Solution Key (chapter 1)
#2. Take
S , 2=. But 2S ∉. If 2S ∈, then there are rational numbers a and b , such that
2=0a ≠ and 0b ≠.) This will lead to
22
423
2a b 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.
#13. (a) Denote the set by S . Take 2()p x x x S =+∈, 2()q x x x S =-+∈.
Then ()()2p x q x x S +=∉. 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 3()1p x x S =+∈, 3()1p x x S =-+∈. Then ()()2p x q x S +=∉. 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 ()1p x x S =+∈, ()1p x x S =-+∈, ()()2p x q x S +=∉. S is not closed under addition. Hence, S is not a subspace.
#15. (c) Denote the set by S . Take ()p x x S =∈. But ()p x x S -=-∉. Thus, the set S is not closed under scalar multiplication. Hence, S is not a subspace.
(e) Denote the set by S . Take ()1p x x S =-∈ ()1q x x S =+∈. But ()()2p x q x x S +=∉. S is not closed under addition. Hence, S is not a subspace.
#17. Since 12{,,,}u v v v i s span ∈ for each i , all combinations of 12,,,u u u r are also in 12{,,,}v v v s span . Thus, 12{,,,}u u u r span is a subspace of 12{,,,}v v v s span . Therefore, 12dim({,,,})u u u r span ≤ 12dim({,,,})v v v s span .
#25. (a) Let 12(,,,)b b b n B = . Then 12(,,,)b b b n AB A A A = .
If AB O =, then b 0i A = for 1,2,,i n = . ()b i N A ∈ for 1,2,,i n = . All lineawr combinations of 12,,,b b b n are also in ()N A . Thus, ()()R B N A ⊂. ()R B is a subspace of ()N A .
If ()R B is a subspace of ()N A , then for each column b i of B , we must have
b 0i A =. Hence,
12(,,,).b b b n AB A A A O ==
(b) By part (a), we know that ()R B is a subspace of ()N A . Thus,
()dim(())dim(())r B R B N A =≤. By the rank-nullity theorem, we obtain that
()()d i m (())(r B r A N A r A n +≤
+= #29. Let
,A B S
∈. Then ()T T T A B A B A B +=+=+, and ()T T kA kA kA ==. S is closed
under addition and scalar multiplication. Thus, S is a subspace of
n n R ⨯
Let ,A B K ∈. Then ()()T T T A B A B A B A B +=+=--=-+, and ()()T T kA kA kA ==-. K is closed under addition and scalar multiplication. Thus, K is a subspace of
n n R ⨯
The proof of n n R S K ⨯=⊕.
Let .n n A R ⨯∈ Then 11()()2
2
T T A A A A A =++-.
1
()2
T A A + is symmetric and 1
()2
T A A - is anti-symmetric. This show that n n R S K ⨯=+. Next, we show that the sum S K + is a direct sum. If A S K ∈⋂, then we have both T A A = and T A A =-. This will imply that A A =-. Thus, A must be the zero matrix. This proves that the sum S K + is a direct sum.
#32. Let ij E denote the matrix whose (,)i j entry is 1, zero elsewhere. For any
()m n ij ij A a C ⨯=∈, where ,ij ij a b are real numbers, A can be written as
11
11
n m n m
ij ij ij ij j i j i A a E b E =====∑∑.
This shows that the matrices
{|1,2,,,1,2,,} ij ij E i m j n == forms a spanning
set for
m n
C
⨯. If
11
11
n m
n m
ij
ij
ij ij j i j i a E
b E O
=====∑∑, then 0ij ij a = for
1,2,,i m
= ,
1,2,,j n = . Thus, we must have 0ij ij a b ==
for 1,2,,i m = , 1,2,,j n = . Therefore,
{|1,2,,,1,2,,} ij ij E i m j n == forms a basis for m n C ⨯. The dimension is 2mn .
