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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 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
= −− − → − − → − − → = − − − = −
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
− − → → → − − −
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= = = −
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 .
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.