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Ashley Robinson Pride Due Date: Saturday, March 26, 2011 Exercise Set 1.4 1) Label the following statements as true or false. a) The zero vector is a linear combination of any nonempty set of vectors. True. b) The span of is . False ( ) { } span 0 ∅= . c) If S is a subset of a vector space V, then span(S) equals the intersection of all subspaces of V that contain S. True. d) In solving a system of linear equations, it is permissible to multiply an equation by any constant. False. You may multiply by an nonzero constant. e) In solving a system of linear equations, it is permissible to add any multiple of one equation to another. True. f) Every system of linear equations has a solution. False. A system of linear equations may have no solution. 2) Solve the following systems of linear equations by the method introduced in this section. a) 1 2 3 1 2 3 4 1 2 3 4 2 2 3 2 3 3 2 5 7 2 3 x x x x x x x x x x x =− + = =− 2 2 3 0 2 1 1 2 1 3 3 3 2 5 7 0 0 1 2 4 1 1 2 1 3 0 0 4 8 13 1 2 3 4 3 4 1 1 2 1 3 2 3 0 0 1 2 4 2 4 0 0 0 0 0 x x x x x x =− + = Let 2 0 x = and 4 0 x = , then 3 4 x = and 1 5 x = . b) 1 2 3 1 2 3 1 2 3 3 7 4 10 3 7 4 10 1 2 1 3 1 2 1 3 2 3 1 2 1 3 3 7 4 10 0 1 1 1 2 1 2 6 2 1 2 6 2 1 2 6 2 2 6 x x x x x x x x x + = + = = 1 2 3 2 1 2 1 3 1 2 1 3 0 1 1 1 0 1 1 1 4 0 3 4 0 0 0 1 3 3 x x x =− =− =−

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Page 1: sec 1.4 - lin combos, sys

Ashley Robinson Pride Due Date: Saturday, March 26, 2011

Exercise Set 1.4

1) Label the following statements as true or false.

a) The zero vector is a linear combination of any nonempty set of vectors.

True.

b) The span of ∅ is ∅ . False ( ) { }span 0∅ = .

c) If S is a subset of a vector space V, then span(S) equals the intersection of all subspaces of V

that contain S.

True.

d) In solving a system of linear equations, it is permissible to multiply an equation by any

constant.

False. You may multiply by an nonzero constant.

e) In solving a system of linear equations, it is permissible to add any multiple of one equation

to another.

True.

f) Every system of linear equations has a solution.

False. A system of linear equations may have no solution.

2) Solve the following systems of linear equations by the method introduced in this section.

a)

1 2 3

1 2 3 4

1 2 3 4

2 2 3 2

3 3 2 5 7

2 3

x x x

x x x x

x x x x

− − = −

− − + = − − − = −

2 2 3 0 2 1 1 2 1 3

3 3 2 5 7 0 0 1 2 4

1 1 2 1 3 0 0 4 8 13

− − − − − − − − − → − − − −

1 2 3 4

3 4

1 1 2 1 32 3

0 0 1 2 42 4

0 0 0 0 0

x x x x

x x

− − − − − − − = − → + =

Let 2 0x = and 4 0x = , then 3 4x = and 1 5x = .

b)

1 2 3

1 2 3

1 2 3

3 7 4 10 3 7 4 10 1 2 1 3 1 2 1 3

2 3 1 2 1 3 3 7 4 10 0 1 1 1

2 1 2 6 2 1 2 6 2 1 2 62 2 6

x x x

x x x

x x x

− + = − − − − + = → − → − → − − − − − − − −− − =

1

2

3

21 2 1 3 1 2 1 3

0 1 1 1 0 1 1 1 4

0 3 4 0 0 0 1 3 3

x

x

x

= −− − → − − → − − → = − − − = −

Page 2: sec 1.4 - lin combos, sys

f)

1 2 3

1 2 3

1 2 3

1 2 3

2 6 1 1 2 6 1 1 2 6 1

2 8 2 1 1 8 0 3 11 10

3 1 1 15 0 5 19 183 15

1 3 10 5 0 1 4 43 10 5

x x x

x x x

x x x

x x x

+ + = − − − + + = − → → → − −+ − = − −+ + = −

1

2

3

1 2 6 13

0 1 4 44

0 0 1 22

0 0 1 2

x

x

x

− = − → = − = − −

3) For each of the following lists of vectors in 3ℝ , determine whether the first vector can be expressed

as a linear combination of the other two.

a) ( ) ( ) ( )2,0,3 , 1,3,0 , 2,4, 1− − – Yes when 1 4a = and 2 3a = −

1 2 2 1 0 4 1 0 4

3 4 0 0 1 3 0 1 3

0 1 3 0 2 6 0 0 0

− → − → − − −

b) ( ) ( ) ( )1,2, 3 , 3,2,1 , 2, 1, 1− − − − – Yes when 1 5a = and 2 8a =

3 2 1 1 1 3 1 0 5

2 1 2 0 1 8 0 1 8

1 1 3 0 1 8 0 0 0

− − − − → − − → − −

c) ( ) ( ) ( )3, 4,1 , 1, 2,1 , 2, 1,1− − − - No

1 0 51 2 3 1 1 1

22 1 4 0 3 2 0 1

31 1 1 0 1 6

0 1 6

− − − − → − → −

4) For each list of polynomials in ( )3P ℝ , determine whether the first polynomial can be expressed as a

linear combination of the other two.

a) 3 3 2 3 23 5, 2 1, 3 1x x x x x x x− + + − + + − - Yes when 1 3a = and 2 2a = −

1 1 1 2 0 6 1 0 3

2 3 0 0 3 6 0 1 2

1 0 3 1 0 3 0 0 0

1 1 5 0 1 2 0 0 0

− − → → − −

− −

b) 3 2 3 2 3 24 2 6, 2 4 1, 3 6 4x x x x x x x x+ − − + + − + + - No!

81 053 1 4 1 4 0 1 4 0

10 16 2 2 0 0 10 0 11 111

1 4 0 3 1 1 0 15 6 20 15

4 1 6 4 1 6 0 0 100 0 10

− − → → → − − −

Page 3: sec 1.4 - lin combos, sys

5) Determine whether the given vector is in the span of S.

a) ( ) ( ) ( ){ }2, 1,1 , 1,0, 2 , 1,1,1S− = − - Yes when 1, 1r s= = −

( ) ( ) ( )1,0, 2 1,1,1 2, 1,1r s+ − = −

1 1 2 1 0 1 1 0 1

0 1 1 0 1 1 0 1 1

2 1 1 0 3 3 0 0 0

− − → − → − −

c) ( ) ( ) ( ){ }1,1,1,2 , 1,0,1, 1 , 0,1,1,1S− = − - No

( ) ( ) ( )1,0,1, 1 0,1,1,1 1,1,1, 2r s− + = −

1 0 1 1 0 1 1 0 1

0 1 1 0 1 1 0 1 1

1 1 1 0 1 2 0 1 2

1 1 2 0 1 1 0 0 0

− − − → → −

e) { }3 2 3 2 22 3 3, 1, 1, 1x x x S x x x x x x− + + + = + + + + + + - Yes

( ) ( ) ( )3 2 2 3 21 1 1 2 3 3r x x x s x x t x x x x+ + + + + + + + = − + + +

1 0 0 1 1 0 0 1

1 1 0 2 0 1 0 3

1 1 1 3 0 0 1 1

1 1 1 3 0 0 0 0

− − →

1, 3, 1r s t= − = =

g) 1 2 1 0 0 1 1 1

, , ,3 4 1 0 0 1 0 0

S

= − − - Yes

1 0 0 1 1 1 1 2

1 0 0 1 0 0 3 4r s t

+ + = − −

1 0 1 1 1 0 1 1 1 0 0 3

0 1 1 2 0 1 0 4 0 1 0 4

1 0 0 3 0 0 1 2 0 0 1 2

0 1 0 4 0 0 1 2 0 0 0 0

→ → − − − −

3, 4, 2r s t= = = −

Page 4: sec 1.4 - lin combos, sys

6) Show that vectors ( )1,1,0 , ( )1,0,1 , and ( )0,1,1 generate 3F .

( ) ( ) ( ) ( )1 2 31,1,0 1,0,1 0,1,1 , ,r s t a a a+ + =

1 1 3 1 3

2 1 2 1 2 3

3 3 3 1 2

1 1 0 1 0 1 1 0 1

1 0 1 0 1 1 0 2 0

0 1 1 0 1 1 0 0 2

a a a a a

a a a a a a

a a a a a

− − − − → − − → − + → − +

1 2 31 2 3

1 2 3 1 2 3

1 2 3 1 2 3

1 1 11 1 11 0 0

2 2 22 2 2

1 1 1 1 1 10 1 0

2 2 2 2 2 2

1 1 1 1 1 10 0 1

2 2 2 2 2 2

r a a aa a a

a a a s a a a

a a a t a a a

= + −+ − − + → = − + − + + = − + +

Since F is a field and 0,1 F∈ and 1 1 F+ ∈ since it is closed under addition, then 2 F∈ . Since F is

a field, 2 has a multiplicative inverse and therefore 1

2F∈ .

9) Show that the matrices 1 0

0 0

, 0 1

0 0

, 0 0

1 0

, and 0 0

0 1

generate ( )2 2M F× .

1 2

3 4

1 0 0 1 0 0 0 0

0 0 0 0 1 0 0 1

a ar s t u

a a

+ + + =

1 1

2 2

3 3

4 4

1 0 0 0

0 1 0 0

0 0 1 0

0 0 1

r s t u a r a

r s t u a s a

r s t u a t a

or s t u a u a

+ + + = = + + + = =

→ + + + = =

+ + + = =

Therefore the matrices generate ( )2 2M F× .

12) Show that a subset W of a vector space V is a subspace of V if and only if ( )span =W W .

Proof ( )⇒ : Given ⊆W V and W is a subspace of V, let x∈W such that 1 1 2 2 n nx a v a v a v= + + +…

for 1 2, , , nv v v ∈W… and 1 2, , , na a a F∈… . Thus ( )spanx∈ W and ( )span⊆W W .

Let ( )spanx∈ W , then x is a linear combination of vectors in W. Therefore x∈W and

( )span ⊆W W . Since ( )span⊆W W and ( )span ⊆W W , then ( )span =W W .

Proof ( )⇐ : Given ⊆W V and ( )span =W W , we must show that W is a subspace of V.

a) 0∈W because ( )0 span∈ W and ( )span =W W

b) Let ( ), spanx y∈ W , then 1 1 2 2 n nx a v a v a v= + + +… and 1 1 2 2 n ny b u b u b u= + + +… for all

,i iv u ∈W and ,i ia b F∈ . Then 1 1 2 2 1 1 2 2n n n nx y a v a v a v b u b u b u+ = + + + + + + +… … and

( )spanx y+ ∈ W and thus x y+ ∈W .

c) Let ( )spanx∈ W such that 1 1 2 2 n nx a v a v a v= + + +… for all iv ∈W and ia F∈ . Then

1 1 2 2 n ncx ca v ca v ca v= + + +… for some c F∈ . Therefore ( )spancx∈ W which implies that

cx∈W .

Therefore W is a subspace of V.

Therefore a subset W of a vector space V is a subspace of V if and only if ( )span =W W .

Page 5: sec 1.4 - lin combos, sys

13) Show that if 1S and 2S are subsets of a vector space V such that 1 2S S⊆ , then ( ) ( )1 2span spanS S⊆ .

In particular, if 1 2S S⊆ and ( )1span S = V , deduce that ( )2span S = V .

Proof: Given 1 2S S⊆ and ( )1span S = V , let ( )1spanx S∈ . Then 1 1 2 2 n nx a v a v a v= + + +… for all

1iv S∈ and ia F∈ . Since ( )1spanx S∈ , 1x S∈ . Also 1 2S S⊆ , and thus 2x S∈ . Then

( )2spanx S∈ . Therefore ( ) ( )1 2span spanS S⊆ and ( )2span S = V .

15) Let 1S and 2S be subsets of a vector space V. Prove that ( ) ( ) ( )1 2 1 2span span spanS S S S⊆∩ ∩ . Give

an example in which ( )1 2span S S∩ and ( ) ( )1 2span spanS S∩ are equal and one in which they are

unequal.

Proof: Let ( )1 2spanx S S∈ ∩ , then 1 1 2 2 n nx a v a v a v= + + +… for all 1 2iv S S∈ ∩ and ia F∈ . Therefore

1x S∈ and 2x S∈ which means ( )1spanx S∈ and ( )2spanx S∈ . Therefore

( ) ( )1 2span spanx S S∈ ∩ and ( ) ( ) ( )1 2 1 2span span spanS S S S⊆∩ ∩ .

Example of Equal: Let ( ){ }1 1,1S = − and ( ){ }2 1,1S = with 2=V ℝ .

Example of Not Equal: Let ( ){ }1 2,0S = and ( ){ }2 1,0S = with 2=V ℝ .

16) Let V be a vector space and S a subset of V with the property that whenever 1 2, , , nv v v S∈… and

1 1 2 2 0n na v a v a v+ + + =… , then, 1 2 0na a a= = = =… . Prove that every vector in the span of S can be

uniquely written as a linear combination of vectors of S.

Proof: Given that S ⊆ V and that whenever 1 1 2 2 0n na v a v a v+ + + =… for 1 2, , , nv v v S∈… . Let

( ), spanx y S∈ such that 1 1 2 2 n nx a x a x a x= + + +… and 1 1 2 2 b ny b y b y b y= + + +… for

, , ,i i i ix y S a b F∈ ∈ . Then

( )0

Since 0

Cancellation Law

x y

x y x x x x

y x x x

y x

y x

− =

− = + − = + −

− + = − +

− = −

=

Therefore the linear combinations in S are unique.