【线性代数基础进阶】二次型

烧灯续昼2002
2,451 阅读3分钟

本文已参与「新人创作礼」活动，一起开启掘金创作之路。

 

注意与特征值、特征向量的联系

 

一、概念、定理

概念

二次型及其矩阵表示

f(x1,x2,x3)=x12+5x22+5x32+2x1x2−6x2x3=x12+x1x2+x1x2+5x22−3x2x3−3x2x3+5x32=x1(x1+x2)+x2(x1+5x2−3x3)+x3(−3x2+5x3)=(x1,x2,x3)(x1+x2x1+5x2−3x3−3x2+5x3)=(x1,x2,x3)(11015−30−35)(x1x2x3)平方项放到对角线,混合项对半分=xTAx \begin{aligned} f(x_{1},x_{2},x_{3})&=x_{1}^{2}+5x_{2}^{2}+5x_{3}^{2}+2x_{1}x_{2}-6x_{2}x_{3}\\ &=x_{1}^{2}+x_{1}x_{2}+x_{1}x_{2}+5x_{2}^{2}-3x_{2}x_{3}-3x_{2}x_{3}+5x_{3}^{2}\\ &=x_{1}(x_{1}+x_{2})+x_{2}(x_{1}+5x_{2}-3x_{3})+x_{3}(-3x_{2}+5x_{3})\\ &=(x_{1},x_{2},x_{3})\begin{pmatrix} x_{1}+x_{2} \\ x_{1}+5x_{2}-3x_{3} \\ -3x_{2}+5x_{3} \end{pmatrix}\\ &=(x_{1},x_{2},x_{3})\begin{pmatrix} 1 & 1 & 0 \\ 1 & 5 & -3 \\ 0 & -3 & 5 \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}\\ &平方项放到对角线,混合项对半分\\ &=x^{T}Ax \end{aligned}

有AT=AA ^{T}=A是二次型的矩阵

 

标准形

2x12+5x22−6x32Λ=(2   5   −6) \begin{gathered} 2x_{1}^{2}+5x_{2}^{2}-6x_{3}^{2}\\ \Lambda=\begin{pmatrix} 2 &  &  \\  & 5 &  \\  &  & -6 \end{pmatrix} \end{gathered}

 

规范形

x12+x22+x32x12+x22−x32x12−x22−x12 \begin{gathered} x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\ x_{1}^{2}+x_{2}^{2}-x_{3}^{2}\\ x_{1}^{2}-x_{2}^{2}\\ -x_{1}^{2} \end{gathered}

 

正惯性指数、负惯性指数

对于标准形

2x12+5x22−6x32正惯性指数为p=2,负惯性指数为q=1x12+x22+x32p=3,q=0 \begin{aligned} 2x_{1}^{2}+5x_{2}^{2}-6x_{3}^{2}\quad &正惯性指数为p=2,负惯性指数为q=1\\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\quad&p=3,q=0 \end{aligned}

非标准形的化成标准形再看

 

二次型的秩

r(f)=r(A) r(f)=r(A)

因为

r(1−20−234045)=3 r \begin{pmatrix} 1 & -2 & 0 \\ -2 & 3 & 4 \\ 0 & 4 & 5 \end{pmatrix}=3

故二次型f=x12+3x22+5x32−4x1x2+8x2x3f=x_{1}^{2}+3x_{2}^{2}+5x_{3}^{2}-4x_{1}x_{2}+8x_{2}x_{3}的秩r(f)=3r(f)=3

 

坐标变换

{x1=c11y1+c12y2+c13y3x2=c21y1+c22y2+c23y3x3=c31y1+c32y2+c33y3∣C∣≠0 \left\{\begin{aligned}&x_{1}=c_{11}y_{1}+c_{12}y_{2}+c_{13}y_{3}\\ &x_{2}=c_{21}y_{1}+c_{22}y_{2}+c_{23}y_{3}\\ &x_{3}=c_{31}y_{1}+c_{32}y_{2}+c_{33}y_{3}\end{aligned}\right.\quad |C|\ne0

则有

(x1x2x3)=(c11c12c13c21c22c23c31c32c33)(y1y2y3) \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}=\begin{pmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}

即可表示为x=cyx=cy

 

注意系数行列式∣C∣|C|一定要不为00

 

合同

如果CTAC=B,其中C可逆C^{T}AC=B,其中C可逆，则称矩阵AA和BB合同，记A≃BA \simeq B

 

xTAx=任取x=Cy(Cy)TA(Cy)=yTCTACy=yTBy \begin{aligned} x^{T}Ax &\overset{任取x=Cy}{=}(Cy)^{T}A(Cy)\\ &=y^{T}C^{T}ACy\\ &=y^{T}By \end{aligned}

其中B=CTACB=C^{T}AC，有

BT=(CTAC)T=CTAT(CT)T=CTAC=B B^{T}=(C^{T}AC)^{T}=C^{T}A^{T}(C^{T})^{T}=C^{T}AC=B

 

性质

  • A≃AA \simeq A

  • 如果A≃BA \simeq B，则B≃AB \simeq A

  • 如果A≃B,B≃CA \simeq B,B \simeq C，则A≃CA \simeq C

 

合同充要条件

A≃B⇔pA=pB,qA=qB A \simeq B \Leftrightarrow p_{A}=p_{B},q_{A}=q_{B}

 

定理（惯性定理）：二次型xTAxx^{T}Ax经坐标变换其正惯性指数和负惯性指数是唯一确定的

 

二、标准形

配方法

定理：对任一nn元二次型f=xTAxf=x^{T}Ax都可以通过（配方法）可逆线性变换x=Cyx=Cy，其中CC为可逆矩阵，化成标准形

 

推论：任一nn元二次型f=xTAxf=x^{T}Ax都存在坐标变换x=Czx=Cz，使ff化为规范形

f=z12+⋯+zp2−zp+12−⋯−zp+q2 f=z_{1}^{2}+\cdots +z_{p}^{2}-z_{p+1}^{2}-\cdots -z ^{2}_{p+q}

 

例：f=y12+3y22−5y32,x=C1yf=y_{1}^{2}+3y_{2}^{2}-5y_{3}^{2},x=C_{1}y

令

{y1=z1y2=13z2y3=15z3 \left\{\begin{aligned}&y_{1}=z_{1}\\ &y_{2}= \frac{1}{\sqrt{3}}z_{2}\\ &y_{3}=\frac{1}{\sqrt{5}}z_{3}\end{aligned}\right.

有

(y1y2y3)=(1   13   15)(z1z2z3) \begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}=\begin{pmatrix} 1 &  &  \\  & \frac{1}{\sqrt{3}} &  \\  &  & \frac{1}{\sqrt{5}} \end{pmatrix}\begin{pmatrix} z_{1} \\ z_{2} \\ z_{3} \end{pmatrix}

则f=z12+z22−z32f=z_{1}^{2}+z_{2}^{2}-z_{3}^{2}，有

x=C1y=C1C2z,C=C1C2 x=C_{1}y=C_{1}C_{2}z,C=C_{1}C_{2}

 

例：f(y1,y2,y3)=4y12−9y22,x=C1yf(y_{1},y_{2},y_{3})=4y_{1}^{2}-9y_{2}^{2},x=C_{1}y

令

{y1=12z1y2=13z2y3=z3 \left\{\begin{aligned}&y_{1}=\frac{1}{2}z_{1}\\ &y_{2}=\frac{1}{3}z_{2}\\ &y_{3}=z_{3}\end{aligned}\right.

则

y=(12   13   1)z y=\begin{pmatrix} \frac{1}{2} &  &  \\  & \frac{1}{3} &  \\  &  & 1 \end{pmatrix}z

有f=(z12−z22)f=(z_{1}^{2}-z_{2}^{2})

 

例：用配方法化二次型，2x12+3x22+5x32+4x1x2−8x2x3−4x3x12x_{1}^{2}+3x_{2}^{2}+5x_{3}^{2}+4x_{1}x_{2}-8x_{2}x_{3}-4x_{3}x_{1}为标准形，并写出所用坐标变换

 

f=2[x12+2x1(x2−x3)]+3x225+x32−8x2x3=2[x12+2x1(x2−x3)+(x2−x3)2]−2(x2−x3)2+3x225+x32−8x2x3=2(x1+x2−x3)2+x22+3x32−4x2x3=2(x1+x2−x3)2+(x22−4x2x3+4x32)−4x32+3x32=2(x1+x2−x3)2+(x2−2x3)2−x32 \begin{aligned} f&=2[x_{1}^{2}+2x_{1}(x_{2}-x_{3})]+3x_{2}^{2}5+x_{3}^{2}-8x_{2}x_{3}\\ &=2[x_{1}^{2}+2x_{1}(x_{2}-x_{3})+(x_{2}-x_{3})^{2}]-2(x_{2}-x_{3})^{2}+3x_{2}^{2}5+x_{3}^{2}-8x_{2}x_{3}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+x_{2}^{2}+3x_{3}^{2}-4x_{2}x_{3}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+(x_{2}^{2}-4x_{2}x_{3}+4x_{3}^{2})-4x_{3}^{2}+3x_{3}^{2}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+(x_{2}-2x_{3})^{2}-x_{3}^{2} \end{aligned}

 

先配方掉x1x_{1}，则后面的式子都没有x1x_{1}，因此最后的系数行列式是上三角形式，一定不为00

 

令

{y1=x1+x2−x3y2=x2−2x3y3=x3∣11−101−2001∣≠0 \left\{\begin{aligned}&y_{1}=x_{1}+x_{2}-x_{3}\\ &y_{2}=x_{2}-2x_{3}\\ &y_{3}=x_{3}\end{aligned}\right.\quad \begin{vmatrix} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{vmatrix}\ne 0

即

{x1=y1−y2−y3x2=y2+2y3x3=y3(x1x2x3)=(1−1−1012001)(y1y2y3) \left\{\begin{aligned}&x_{1}=y_{1}-y_{2}-y_{3}\\ &x_{2}=y_{2}+2y_{3}\\ &x_{3}=y_{3}\end{aligned}\right.\quad \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}

有f=2y12+y22−y32f=2y_{1}^{2}+y_{2}^{2}-y_{3}^{2}

 

例：用配方法化成二次型，f(x1,x2,x3)=2x1x2+4x1x3f(x_{1},x_{2},x_{3})=2x_{1}x_{2}+4x_{1}x_{3}为标准形，并写出所用的坐标变换

 

对于没有平方项的，先做辅助坐标变换

 

令

{x1=y1+y2x2=y1−y2x3=y3 \left\{\begin{aligned}&x_{1}=y_{1}+y_{2}\\&x_{2}=y_{1}-y_{2}\\&x_{3}=y_{3}\end{aligned}\right.

 

该变换既能配方又能保证系数行列式不为00

 

得

f=2(y1+y2)(y1−y2)+4(y1+y2)y3=2y12−2y22+4y1y2+4y2y3=2(y12+2y1y3+y32)−2y22+4y2y3−2y32=2(y1+y3)2−2(y2−y3)2 \begin{aligned} f&=2(y_{1}+y_{2})(y_{1}-y_{2})+4(y_{1}+y_{2})y_{3}\\ &=2y_{1}^{2}-2y_{2}^{2}+4y_{1}y_{2}+4y_{2}y_{3}\\ &=2(y_{1}^{2}+2y_{1}y_{3}+y_{3}^{2})-2y_{2}^{2}+4y_{2}y_{3}-2y_{3}^{2}\\ &=2(y_{1}+y_{3})^{2}-2(y_{2}-y_{3})^{2} \end{aligned}

令

{z1=y1+y3z2=y2−y3z3=y3即{y1=z1−z3y2=z2+z3y3=z3 \left\{\begin{aligned}&z_{1}=y_{1}+y_{3}\\&z_{2}=y_{2}-y_{3}\\&z_{3}=y_{3}\end{aligned}\right.\quad 即\left\{\begin{aligned}&y_{1}=z_{1}-z_{3}\\&y_{2}=z_{2}+z_{3}\\&y_{3}=z_{3}\end{aligned}\right.

则f=2z12−2z22f=2z_{1}^{2}-2z_{2}^{2}

x=C1y,y=C2z⇒x=Cz,C=C1C2 x=C_{1}y,y=C_{2}z \Rightarrow x=Cz,C=C_{1}C_{2}

其中

C=(1101−10001)(10−1011001)=(1101−1−2001) C=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & -2 \\ 0 & 0 & 1 \end{pmatrix}

 

正交变换法

定理：对任一nn元二次型f(x1,x2,⋯ ,xn)=xTAxf(x_{1},x_{2},\cdots,x_{n})=x^{T}Ax存在正交变换x=Qyx=Qy，使ff化为标准形

 

由于AA是实对称矩阵，则存在正交矩阵QQ

Q−1AQ=Λ Q^{-1}AQ=\Lambda

令x=Qyx=Qy，QQ为正交矩阵，Q−1=QTQ^{-1}=Q^{T}，有

f=xTAx=(Qy)TA(Qy)=yTQTAQy=yTQ−1AQy=yTΛy=λ1y12+λ2y22+⋯+λnyn2 \begin{aligned} f=x^{T}Ax&=(Qy)^{T}A(Qy)\\ &=y^{T}Q^{T}AQy\\ &=y^{T}Q^{-1}AQy\\ &=y^{T}\Lambda y\\ &=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\cdots +\lambda_{n}y_{n}^{2} \end{aligned}

 

例：二次型x12+3x22+x32+2ax1x2+2x1x3+2x2x3x_{1}^{2}+3x_{2}^{2}+x_{3}^{2}+2a x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}经正交变换x=Pyx=Py化成标准形y12+4y22y_{1}^{2}+4y_{2}^{2}，则a=()a=()

 

二次型xTAxx^{T}Ax经正交变换化为标准形y12+4y22y_{1}^{2}+4y_{2}^{2}，那么标准形中平方项的系数1,4,01,4,0就是AA的特征值

∣0E−A∣=−∣1a1a31111∣=−∣0a−10a31111∣=(a−1)2=0 |0E-A|=-\begin{vmatrix} 1 & a & 1 \\ a & 3 & 1 \\ 1 & 1 & 1 \end{vmatrix}=-\begin{vmatrix} 0 & a-1 & 0 \\ a & 3 & 1 \\ 1 & 1 & 1 \end{vmatrix}=(a-1)^{2}=0

因此a=1a=1。用λ=1\lambda=1或44也同理，这里用λ=1\lambda=1做演示

∣E−A∣=∣0−a−1−a2−1−1−10∣=∣0−a−1−aa−2−1−100∣=2−2a=0 |E-A|=\begin{vmatrix} 0 & -a & -1 \\ -a & 2 & -1 \\ -1 & -1 & 0 \end{vmatrix}=\begin{vmatrix} 0 & -a & -1 \\ -a & a-2 & -1 \\ -1 & 0 & 0 \end{vmatrix}=2-2a=0

 

例：二次型f(x1,x2)=xTAxf(x_{1},x_{2})=x^{T}Ax经正交变换x=Qyx=Qy化为标准形y12+3y22y_{1}^{2}+3y_{2}^{2}，若QQ的第一列是(12,12)T(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T}，则Q=()Q=()

 

正交变换x=Qyx=Qy

xTAx=y12+3y22 x^{T}Ax=y_{1}^{2}+3y_{2}^{2}

则AA的特征值为1,31,3，Q=(γ1,γ2)Q=(\gamma_{1},\gamma_{2})分别是1,31,3的特征向量，设λ=3\lambda=3的特征向量α=(x1,x2)T\alpha=(x_{1},x_{2})^{T}。又因为AA是实对称矩阵，不同特征值对应特征向量正交，有

12x1+12x2=0 \frac{1}{\sqrt{2}}x_{1}+ \frac{1}{\sqrt{2}}x_{2}=0

基础解系(1,−1)T(1,-1)^{T}，单位化(12,−12)T(\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}})^{T}，则

Q=(121212−12) Q=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}

 

例：用正交变换化二次型f(x1,x2,x3)=x12−x22+2x32+4x2x3f(x_{1},x_{2},x_{3})=x_{1}^{2}-x_{2}^{2}+2x_{3}^{2}+4x_{2}x_{3}为标准形，写出所用坐标变换

 

二次型矩阵

A=(1000−12022) A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 2 \\ 0 & 2 & 2 \end{pmatrix}

有

∣λE−A∣=∣λ−1000λ+1−20−2λ−2∣=(λ−1)(λ−3)(λ+2) |\lambda E-A|=\begin{vmatrix} \lambda-1 & 0 & 0 \\ 0 & \lambda+1 & -2 \\ 0 & -2 & \lambda-2 \end{vmatrix}=(\lambda-1)(\lambda-3)(\lambda+2)

AA的特征值1,3,−21,3,-2

 

如果只要求标准形到这里就可以了，标准形为f=y12+3y22−2y32f=y_{1}^{2}+3y_{2}^{2}-2y_{3}^{2}

 

由(E−A)x=0(3E−A)x=0(−2E−A)x=0\begin{aligned} (E-A)x&=0\\(3E-A)x&=0\\(-2E-A)x&=0\end{aligned}，得α1=(1,0,0)Tα2=(0,1,2)Tα3=(0,−2,1)T\begin{aligned} \alpha_{1}&=(1,0,0)^{T}\\\alpha_{2}&=(0,1,2)^{T}\\ \alpha_{3}&=(0,-2,1)^{T}\end{aligned}，特征值不同特征向量已经正交，单位化

γ1=(100),γ2=15(012),γ3=15(0−21) \gamma_{1}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\gamma_{2}=\frac{1}{\sqrt{5}}\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix},\gamma_{3}=\frac{1}{\sqrt{5}}\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}

令Q=(γ1,γ2,γ3)Q=(\gamma_{1},\gamma_{2},\gamma_{3})，经

(x1x2x3)=(100015−2502515)(y1y2y3) \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{5}} & - \frac{2}{\sqrt{5}} \\ 0 & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix}

则

xTAx=yT(QTAQ)y=yTΛy=y12+3y22−2y32 x^{T}Ax=y^{T}(Q^{T}AQ)y=y^{T}\Lambda y=y_{1}^{2}+3y^{2}_{2}-2y_{3}^{2}

 

例：已知二次型f(x1,x2,x3)=5x12+5x22+cx32−2x1x2+6x1x3−6x2x3f(x_{1},x_{2},x_{3})=5x_{1}^{2}+5x_{2}^{2}+c x_{3}^{2}-2x_{1}x_{2}+6x_{1}x_{3}-6x_{2}x_{3}的秩为22，求cc，求二次型ff的p,qp,q

 

二次型矩阵

A=(5−13−15−33−3c) A=\begin{pmatrix} 5 & -1 & 3 \\ -1 & 5 & -3 \\ 3 & -3 & c \end{pmatrix}

由r(f)=2⇔r(A)=2r(f)=2\Leftrightarrow r(A)=2

∣A∣=∣5−13−15−33−3c∣=∣4−1306−60−3c∣=24(c−3) |A|=\begin{vmatrix} 5 & -1 & 3 \\ -1 & 5 & -3 \\ 3 & -3 & c \end{vmatrix}=\begin{vmatrix} 4 & -1 & 3 \\ 0 & 6 & -6 \\ 0 & -3 & c \end{vmatrix}=24(c-3)

因此c=3c=3，代入原式，有

∣λE−A∣=∣λ−51−31λ−53−33λ−3∣=λ(λ−4)(λ−9) |\lambda E-A|=\begin{vmatrix} \lambda-5 & 1 & -3 \\ 1 & \lambda-5 & 3 \\ -3 & 3 & \lambda-3 \end{vmatrix}=\lambda(\lambda-4)(\lambda-9)

因此p=2,q=0p=2,q=0

 

三、正定二次型

定义：任取x≠0x \ne 0，恒有f(x1,x2,x3)=xTAx>0f(x_{1},x_{2},x_{3})=x^{T}Ax>0，则称二次型ff为正定二次型，二次型矩阵AA称为正定矩阵

 

二次型要想是正定矩阵则二次型系数必须全部大于00

 

定理：经坐标变换不改变二次型的正定性

 

定理（正定的充分必要条件）：

xTAxx^{T}Ax正定

⇔p=n\Leftrightarrow p=n

⇔A≃E\Leftrightarrow A \simeq E（即存在可逆CC，使CTAC=EC^{T}AC=E，也可描述为A=DTD,D可逆A=D^{T}D,D可逆）

⇔A\Leftrightarrow A的特征值全大于00

⇔A\Leftrightarrow A的顺序主子式全大于00

 

定理（正定的必要条件）：

  • 矩阵AA主对角线元素全大于00

  • ∣A∣>0|A|>0

 

例：判断二次型f(x1,x2,x3)=2x12+5x22+5x32+4x1x2−4x1x3−8x2x3f(x_{1},x_{2},x_{3})=2x_{1}^{2}+5x_{2}^{2}+5x_{3}^{2}+4x_{1}x_{2}-4x_{1}x_{3}-8x_{2}x_{3}的正定性

 

顺序主子式

A=(22−225−4−2−45) A=\begin{pmatrix} 2 & 2 & -2 \\ 2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}

Δ1=2>0,Δ2=∣2225∣,Δ3=∣A∣=10>0\Delta_{1}=2>0,\Delta_{2}=\begin{vmatrix}2 & 2 \\ 2 & 5\end{vmatrix},\Delta_{3}=|A|=10>0，因此正定

 

特征值

∣λE−A∣=∣λ−2−22−2λ−5424λ−5∣=(λ−1)2(λ−10) |\lambda E-A|=\begin{vmatrix} \lambda-2 & -2 & 2 \\ -2 & \lambda-5 & 4 \\ 2 & 4 & \lambda-5 \end{vmatrix}=(\lambda-1)^{2}(\lambda-10)

特征值1,1,101,1,10，因此正定

 

配方法

f=2(x12+2x1x2−2x1x3)+5x22+5x32−8x2x3=2[x12+2x1(x2−x3)+(x2−x3)2]+5x22+5x32−8x2x3−2(x2−x3)2=2(x1+x2−x3)2+3x22−4x2x3+3x32=2(x1+x2−x3)2+3[x22−43x2x3+(23x3)2]+3x32−43x32=2(x1+x2−x3)2+3(x2−23x3)2+53x32=2y12+3y22+53y32 \begin{aligned} f&=2(x_{1}^{2}+2x_{1}x_{2}-2x_{1}x_{3})+5x_{2}^{2}+5x_{3}^{2}-8x_{2}x_{3}\\ &=2[x_{1}^{2}+2x_{1}(x_{2}-x_{3})+(x_{2}-x_{3})^{2}]+5x_{2}^{2}+5x_{3}^{2}-8x_{2}x_{3}-2(x_{2}-x_{3})^{2}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+3x_{2}^{2}-4x_{2}x_{3}+3x_{3}^{2}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+3[x_{2}^{2}- \frac{4}{3}x_{2}x_{3}+(\frac{2}{3}x_{3})^{2}]+3x_{3}^{2}- \frac{4}{3}x_{3}^{2}\\ &=2(x_{1}+x_{2}-x_{3})^{2}+3(x_{2}- \frac{2}{3}x_{3})^{2}+ \frac{5}{3}x_{3}^{2}\\ &=2y_{1}^{2}+3y_{2}^{2}+ \frac{5}{3}y_{3}^{2} \end{aligned}

因此正定

avatar
文章
阅读
粉丝