# Homework Help: 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).