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Another exercise on vector space

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine if the following set is a vector space under the the given operations.

    The set V of all pairs of real numbers of the form (1,x) with the operations:

    (1, y) + (1, y') = (1, y + y')

    k(1, y) = (1, ky)


    2. Relevant equations


    3. The attempt at a solution

    Axiom 4: There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u.

    u = (1, y)

    (1, y) + (1, 0) = (1, y+ 0) = (1, u) = u

    May I consider (1,0) as the zero vector??? I have a doubt if the zero vector 0 has to be (0,0) always or the zero can be defined as any vector (in this case (1,0) ) that remains the vector untouched under addition.

    Axim 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

    u = (1,x)
    -u = (-1)(1,x) = (1, -x) by definition of scalar multiplication given for the set V. (Is this correct?)

    then u + (-u) = (1,x) + (-1)(1,x) = (1,x) + (1,-x) = (1,x+(-x)) = (1,0) = 0

    in this case (1,0) is the zero vector.
     
  2. jcsd
  3. Aug 14, 2012 #2

    Simon Bridge

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    In this case (0,0) cannot be the zero vector, since (0,0) is not in V.

    yep - defn was k(1,y)=(1,ky) so put k=-1 ... but just like "0" does not need to be (0,0), so -u need not be literally "-1.u". It's a notation. However:

    ... well done. You have discovered that the negative vector [-u] is the same as -1.u

    You'd normally want to say:

    if v=(1,y) is the negative of u=(1,x), then u+v=0

    observe: u+v=0 iff y=-x, therefore: -u=(1,-x)=-1.(1,x)=-1.u
     
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