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Determining vector spaces

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    The 10 axioms:
    1. If u and v are objects in V, then u + v is in V
    2. u + v = v + u
    3. u + (v+w) = (u+v)+w
    4. There is an object 0 in V, called a zero factor for V, such that 0+u = u+0 = u for all u in V
    5. For each u in V, there is an object -u in V, called a negative of u, such that u+(-u)=(-u)+u = 0
    6. If k is any scalar and u is any object in V, then ku is in V
    7. k(u + v) = ku + kv
    8. (k + m)u = ku + mu
    9. k(mu) = (km)(u)
    10. 1u = u
    3. The attempt at a solution
    First of all, are u, v, and w just (x_{1},y_{1}), (x_{2},y_{2}), (x_{3},y_{3}) and so on? If so, am I going about this the right way?

    Axiom 1. Since x,y[tex]\in[/tex]R, then any (x,y)[tex]\in[/tex] thus the addition would hold. (Not really sure how to explain this one well)
    Axiom 2. [tex](x_{1},y_{1})+(x_{2},y_{2}) = (x_{1}+x_{2}+_{1},y_{1}+y_{2}+_{1}) = (x_{2},y_{2})+(x_{1},y_{1}) = (x_{2}+x_{1}+_{1},y_{2}+y_{1}+_{1})[/tex]
    Axiom 3.

    [tex]u+(v+w) = (x_{1},y_{1})+(x_{2}+x_{3}+1,y_{2}+y_{3}+1) = (x_{1}+(x_{2}+x_{3}+1)+1,y_{1}+(y_{2}+y_{3}+1)+1) = (x_{1}+x_{2}+x_{3}+_{2},y_{1}+y_{2}+y_{3}+2)[/tex]
    [tex](u+v)+w = (x_{1}+x_{2}+1,y_{1}+y_{2}+1) + (x_{3},y_{3}) = ((x_{1}+x_{2}+_{1})+x_{3}+1,(y_{1}+y_{2}+1)+y_{3}+1) = (x_{1}+x_{2}+x_{3}+_{2},y_{1}+y_{2}+y_{3}+2)[/tex]

    and so on for the rest...

    Is this the correct approach? It seems that it is to me, but I may be way off here.
    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 23, 2010 #2


    Staff: Mentor

    That would be zero vector, not factor.
    Yes, u, v, and w and just ordered pairs.
    You left something out. ... then any (x, y) [itex]\in[/itex] V, so V is closed under addition.
    Yes, this is the right approach. I have my suspicions, though, that V is not a vector space.
  4. Mar 23, 2010 #3
    Thanks for the tip and clarification :]

    I think you are correct, unless I made a mistake. Using this approach I found that all axioms were satisfied except axiom 5. Thoughts on this?
  5. Mar 23, 2010 #4


    Staff: Mentor

    Axiom 4 is also violated. There might be others, but I didn't check all of them.
  6. Mar 23, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
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    Both axioms hold, actually. The zero vector would be 0=(-1,-1), and the additive inverse would be -(x,y) = (-x-2,-y-2). I would have thought one of axioms 7-10 would fail, but I think they all work out.
  7. Mar 23, 2010 #6


    Staff: Mentor

    Live and learn. Thanks, Vela.
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