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Linear Algebra: Show it's a vector space question

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Define V =R with vector addition a+b=ab and scalar multiplication za=a^z.
    Show that V is a vector space.




    2. Relevant equations
    a+b=ab, za=a^z


    3. The attempt at a solution

    I was able to check all the axioms but one, the additive inverse axiom where for all v in V there exists a... -v in V such that v+(-v)= 0. So far I have this: a+0=a(0)=0. In this case I believe 0 is the additive inverse.
     
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  3. Jul 2, 2009 #2

    Office_Shredder

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    a(0) = 0 =/= a. So 0 is not the additive identity.

    For the purposes of this thread, I propose using + and * to represent the addition and multiplication in the vector space (as opposed to canonical addition and multiplication in R) in order to avoid confusion
     
  4. Jul 2, 2009 #3

    HallsofIvy

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    While there is a single additive identity (and it is NOT 0), each member of V has a different additive inverse. The definition of "additive identity" (I will call it "O" for the moment) is that "a+ O= a". Here, "addition" of vectors is defined as the product of the numbers: "a+ b= ab" above. Therefore, "a+ O"= aO= a. What number is O?

    However, I see a distince problem with this as stated. What is the additive inverse of the number 0? Also, if scalar multiplication is defined by za= az, what is (1/2)(-4)? Are you sure you were not supposed to take V= R+, the set of positive real numbers?
     
  5. Jul 7, 2009 #4
    yeah I'm sure. would I be able to use 1/a for O then?
     
  6. Jul 7, 2009 #5

    HallsofIvy

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    No, of course, not! There must be a single number, O, such that a"+" O= a. The additive identity cannot depend on what a is. Here, "+" is defined as "times: a"+" O= aO= a so O= 1. In that case, the additive inverse, "-a", must be such that a"+" ("-a")= O which now is interpreted as a("-a")= 1 or "-a"= 1/a. As I said in my first response, "While there is a single additive identity (and it is NOT 0), each member of V has a different additive inverse". The additive inverse of a is 1/a. That does NOT exist for a= 0 so this does NOT define a vector space.

    If you restrict the set of vectors to the positive numbers only, then you have a vector space.
     
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