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Showing that there is no such zero vector

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Let V denote the set of ordered pairs of real numbers. If (x, y) and (w, z) are elements of V and a is a real scalar, define (x, y) + (w, z) = (x + w, yz) and a(x, y) = (ax, y). Is V a vector space?

    2. Relevant equations


    3. The attempt at a solution
    Going through the vector space axioms everything is fine until we need to show that a zero vector exists.
    Specifically, it is clearly evident that no such zero vector exists. However, how do we prove that no such zero vector exists, and hence it is not a vector space? For example, we can easily show that (0, 0) and (1, 1) do not work. However, this does not show that now pair will work (although clearly no pair will work).
     
    Last edited: Oct 9, 2016
  2. jcsd
  3. Oct 9, 2016 #2

    Mark44

    Staff: Mentor

    How is it clearly evident that no such zero vector exists?

    I think that I would start by assuming that <u, v> is a zero vector in V. Then for an arbitrary vector <x, y> it must be true that <x, y> + <u, v> = <x, y>.
     
  4. Oct 9, 2016 #3

    PeroK

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    You just have to find a proof of what is "clearly evident".
     
  5. Oct 9, 2016 #4
    So (x, y) + (u, v) = (x, y) ==> (x + u, vy) = (x, y)

    therefore, u = 0, v = 1. So would (0, 1) be the zero vector in this vector space?
     
  6. Oct 9, 2016 #5

    PeroK

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    Can you confirm the rule for vector addition as this is very different from what you have in the original post.
     
  7. Oct 9, 2016 #6
    Oops, I did mess up in the original post, sorry. The rule for vector addition is (x, y) + (w, z) = (x + w, yz).
     
  8. Oct 9, 2016 #7

    Mark44

    Staff: Mentor

    Looks like you found a zero vector, even though it was "clearly evident" that no such vector exists.
     
  9. Oct 9, 2016 #8
    Yes yes, I have learned my lesson. However, now let's look at if an inverse vector always exists.

    ##(x, y) + (u, v) = (0, 1) \implies (x + u, yv) = (0, 1) \implies u = -x,~~ v = 1/y##

    Does this show that an inverse vector always exists?
     
  10. Oct 9, 2016 #9

    Mark44

    Staff: Mentor

    No, it doesn't. There are lots of (i.e., an infinite number of) vectors that don't have additive inverses.
     
  11. Oct 9, 2016 #10
    So any vector of the form (a, 0) doesn't have an inverse, so thus this is not a vector space?
     
  12. Oct 9, 2016 #11

    Mark44

    Staff: Mentor

    What do you think? Does this particular axiom allow for exceptions?
     
  13. Oct 9, 2016 #12
    No, since there must exist an inverse for each element in the vector space, and obviously (a, 0) is an element (as it is a 2-tuple).
     
  14. Oct 9, 2016 #13

    Mark44

    Staff: Mentor

    There you go. And since V fails this axiom, it is not a vector space. It's possible that there are other axioms among the scalar multiplication bunch that fail, but since you found that some elements of V don't have additive inverses, that's enough to say that V isn't a vector space. It might be useful, though, to see if some of the scalar multiplcation axioms fail -- might be good practice for a test.
     
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