線代矩陣的相似和相合

例2設(shè)A A能否對(duì)角化？若能對(duì)角化,則求出可逆矩 P使P1AP為對(duì)角陣解IA

所以A1213 zeng,ShanghaiUniversity,linear1

2 01IAx zeng,ShanghaiUniversity,linear將32代入IAx0,得方程組的基礎(chǔ)3 由于1,2,3線性無關(guān) 所以A可對(duì)角化2 1 0 0 zeng,ShanghaiUniversity,linear 1 若令P

P1AP

即矩陣P的列向量和對(duì)角矩陣中特征值的位置 zeng,ShanghaiUniversity,linearr(IA)n zeng,ShanghaiUniversity,linearA與B相似,則det(A)若A與B相似,且A可逆,則B也可逆,且A1B1相似A與B則kA與kB相似k為常數(shù)若A與B相似而fx)是一多項(xiàng)式,則fA)與f(B)相似. zeng,ShanghaiUniversity,linear zeng,ShanghaiUniversity,linear定理1對(duì)稱實(shí)矩陣的特征值為實(shí)數(shù)證明設(shè)復(fù)數(shù)為對(duì)稱矩陣A,復(fù)向量x為對(duì)應(yīng)的特征向量, Axx,x用表示的共軛復(fù)數(shù)

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

xTAxxTAxxTx xT(xTAT)xAxT x

i所 xTx∑xx∑ i

即,由此可得是實(shí)數(shù) zeng,ShanghaiUniversity,linear(iIA)x是實(shí)系數(shù)方程組,由iIA0知必有實(shí)的基礎(chǔ)解系,從而對(duì)應(yīng)的特征向量可以取實(shí)向量. zeng,ShanghaiUniversity,linear p2是對(duì)應(yīng)的特征向量,若12,則p1與p2正交.1p1Ap1,2p2Ap2,12p

p

T

T

pT

于是

pT pT

T

pTp

pT ∵12

pT

0.即p與

正交 zeng,ShanghaiUniversity,linear 設(shè)A為n階對(duì)稱矩陣,是的特征方程的r重根,則矩陣IA的秩r(IA)nr,從而對(duì)應(yīng)特征值恰有r個(gè)線性無關(guān)的特征向量. 設(shè)A為n階對(duì)稱矩陣,則必有正交矩陣P,使P1AP,其中是以的n 個(gè)特征值為對(duì)角證明設(shè)A的互不相等的特征值為1,2 1, ,

(1r2 rs理3(如上)可得： zeng,ShanghaiUniversity,linear對(duì)應(yīng)特征值i(i1,2, ,s),恰有ri個(gè)線性無

rsn知故這n個(gè)單位特征向量?jī)蓛烧?P1APP1P其中對(duì)角矩陣的對(duì)角元素含r1個(gè)1, zeng,ShanghaiUniversity,linear由iIAx0,求出的特征向量將特征向量正交化將特征向量單位化 zeng,ShanghaiUniversity,linear例1對(duì)下列各實(shí)對(duì)稱矩陣，分別求出正交矩陣P，使P1AP為對(duì)角陣.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是屬于A的3個(gè)不同特征值123的特征向量,故它們必兩兩正交 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與3 所以123兩兩正交再將單位化,令

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

121212 12121220200P1AP040.004 zeng,ShanghaiUniversity,linear 求正交矩陣 PTAP為對(duì)角矩陣解|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為對(duì)角矩陣211142111411a11 112141，得a

|IA

31當(dāng)4時(shí)，有特征向 131當(dāng)1

1(1,1,0)T,

1(1,2261 226122

2

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

A

，則稱A為 證設(shè)AC,

A由于TAAT)T

A(A)T T ()T

，所以T

，即

zeng,ShanghaiUniversity,linear判斷下列兩矩陣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矩陣1,

P1AP

det(IB)(即B與A有相同的特征值 zeng,ShanghaiUniversity,linear對(duì)應(yīng)特征值2 n0,有n1個(gè)線性無關(guān)的特征向量,故存在可逆矩陣P2,使得PPBP BP, PPBP BP,P P

AP1 P PP

1AP1P

21故A與B相似21 zeng,ShanghaiUniversity,linear定義 含有n個(gè)變量x1,x2 ,xn的二次齊次函fx,x ,xax2 x2 2a12x1x2

2an1,n當(dāng)aij是復(fù)數(shù)時(shí),f稱為復(fù)二次 當(dāng)aij是實(shí)數(shù)時(shí),f稱為實(shí)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 則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為對(duì)稱矩陣 zeng,ShanghaiUniversity,linear在二次型的矩陣表示中，一個(gè)二次型，就唯一地確定一個(gè)對(duì)稱矩陣；反之，一個(gè)對(duì)對(duì)稱矩陣A叫做二次型f的矩陣f叫做對(duì)稱矩陣A的二次型對(duì)稱矩陣Af的秩 zeng,ShanghaiUniversity,linear例 fx22x23x2

6x2解

1, 2,

a21

a31 A zeng,ShanghaiUniversity,linear 2xc21y1c22y2 2

cn2

記Ccij則上述可逆線性變換可x zeng,ShanghaiUniversity,linearfxTAx,fxTAxCyT

yTCTACBPT B，則 B，且 zeng,ShanghaiUniversity,linear二次型經(jīng)可逆變換xCy后其秩不變但的矩陣由A變?yōu)锽CTAC要使二次型f經(jīng)可逆變 xCy變成標(biāo)準(zhǔn)形

yTCTACyky2+ky2 kyy y

yk yk ykyn也就是要使CTAC成為對(duì)角矩陣 zeng,ShanghaiUniversity,linear使P1AP,即PTAP.把此結(jié)論應(yīng)用于二次型,f n xx i,jxPy使f化為標(biāo)準(zhǔn)fy2y2 y2 其中1,2 zeng,ShanghaiUniversity,linear將二次型表成矩陣形式fxTAx，求出 求出對(duì)應(yīng)于特征值的特征向量1，2 ,n將特征向量1,2 ,n正交化,單位化, ,n,記C ,n 作正交變換xCy,則得f的標(biāo)準(zhǔn)形f y2+ y zeng,ShanghaiUniversity,linear例2f17x214x214x24x

4x1

8x2通過正交變換xPy化成標(biāo)準(zhǔn)形解1 A IA

2 zeng,ShanghaiUniversity,linear 19, 將19代入IAx0,1(12,1,1)T1 將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將正交向量組單位化，得正交矩陣

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

zeng,ShanghaiUniversity,linear例3求一個(gè)正交變換xPyf2x1x22x1x32x1x42x2x32x2x42x3化為標(biāo)準(zhǔn)形解 二次型的矩陣為A

zeng,ShanghaiUniversity,linear11111計(jì)算特征多項(xiàng)式把二,三,四列都加到第一列上, 111IA(1) 111把二,三,四行分別減去第一行, zeng,ShanghaiUniversity,linear12200IA(-1)00 (

(1)2(223)(3)(于是A的特征值為13,234當(dāng)13時(shí),解方程(3IA)x zeng,ShanghaiUniversity,linear 得基礎(chǔ)解系

,單位化即得p 1 1

1當(dāng)234時(shí),解方程(IA)x ,

1 zeng,ShanghaiUniversity,linear 01單位化即得

,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當(dāng)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（負(fù)）系數(shù)的二次項(xiàng)的項(xiàng)數(shù)也是確定設(shè)有實(shí)二次型fxTAx,它的秩為r,xfky2ky2 ky2

ky2k

- - -212r diag(k 212r

fz2z2 z2

zeng,ShanghaiUniversity,linear

I r 其中rrA),且由唯一確定，分別稱prp為的 zeng,ShanghaiUniversity,linear 設(shè)有實(shí)二次型f(x)xTAx,如果對(duì)任x0都有fx0f00,則稱f為正定二次型,A是正定的;x0都有fx0,f,并A是負(fù)定的fx24y216zfx23x2

zeng,ShanghaiUniversity,linearAAA的正慣性指數(shù)為(4)存在n階可逆的矩陣PRnn,使APT zeng,ShanghaiUniversity,linear 設(shè)可逆變換xCyfxfCy

n ∑ky2 則yC-1x

ii0i

給x fx

kki

2 ）（ks0,則當(dāng)yes(單位坐標(biāo)向量 Ces

fCesks這與f為正定 ki0i , zeng,ShanghaiUniversity,linear定

0，r , zeng,ShanghaiUniversity,linear設(shè)A為正定實(shí)對(duì)稱陣,則ATA1A均為定矩陣若AB均為n階正定矩陣,則AB也是正定矩陣. zeng,ShanghaiUniversity,linear例1fx,x,x5x x25x24x1x2

8x

4x

fx1,x2,x3的矩陣

5 1

1 zeng,ShanghaiUniversity,linear例 判別二次fx,x,x2x

4

5x24x

用特征值判別法

A 令I(lǐng)

0?

1, 4, zeng,ShanghaiUniversity,linear例 判別二次f5x26y24z24xy4的正定性 解f的矩陣 A 5