本文已参加「新人创造礼」活动，一同开启创造之路。
留意与特征值、特征向量的联络
一、概念、定理
概念
二次型及其矩阵表明
f(x1,x2,x3)=x12+5×22+5×32+2x1x2−6x2x3=x12+x1x2+x1x2+5×22−3x2x3−3x2x3+5×32=x1(x1+x2)+x2(x1+5×2−3×3)+x3(−3×2+5×3)=(x1,x2,x3)(x1+x2x1+5×2−3×3−3×2+5×3)=(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}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
有AT=AA ^{T}=AAT=A是二次型的矩阵
规范形
2×12+5×22−6×32=(25−6)\begin{gathered} 2x_{1}^{2}+5x_{2}^{2}-6x_{3}^{2}\\ \Lambda=\begin{pmatrix} 2 & & \\ & 5 & \\ & & -6 \end{pmatrix} \end{gathered}2x12+5x22−6x32=⎝⎛25−6⎠⎞
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}x12+x22+x32x12+x22−x32x12−x22−x12
正惯性指数、负惯性指数
关于规范形
2×12+5×22−6×32正惯性指数为p=2,负惯性指数为q=1×12+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}2x12+5x22−6x32x12+x22+x32正惯性指数为p=2,负惯性指数为q=1p=3,q=0
非规范形的化成规范形再看
二次型的秩
r(f)=r(A)r(f)=r(A)r(f)=r(A)
因为
r(1−20−234045)=3r \begin{pmatrix} 1 & -2 & 0 \\ -2 & 3 & 4 \\ 0 & 4 & 5 \end{pmatrix}=3r⎝⎛1−20−234045⎠⎞=3
故二次型f=x12+3×22+5×32−4x1x2+8x2x3f=x_{1}^{2}+3x_{2}^{2}+5x_{3}^{2}-4x_{1}x_{2}+8x_{2}x_{3}f=x12+3x22+5x32−4x1x2+8x2x3的秩r(f)=3r(f)=3r(f)=3
坐标改换
{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⎩⎪⎪⎨⎪⎪⎧x1=c11y1+c12y2+c13y3x2=c21y1+c22y2+c23y3x3=c31y1+c32y2+c33y3∣C∣=0
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}⎝⎛x1x2x3⎠⎞=⎝⎛c11c21c31c12c22c32c13c23c33⎠⎞⎝⎛y1y2y3⎠⎞
即可表明为x=cyx=cyx=cy
留意系数行列式∣C∣|C|∣C∣必定要不为000
合同
假如CTAC=B,其间C可逆C^{T}AC=B,其间C可逆CTAC=B,其中C可逆，则称矩阵AAA和BBB合同，记A≃BA \simeq BA≃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}xTAx=任取x=Cy(Cy)TA(Cy)=yTCTACy=yTBy
其间B=CTACB=C^{T}ACB=CTAC，有
BT=(CTAC)T=CTAT(CT)T=CTAC=BB^{T}=(C^{T}AC)^{T}=C^{T}A^{T}(C^{T})^{T}=C^{T}AC=BBT=(CTAC)T=CTAT(CT)T=CTAC=B
性质
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A≃AA \simeq AA≃A
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假如A≃BA \simeq BA≃B，则B≃AB \simeq AB≃A
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假如A≃B,B≃CA \simeq B,B \simeq CA≃B,B≃C，则A≃CA \simeq CA≃C
合同充要条件
A≃B⇔pA=pB,qA=qBA \simeq B \Leftrightarrow p_{A}=p_{B},q_{A}=q_{B}A≃B⇔pA=pB,qA=qB
定理（惯性定理）：二次型xTAxx^{T}AxxTAx经坐标改换其正惯性指数和负惯性指数是仅有确认的
二、规范形
配办法
定理：对任一nnn元二次型f=xTAxf=x^{T}Axf=xTAx都可以通过（配办法）可逆线性改换x=Cyx=Cyx=Cy，其间CCC为可逆矩阵，化成规范形
推论：任一nnn元二次型f=xTAxf=x^{T}Axf=xTAx都存在坐标改换x=Czx=Czx=Cz，使fff化为规范形
f=z12+⋯+zp2−zp+12−⋯−zp+q2f=z_{1}^{2}+\cdots +z_{p}^{2}-z_{p+1}^{2}-\cdots -z ^{2}_{p+q}f=z12+⋯+zp2−zp+12−⋯−zp+q2
例：f=y12+3y22−5y32,x=C1yf=y_{1}^{2}+3y_{2}^{2}-5y_{3}^{2},x=C_{1}yf=y12+3y22−5y32,x=C1y
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.⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧y1=z1y2=31z2y3=51z3
有
(y1y2y3)=(11315)(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}⎝⎛y1y2y3⎠⎞=⎝⎛13151⎠⎞⎝⎛z1z2z3⎠⎞
则f=z12+z22−z32f=z_{1}^{2}+z_{2}^{2}-z_{3}^{2}f=z12+z22−z32，有
x=C1y=C1C2z,C=C1C2x=C_{1}y=C_{1}C_{2}z,C=C_{1}C_{2}x=C1y=C1C2z,C=C1C2
例：f(y1,y2,y3)=4y12−9y22,x=C1yf(y_{1},y_{2},y_{3})=4y_{1}^{2}-9y_{2}^{2},x=C_{1}yf(y1,y2,y3)=4y12−9y22,x=C1y
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.⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧y1=21z1y2=31z2y3=z3
则
y=(12131)zy=\begin{pmatrix} \frac{1}{2} & & \\ & \frac{1}{3} & \\ & & 1 \end{pmatrix}zy=⎝⎛21311⎠⎞z
有f=(z12−z22)f=(z_{1}^{2}-z_{2}^{2})f=(z12−z22)
例：用配办法化二次型，2×12+3×22+5×32+4x1x2−8x2x3−4x3x12x_{1}^{2}+3x_{2}^{2}+5x_{3}^{2}+4x_{1}x_{2}-8x_{2}x_{3}-4x_{3}x_{1}2x12+3x22+5x32+4x1x2−8x2x3−4x3x1为规范形，并写出所用坐标改换
f=2[x12+2×1(x2−x3)]+3×225+x32−8x2x3=2[x12+2×1(x2−x3)+(x2−x3)2]−2(x2−x3)2+3×225+x32−8x2x3=2(x1+x2−x3)2+x22+3×32−4x2x3=2(x1+x2−x3)2+(x22−4x2x3+4×32)−4×32+3×32=2(x1+x2−x3)2+(x2−2×3)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}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
先配方掉x1x_{1}x1，则后面的式子都没有x1x_{1}x1，因而最后的系数行列式是上三角方式，必定不为000
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⎩⎪⎪⎨⎪⎪⎧y1=x1+x2−x3y2=x2−2x3y3=x3∣∣∣∣∣∣∣100110−1−21∣∣∣∣∣∣∣=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}⎩⎪⎪⎨⎪⎪⎧x1=y1−y2−y3x2=y2+2y3x3=y3⎝⎛x1x2x3⎠⎞=⎝⎛100−110−121⎠⎞⎝⎛y1y2y3⎠⎞
有f=2y12+y22−y32f=2y_{1}^{2}+y_{2}^{2}-y_{3}^{2}f=2y12+y22−y32
例：用配办法化成二次型，f(x1,x2,x3)=2x1x2+4x1x3f(x_{1},x_{2},x_{3})=2x_{1}x_{2}+4x_{1}x_{3}f(x1,x2,x3)=2x1x2+4x1x3为规范形，并写出所用的坐标改换
关于没有平方项的，先做辅佐坐标改换
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.⎩⎪⎪⎨⎪⎪⎧x1=y1+y2x2=y1−y2x3=y3
该改换既能配方又能确保系数行列式不为000
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}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
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.⎩⎪⎪⎨⎪⎪⎧z1=y1+y3z2=y2−y3z3=y3即⎩⎪⎪⎨⎪⎪⎧y1=z1−z3y2=z2+z3y3=z3
则f=2z12−2z22f=2z_{1}^{2}-2z_{2}^{2}f=2z12−2z22
x=C1y,y=C2z⇒x=Cz,C=C1C2x=C_{1}y,y=C_{2}z \Rightarrow x=Cz,C=C_{1}C_{2}x=C1y,y=C2z⇒x=Cz,C=C1C2
其间
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}C=⎝⎛1101−10001⎠⎞⎝⎛100010−111⎠⎞=⎝⎛1101−100−21⎠⎞
正交改换法
定理：对任一nnn元二次型f(x1,x2,⋯ ,xn)=xTAxf(x_{1},x_{2},\cdots,x_{n})=x^{T}Axf(x1,x2,⋯,xn)=xTAx存在正交改换x=Qyx=Qyx=Qy，使fff化为规范形
因为AAA是实对称矩阵，则存在正交矩阵QQQ
Q−1AQ=Q^{-1}AQ=\LambdaQ−1AQ=
令x=Qyx=Qyx=Qy，QQQ为正交矩阵，Q−1=QTQ^{-1}=Q^{T}Q−1=QT，有
f=xTAx=(Qy)TA(Qy)=yTQTAQy=yTQ−1AQy=yTy=1y12+2y22+⋯+nyn2\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}f=xTAx=(Qy)TA(Qy)=yTQTAQy=yTQ−1AQy=yTy=1y12+2y22+⋯+nyn2
例：二次型x12+3×22+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}x12+3x22+x32+2ax1x2+2x1x3+2x2x3经正交改换x=Pyx=Pyx=Py化成规范形y12+4y22y_{1}^{2}+4y_{2}^{2}y12+4y22，则a=()a=()a=()
二次型xTAxx^{T}AxxTAx经正交改换化为规范形y12+4y22y_{1}^{2}+4y_{2}^{2}y12+4y22，那么规范形中平方项的系数1,4,01,4,01,4,0便是AAA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∣0E−A∣=−∣∣∣∣∣∣∣1a1a31111∣∣∣∣∣∣∣=−∣∣∣∣∣∣∣0a1a−131011∣∣∣∣∣∣∣=(a−1)2=0
因而a=1a=1a=1。用=1\lambda=1=1或444也同理，这儿用=1\lambda=1=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∣E−A∣=∣∣∣∣∣∣∣0−a−1−a2−1−1−10∣∣∣∣∣∣∣=∣∣∣∣∣∣∣0−a−1−aa−20−1−10∣∣∣∣∣∣∣=2−2a=0
例：二次型f(x1,x2)=xTAxf(x_{1},x_{2})=x^{T}Axf(x1,x2)=xTAx经正交改换x=Qyx=Qyx=Qy化为规范形y12+3y22y_{1}^{2}+3y_{2}^{2}y12+3y22，若QQQ的榜首列是(12,12)T(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})^{T}(21,21)T，则Q=()Q=()Q=()
正交改换x=Qyx=Qyx=Qy
xTAx=y12+3y22x^{T}Ax=y_{1}^{2}+3y_{2}^{2}xTAx=y12+3y22
则AAA的特征值为1,31,31,3，Q=(1,2)Q=(\gamma_{1},\gamma_{2})Q=(1,2)分别是1,31,31,3的特征向量，设=3\lambda=3=3的特征向量=(x1,x2)T\alpha=(x_{1},x_{2})^{T}=(x1,x2)T。又因为AAA是实对称矩阵，不同特征值对应特征向量正交，有
12×1+12×2=0\frac{1}{\sqrt{2}}x_{1}+ \frac{1}{\sqrt{2}}x_{2}=021x1+21x2=0
基础解系(1,−1)T(1,-1)^{T}(1,−1)T，单位化(12,−12)T(\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}})^{T}(21,−21)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}Q=(212121−21)
例：用正交改换化二次型f(x1,x2,x3)=x12−x22+2×32+4x2x3f(x_{1},x_{2},x_{3})=x_{1}^{2}-x_{2}^{2}+2x_{3}^{2}+4x_{2}x_{3}f(x1,x2,x3)=x12−x22+2x32+4x2x3为规范形，写出所用坐标改换
二次型矩阵
A=(1000−12022)A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 2 \\ 0 & 2 & 2 \end{pmatrix}A=⎝⎛1000−12022⎠⎞
有
∣E−A∣=∣−1000+1−20−2−2∣=(−1)(−3)(+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)∣E−A∣=∣∣∣∣∣∣∣−1000+1−20−2−2∣∣∣∣∣∣∣=(−1)(−3)(+2)
AAA的特征值1,3,−21,3,-21,3,−2
假如只要求规范形到这儿就可以了，规范形为f=y12+3y22−2y32f=y_{1}^{2}+3y_{2}^{2}-2y_{3}^{2}f=y12+3y22−2y32
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}(E−A)x(3E−A)x(−2E−A)x=0=0=0，得1=(1,0,0)T2=(0,1,2)T3=(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}123=(1,0,0)T=(0,1,2)T=(0,−2,1)T，特征值不同特征向量已经正交，单位化
1=(100),2=15(012),3=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}1=⎝⎛100⎠⎞,2=51⎝⎛012⎠⎞,3=51⎝⎛0−21⎠⎞
令Q=(1,2,3)Q=(\gamma_{1},\gamma_{2},\gamma_{3})Q=(1,2,3)，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}⎝⎛x1x2x3⎠⎞=⎝⎛100051520−5251⎠⎞⎝⎛y1y2y3⎠⎞
则
xTAx=yT(QTAQ)y=yTy=y12+3y22−2y32x^{T}Ax=y^{T}(Q^{T}AQ)y=y^{T}\Lambda y=y_{1}^{2}+3y^{2}_{2}-2y_{3}^{2}xTAx=yT(QTAQ)y=yTy=y12+3y22−2y32
例：已知二次型f(x1,x2,x3)=5×12+5×22+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}f(x1,x2,x3)=5x12+5x22+cx32−2x1x2+6x1x3−6x2x3的秩为222，求ccc，求二次型fff的p,qp,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}A=⎝⎛5−13−15−33−3c⎠⎞
由r(f)=2⇔r(A)=2r(f)=2\Leftrightarrow r(A)=2r(f)=2⇔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)∣A∣=∣∣∣∣∣∣∣5−13−15−33−3c∣∣∣∣∣∣∣=∣∣∣∣∣∣∣400−16−33−6c∣∣∣∣∣∣∣=24(c−3)
因而c=3c=3c=3，代入原式，有
∣E−A∣=∣−51−31−53−33−3∣=(−4)(−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)∣E−A∣=∣∣∣∣∣∣∣−51−31−53−33−3∣∣∣∣∣∣∣=(−4)(−9)
因而p=2,q=0p=2,q=0p=2,q=0
三、正定二次型
界说：任取x≠0x \ne 0x=0，恒有f(x1,x2,x3)=xTAx>0f(x_{1},x_{2},x_{3})=x^{T}Ax>0f(x1,x2,x3)=xTAx>0，则称二次型fff为正定二次型，二次型矩阵AAA称为正定矩阵
二次型要想是正定矩阵则二次型系数必须悉数大于000
定理：经坐标改换不改变二次型的正定性
定理（正定的充分必要条件）：
xTAxx^{T}AxxTAx正定
⇔p=n\Leftrightarrow p=n⇔p=n
⇔A≃E\Leftrightarrow A \simeq E⇔A≃E（即存在可逆CCC，使CTAC=EC^{T}AC=ECTAC=E，也可描述为A=DTD,D可逆A=D^{T}D,D可逆A=DTD,D可逆）
⇔A\Leftrightarrow A⇔A的特征值全大于000
⇔A\Leftrightarrow A⇔A的次序主子式全大于000
定理（正定的必要条件）：
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矩阵AAA主对角线元素全大于000
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∣A∣>0|A|>0∣A∣>0
例：判断二次型f(x1,x2,x3)=2×12+5×22+5×32+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}f(x1,x2,x3)=2x12+5x22+5x32+4x1x2−4x1x3−8x2x3的正定性
次序主子式
A=(22−225−4−2−45)A=\begin{pmatrix} 2 & 2 & -2 \\ 2 & 5 & -4 \\ -2 & -4 & 5 \end{pmatrix}A=⎝⎛22−225−4−2−45⎠⎞
1=2>0,2=∣2225∣,3=∣A∣=10>0\Delta_{1}=2>0,\Delta_{2}=\begin{vmatrix}2 & 2 \\ 2 & 5\end{vmatrix},\Delta_{3}=|A|=10>01=2>0,2=∣∣∣∣∣2225∣∣∣∣∣,3=∣A∣=10>0，因而正定
特征值
∣E−A∣=∣−2−22−2−5424−5∣=(−1)2(−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)∣E−A∣=∣∣∣∣∣∣∣−2−22−2−5424−5∣∣∣∣∣∣∣=(−1)2(−10)
特征值1,1,101,1,101,1,10，因而正定
配办法
f=2(x12+2x1x2−2x1x3)+5×22+5×32−8x2x3=2[x12+2×1(x2−x3)+(x2−x3)2]+5×22+5×32−8x2x3−2(x2−x3)2=2(x1+x2−x3)2+3×22−4x2x3+3×32=2(x1+x2−x3)2+3[x22−43x2x3+(23×3)2]+3×32−43×32=2(x1+x2−x3)2+3(x2−23×3)2+53×32=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}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[x22−34x2x3+(32x3)2]+3x32−34x32=2(x1+x2−x3)2+3(x2−32x3)2+35x32=2y12+3y22+35y32
因而正定
