Download Lab 6 Solutions 4.1 a. Additive inverse b. Transitive

Lab 6 Solutions
4.1
a.
b.
c.
d.
e.
f.
Additive inverse
Transitive property of equality
Add v to both sides
Associative property of addition
Commutative property and Additive inverse
Additive identity
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4.2
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4.2 #47. Prove that 0v = 0 for any element v of a vector space 𝕍.
0v
0v
0v
0v + (0v)
0v + (0v)
0
0
=
=
=
=
=
=
=
0v
(0 + 0)v
0v + 0v
(0v + 0v) + (0v)
0v + (0v + (0v))
0v + 0
0v
Reflexive property of equality
Additive identity in ℝ
Distributive property
Add 0v to both sides
Associative property
Additive inverses
Additive identity in 𝕍
4.2 #48
Prove that (–1)v = –v.
Distributive property
Additive inverses in R
Theorem 4.4(1), i.e. 0v = 0
Scalar identity
Add –v to both sides
Associative property
Additive inverse
Commutative property
Additive identity
(–1)v + (1)v = (–1 + 1)v
(–1)v + (1)v = 0v
(–1)v + (1)v = 0
(–1)v + v = 0
((–1)v + v) + (–v) = 0 + (–v)
(–1)v + (v + (–v)) = 0 + (–v)
(–1)v + 0 = 0 + (–v)
(–1)v + 0 = (–v) + 0
(–1)v = –v
4.3 #48
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To show that  = { f ∈ C[0, 1] such that
∫ f (x)dx = 0 } is a subspace of C[0, 1], we need to
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show that  is not empty and  is closed under vector addition and scalar multiplication.
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 is not empty because the zero function z(x) = 0 is in :
∫ z(x)dx = 0
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 is closed under addition: if f ∈  and g ∈  then the new function f + g ∈  because
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∫ [ f + g](x)dx =
0
1
∫
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f (x) + g(x)dx =
0
∫
0
1
f (x)dx + ∫ g(x)dx = 0 + 0 = 0
0
Finally,  is closed under scalar multiplication: if c is a scalar and f ∈  then the new function
cf ∈  because
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1
1
0
0
0
∫ [cf ](x)dx = ∫ c f (x)dx = c ∫ f (x)dx = c0 = 0
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4.3 #52
Prove that W is a subspace.
4.3 #58
a) To show that  +  a subspace of  when  and  are subspaces of , we need to show that
 +  is not empty and  +  is closed under vector addition and scalar multiplication.
 +  is not empty because 0 ∈  and 0 ∈  so 0 = 0 + 0 ∈  + .
 +  is closed under addition: if u1 ∈  +  and u2 ∈  +  then u1 = v1 + w1 and
u2 = v2 + w2 for some v1, v2 ∈  and w1, w2 ∈ . So
u1 + u2 = (v1 + w1) + (v2 + w2) = (v1 + v2) + (w1 + w2) ∈  + 
because  and  are closed under addition.
Finally,  +  is closed under scalar multiplication: if c is a scalar and u ∈  +  then
u1 = v1 + w1 for some v ∈  and w ∈ . So
cu = c(v + w) = cv1 + cv2 ∈  + 
because  and  are closed under scalar multiplication.
b) If  = {(x, 0): x is a real number} and  = {(0, y): y is a real number} then
 +  = {(x, y): x and y are a real numbers} so  +  = 2.
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