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Determining if set is a real vector space

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data
    The set R^2 with addition defined by <x,y>+<a,b>=<x+a+1,y+b>, and scalar multiplication defined by r<x,y>= <rx+r-1,ry>.

    The answer in the back of the book says it is a vector space, but I am having trouble proving that 0+v=v and v+(-v)=0


    2. Relevant equations



    3. The attempt at a solution
    My guess is that you take v+(-1)v=0, applying the above definition of scalar multiplication to (-1)v, which gives 0=<-1,0>. Using that definition of 0, then 0+v=v, but I'm not sure if any of that is correct.
     
  2. jcsd
  3. Jun 21, 2009 #2
    The vector 0 is defined as a vector z such that z + v = v for each v in V. It is an axiom that this vector exists, so in showing that a space is a vector space, you must show that there exists such a vector. The easiest way is to directly use your formula for the sum of two vectors:
    <x, y> + <z1, z2> = <x + z1 + 1, y + z2> = <x, y>
    and solve for z, which must not depend on x or y.
     
  4. Jun 21, 2009 #3

    dx

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    Argh! Sorry, I should read more carefully..
     
  5. Jun 21, 2009 #4

    HallsofIvy

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    In this vector space, the "0 vector" is not <0, 0>.

    Any easy way of determining what the 0 vector, [itex]\vec{0}[/itex], is is to use the fact, true in any vector space, that [itex]0\vec{v}= \vec{0}[/itex]. (That follows from the "distributive law": [itex]a\vec{v}= (0+ a)\vec{v}= 0\vec{v}+ a\vec{v}[/itex].)

    Here, r<x,y>= <rx+r-1,ry>. What do you get if r= 0?
     
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