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Help with vector spaces

  1. Aug 5, 2012 #1
    1. The problem statement, all variables and given/known data
    let [tex]S={(a_1,a_2):a_1,a_2 \in \mathbb{R}}[/tex] For [tex] (a_1,a_2),(b_1,b_2)\in{S}[/tex] and [tex]c\in\mathbb{R}[/tex] define [tex](a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2-b_2)[/tex] and [tex]c(a_1,a_2)=(ca_1,ca_2)[/tex].
    show that this is not a vector space


    2. Relevant equations
    vector space axioms


    3. The attempt at a solution
    this isn't an exercise in the book, but an example from the book that states that commutativity and associativity of addition and the distributive law all fail, so this in fact is not a vector space
    i tried working these out and i think i got commutativity one right
    because then you have [tex](a_1+b_1,a_2-b_2)[/tex] does not equal [tex](b_1+a_1,b_2-a_2)[/tex] is this correct?
    i got stuck on associativity, i worked it out but to me it seems that it does in fact hold true
    haven't check the distributive law though
    the book im using is linear algebra by friedberg, insel and spence, second edition
     
  2. jcsd
  3. Aug 5, 2012 #2

    HallsofIvy

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    Yes, that is correct.

    "Associativity of addition" would require that ((a1, b1)+ (a2, b2))+ (a3, b3)= (a1+ b1)+ ((a2, b2)+ (a3, b3)). What do you get for each of those?

    Distributivity requires that (a1, b1)((a2, b2)+ (a3, b3))= (a1,b1)(a2,b2)+ (a1,b1)(a3,b3). What do you get for each of those?
     
  4. Aug 5, 2012 #3
    nevermind i just checked my work again and i think i figured it out
    for associativity we end up with
    [tex](a_1+b_1+c_1,a_2-b_2-c_2)[/tex] on the left side while the right side gives us
    [tex](a_1+b_1+c_1,a_2-b_2+c_2)[/tex]
    which aren't equal, so this shows associativity of addition fails right?

    im going to work on the distributive law, if i get stuck on that ill post back on this thread for help

    first time working with vector spaces, so im just trying to make sure i get this right before i actually start my linear algebra class in the fall haha
     
  5. Aug 5, 2012 #4

    HallsofIvy

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    Yes, that is correct.

     
  6. Aug 5, 2012 #5
    thanks hallsofivy

    looking back into my book im not even sure if its the distributive law that fails
    my book says VS8 fails which is for a,b of elements in F(field) and each element x in V
    (a+b)x=ax+bx, is this the distributive law? it doesnt look like the one you posted hallsofivy

    anyways i checked it using x,y as the elements of the field
    [tex](x+y)(a_1,a_2)=(xa_1+ya_1,xa_2+ya_2)[/tex] and [tex]x(a_1,a_2)+y(a_1,a_2)=(xa_1,xa_2)+(ya_1,ya_2)[/tex]
    both sides are not equal so it fails that axiom
     
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