1、1习 习 习 习 题 题 题 题 一 一 一 一1. 设 为的任一特征值 , 则因 22 为 A = A22 O 的特征值 , 故 022 = . 即 = 0 或 2.2. A B , C D 时 , 分别存在可逆矩阵 P 和 Q , 使得 P 1 A P = B , Q 1 C Q = D . 令T = QO OP则 T 是可逆矩阵 , 且 T 1 CO OA T = QO OPCO OAQO OP 11 = DO OB3. 设 ix 是对应于特征值 i 的特征向量 , 则 A ix = i ix , 用 1A 左乘得 iii xAx 1= . 即iii xxA11 =故 1i 是 A 的特
2、征值 , i = 1,2, , n .4. ( 1) 可以 . AE = )2) (1) (1( + , =P 104003214, =2111 A PP .( 2) 不可以 . ( 3) =110101010P , =1221 A PP .5. ( 1) A 的特征值是 0, 1, 2. 故 A = ( b a ) 2 = 0. 从而 b = a . 又11111=aaaaAI = )223( 22 + a 将 = 1, 2 代入上式求得 A = 0.( 2) P = 101010101.6. AI = )1()2( 2 + , A 有特征值 2, 2, 1. = 2 所对应的方程组 ( 2
3、 I A ) x = 0 有解向量 p 1 = 041, p 2 = 401 = 1 所对应的方程组 ( I + A ) x = 0 有解向量 p 3 = 001令 P = ( p ,1 p ,2 p 3 ) = 140004111, 则 P 1 = 4416414030121 . 于是有A 1 0 0 = P 1221 0 01 0 0P 1 = 124122440230121224311 0 01 0 01 0 01 0 01 0 01 0 01 0 0.27 ( 1) AI = )1(2 + = D 3 ( ) , I A 有 2 阶子式 1721 11 = 4 4 不是 D 3 ( )
4、 的因子 , 所以 D 2 ( ) = D 1 ( ) = 1, A 的初等因子为 1, 2 . A 的 Jord a n标准形为 J = 000100001设 A 的相似变换矩阵为 P = ( p 1 , p 2 , p 3 ) , 则由 A P = P J 得 =23211pA pA ppA p0 解出 P = 241231111;( 2) 因为 ) ,2()1()( 23 = D 1)()( 12 = DD ，故 A J = 200010011设变换矩阵为 P = ( 321 , ppp ) , 则 =+=33212112 pA pppA ppA p P = 502513803( 3)
5、) ,2()1()( 23 += AID ,1)(2 += D 1)(1 =D . A 的不变因子是,11 =d ,12 += d )2) (1(3 += d A J = 211因为 A 可对角化，可分别求出特征值 1 ， 2 所对应的三个线性无关的特征向量：当 = 1 时，解方程组 ,0)( =+ xAI 求得两个线性无关的特征向 ,1011 =p =0122p当 = 2 时，解方程组 ,0)2( = xAI 得 =1123p , P = 101110221( 4) 因 +=41131621 AI 2)1(11 , 故 A J = 10111设变换矩阵为 P = ),( 321 ppp ,
6、则 +=3232211ppA ppA ppA p21 , pp 是线性方程组 0= xAI )( 的解向量 ， 此方程仴的一般解形为 p = +tsts 3取 =0111p ,=1032p 为求滿足方程 23)( ppAI = 的解向量 3p , 再取 ,2 pp = 根据3tsts3113113622 tstss00033000311由此可得 s = t , 从而向量 T3213 ),( xxx=p 的坐标应満足方程 sxxx =+ 321 3取 T3 )0,0,1( =p , 最后得 P = 0100011318. 设 f ( ) = 432 2458 + . A 的最小多项式为 12)(
7、 3 += Am , 作带余除法得 f( ) = ( 149542 235 + ) , )( Am + 103724 2 + , 于是f ( A ) = IAA 103724 2 + = 346106195026483.9. A 的最小多项式为 76)( 2 += Am , 设 f ( ) = 3 72 91 91 22 234 + , 则f ( ) = )()52( 2 Am+ + 2+ . 于是 f ( A ) 1 = 1)2( + IA . 由此求出 f ( A ) 1 = 32 1723110. ( 1) I A = +41131621标准形 2)1(00010001 , A 的最小多
8、项式为 2)1( ;2) )1) (1( + ; ( 3) 2 .11. 将方程组写成矩阵形式 :=321321188034011ddddddxxxtxtxtx, =321xxxx ,=txtxtxtdddddddd321x , A =188034011则有 J = P A P 1 = 100010011, . 其中 P = 124012001.令 x=P y , 将原方程组改写成 : ,dd Jyy =t 则=+=3321211ddddddytyyytyyty解此方程组得 : y 1 = C 1 e t + C 2 T e t , y 2 = C 2 e t , y 3 = C 3 e t
9、. 于是x = P y = +tttttttc)t(cc)t(cctccee24e4e12e2ee3212121 .12. ( 1) A 是实对称矩阵 . AI = 2)1) (10( , A 有特征值 10, 2, 2.当 = 10 时 . 对应的齐次线性方程组 ( 10I A ) x = 0 的系数矩阵4542452228 000110102由此求出特征向量 p 1 = ( 1, 2, 2)T , 单位化后得 e 1 = ( 32,32,31 ) T .当 = 1 时 , 对应的齐次线性方程组 ( I A ) x = 0 的系数矩阵442442221 000000221由此求出特征向量 p
10、 2 = ( 2, 1, 0)T , p 3 = ( 2, 0, 1)T . 单位化后得 e 2 = ( 0,51,52 ) T ,e 3 = ( 53 5,53 4,53 2 ) T . 令 U =535032 5345132 5325231, 则 U 1 A U = 1110.( 2) A 是 H e r mi tH H H 矩阵 . 同理可求出相似变换矩阵U =212121 2i2i2i 21210, U 1 A U = 220.13. 若 A 是 H e r mi tH H H 正定矩阵 , 则由定理 1.24可知存在 n 阶酉矩阵 U , 使得U H A U =n21, i 0, I
11、 = 1, 2, , n . 于是 A = Un21U H= Un21U H Un21U H 令B = Un21U H则 A = B 2 . 反之 , 当 A = B 2 且 B 是 H e r mi tH H H 正定矩阵时 , 则因 H e r mi tH H H 正定矩阵的乘积仍为 H e r miH H H t正定矩阵 , 故 A 是 H e r mi tH H H 正定的 .14. ( 1) ( 2). 因 A 是 H e r mi tH H H 矩阵 , 则存在酉矩阵 U , 使得 U H A U = d i ag( n , 21 )令 x = U y , 其中 y = e k .
12、 则 x 0. 于是 x H A x = y H ( U H A U ) y = k 0 ( k = 1, 2, , n ) .5( 2) ( 3). A = U d i ag( n , 21 ) U H = U d i ag( n , 21 ) d i ag( n , 21 ) U H令 P = d i ag( n , 21 ) U H , 则 A = P H P .( 3) ( 1). 任取 x 0, 有 x H A x = x H P H P x = 22P x 0.习 习 习 习 题 题 题 题 二 二 二 二1. 1x = 01i42i1 + = 7+ 2 , 2x = 1i )4i
13、 (4)2(i )1i ) (1( 2 + = 23 ,x = max 1i42i1 , + = 4.2. 当 x 0 时 , 有 x 0; 当 x 0 时 , 显然有 x = 0. 对任意 C , 有x = xnkkknkkk = = 1212 .为证明三角不等式成立 , 先证明 M i n k ows k i 不等式 :设 1 p , 则对任意实数 x k , y k ( k = 1, 2, , n ) 有pnkpkk yx11)( =+ =+ nkppknkppk yx1111)()(证 当 p = 1 时，此不等式显然成立 . 下设 p 1, 则有=+nkpkk yx1 = + nkp
14、kkknkpkkk yxyyxx1111对上式右边的每一个加式分别使用 H H H H lder不等式 , 并由 ( p 1)q = p , 得=+nkpkk yx1 qnkqpkkpnkpkqnkqpkkpnkpk yxyyxx11)1(1111)1(11)()()()( =+= qnkpkkpnkpkpnkpk yxyx111111) ()() ( =+再用 qnkpkk yx11)( =+ 除上式两边 , 即得 M i n k ows k i 不等式 .现设任意 y = ( n , 21 ) T C n , 则有=+=+ nkkkkyx12 = =+nkkkk12)( =+nkkkkk1
15、2)( =+ nkjknkkk1212 ()( = yx + .3. ( 1) 函数的非负性与齐次性是显然的 , 我们只证三角不等式 . 利用最大函数的等价定义 :max(A , B ) = )(21 baba +max( ), ba yxyx + max( bbaa yxyx + , )= )(21 bbaababa yxyxyyxx + )(21 babababa yyxxyyxx += )(21)(21 babababa yyyyxxxx += max( ba xx , ) + max( ba yy , )6( 2) 只证三角不等式 .k 1 ayx + + k 2 byx + k 1
16、ax + k 1 ay + k 2 bx + k 2 by= ( k 1 ax + k 2 bx ) + ( k 1 ay + k 2 by ) .4. 218132i453i11m+=+=A ;66132i453i1 222222F =+=A ; 15m =A ;=1A 列和范数 ( 最大列模和 ) = 27 + ; A = 行和范数 ( 最大行模和 ) = 9 ;5. 非负性 : A O 时 S 1 A S O , 于是 m1 A SSA = 0. A = O 时 , 显然 A = 0;齐次性 : 设 C , 则 = m1 )( SASA m1 A SS = A ;三角不等式 : m11m
17、1 )( B SSA SSSBASBA +=+=+BAB SSA SS +=+ m1m1 ;相容性 : m11m1 )( B SA S SSSA BSA B = m1m1 B SSA SS = A B .6. 因为 I n O , 所以 nI 0.从而利用矩阵范数的相容性得 :nnn III = nI nI , 即 nI 1.7 . 设 A = ( A i j ) C nn , x = T21 ),( n C n , 且 A = i jji a,m a x , 则 =i kki kA x a1 i kki ka = k ii kk a n A kk = mA 1x ; =i kki kA x2
18、2 a i k ki ka2 = i kka22 = n A 2x n A =mA 2x .8. 非负性与齐次性是显然的 , 我们先证三角不等式和相容性成立 . A = ( a i j ) , B = ( b i j ) C nm ,C = ( c s t ) C ln 且 A = i jji a,m a x , B = i jji a,m a x , C = s tts c,m a x . 则MBA + =maxm , n i ji jji ba +,m a x maxm , n )(m a x, i ji jji ba + max m , n ( A+ B )= maxm , n A +m
19、axm , n B = MM BA + ;MA C = maxm m m m , l k k ti kti ca,m a x maxm m m m , n m a x, k k ti kti ca maxm m m m , n m a x 22, k k tk i kti ca( M i n k ows k i 不等式 )= maxm m m m , n n A C maxm m m m , n maxn , l A C = MM CA .下证与相应的向量范数的相容性 .设 x = T21 ),( n C n , d = km a x k , 则有 =i kki kaA x 1 i kki k
20、a = k ii kk a )( kkna = n A kk maxm m m m , n A kk = 1M xA ;72A x = i k ki ka2 i kki ka2)( i k kki ka )(22 ( H H H H l d e r 不等式 )= kki ki ka22 mn A2x maxm m m m , n A 2x = 2M xA ;m a x1= nkki kiA x a =nkki ki a1m a x m ax 22 k kk i kia m a x 22 ndnai= n A D maxm m m m , n A D = xA M .9. 只证范数的相容性公理及
21、与向量 2 范数的相容性 . 设 A = ( a i j ) C nm , B = ( b s t ) C ln ,x = T21 ),( n C n 且 A = i jji a,m a x , B = s tts b,m a x , 则= nkk ti kltm m m miA B11,1Gm a x baml m a x, k tki kti baml m a x 22, kk tki kti baml ( M i n k ows k i 不等式 ) ml n ab= ) ( bnlamn = GG BA . = = m m m minkki kA x1212a i kki ka2)( i
22、 k kki ka )(22 （ H H H H l d e r 不等式） i kkna )(22 = mn A2x= 2G xA .10. 利用定理 2.12得 122H2 = nIUUU .11.A 1 =0110211214321c ond 1 ( A ) = 225255111 =AA ; c ond ( A ) = 10251 = AA .12 设 x 是对应于 的特征向量 , 则 A xx m m m mm m m m = . 又设 v 是 C n 上与矩阵范数 相容的向量范数 ,那么 vm m m mvm m m mvm m m m xAxx = vm m m m xA因 vx
23、0, 故由上式可得 m m m m m m m mA m m m m m m m mA .习 习 习 习 题 题 题 题 三 三 三 三1. 2cc ) (2( += AI , 当 c =)( 1 时 , 根据定理 3.3, A 为收敛矩阵 .2. 令 S )N( = =N0)(kkA , )(l i m NN S+ = S , 则 0)(l i ml i m)()()( =+ + kkkkk SSA .8反例 : 设 A )( k =k0001k , 则因 + =01k k发散 , 故 + =0)(kkA 发散 , 但 )(l i m kk A+ = O .3. 设 A = 6.03.0 7
24、.01.0 , 则 )( A =A 行和范数 = 0.9 1, 根据定理 3.7,+= 0 6.03.07.01.0kk= ( I A ) 1 = 93 7432 .4. 我们用用两种方法求矩阵函数 e A :相似对角化法 . 22 a += AI , a-a i,i=当 = i a 时 , 解方程组 ( i a A ) x = 0 , 得解向量 p 1 = ( i , 1)T .当 = i a 时 , 解方程组 ( i a + A ) x = 0 , 得解向量 p 2 = ( i , 1)T . 令P = 11 ii , 则 P 1 = i1 i1i21 , 于是 e A = P aa i0
25、 0i P 1 = aa a-a c oss i n s i nc os .利用待定系数法 . 设 e = ( 2 + a 2 ) q ( ) + r ( ) , 且 r ( ) = b 0 + b 1 , 则由=+ aaabbabbi10i10eiei b0 = c os a , b 1 = a1 s i n a . 于是e A = b 0 I + b 1 A = c osa 11 + a1 s i n a a a = aa aa c oss i n s i nc os .后一求法显然比前一种方法更简便 , 以后我们多用待定系数法 . 设f ( ) = c os , 或 s i n 则有=+
26、a-abbaabbs i n iis i n ii1010 与=+aabbaabbic osiic osi1010由此可得 = aabbs i n ii010 与=0ic os10bab故 ( a2i s i n i a ) A = 0i s i n i i s i n i0 a a = s i n A 与 ( c osi a ) I = aa c o s i0 0c o s i = c osA .5. 对 A 求得P = 013013111, P 1 = 24633011061 , P 1 A P = 211根据 p 69 方法二 ,e A t = P d i ag(e t ,e t ,e
27、t2 ) P 1 = +tttttttttttttte3e3e3e30e3e3e3e30ee3e2ee3e4e661222 t9s i n A = P d i ag(s i n ( 1),s i n 1,si n 2) P 1 = 01s i n601s i n6001s i n42s i n21s i n22s i n42s i n616. D 3 ( ) =10010011= 2)1( , D 2 ( ) = D 1 ( ) = 1, A J = 000010011. 现设r ( , t ) = b 0 + b 1 + b 2 2 , 则有=+=+1e2e021210btbbbbbtt b
28、 0 = 1, b 1 = 2e t t e t 2, b 2 = t e t e t + 1. 于是e tA = r ( A , t ) = b 0 I + b 1 A + b 2 A 2 = I + ( 2e t t e t 2) 100100011+ ( t e t e t + 1) 100100111= +tte001e101ee1eettttt同理 , 由=+=+1s i n2c os021210bttbbtbbb b 0 = 1, b 1 = t s i n t + 2cost 2, b 2 = 1 t s i n t c os t . 将其代入c os A t = b 0 I +
29、 b 1 A + b 2 A 2 , 求出 c os A t = tttttttc os001c os10c oss i n11c osc os7. 设 f ( A ) = + =0kk Aka , S N = =Nkk A0ka . 则 f ( A ) = NN S+ l i m 并且由于( S N ) T = T0)( =Nkkk Aa = =Nkkk A0T )(a 所以 , f ( A T ) = T)(l i m NN S+ = f ( A )T .8, ( 1) 对 A 求得P =1111, P 1 = P , J =1111111则有e tA = Pttttttttttttttt
30、teeee2eee6e2e232eP 1 =ttttttttteee2e60eee200ee000e232ttttttt10s i n A t = Ptttttttttttttttts i nc oss i ns i n2c oss i nc os6s i n2c oss i n232P 1 =tttttttttttttttts i nc oss i n2c os6s i nc oss i n2s i nc oss i n232c os A t = Pttttttttttttttttc oss i nc osc os2s i nc oss i n6c os2s i nc os232P =ttt
31、tttttttttttttc oss i nc os2s i n60c oss i nc os200c oss i n000c os232( 2) 对 A 求出P = P 1 =0100100000100001, J =010212则有e A t = P11eee222tttttP 1 =100010000e000ee222ttttts i n A t = P002s i n2c os2s i ntttttP 1 =0000000002s i n0002c os2s i ntttttc os A t = P1012c os2s i n2c osttttP 1 =10000100002c os0
32、002s i n2c ostttt9. ( 1) s i n 2 A + c os 2 A = )e( ei21 ii AA 2 = )( e21 ii AA e + 2= )eee( e41)eee( e41 i2i2i2i2 OOAAOOAA + = e O = I( 2) s i n ( A + 2 I ) = s i n A c os ( 2 I ) + c os A s i n ( 2 I )= s i n A I !21 ( 2 I ) 2 + !41 ( 2 I ) 4 + c os A 2 I !31 ( 2 I ) 3 + !51 ( 2 I ) 5 = s i n A 1
33、!21 ( 2 ) 2 + !41 ( 2 ) 4 I + c os A 2 !31 ( 2 ) 3 + !51 ( 2 ) 5 I= s i n A c os2 + c os A s i n 2 （ 3 ） 的证明同上 .( 4) 因为 A （ 2 i I ） = ( 2 i I ) A , 所以根据定理 3.10可得e IA i2+ = e A e I i2 = e A I + ( 2 I ) + !21 ( 2 i I ) 2 + !31 ( 2 i I ) 3 + 11= e A 1 !21 ( 2 ) 2 + !41 ( 2 ) 4 + i 2 !31 ( 2 ) 3 + !51 (
34、 2 ) 5 I= e A c os 2 + i s i n 2 I = e A此题还可用下列方法证明 :e IA i2+ = e A e Ii2 = e A Pi2i2 i2eee P1 = e A P I P 1 = e A用同样的方法可证 : e IA i2 = e A e I i2 .10. A T = A , 根据第 7 题的结果得 ( e A ) T = e TA = e A , 于是有e A ( e A ) T = e A e TA = e AA = e O = I11. 因 A 是 H e r m( iH H H A ) H = i A H = i A , 于是有e Ai (
35、e Ai ) H = e Ai e Ai = e O = I12. 根据定理 3.13, A 1 tt Aedd = e A t , 利用定理 3.14得 t A0 de = t AA0 1 dedd = A 1 dedd0 At = A 1 ( e A t I ) .13. tdd A ( t ) = tt tt s i nc os c oss i n , tdd ( d e t A ( t ) ) = tdd ( 1)= 0, d e t ( tdd A ( t ) ) = 1,A 1 ( t ) = tt tt c oss i n s i nc os , tdd A 1 ( t ) =
36、tt tt s i nc os c oss i n14. t A0 d)( =00d30de2deddede002002002tttttt=+002301ee1 311ee)1( e212232ttt ttttt15. 取 m m m m = 2, A ( t ) = ttt02 , 则A 2 ( t ) = +2 2340 t ttt , tdd ( A ( t ) ) 2 = +t ttt 20 234 23 2 A ( t ) tdd A ( t ) = +t ttt 20 224 23 .困为 += 21 ) () (dd)() () (dd) ()()(dd) (dd m m m m
37、m m m m AAAAAAAAA ttttttttttttt m+ )(dd) ( 1 ttt AA m m m m 所以当 ( tdd A ( t ) ) A ( t ) = A ( t ) tdd A ( t ) 时 , 有)(dd) ()(dd) ()(dd) () (dd 111 ttttttttttt AAAAAAA m m m mm m m mm m m mm m m m += = m m m m A ( t ) )(dd1 tt Am m m m 16. ( 1) 设 B = ( i jb ) nm m m m , X = ( i j ) m m m mn , 则 B X =
38、( =nkk ji k1b ) m m m mm m m m , 于是有t r ( B X ) = =+ nkk m m m mm km m mnkk jj knkkk11111 bbb 12i jB X )t r ( =j ib ( i = 1,2, , n ; j = 1,2, , m m m m ) =m nm m mnm m m mB XXbbbb111 1)( t r (dd = TB由于 B X 与 TTT)( BXB X = 的迹相同，所以TTT ) )( t r (d d) )( t r (d d BB XXBXX =( 2) 设 A = ( i ja ) nn , f = t
39、 r ( A XX T ) , 则有=n m m m mm m m mnX111 1T ， A X =kk m m m mn kkkn kk m m m mkkkkaaaa1111f = +l kk m m m ml kl m m m ml kk jl kl jl kkl kl aaa 11) ()( += =kk jl kl k i jl jk jl ki jl jl kk jl kl ji ji j aaaf = +kl jl ikk ji k aam m m mni jX =ffdd = XAAXAA X )( TT +=+17. 设 A = ( i ja ) m m m mn , 则 F ( x ) = ( =nkk nknkknkkk1211 , aaa1k ) , 且AdFFFxFn nnnnnn=aaaaaaaaa2122 22 111 21 121ddddddd18.( ) =ttttttttttttttttttA tA t A222222222e4e3e3e6e3e6e2eee4ee2e2eee2ee4ee在上式中令 t = 0, 则有 A = =133131113e OA19.
