1、1Solution Key (chapter 1)#2. Take , . But . If , then there are rational numbers a and 2S2S2b, such that .( It is clear that and .) This will lead to 3ab0ab46The right hand is a rational number and the left hand side is an irrational number. This is impossible. Thus, S is not closed under multiplica
2、tion. Hence, S is not a field.#13. (a) Denote the set by S. Take , . 2()pxS2()qxThen . S is not closed under addition. Hence, S is not a subspace. ()2pxqx(Or: The set S does not contain the zero polynomial, hence, is not a subspace.)(b) Denote the set by S.Take , . Then . S is not closed under 3()1p
3、x3()1px()2pxqaddition. Hence, S is not a subspace.(Or: The set S does not contain the zero polynomial, hence, is not a subspace.)(d) Denote the set by S. Take , , . S ()1pxS()1pxS()2pxqis not closed under addition. Hence, S is not a subspace.#15. (c) Denote the set by S. Take . But . Thus, the set S
4、 is not ()px()pxSclosed under scalar multiplication. Hence, S is not a subspace. (e) Denote the set by S. Take . But . ()1()1q()2pxqxS is not closed under addition. Hence, S is not a subspace.#17. Since for each , all combinations of are also in 12,uvi sspan i 12,ur. Thus, is a subspace of . Therefo
5、re, 12,vsspan 12,ur 12,vsspan.12dim(,)ur 12dim(,)vsspan#25. (a) Let . Then .12(,)bnB 12(,)bnABAIf , then for . for . All lineawr AO0iA,in (iN1,2incombinations of are also in . Thus, . is a subspace of 12,bn ()(RB). ()N2If is a subspace of , then for each column of B, we must have . ()RB()NAbi b0iAHe
6、nce, 12(,.bnAO(b) By part (a), we know that is a subspace of . Thus, ()RB()N. By the rank-nullity theorem, we obtain that ()dim()i()rBRNA(rrn#29. Let . Then , and . S is closed under ,AS()TTBAB()TkAkaddition and scalar multiplication. Thus, S is a subspace of nRLet . Then , and . K is ,BK()()TTA()()
7、TkAclosed under addition and scalar multiplication. Thus, K is a subspace of nRThe proof of .nRSLet Then . is symmetric and .A1()()2TTA1()2TAis anti-symmetric. This show that . 1()2T nRSNext, we show that the sum is a direct sum. If , then we have both SKKand . This will imply that . Thus, A must be
8、 the zero matrix. TATAAThis proves that the sum is a direct sum.#32. Let denote the matrix whose entry is 1, zero elsewhere. For any ijE(,)ij, where are real numbers, A can be written as(1)mnijijAabC,ijab.11nmij ijj jAEEThis shows that the matrices forms a spanning set ,|,2,12, ijij jn for . If , then for , mnC11nmnmij ijj jabO0ijijab1,2im. Thus, we must have for , . Therefore, 1,2j 0ijija1,2m ,nforms a basis for . The dimension is .,|,12, ijijEjn nC2
