南航双语矩阵论第一章习题答案2016年版.pdf

1 Solution Key (chapter 1) Exercise 2. The main point is to show that this set is not closed under multiplication. Take2S,2 22. But2S. If2S, then there are rational numbers a and b, such that 223ab.( It is clear that0a and0b .) This will lead to 22 423 6 2 ab ab The right hand is a rational number and the left hand side is an irrational number. This is impossible. Thus, S is not closed under multiplication. Hence, S is not a field. Exercise 7. zxyx )()()()(zxxyxx zxxyxx)()( z0y0 zy Exercise 12 It is a vector space. A1： A2：, Hence, A3：The existence of the zero element . The zero elementmust satisfy that for any, That is for any，. We obtain that the zero element is A4:The existence of additive inverse. For each，its additive inverse is, since . (Note thatis the zero element of) M1： M2： M3： 2 M4： Exercise 13. (a) No, it is not a subspace. Denote the set by S.Take 2 ( )p xxxS, 2 ( )q xxxS . Then ( )( )2p xq xxS. S is not closed under addition. Hence, S is not a subspace. (Or:The set S does not contain the zero polynomial, hence, is not a subspace.) (b) Denote the set by S. (b) Take 3 ( )1p xxS , 3 ( )1p xxS . Then( )( )2p xq xS. S is not closed under addition. Hence, S is not a subspace. (Or:The set S does not contain the zero polynomial, hence, is not a subspace.) (c) Yes, it is a subspace. Check that this set is closed under addition and scalar multiplication. (d) No, it is not a subspace. Denote the set by S. Take( )1p xxS ,( )1p xxS ,( )( )2p xq xS. S is not closed under addition. Hence, S is not a subspace. Exercise 15. (a) Yes, it is a subspace. Check that this set is closed under addition and scalar multiplication. (b) Yes, it is a subspace. Check that this set is closed under addition and scalar multiplication. (c) Denote the set by S. Take( )p xxS. But( 1) ( )p xxS . Thus, the set S is not closed under scalar multiplication. Hence, S is not a subspace. (d) Yes, it is a subspace. Check that this set is closed under addition and scalar multiplication. (e) Denote the set by S. Take( )1p xxS ( )1q xxS . But( )( )2p xq xxS. S is not closed under addition. Hence, S is not a subspace. Exercise 17. 3 Since 12 ,uv vv is spanfor eachi, all combinations of 12 ,u uurare also in 12 ,v vvsspan. Thus, 12 ,u uurspanis a subspace of 12 ,v vvsspan. Therefore, 12 dim(,)u uurspan 12 dim(,)v vvsspan. Exercise 19 By Taylor expansion formula ( ) 1 1 0 (1) ( )32(1) ! j n nj j f f xxx j 21 2(1)(2)2(1)(2)() 5 12(1) (1)(1)(1)2(1) 2! jn nnnnnj nxxxx j The coordinate vector is 2(1)(2)2(1)(2)() 5,2(1),2) 2! T nnnnnj n j （ Exercise 22 Use the definition of the transition matrix. 111 011 001 Exercise 25. (b) Let 12 (,)b bbnB . Then 12 (,)bbbnABAAA. IfABO, thenb0 i Afor1,2,in.( )biN Afor1,2,in. All linear combinations of 12 ,b bbnare also in( )N A. Thus,( )( )R BN A.( )R Bis a subspace of( )N A. If( )R Bis a subspace of( )N A, then for each columnbiof B, we must have b0 i A. Hence, 12 (,).bbbnABAAAO (b) By part (a), we know that( )R Bis a subspace of( )N A. Thus, 4 ( )dim( ( )dim( )r BR BN A. By the rank-nullity theorem, we obtain that ( )( )dim( )( )r Br AN Ar An Exercise 27 (a) Suppose that0xC. Then0)(xBA. Since the column vectors of A are linearly independent, we must have0xB. Since the column vectors of B are linearly independent, we must have0x. This shows that the column vectors of C are linearly independent. (b) TTT ABAB)( TT BA ，have linearly independent columns By part (a), T AB)(have linearly independent columns. Thus, AB have linearly independent rows. Exercise 29. Let,A BS. Then()T TT ABABAB, and()T T kAkAkA. S is closed under addition and scalar multiplication. Thus, S is a subspace of n n R Let,A BK. Then()() TTT ABABABAB , and()() TT kAkAkA . K is closed under addition and scalar multiplication. Thus, K is a subspace of n n R The proof of n n RSK . Let. n n AR Then 11 ()() 22 TT AAAAA. 1 () 2 T AAis symmetric and 1 () 2 T AAis anti-symmetric. This show that n n RSK . Next, we show that the sumSKis a direct sum. IfASK, then we have both T AAand T AA . This will imply thatAA . Thus, A must be the zero matrix. This proves that the sumSKis a direct sum. Exercise 32. Let ij Edenote the matrix whose( , )i jentry is 1, zero elsewhere. ij Fdenote the matrix whose( , )i jentry is1, zero elsewhere. For any(1) m n ijij AabC , where, ijij a bare real numbers, A can be written as 5 1111 nmnm ijijijij jiji Aa Eb F . This shows that the matrices,|1,2,1,2, ijij EFim jnforms a spanning set for m n C . If 1111 nmnm ijijijij jiji a Eb FO , then10 ijij abfor1,2,im, 1,2,jn. Thus, we must