Axiom 6 Vector Space

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  • #1
Dustinsfl
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{∀ x ϵ ℝ+ : x>0}
Define the operation of scalar multiplication, denoted ∘, by α∘x = x^α, x ϵ ℝ+ and α ϵ ℝ.
Define the operation of addition, denoted ⊕, by x ⊕ y = x·y, x, y ϵ ℝ+.
Thus, for this system, the scalar product of -3 times 1/2 is given by:
-3∘1/2 = (1/2)^-3 = 8 and the sum of 2 and 5 is given by:
2 ⊕ 5 = 2·5 = 10.
Is ℝ+ a vector space with these operations? Prove your answer.

6. (α ⊕ β)∘x = x^(α ⊕ β) = x^(α·β) = x^(β·α)
Am on the right path for this axiom?
 

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  • #2
LCKurtz
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6. (α ⊕ β)∘x = x^(α ⊕ β) = x^(α·β) = x^(β·α)
Am on the right path for this axiom?

What axiom? Are you trying to prove or disprove:

(α ⊕ β)∘x = (α∘x) ⊕ (β∘x), which would be better written as

(x ⊕ y)∘a = (x∘a) ⊕ (y∘a)

If so, you need to continue until you get the right side to come out or show that it doesn't. I would also suggest you be careful to keep consistent notation. Your vectors are in R+ and your scalars are in R. You have already reversed the notation from what is given i.e., x, y in R+ and α, β in R. (Why not use a, b instead to save typing)?
 
  • #3
Dustinsfl
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Alpha and beta are simply scalars. The problem allows the scalars to be in all real where x only exists in real > 0. This is vector space axiom 6 which as stated says (α ⊕ β)∘x = α·x ⊕ β·x. The issue I was caught up on is how scalar multiplication is defined. It says scalar multiplication becomes the exponent. Am I able to ignore that and just distribute x?
 
  • #4
LCKurtz
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Alpha and beta are simply scalars. The problem allows the scalars to be in all real where x only exists in real > 0. This is vector space axiom 6 which as stated says (α ⊕ β)∘x = α·x ⊕ β·x. The issue I was caught up on is how scalar multiplication is defined. It says scalar multiplication becomes the exponent. Am I able to ignore that and just distribute x?

Apparently you didn't understand what I said in my post. You are getting your notation all mixed up. If alpha and beta are scalars in R, then α ⊕ β doesn't make sense because the ⊕ operation is defined on the vectors which are the scalars in R+.
 
  • #5
Dustinsfl
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So alpha and beta can be taking as normal addition then?
 
  • #6
LCKurtz
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So alpha and beta can be taking as normal addition then?

Not sure what that sentence means. Proving something like (x ⊕ y)∘a = (x∘a) ⊕ (y∘a) will not involve the ordinary + operation.
 
  • #7
vela
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The axiom you're trying to show holds is:

For [itex]a, b \in R[/itex] and [itex]\vec{x} \in R^+[/itex],

[tex](a+b)\circ \vec{x} = a\circ \vec{x} \oplus b \circ \vec{x}[/tex]

where:

+ is the normal addition of real numbers
[itex]\oplus[/itex] is the vector addition defined above
[itex]\circ[/itex] is the scalar multiplication of a vector defined above

It's kind of confusing because the + symbol is typically used to denote both field addition and vector addition, but they are different operations. One of the goals of this problem, I'm guessing, is to get you to realize this distinction.
 

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