矩阵论习题
(方保镕编著)习题 5
习题 5
(k)(k)1. 证:由于,再利用矩阵范数的性质 limA,A,limA,A,0,,,,kk
(k)(k) ~ A,A,A,A
()k(k)所以有~即. limA,AlimA,A,0,,,,kk
2. 证:因为
(k)(k)(k)(k) (,A,,B),(,A,,B),(,A,,A),(,B,,B)kkkk
kkkk()()()(),,A,,A,,B,,B,,A,,A,,A,,A,,B,,B,,B,,Bkkkkkkkk
kk()(). ,,A,A,,,,A,,B,B,,,,Bkkkk
(k)(k)利用~~以及lim,,,,0~limA,A,0limB,B,0k,,,,,,kkk
lim,,,,0~~有界~知 ,,kkk,,k
(k)(k)lim,(A,,B,(,A,,B),0kk,,k
(k)(k)lim,(A,,B),,A,,B故有. kk,,k
3. 解:,1,由于~从而是收敛矩阵. A,0.9A1
515,2,由于的特征值为~~故~,,,,,,(A),,1A12626故的收敛矩阵. A
,,,,,a,,2a4. 解:由于的特征值为~于是~故,(A),2aA231
111a,,,a,,(A),1当即或时~为收敛矩阵. A222
1
,NN(k)()()()NkkS,AS,PAQ,P(A)Q5. 证:记~ ,于是 ,,,k0,0,0,kk
N,,(k)(k)(k)(N)PAQ,limS,P(limA)Q,PSQ,P(A)Q,,,,,NN,,,k0k0k0,,
,,,(k)(k)(k)PAQAA即也收敛.如果绝对收敛~则收敛.又由于,,,k,0k,0k,0
(k)(K)(k),其中是与无关的正数~由比较判别法知PAQ,PAQ,,Ak
,,(k)(k)PAQPAQ收敛~故也绝对收敛. ,,k,0k,0
17,,6. 解:,1, 设 A,,,,1,3,,
,1kx,,,,,2可求得的特征值为~所以~由幂级数的收,(A),2A,122k,1k敛半径为
2ak(,1)kr. ,lim,lim,12k,,k,,akk,1
,1kA因 , 知矩阵幂级数发散. ,(A),2,r,2kk,1
1,8,,,,,3,,5,2,设~可求得的特征值为~~所以BB,12,,,21,,
,kk,,x,B,5.又因幂级数的收敛半径 ,k6,0k
k,1ak6kr ,lim,lim,6kk,,k,,ak,16k,1
,kkB即有,(B),r~故矩阵幂级数绝对收敛. ,k6k,0
2
,0.10.7,,kA7. 解:设~由于~故矩阵幂级数收A,0.9,1A,,,,,0.30.6k,0,,
47,,2,1敛~且其和为. (),,IA,,393,,
8. 证:因~所以有 A(2,jI),(2,jI)A
111,,A234A,2,jIA2,jI,,,,,,?eI(,2jI)(2,jI)(2,jI)(2,jI) e,ee,,2!3!41,,
,,1111,,,,A2435 ,,,,?,,,,?e1(2,)(2,)j2,(2,)(2,)I,,,,,,2!4!3!5!,,,,,,
AA,,ecos2,,jsin2,I,e=.
又因~所以有 A(2,I),(2,I)A
sin(A,2,I),sinAcos(2,I),cosAsin(2,I)
1111,,,,2435,,,?,,,,?sinAI(,2I)(,2I)cosA2,I(2,I)(2,I)= ,,,,2!4!3!5!,,,,
1111,,,,2435sinA1,(,2),(,2),?I,cosA2,,(2,),(2,),?I= ,,,,2!4!3!5!,,,,= sinAcos2,,cosAsin2,,sinA
9. 证:因为
TT11,,A23e,I,A,A,A,?,,,,2!3!,,
T2311TTTA,A,,,,,I,A,A,A,?,e,e2!3!
TTAAAAA,A0A所以有.故为正交阵. e,,ee,e,e,e,e,I
HHHjAjA,,jA,jA10. 证:因为, 于是有 ,,,,e,e,e,e
3
HjAjAjA,jA0 ,,ee,ee,e,IjA故为酉阵. e
3211. 解 :,1, , ,,f,,,I,A,,,,
,2,由Cayley-Hamilton定理知
3232,,fA,A,A,0 ~即A=A
从而有
4332A=A?A=A=A
5432A=A?A=A=A
…………………
111A23n故 e,I,A,A,A,?,A,?2!3!n!
111,,2,I,A,A,,?,,?,,2!3!n! ,,
2,,,I,A,e,2A
111k352k,1sin,,1A,A,A,A,?,,A,? ,,3!5!21!k,
,,111k2,,,A,A,,,?,,,,1,?,,3!5!2,,k,1! ,,
2,,,A,sin1,1A
12. 解: ,,,,,,,I,A,,,1,,1,,2,0
,,,1,,1,,2求得A的特征值为~~~于是存在可逆阵 312
1,110,11,,,,1,,,,,1CC,,310,033 ~ ,,,,6,,,,310642,,,,
4
,1,,
,,,1CAC,1使得.再根据矩阵函数值公式为 ,,
,,2,,
22,12,1,,6e4e,3e,e2e,3e,e
1,,,112,1,1,1AeCdiage,e,eC03e3e3e3e,,,,,, ,,6,1,1,,03e3e3e3e,,,,
tA,tt2t,1,,e,Cdiage,e,eC
ttt,ttt,t222,,6e4e,3e,e2e,3e,e
1,,t,tt,t03e3e3e3e,,, ,,6t,tt,t,,03e3e3e3e,,,,
,1,,,,sinA,Cdiagsin,1,sin1,sin2C
sin24sin2,2sin12sin2,4sin1,,1,,006sin1 = ,,6,,06sin10,,
11,,
,,11,1,,13. 解:,1,对A求得C~使得CAC,,J~所以有 ,,11
,,1,,
0000,,
,,1000,,1,1,,,,,lnln ACJC,1002,,
11,,,10,,32,,
J1121,,,,,,1,2,~其中~ A,J,J,21,,,,,,J02012,,,,,,
于是有
1,,01,,ln2,,ln,J ~ ln,J122,,00,,,,0ln2,,
5
1,,ln200,,2lnJ,,,,1ln200 . ln,,A,,,,lnJ,2,01,,
,,0,,
,t,2t,t,2t,,2eeee,,At14. 解:,1,, e,,,,t,2t,t,2t2e2ee2e,,,,,,
,35,2011,11514,,,,,,t,2t3teee,,,,,,Ate,,3,52,01,1,514,2,, ,,,,,,61510,,,,,,3,520,1414514,,,,,,
1,,22,,,22121tttt,,,,2,,At222,2t,,,,,,,4421,3,, ettttte,,2,,,,,,,,,,854232121tttttt
,,,,
1,51063050,,,,,,
,,,,,,At,2t,2t,3t000000010e,e,te,e,4,. ,,,,,,
,,,,,,021000020,,,,,,,
15. 解:
cosA,I,cos1,1A,sinA,sin1A,,,,,
,1, 2A,,e,I,e,1A
2A,,,,cosA,cos1IsinA,sin1Ie,eI,2,~~,
2Ae,I,3,~~. cosA,IsinA,A
5I,AA,I1000100016. 解:,1,, A,,544
3e,1e,3e,1,,
,,A3333e,ee,,e,,2,, ,,
,,3113e,e,,e,,
6
,,,33,,,,A666arcsin,,3,, ,,4,233,,,,,363,,
15,8,,15,12,4,. ,,I,A,,A,A,,,8172110,,
,sintcost,,d17. 解: ,,At,,,,cost,sintdt,,
,sint,cost,,ddd,1. ,,,,,,At,,At,0,At,1,,cost,sintdtdtdt,,
2,,tt18. 解:时取~则 ,,At,m,2,,0t,,
43232,,,,tt,t4t3t,2td22 ,,,, At,,At,,,,,2dt0t02t,,,,
32,,4t2t,2td,,,, 2AtAt,,,dt02t,,
dd2可见~,,,,. ,,At,2AtAtdtdt
19. 解:两边对t求导数~得
5cos5t,3cost10cos5t,2cost5cos5t,cost,,1,,AcosAt,5cos5t,cost10cos5t,2cost5cos5t,cost ,,4,,5cos5t,cost10cos5t,2cost5cos5t,3cost,,令t=0~并注意到~得 cos0,I
221,,
,,A,131. ,,
,,122,,
7
20. 解:这是数量函数对矩阵变量的导数.设,,~则A,aijmxn
mn22T,,a,trAA=. ,,fA,A,,stF,,11st
,f,,,2ai,1,2,?,m;j,1,2,?,n又因为~所以 ij,aij
,,df,f,,,,,,2a,2A . ijm,n,,dA,aij,,m,n
dTTYX,yx,yx,?,yx21. 解:由于~再由~,,XAX,2AX1122nndX
ddfxdc,,T知~而~因此. ,,YX,Y,0,2AX,YdXdxdX
n,,,,,,22. 证:,1,设~则~于是B,b,X,xBXbx,,,,ijijikkjm,nn,mk,1,,m,m有
nnn
,,trBX,bx,?,bx,?,bx ,,,11kkjkkjmkkm,,,111kkk
,tBX,,r ,,,bi,1,2,?,n;j,1,2,?,mji,xij
b?b,,11m1d,,T,,,,trBX,??,B ,,dX,,b?b1nmn,,
TTT注意到BX与,BX,=XB有相同的迹~所以
ddTTT ,,,, ,,,,trXB,trBX,BdXdX
T,,,,A,a,X,x,f,tr,,XAX,2,设 ijij,,nnnm
则有
8
nn,,axax?,,1kk11kkm,,xx?,,11n1k,1k,1,,,,T ,,XAX??,??,,,,nn,,,,xx?1mnm,,axax?,,nkk1nkkm,,k,1k,1,,
nnnnnn
f,xax,?,xax,?,xax ,,,,,,11eekkejekkjemekkm,1,,11,1,1,1ekekek
nn,,f,, ,xax,,ejekkj,,,,xxe,1k,1,,ijij
nnn,,,x,,,,,ej, = axxax,,,,,,,,,ekkjejekkj,,xx,,11,1,,,,ekk,,ijij,,
nn
ax,ax = ,,jkkjekej,1,1kk
,,df,fTT,,,,,,AX,AX,A,AX.,,dX,xij,,n,m
TT,,,,f,X,uAX,u23. 证:设~因为A,A~所以
TTT,,f,XAX,2AuX,uAu
df利用第21题的结果可得,,. ,2AX,2Au,2AX,udX
aA,,24. 证:设~记的代数等子式为~将detA按第行A,aiijijijn,n
展开~得
detA,aA,?,aA,?,aA~ i1i1ijijinin
,f,,,Ai,j,1,2,?,n所以~从而有 ij,aij
dfTTT,1,1,,,,,, ,,,, ,A,adjA,detAA,detAAij,nndA
其中adjA是A的伴随矩阵.
9
T25. 解:设,,~,,.由于ABA的第k行第k列元素B,bA,aijijn,nn,m
nn
aba为~所以 ,,sksttk,,11st
mnn,,T ,,,,fAtrABAaba,,,,,,,sksttk,,,111,,kst
mnnnn,,,aba,aba,,,,,,,sksttksjsttjk,,,111s,,11tst,,,,k,j
mnnn,, ,aba,aba,,,,,,sksttk1j1ttjk,,11s,,11tt,,,,k,j
nn
,aba,?,aba,,ijittjnjnttjt,1t,1故
n,f,,,ab,?,ab,ba,ab,ab,,,11,1,,1,,1,,1,jiijiiittjijijijii,a,1t,,ij nn
,?,ab,ba,ba,,njniittjbisj,1,1ts最后得
nn,,df,f,,,,,,,,ba,ba,,,,,ittj,sisj,,dA,a ts,1,,1,,,ijnmnm,,,,nm,
T,BX,BX
dfTf,ABA特别地~当B是对称矩阵时~,当A为列向量时~~,2BAdA
dfT且. ,BA,BAdA
T,,26. 解:设 ~, 由于 A,a,,X,x,x,?,xij12nm,n
Tnn,,F(X),AX,ax,?,ax,,,,1kkmkk,,k,1k,1 Tnn,,TG(X),(AX),ax,?,ax,,,,,1kkmkkk,1k,1,,
所以
10
,F,GT,a,?,a,,a,?,a,,,,,11imiimi,x,xii T,,dF,F,FT,,,,故,,?,,a,?,a,?,a,?,a1111mnmn,,dX,x,x1n,,
a?a,,111n,,dF,F,F,,,,,,?,,??,A T,,,,,x,xdX1n,,,,a?am1mn,,
a?a,,T11m1,,dG,G,G,,T,,,,?,,??,A ,,,,dX,x,xn1,,,,a?a1nmn,,
27. 解:因为
2TTTTTTT ,,,,,,fx,AX,b,AX,bAX,b,XAAX,XAb,bAX,bb2
故由上两题的结果得
dfTTTTTT ,,,,,2AAX,Ab,bA,2AAX,AbdX
T,,,f,f,fT1,,,,28. 解:因为,,?,n,a,?,a,, 1inii,,,x,x,xiii,,
故有
T,,dfff,,T,,,,,,?,,,,?,,. 1n,,dXxx,,1n,,
11,,2tt3ect1ectc,,,,,,111213,,23,,,t2tA,,tdt,,e,ce,cc29. 解: 212223,,,32,,t,ccc313233,,2,,
11
11,,2,11,,e,,23,,,12 ,,,,,110Atdtee,,,3,,002,,,,
22224tt,,etet2t,,22d,,,t2tAxdx2te2e0 ,,, ,,,,,0dt,,2,,3t00,,
,12,,1,3i,,30. 解:设~A的特征值为~相应的两个特A,,,2,21,,
征向量为
11,,,,
,,,,1,3i1,3i,, ,~,
,,,,22,,,,作矩阵
11,,
,,1,3i1,3iP,
,,22,,
ix利用欧拉公式~则 e,cosx,isinx
it3,,e0At,1,ePP,,it,30e,,,,
12,, ,cos3tsin3tsin3t,,33,,,,21,,sin3tcos3t,sin3t,,33,,故
2,,tsin3,,0,,3At,,xt,e,,,. ,,11,,,,t,tcos3sin3,,3,,
12
35,,31. 解:设~它的特征值为~对应的两个线,,3,5iA,,,,53,,
1i,,TT,,,,,,1,i,,,i,1性无关的特征向量为~作可逆矩阵~从而有 P,,,i1,,
,35it,,11,cos5sin5,,iitt0,,,,,,e1A3t ,,ee,,,,,,,,,,,35it1,1,sin5cos5iitt20e,,,,,,,,
故
,,t0,,e,,,AtAt,,,,Xt,e,ed,,,,,,,010,,,,
t,,sin5tcos5tcos5,sin5tsin5,,,,3t3t4,, ,e,eed,,,,,,0cos5t,sin5tcos5,cos5tsin5,,,,,,
,,sin5tat,,,,3t3t,e,e,其中,,,,,,cos5tbt,,,,
,4t,,e4,,,,at4cos5t5sin5tcos5t,,,,,,4141,,
,4t,,e5,,4sin5t,5cos5t,sin5t,,,,4141,,
,4t,,e4,,,,bt,,,4cos5t,5sin5t,sin5t,,4141,,
,4t,,e5,,,,4sin5t,5cos5t,cos5t,,4141,,
,,,21011,,,,
,,,,,,Ab,,t,,420,,2,,x0,1 32. 解: 设 ~~ ,,,,,,t,,,,,,101,1e,1,,,,,,
32,,,,,,,det,I,A= , 由C-H定理知 ,,,
324252 A=A~A=A~A=A~……~
从而有
111234AteIAtAtAtAt,,,,,,,, ,,,,,,?2!3!4!
13
111,,2342,I,tA,t,t,t,A?,,2!3!4!,,
1,2tt0 ,,
,,t2,I,tA,e,1,tA,,4tt0,,,,ttt,,12teet1e,,,,,,故
,,1,,tt,,,,At,A,At,,()()()()2xt,exo,eb,d,,exo,dt ,,,,,,00,,,,0,,,,
,,1t1,,,,,,
,,,,,,,,Ate1,2t,1 = . ,,,,,,,,t,,,,,,,,10(t,1)e,,,,,,,,
14