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Vector Space Properties

  1. Jul 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Let V = {RXR} with addition defined as (a1,a2) + (b1,b2) = (a1 + b1, a2b2)

    Show the vector space condition, for each x in V there exists a y in V such that x + y = 0, fails for the V defined above.

    2. Relevant equations



    3. The attempt at a solution

    a + d = (a1,a2) + (d1,d2) = (a1 + d1, a2d2), so let d1 = -a1 and d2 = 0, therefore (a1 - a1, 0*a2) = (0,0) = 0

    Seems to me that the condition holds. What have I missed?
     
  2. jcsd
  3. Jul 17, 2012 #2

    jbunniii

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    I agree that it does hold. However, there is another vector space axiom that is not satisfied. Can you see which one?
     
  4. Jul 17, 2012 #3

    HallsofIvy

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    When you write "x+ y= 0", exacty what do you mean by "0"? You have, apparently, assumed that it meant (0, 0) but there "vector space condition" requiring that, for each x, there exist y such that x+y= (0, 0).

    There is a vector space condition that there exist a "zero vector", 0, such that x+ 0= 0 and a condition that, given any x, there exist a y such that x+ y= 0. (The existence of an "additive identity" and the existence of "additive inverses".)

    Since we are defining (a1, a1)+ (b1, b2)= (a1+ b2, a2b2), (0, 0) is NOT the "additive identiy": (a1, a2)+ (0 , 0)= (a1, 0), NOT (a1, a2).

    What is the additive identity for this space?
     
  5. Jul 17, 2012 #4
    Thanks for the replies.

    (a1, a2)+ (0, 1)= (a1+ 0, a2*1) = (a1, a2) is the additive identity so 0 = (0,1)

    (a1,a2) + (b1,b2) = (a1+ b1, a2b2) = (0, 1) means b1 = -a1 and b2 = 1/a2... still missing something.
     
  6. Jul 17, 2012 #5

    HallsofIvy

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    So what would be the additive inverse of (a, 0)?
     
  7. Jul 17, 2012 #6
    (-a,1/0)!!!

    Haha thanks, appreciate the help.
     
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