Suppose ₁ and ₂ both have the property of being an additive inverse of v. That is, for this arb particular vector v, u₁ + v = 0 and u₂+ v = 0. Supply a reason for each line of the string of proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1. v is th nverse of v]. U₁ = U₁ + 0 = U₁ + (U₂ + v) = U₁ + (v + U₂) vector space property 4: property of the additive identity vector space property 4: property of the additive identity vector space property 2: vector addition is commutative X
Suppose ₁ and ₂ both have the property of being an additive inverse of v. That is, for this arb particular vector v, u₁ + v = 0 and u₂+ v = 0. Supply a reason for each line of the string of proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore -1. v is th nverse of v]. U₁ = U₁ + 0 = U₁ + (U₂ + v) = U₁ + (v + U₂) vector space property 4: property of the additive identity vector space property 4: property of the additive identity vector space property 2: vector addition is commutative X
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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This is a selection-choice question， i only need u1+(u2+V) and 0+u2.
![Suppose u₁ and u2 both have the property of being an additive inverse of v. That is, for this arbitrary but
particular vector v, u₁ + v = 0 and u₂ + v = 0. Supply a reason for each line of the string of equalities
proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore −1 . v is the additive
inverse of v].
U₁ = U₁ +0
= U₁ + (U₂ + v)
= U₁ + (v + U₂)
(U₁ + v) + U₂
=
= 0 + U₂
=
= U2
vector space property 4: property of the additive identity
vector space property 4: property of the additive identity
vector space property 2: vector addition is commutative
vector space property 3: vector addition is associative
vector space property 7: property of the multiplicative identity, 1 X
vector space property 4: property of the additive identity](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad61a16a-ccb3-4100-87e8-33e1dead9976%2F111db03f-44c2-4bc2-a3b4-1f3b18cf9c45%2Fnko8tkl_processed.png&w=3840&q=75)
Transcribed Image Text:Suppose u₁ and u2 both have the property of being an additive inverse of v. That is, for this arbitrary but
particular vector v, u₁ + v = 0 and u₂ + v = 0. Supply a reason for each line of the string of equalities
proving that u₁ = u₂, thus proving that additive inverses are unique [and therefore −1 . v is the additive
inverse of v].
U₁ = U₁ +0
= U₁ + (U₂ + v)
= U₁ + (v + U₂)
(U₁ + v) + U₂
=
= 0 + U₂
=
= U2
vector space property 4: property of the additive identity
vector space property 4: property of the additive identity
vector space property 2: vector addition is commutative
vector space property 3: vector addition is associative
vector space property 7: property of the multiplicative identity, 1 X
vector space property 4: property of the additive identity
Transcribed Image Text:Select an answer
assumption of the claim
vector space property 1: closure under addition
vector space property 2: vector addition is commutative
vector space property 3: vector addition is associative
✓ vector space property 4: property of the additive identity
vector space property 5: existence of additive inverses
vector space property 6: closure under scalar multiplication
vector space property 7: property of the multiplicative identity, 1
vector space property 8: distribution of scalars over vectors
vector space property 9: distribution of vectors over scalars
vector space property 10: associativity of scalar multiplication
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