right\rbrack

=2(y1−y3)2−2(y2−2y3)2+6y32= 2(y_{1} - y_{3})^{2} - 2(y_{2} - 2y_{3})^{2} + 6y_{3}^{2}

四、化为规范形

令{z1=y1−y3,z2=y2−2y3,z3=y3⇒f=2z12−2z22+6z32.\left\{ \begin{matrix} z_{1} = y_{1} - y_{3}, \\ z_{2} = y_{2} - 2y_{3}, \\ z_{3} = y_{3} \end{matrix} \right.\ \quad \Rightarrow \quad f = 2z_{1}^{2} - 2z_{2}^{2} + 6z_{3}^{2}.

再令{w1=2z1,w2=2z2,w3=6z3⇒f=w12−w22+w32.\left\{ \begin{matrix} w_{1} = \sqrt{2}z_{1}, \\ w_{2} = \sqrt{2}z_{2}, \\ w_{3} = \sqrt{6}z_{3} \end{matrix} \right.\ \quad \Rightarrow \quad f = w_{1}^{2} - w_{2}^{2} + w_{3}^{2}.

此为规范形。

五、求变换矩阵

我们要求可逆矩阵CC，使得x=Cwx = Cw。

1.从ww表示yy

由{w1=2z1=2(y1−y3),w2=2z2=2(y2−2y3),w3=6z3=6y3,\left\{ \begin{matrix} w_{1} = \sqrt{2}z_{1} = \sqrt{2}(y_{1} - y_{3}), \\ w_{2} = \sqrt{2}z_{2} = \sqrt{2}(y_{2} - 2y_{3}), \\ w_{3} = \sqrt{6}z_{3} = \sqrt{6}y_{3}, \end{matrix} \right.\

解得{y1=12w1+y3=12w1+16w3y2=12w2+2y3=12w2+26w3y3=16w3\left\{ \begin{matrix} y_{1} = \frac{1}{\sqrt{2}}w_{1} + y_{3} = \frac{1}{\sqrt{2}}w_{1} + \frac{1}{\sqrt{6}}w_{3} \\ y_{2} = \frac{1}{\sqrt{2}}w_{2} + 2y_{3} = \frac{1}{\sqrt{2}}w_{2} + \frac{2}{\sqrt{6}}w_{3} \\ y_{3} = \frac{1}{\sqrt{6}}w_{3} \end{matrix} \right.\

2.从yy表示xx

由{x1=y1+y2,x2=y1−y2,x3=y3,\left\{ \begin{matrix} x_{1} = y_{1} + y_{2}, \\ x_{2} = y_{1} - y_{2}, \\ x_{3} = y_{3}, \end{matrix} \right.\

代入得

x1=(12w1+16w3)+(12w2+26w3)=12(w1+w2)+36w3x_{1} = \left( \frac{1}{\sqrt{2}}w_{1} + \frac{1}{\sqrt{6}}w_{3} \right) + \left( \frac{1}{\sqrt{2}}w_{2} + \frac{2}{\sqrt{6}}w_{3} \right) = \frac{1}{\sqrt{2}}(w_{1} + w_{2}) + \frac{3}{\sqrt{6}}w_{3}

x2=(12w1+16w3)−(12w2+26w3)=12(w1−w2)−16w3x_{2} = \left( \frac{1}{\sqrt{2}}w_{1} + \frac{1}{\sqrt{6}}w_{3} \right) - \left( \frac{1}{\sqrt{2}}w_{2} + \frac{2}{\sqrt{6}}w_{3} \right) = \frac{1}{\sqrt{2}}(w_{1} - w_{2}) - \frac{1}{\sqrt{6}}w_{3}

x3=16w3x_{3} = \frac{1}{\sqrt{6}}w_{3}

3.写出矩阵形式

[x1x2x3]=[12123612−12−160016][w1w2w3]=Cw.\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{3}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ 0 & 0 & \frac{1}{\sqrt{6}} \end{bmatrix}\begin{bmatrix} w_{1} \\ w_{2} \\ w_{3} \end{bmatrix} = Cw.

六、最终答案

规范形：f=w12−w22+w32f = w_{1}^{2} - w_{2}^{2} + w_{3}^{2}

所用变换矩阵：C=[12123612−12−160016]C = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{3}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} \\ 0 & 0 & \frac{1}{\sqrt{6}} \end{bmatrix}

满足x=Cwx = Cw，且f=w12−w22+w32f = w_{1}^{2} - w_{2}^{2} + w_{3}^{2}。