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Vector space, abelian groups

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data

    The set ℝ^2 with vector addiction forms an abelian group.

    a ∈ ℝ,

    x = [tex]\binom{p}{q}[/tex]

    we put: a ⊗ x = [latex]\binom{ap}{0}[/latex] ∈ ℝ^2; this defines scalar multiplication

    ℝ × ℝ^2 → ℝ^2
    (p, x) → (p ⊗ x)

    of the field ℝ on ℝ^2.

    Determine which of the axioms defining a vector space hold for this abelian group ℝ^2 with this scalar multiplication.

    2. Relevant equations


    3. The attempt at a solution
    a ⊗ x = [tex]\binom{ap}{0}[/tex]

    i have no idea where to begin. We'd have to look at the axioms for the vector space and go from there? which ones would i want to prove are false?
     
  2. jcsd
  3. Feb 18, 2015 #2

    Dick

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    Use the "tex" tag instead of "latex". And yes, you'd look at all of the vector space axioms. You might want to pay particular attention to the ones involving the scalar product.
     
  4. Feb 18, 2015 #3
    i see that you have to 'use' the axioms, but how do i 'use' them?
     
  5. Feb 18, 2015 #4

    HallsofIvy

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    You don't "use" the axioms, you determine whether or not the axioms are true. What are those axioms and what are they in terms of the definitions given here?
     
  6. Feb 19, 2015 #5
    a ⊗ ([tex]\binom{p}{q}\binom{r}{s}[/tex]) = a ⊗ [tex]\binom{pr}{qs}[/tex] = [tex]\binom{apr}{0}[/tex]

    (a ⊗ [tex]\binom{p}{q})\binom{r}{s}[/tex]) = [tex]\binom{ap}{0}\binom{r}{s}[/tex] = [tex]\binom{apr}{0}[/tex]


    is this correct? im trying to prove multiplicative distributuvuty.
     
  7. Feb 19, 2015 #6

    HallsofIvy

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    "multiplicative distributivity" means that a(u+ v)= au+ av for scalar a and vectors u and v. I don't see any addition of vectors in that. In fact, you have not stated what the vectors are and what the scalar is. Perhaps you mean "associativity"? That a(bv)= (ab)v for scalars a and b and vector v?

    (Using "itex" instead of "tex" lets you keep everything on one line.)
     
  8. Feb 19, 2015 #7

    Dick

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    Where did you find that axiom? In a vector space multiplication of vectors is not generally defined. I'll give you a big hint. Can you find an axiom that would tell you what 1⊗##\binom{p}{q}## should be? State it. That's the scalar product of 1 with a vector.
     
  9. Feb 19, 2015 #8
    I think i understand now. I got help irl. It's a bit funky because you have to assume their multiplication is true, when it appears not to be.

    There is no vector multiplication axiom in a vector space (you can't multiply vecturs, duh). The first distributivity axiom is as follows:

    a, b ∈ Field
    u, v ∈ vector space

    1st law: v(a + b) = va + vb

    proof as in the question...

    prove: [itex]\binom{p}{q}[/itex]⊗(a + b) = [itex]\binom{p(a+b)}{0}[/itex]

    proof:

    compare with [itex]\binom{p}{q}[/itex]⊗(a+b) = a⊗[itex]\binom{p}{q}[/itex] + b⊗ [itex]\binom{p)}{q}[/itex]

    = [itex]\binom{ap}{0}[/itex] + [itex]\binom{bp}{0}[/itex]

    = [itex]\binom{ap + bp)}{0}[/itex]

    = [itex]\binom{p(a+b)}{0}[/itex]

    therefore the first law of distributivity holds for this multiplication.

    Am I going along the right lines? And the ⊗ represents the "interaction" (or binary operation) between the field and the vector space.
     
  10. Feb 19, 2015 #9

    Dick

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    Yes, you are getting the hang of it. And, yes, that proves that axiom. Now look through the list of axioms and see if you can find one that's not going to be true. There is one.
     
  11. Feb 19, 2015 #10
    I think I found the one that doesn't work, it's the additive inverse.

    0 in Field.
    v in Vector space.

    0 + v = v


    0 ⊗ [itex]\binom{p}{q}[/itex] = [itex]\binom{0p}{0}[/itex] = [itex]\binom{0}{0}[/itex].

    Am I correct? thanks. From what I can see, I think this is the only false axiom.

    EDIT: I might be wrong, since im adding an element from the field (zero) into the vector space: 0 + v = v, or perhaps the zero is meant to be the zero vector (?)./
     
  12. Feb 19, 2015 #11
    Actually, I think I know now. It's the v + (-v) = 0.

    [itex]\binom{p}{q}[/itex] + (-1)⊗[itex]\binom{p}{q}[/itex]

    = [itex]\binom{p}{q}[/itex] + [itex]\binom{-1p}{0}[/itex]

    = [itex]\binom{p - p}{q}[/itex]

    [itex]\binom{0}{q}[/itex] =/= [itex]\binom{0}{0}[/itex]

    so not true.
     
  13. Feb 19, 2015 #12

    Dick

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    You're getting closer and that is a valid problem but how do you know (-1)⊗v should be (-v)? I actually don't know exactly what your axiom list looks like. There are variations. Do you have an axiom that says (1)⊗v=v?
     
  14. Feb 20, 2015 #13
    here are the axioms that i used:

    http://mathworld.wolfram.com/VectorSpace.html

    and yes, there is a 1v = v, which i proved rather simply (within the question) on my own. Therefore, -1v must equal -v, so my proof from above should be correct (?)/

    v + (-v) =
    v + (-1)*v


    ==> v + (-1)⊗v
     
  15. Feb 20, 2015 #14

    Dick

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    Yes, it's correct. It's just a little roundabout. Why don't you just show 1⊗v=v isn't generally true? That's enough. And that's what's REALLY going wrong. Not the additive inverse. v+(-v)=0 is correct. It's just that (-1)⊗v isn't equal to -v.
     
    Last edited: Feb 20, 2015
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