Is it a subspace or not?

1. Feb 28, 2008

karnten07

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. Feb 28, 2008

e(ho0n3

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.

3. Feb 28, 2008

karnten07

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.

4. Feb 28, 2008

karnten07

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?

5. Feb 28, 2008

karnten07

Oh does it mean that the scalars that can be applied are only the two elements 0 and 1 of F2?

6. Feb 28, 2008

e(ho0n3

7. Feb 28, 2008

karnten07

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??

8. Feb 28, 2008

e(ho0n3

Yes, 0 and 1 are the only allowed scalars. As long as F2 is a field, the vector space over it is well-defined.

9. Feb 28, 2008

karnten07

Great, thanks for clearing that up.

10. Feb 28, 2008

karnten07

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).