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Subspace question

  • #1

Homework Statement



Let X=ℝ3 and let V={(a,b,c) such that a2+b2=c2}. Is V a subspace of X? If so, what dimensions?

Homework Equations



A vector space V exists over a field F if V is an abelian group under addition, and if for each a ∈ F and v ∈ V, there is an element av ∈ V such that all of the following conditions apply for all a,b ∈F and all u,v ∈V -- a(u + v) = au + av, (a + b)v = av + bv, a(bv)= (ab)v, and 1 × v = v.

The Attempt at a Solution



a2+b2=c2 is abelian under addition because real numbers are commutative under that operation. I believe we would not have to concern ourselves with multiplication because we are not multiplying, and we would not have to concern ourselves with associativity under multiplication for the same reason. If we multiply this my the multiplicative identity, we get 1 × (a2+b2=c2)=a2+b2=c2. So, we are good for a vector space.

V is a subspace of X because it is nonempty, and closed under linear operations -- our friend Pythagoras showed that is true several millennia ago. Addition and multiplication are closed under ℝ, so we are good to go. Also, it passes through the origin, 0^2 + 0^2 = 0^2, obviously enough. Therefore, I have is as a subspace existing in three dimensions.

Thank you all for your time,

SY
 

Answers and Replies

  • #2
33,271
4,976

Homework Statement



Let X=ℝ3 and let V={(a,b,c) such that a2+b2=c2}. Is V a subspace of X? If so, what dimensions?

Homework Equations



A vector space V exists over a field F if V is an abelian group under addition, and if for each a ∈ F and v ∈ V, there is an element av ∈ V such that all of the following conditions apply for all a,b ∈F and all u,v ∈V -- a(u + v) = au + av, (a + b)v = av + bv, a(bv)= (ab)v, and 1 × v = v.

The Attempt at a Solution



a2+b2=c2 is abelian under addition because real numbers are commutative under that operation.
This doesn't make any sense. The elements of V are vectors in ##\mathbb{R}^3## of the form <x, y, z> such that ##x^2 + y^2 = z^2##. x, y, and z are all real numbers, it's true.
SYoungblood said:
I believe we would not have to concern ourselves with multiplication because we are not multiplying, and we would not have to concern ourselves with associativity under multiplication for the same reason.
Since the underlying space is ##\mathbb{R}^3##, addition is the usual vector addition, and scalar multiplication is the usual scalar multiplication.
Since V is a subset of ##\mathbb{R}^3## you don't have to verify all of the axioms for a vector space. What you do need to verify is that for vectors in set V,
  • The 0 vector is an element of V
  • V is closed under vector addtion. I.e., if ##u \in V## and ##w \in V## is ##u + w \in V##?
  • V is closed under scalar multiplication. I.el, if ##u \in V## and c is a scalar (a real) is ##cv \in V##?
SYoungblood said:
If we multiply this my the multiplicative identity, we get 1 × (a2+b2=c2)=a2+b2=c2. So, we are good for a vector space.
No, that's not how it works. See above.
SYoungblood said:
V is a subspace of X because it is nonempty, and closed under linear operations -- our friend Pythagoras showed that is true several millennia ago. Addition and multiplication are closed under ℝ, so we are good to go. Also, it passes through the origin, 0^2 + 0^2 = 0^2, obviously enough. Therefore, I have is as a subspace existing in three dimensions.

Thank you all for your time,

SY
 
  • #3
I got it, this and breaking out my old LA text helped. Thank you.

SY
 

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