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Is it a subspace or not?

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Let (F2) ={0,1} denote the field with 2 elements.

    i) Let V be a vector space over (F2) . Show that every non empty subset W of V which is closed under addition is a subspace of V.

    ii) Write down all subsets of the vector space (F2)^2 over (F2) and underline those subsets which are subspaces.

    2. Relevant equations



    3. The attempt at a solution

    For i.) do i need to show that it is closed under scalar multiplication also? I don't understand how it is because for example 6x1=6 which is not of F2??
     
    Last edited: Feb 28, 2008
  2. jcsd
  3. Feb 28, 2008 #2
    To show that W is a subspace of V, it suffices to show that for two scalars in your field, a and b, and two vectors in W, v and w, av + bw is in W.

    6 x 1 = 6 is certainly not in F2, but it is in V (assuming V is R).

    I think your confusion is that you're letting V = F2.
     
  4. Feb 28, 2008 #3
    Yes you are right, thats what i was kind of thinking. Could anyone explain to me what it means that V is a vector space over F2? please.
     
  5. Feb 28, 2008 #4
    For part ii.) i have that F2^2 = {(0,0),(0,1),(1,0),(1,1)}

    So i assume that 0,0 is a subspace because it is the zero subspace. Do i say that (0,1) is a subspace also as this is the vector space of F2 itself?? Are (1,0) and (1,1) subspaces also?
     
  6. Feb 28, 2008 #5
    Oh does it mean that the scalars that can be applied are only the two elements 0 and 1 of F2?
     
  7. Feb 28, 2008 #6
  8. Feb 28, 2008 #7
    Yes, i have read about vector spaces and it does seem to me that in this case the vector space is over F2 and F2 only consists of 2 elements. So does this mean only 0 and 1 are the scalars that this vector space deals with, so to speak??
     
  9. Feb 28, 2008 #8
    Yes, 0 and 1 are the only allowed scalars. As long as F2 is a field, the vector space over it is well-defined.
     
  10. Feb 28, 2008 #9
    Great, thanks for clearing that up.
     
  11. Feb 28, 2008 #10
    So since 0 and 1 are the only scalars used on V, then i could simultaneously prove it is closed under scalar multiplication and show the existence of a multiplicative identity element (1).
     
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