View Full Version : Subspaces
karnten07
Feb28-08, 07:18 PM
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??
e(ho0n3
Feb28-08, 07:28 PM
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.
karnten07
Feb28-08, 07:32 PM
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.
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.
karnten07
Feb28-08, 07:52 PM
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?
karnten07
Feb28-08, 08:20 PM
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.
Oh does it mean that the scalars that can be applied are only the two elements 0 and 1 of F2?
e(ho0n3
Feb28-08, 08:25 PM
It seems to me that you have very little knowledge of what vector spaces are. Perhaps this will help: http://en.wikipedia.org/wiki/Vector_space.
karnten07
Feb28-08, 08:31 PM
It seems to me that you have very little knowledge of what vector spaces are. Perhaps this will help: http://en.wikipedia.org/wiki/Vector_space.
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??
e(ho0n3
Feb28-08, 08:36 PM
Yes, 0 and 1 are the only allowed scalars. As long as F2 is a field, the vector space over it is well-defined.
karnten07
Feb28-08, 08:44 PM
Yes, 0 and 1 are the only allowed scalars. As long as F2 is a field, the vector space over it is well-defined.
Great, thanks for clearing that up.
karnten07
Feb28-08, 08:51 PM
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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