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Set R^(2) with the usual vector addition forms an abelian group

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    the set R^(2) with the usual vector addition forms an abelian group. For a belongs to R and x=(x1,x2) belongs to R^(2) we put a *x :=(ax1,0),this defines a scalar multiplication R*R^2 ---R^2 (a,x)---a*x.
    determine which of the axioms defining a vector space hold for the abelian group R^2 with the scalar multiplication

    2. Relevant equations

    3. The attempt at a solution
    I know that a*x1=ax1 but a*x2=0?? It confused me . And how to use an axiom to define it ?
  2. jcsd
  3. Feb 25, 2009 #2


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    Re: vectors

    You don't "use an axiom to define it"- it is already defined. And "a*x1" doesn't mean anything- a*(x1, x2) is defined as (ax1, 0). You need to show that this still obeys the axioms defining a vector space. What are those axioms?
  4. Feb 25, 2009 #3
    Re: vectors

    oh,i see . i have to use ax1 and 0 to show 1st and 2nd distributivity law,identity elements and the compatibility are all exist ?
    Last edited: Feb 25, 2009
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