矩阵论3_λ矩阵与矩阵的Jordan标准型.pdf

《矩阵论3_λ矩阵与矩阵的Jordan标准型.pdf》由会员分享，可在线阅读，更多相关《矩阵论3_λ矩阵与矩阵的Jordan标准型.pdf（29页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、1n?JordanIO/3.1?“3.1.1?P?n?K?i/“L?“f()=ann+an1n1+a1+a0?ai P(i=0,1,n)?P?“?P?“?NP?Pn?f()?gP?(f()=n;ean=1Kf()?“?“?!?!?!?!9?$?5 n3.1.1?f(),g()P?g()6=0K?3?“q(),r()P?f()=q()g()+r()?r()=0r()6=0?(r()?A?m?A()?g?C?u3A()m?A?n?P(i,j)1=P(i,j),P(i(k)1=P(i(k1),P(i,j()1=P(i,j(),3.2.5?A(),B()PmnXJA()Lk?g?C?z?B()K?A(

2、)?B()?-P?A()=B()n3.2.2?A(),B()PmnKA()?B()?-?7?3m?P1(),Pl()?n?Q1(),Qt()?A()=Pl()P1()B()Q1()Qt()n3.3.5?A(),B()PmnKA()?B()?-?7?3?_?P()Pmm?Q()Pnn?A()=P()B()Q()n3.3.4?A()PnnKA()?_?7?A()?L?X?3.2.3?3?-e?IO/?n3.2.1?A()=(aij()?a11()6=0?A()?k?U?a11()?K?3?A()?-?B()=(bij()?b11()6=0?(b11()1,j 1)?a1j()=()a11()a11

3、()()a11().ai1()aij().=j+1()a11()0.ai1()aij()()ai1().=1+j(1)a11()0.aij()+1 ()ai1()aij()()ai1().?z?1 n3.2.3?A()=(aij()Pmn,rank(A()=rKA()?-uSmithIO/d1()d2().dr()0.0 Pmn?di()(i=1,r)1?“?di()|di+1()(i=1,r 1)y?r 0?a11()6=0d?n3.2.1?A()=B()=(bij()b11()1?“b11()|bij()(i=1,m;j=1,n)B()=?d1()A1()?d1()=b11()d1()?A

4、1()?k?eA1()6=0A1()E?LkB()=d1()d2()A2()d1(),d2()?1?“?d1()|d2()d2()?A2()?k?UY?L?SmithIO/n3.3.2?A()?SmithIO/?.(?1?“f5y)3.2.6?A()Pmn?SmithI O/?0?d1()d2()dr()?A()?Cf n3.3.3?A(),B()PmnKA()?B()?-?7?k?Cf 3.2.2+1 22+122=1+3(1)1 201 223+1(1)=1 20002=2+1(2)3+1()1 000 0 0 2 3(1)=3+2(1)1 000 00 0 2+3.3?f?A()?Cf?

5、d1(),d2(),dr()3E?S?)?g“?d1()=(1)e11(2)e12(s)e1sd2()=(1)e21(2)e22(s)e2sdr()=(1)er1(2)er2(s)ers?di()|di+1()(i=1,r 1)0 e11 e21 er10 e12 e22 er20 e1s e2s ers 3.3.2(j)eij(i=1,r;j=1,s)?A()?f X e?A()?Cf?d1()=1d2()=(1)d3()=(1)2(+1)2d4()=2(1)3(+1)3(2)KA()?f?,2,1,(1)2,(1)3,(+1)2,(+1)3,2?L5XJ?r?f?(?Cf?r(?Cf?g“

6、?f?g?p?73dr()?)?ggp?73dr1()?)Xd?X5 6?A()?4?f,2,1,(1)2,(1)3,(+i)3,(i)3K?Cfd4()=2(1)3(+i)3(i)3=2(1)3(2+1)3d3()=(1)2d2()=(1)d1()=1SmithIO/A()=10000 00(1)000 000(1)200 00002(1)3(2+1)30 000000 0 n3.3.6?A(),B()PmnKA()?B()?-?7?k?f n3.3.7?A()=?B()C()?KB()?C()?f?NA()?f n3.3.8e?A()?-u?A()=A1()A2().At()KA1(),A

7、2(),At()?f?N?A()?f3.4?q?n3.4.1A,B CnnXJ?3P,Q Cnn?I A=P(I B)QKA?B?qy?I A=PQ PBQ?X?PQ=I,A=PBQddQ=P1,A=PBP1 n3.4.1n?A?B?q?7?I A?I B?-n 3.4.2n?A?B?q?7?IA?IB k?Cf n 3.4.3n?A?B?q?7?I A?I Bk?f3.5?JordanIO/3.5.1/G?Ji=i1.i1i Cnini?Jordan?i?EdeZ?Jordan?|?Jordan/?X?1 1144 14i1i1i Ji?k?niA?iAuA?i=k?5?A?Ji?kw?L“

8、Jpi=fp(i)f0p(i)12!f00p(i)1(ni 1)!fni1p(i).fp(i)f0p(i).12!f00p(i).0.f0p(i)fp(i),p=1,2,?fp()=pI Ji=11.(i)ni PniniJi?Cf?d1()=dni1()=1,dni()=(i)nil?Ji?f?(i)ni?J=J1J2.JsJordan/?Ji CniniJordanJ?A?I J=I J1I J2.I Jsdn3.3.8?Jordan/?J?f?(1)n1,(2)n2,(s)ns n3.5.1?A Cnn,KA?Jordan/?q?Jordan/?Jordan?gS?A?y?A?f?(1)

9、n1,(2)n2,(s)ns,(1)?1,s?Uk?n1,ns?Uk?z?f(i)niAu?Jordan Ji=i10i1.10i Cnini,i=1,s?Jordan?Jordan/?J=diag(J1,J2,Js)(2)?f?(1)?J?A k?f J?A?qJordan/?(2)?A?JordanIO/ek,?Jordan/?J0?A?qK J0?A k?fd J0?J?Jordan?gS?y?5 n3.5.3?A Cnn,KA?q?7?A?f?g?3.5.1?A=1 2 61031 1 4?JordanIO/)?I A=+1 261311 4=1 1(1)2?f 1,(1)2Jorda

10、nIO/J=11 11 3.5.23.5.1?A3?_?P?P1AP=J)?PP=p1,p2,p3,KAp1,Ap2,Ap3=p1,p2,p31 0 00 1 10 0 1?Ap1=p1,Ap2=p2,Ap3=p2+p3p1,p2A?1?5A?)(I A)x=0?5A?=?1 1 0?T,=?3 0 1?T?p1=?e?p2=e?p3?|)(I A)p3=I 1 2 61031 1 4p3=2 2 61 1 31 1 3p3=301=p2p2?A?y(I A)p2=0p2?p1?5?(I A)p3=p2k)dp2=k1+k2=?k1+3k2k1k2?T,k26=0?|2 2 61 1 31 1

11、 3p3=k1 3k2k1k2k)Nw?k1=k2?|k)?k1=k2=1?p2=?2 1 1?T,p3=?2 0 1?T,P=1 2 211 001 1 A CnnK?3n?_?P?P1AP=J=J1J2.Js?Ji Cnini?JordanPP=?P1P2 Ps?Pi Cnniu?AP1AP2 APs?=?P1J1P2J2 PsJs?“?APi=PiJi,i 1,2,sPPi=hp(i)1,p(i)2,p(i)niiK?“?i=1,2,sAp(i)1=ip(i)1Ap(i)2=ip(i)2+p(i)1Ap(i)ni=ip(i)ni+p(i)ni1p(i)1?AAuA?i?A?dp(i)1?

12、g?p(i)2,p(i)ni.A?p(i)1?A?yp(i)2?aq/p(i)2?p(i)2?=?A?yp(i)3?ga?p(i)1,p(i)2,p(i)ni?53.6Caylay-Hamiltonn?“n3.6.1(Caylay-Hamiltonn)A Cnn,f()=det(I A)=n+a1n1+a2n2+an1+anA?A?“Kf(A)=An+a1An1+a2An2+an1A+anI=0y?(I A)=C1n1+C2n2+Cn1+Cn(I A)(I A)=f()I(IA)(C1n1+C2n2+Cn1+Cn)=In+a1In1+an1I+anIC1=I,C2 AC1=a1I,Cn ACn

13、1=an1I,ACn=anI0=AnC1+An1(C2 AC1)+A(Cn ACn1)ACn=An+a1An1+an1A+anI=f(A)3.6.1?A CnnXJ?3?“()?(A)=0,K()?A?z?“3.6.2A Cnn?kz?“g?$?1?“?A?“n3.6.3?q?k?“y?A?B?qK?3?P?B=P1APBk=P1AkP,k=1,2,u?“g()=g0m+g1m1+gm1+gm?kg(B)=g0Bm+g1Bm1+gm1B+gmI=g0P1AmP+g1P1Am1P+gm1P1AP+gmP1P=P1g(A)P?A?B k?z?“l?k?“.n3.6.4?A=diag(A1,As)?“?u?“?“y?Ai?“?mi()(i=1,s)du?“()(A)=(A1.As)=(A1).(As)XJ()?A?z?“K()7?Ai(i=1,s)?z?“l?mi()|()(i=1,s)d()?m1(),ms()?“?L5XJ()?m1(),ms()?“K(Ai)=0(i=1,s)l?(A)=0dA?“?m1(),ms()?“g?$?=?“n3.6.5?A Cnn,K A?“?A?1 n?Cf dn().:A=3 1 10 201 11I A=3110 2011 1=1000 2000(2)2?“?13?Cfd3()=(2)2 n3.6.6n?A?qu?7?A?“m()vk: