南航双语矩阵论matrixtheory第一章部分题解.doc

Solution Key (chapter 1)#2. Take , . But . If , then there are rational numbers a and b, such that .( It is clear that and .) This will lead to 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.#13. (a) Denote the set by S. Take , . Then . 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.Take , . Then . 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.)(d) Denote the set by S. Take , , . S is not closed under addition. Hence, S is not a subspace.#15. (c) Denote the set by S. Take. But . Thus, the set S is not closed under scalar multiplication. Hence, S is not a subspace. (e) Denote the set by S. Take . But . S is not closed under addition. Hence, S is not a subspace.#17. Since for each , all combinations of are also in . Thus, is a subspace of . Therefore, .#25. (a) Let . Then . If , then for . for . All lineawr combinations of are also in. Thus, . is a subspace of . If is a subspace of , then for each column of B, we must have . Hence, (b) By part (a), we know that is a subspace of . Thus, . By the rank-nullity theorem, we obtain that #29. Let . Then , and . S is closed under addition and scalar multiplication. Thus, S is a subspace of Let . Then , and . K is closed under addition and scalar multiplication. Thus, K is a subspace of The proof of . Let Then . is symmetric and is anti-symmetric. This show that . Next, we show that the sum is a direct sum. If , then we have both and . This will imply that . Thus, A must be the zero matrix. This proves that the sum is a direct sum.#32. Let denote the matrix whose entry is 1, zero elsewhere. For any , where are real numbers, A can be written as . This shows that the matrices forms a spanning s