线代矩阵的相似和相合

例2设A A能否对角化？若能对角化,则求出可逆矩 P使P1AP为对角阵解IA

所以A1213 zeng,ShanghaiUniversity,linear1

2 01IAx zeng,ShanghaiUniversity,linear将32代入IAx0,得方程组的基础3 由于1,2,3线性无关 所以A可对角化2 1 0 0 zeng,ShanghaiUniversity,linear 1 若令P

P1AP

即矩阵P的列向量和对角矩阵中特征值的位置 zeng,ShanghaiUniversity,linearr(IA)n zeng,ShanghaiUniversity,linearA与B相似,则det(A)若A与B相似,且A可逆,则B也可逆,且A1B1相似A与B则kA与kB相似k为常数若A与B相似而fx)是一多项式,则fA)与f(B)相似. zeng,ShanghaiUniversity,linear zeng,ShanghaiUniversity,linear定理1对称实矩阵的特征值为实数证明设复数为对称矩阵A,复向量x为对应的特征向量, Axx,x用表示的共轭复数

x表示x的 AxAx= Axx zeng,ShanghaiUniversity,linear及xT

xTAxxTAxxTx xT(xTAT)xAxT x

i所 xTx∑xx∑ i

即,由此可得是实数 zeng,ShanghaiUniversity,linear(iIA)x是实系数方程组,由iIA0知必有实的基础解系,从而对应的特征向量可以取实向量. zeng,ShanghaiUniversity,linear p2是对应的特征向量,若12,则p1与p2正交.1p1Ap1,2p2Ap2,12p

p

T

T

pT

于是

pT pT

T

pTp

pT ∵12

pT

0.即p与

正交 zeng,ShanghaiUniversity,linear 设A为n阶对称矩阵,是的特征方程的r重根,则矩阵IA的秩r(IA)nr,从而对应特征值恰有r个线性无关的特征向量. 设A为n阶对称矩阵,则必有正交矩阵P,使P1AP,其中是以的n 个特征值为对角证明设A的互不相等的特征值为1,2 1, ,

(1r2 rs理3(如上)可得： zeng,ShanghaiUniversity,linear对应特征值i(i1,2, ,s),恰有ri个线性无

rsn知故这n个单位特征向量两两正交.P1APP1P其中对角矩阵的对角元素含r1个1, zeng,ShanghaiUniversity,linear由iIAx0,求出的特征向量将特征向量正交化将特征向量单位化 zeng,ShanghaiUniversity,linear例1对下列各实对称矩阵，分别求出正交矩阵P，使P1AP为对角阵.20400(1)A1,(2)A03100013解(1)第一步求A IA

14,21,3 zeng,ShanghaiUniversity,linear第二 由iIAx0,求出的特征向14,由4IAx21,由IAx0,

1 22 zeng,ShanghaiUniversity,linear3122 由于1,2,3是属于A的3个不同特征值123的特征向量,故它们必两两正交 iii zeng,ShanghaiUniversity,linear2222得12,21,123 232 P

1,3 1 P1AP zeng,ShanghaiUniversity,linear (2)A IA

12,23

012,由2IAx0,

1 234,由4IAx0, zeng,ShanghaiUniversity,linear12 0

3 1

2与3 所以123两两正交再将单位化,令

i1,2,3 i 120,120,312.012 zeng,ShanghaiUniversity,linear

121212 12121220200P1AP040.004 zeng,ShanghaiUniversity,linear 求正交矩阵 PTAP为对角矩阵解|IA

1 1

13 3

1(1,1,0)T,

1(1,21 2122

2

26 P[1,2,326 zeng,ShanghaiUniversity,linear求

2A11

2

使得PTAP为对角矩阵211142111411a11 112141，得a

|IA

31当4时，有特征向 131当1

1(1,1,0)T,

1(1,2261 226122

2

1所以P1,2,3 zeng,ShanghaiUniversity,linear zeng,ShanghaiUniversity,linearA

A

，则称A为 证设AC,

A由于TAAT)T

A(A)T T ()T

，所以T

，即

zeng,ShanghaiUniversity,linear判断下列两矩阵AB是否相似1 A 1

B 1 zeng,ShanghaiUniversity,linear 因det(IA)(n)()n1,的特征值1n,2 n矩阵1,

P1AP

det(IB)(即B与A有相同的特征值 zeng,ShanghaiUniversity,linear对应特征值2 n0,有n1个线性无关的特征向量,故存在可逆矩阵P2,使得PPBP BP, PPBP BP,P P

AP1 P PP

1AP1P

21故A与B相似21 zeng,ShanghaiUniversity,linear定义 含有n个变量x1,x2 ,xn的二次齐次函fx,x ,xax2 x2 2a12x1x2

2an1,n当aij是复数时,f称为复二次 当aij是实数时,f称为实二次型 zeng,ShanghaiUniversity,linear fky2ky fx,x,x2x24x25x24x fx1,x2,x3x1x2x1x3x2fx,x,xx24x24x2 zeng,ShanghaiUniversity,linearfx,x ,x

x2

x2 2a12x1x22a13x1x3 2an1,n取ajiaij 则2aijxixjaijxixjajixjxif

x2 xx x xx

x2 x 2 xx xx n i, zeng,ShanghaiUniversity,linearf

x2 xx xxx x xxx x x2 2 xx xx n x1(a11x1a12x2 a1nxnx2(a21x1a22x2 xn(an1x1an2x2 a11x1a12x2 a1n(x,x ,x

ax x

2

an2

zeng,ShanghaiUniversity,linear

A

，x fxTx,其中A为对称矩阵 zeng,ShanghaiUniversity,linear在二次型的矩阵表示中，一个二次型，就唯一地确定一个对称矩阵；反之，一个对对称矩阵A叫做二次型f的矩阵f叫做对称矩阵A的二次型对称矩阵Af的秩 zeng,ShanghaiUniversity,linear例 fx22x23x2

6x2解

1, 2,

a21

a31 A zeng,ShanghaiUniversity,linear 2xc21y1c22y2 2

cn2

记Ccij则上述可逆线性变换可x zeng,ShanghaiUniversity,linearfxTAx,fxTAxCyT

yTCTACBPT B，则 B，且 zeng,ShanghaiUniversity,linear二次型经可逆变换xCy后其秩不变但的矩阵由A变为BCTAC要使二次型f经可逆变 xCy变成标准形

yTCTACyky2+ky2 kyy y

yk yk ykyn也就是要使CTAC成为对角矩阵 zeng,ShanghaiUniversity,linear使P1AP,即PTAP.把此结论应用于二次型,f n xx i,jxPy使f化为标准fy2y2 y2 其中1,2 zeng,ShanghaiUniversity,linear将二次型表成矩阵形式fxTAx，求出 求出对应于特征值的特征向量1，2 ,n将特征向量1,2 ,n正交化,单位化, ,n,记C ,n 作正交变换xCy,则得f的标准形f y2+ y zeng,ShanghaiUniversity,linear例2f17x214x214x24x

4x1

8x2通过正交变换xPy化成标准形解1 A IA

2 zeng,ShanghaiUniversity,linear 19, 将19代入IAx0,1(12,1,1)T1 将2318代入IAx (2,1,0), (2,0,1)

, , ,

,T T 1 (12,1,1) 2T

(2,1,0)3(25,45,1) zeng,ShanghaiUniversity,linear将正交向量组单位化，得正交矩阵

ii1511515.2055P215.205 1 2 ,3 zeng,ShanghaiUniversity,linear 1 215,3205且 f9y218y2

zeng,ShanghaiUniversity,linear例3求一个正交变换xPyf2x1x22x1x32x1x42x2x32x2x42x3化为标准形解 二次型的矩阵为A

zeng,ShanghaiUniversity,linear11111计算特征多项式把二,三,四列都加到第一列上, 111IA(1) 111把二,三,四行分别减去第一行, zeng,ShanghaiUniversity,linear12200IA(-1)00 (

(1)2(223)(3)(于是A的特征值为13,234当13时,解方程(3IA)x zeng,ShanghaiUniversity,linear 得基础解系

,单位化即得p 1 1

1当234时,解方程(IA)x ,

1 zeng,ShanghaiUniversity,linear 01单位化即得

,p

,p 112 120

112112 22 1 1 22x2 1

1 3 1 3

1 x4 1

1 1212 f3y2y2y21212 zeng,ShanghaiUniversity,linear f2x23x23x 032.02300|I 0 zeng,ShanghaiUniversity,linear当2

(2IA)x000000000010000030010000000000110,P0.00 (5IA)x

zeng,ShanghaiUniversity,linear /2 1/1/1/ (5IA)x001001000 0220110000000 2 1 1/ zeng,ShanghaiUniversity,linear 22P 01/2-220

xPy

f2y25y2y2. zeng,ShanghaiUniversity,linear五 次型的秩,正（负）系数的二次项的项数也是确定设有实二次型fxTAx,它的秩为r,xfky2ky2 ky2

ky2k

- - -212r diag(k 212r

fz2z2 z2

zeng,ShanghaiUniversity,linear

I r 其中rrA),且由唯一确定，分别称prp为的 zeng,ShanghaiUniversity,linear 设有实二次型f(x)xTAx,如果对任x0都有fx0f00,则称f为正定二次型,A是正定的;x0都有fx0,f,并A是负定的fx24y216zfx23x2

zeng,ShanghaiUniversity,linearAAA的正惯性指数为(4)存在n阶可逆的矩阵PRnn,使APT zeng,ShanghaiUniversity,linear 设可逆变换xCyfxfCy

n ∑ky2 则yC-1x

ii0i

给x fx

kki

2 ）（ks0,则当yes(单位坐标向量 Ces

fCesks这与f为正定 ki0i , zeng,ShanghaiUniversity,linear定

0，r , zeng,ShanghaiUniversity,linear设A为正定实对称阵,则ATA1A均为定矩阵若AB均为n阶正定矩阵,则AB也是正定矩阵. zeng,ShanghaiUniversity,linear例1fx,x,x5x x25x24x1x2

8x

4x

fx1,x2,x3的矩阵

5 1

1 zeng,ShanghaiUniversity,linear例 判别二次fx,x,x2x

4

5x24x

用特征值判别法

A 令I

0⇒

1, 4, zeng,ShanghaiUniversity,linear例 判别二次f5x26y24z24xy4的正定性 解f的矩阵 A 5