Vector space basis proof

1. Dec 1, 2011

flyerpower

1. The problem statement, all variables and given/known data
Let S be any non-empty set, F be a field and V={ f : S -> F such that f(x) = 0 } be a vector space over F.
Let f[sub k] (x) : S -> F such that f[sub k] (x) = 1 for k=x, otherwise f[sub k] (x) = 0.

Prove that the set { f [sub k] } with k from S is a basis for the vector space V.

3. The attempt at a solution

I tried to sketch something but i am not sure i'm on the right path.

So, given B={ f [sub k] }, k from S, it is a basis for V if and only if B spans any vector from V and B is linearly independent.

Let g : S -> F be a vector from V, then g(x)=0 and a some scalars from F with i >= 1.

Then B spans g if and only if g = sum ( a * f ).
But g(x) = 0 so 0 = sum ( a * f ), so the vectors f are linearly independent.

So i'd say B is a basis for the vector V, but i'm not sure it's correct because i didn't make use of the definition of the function f.

Last edited: Dec 1, 2011
2. Dec 1, 2011

jgens

I am a little confused by your notation. Which is the correct V:
• V = {f:S → F | f = 0}
• V = {f:S → F | f(x0) = 0 for some x0 in S}

3. Dec 1, 2011

flyerpower

It's not specified, so i guess it's f(x) = 0 for all x in S.

4. Dec 1, 2011

jgens

I didn't think of this at first, but that seems to be problematic. For {fk} to be a basis, it needs to be a subset of V. But since V contains only the function that is 0 everywhere and fk is not the zero function, this is a contradiction.

Are you sure that's what is meant by V?

5. Dec 1, 2011

flyerpower

This is what concerned me too.
Honestly i don't quite understand the definition of V as it doesn't say anything clear about x in f(x), but, actually i think V is defined such that f(x)=0 for a finite number of elements in S.

6. Dec 1, 2011

jgens

I think I have it figured out. Use V = {f:S → F | f(x) = 0 for all but finitely many x}. Can you show that {fk} are a basis for V?

7. Dec 1, 2011

flyerpower

Well, the definition of V doesn't change the situation, the problem is that i don't know the dimension of V, is it finite?

8. Dec 1, 2011

jgens

The dimension of V does not matter. With this definition of V you can show that {fk} is a basis. And your proof earlier isn't quite right, so you'll need to improve that for this.

9. Dec 1, 2011

flyerpower

Ok, thank you, i'll take one more ride :).

10. Dec 2, 2011

flyerpower

I don't understand why V is the set of all functions such that f(x)=0 for a finite number of S.

For example if S={1,2}, does that mean that a vector in S is (f(1),f(2))? with f(1)=0, f(2)=b, with b in F. Or
f(1)=a, f(2)=b, with a,b in F.

The basis is {(f[sub 1](1),f[sub 1](2)),f[sub 2](1),f[sub 2](2))} = {(1,0),(0,1)}.
But i don't understand why f(x) must be 0 for some arbitrary points in S.

11. Dec 2, 2011

jgens

You are confused because you are not reading things correctly. If you use V = {f:S → F | f(x) = 0 for all but finitely many x} then the {fk} constructed in the posts above are a basis for V.

As for your particular objection, note that if S is finite and f is non-zero everywhere, then f(x) = 0 except at a finite number of points.

Last edited: Dec 2, 2011
12. Dec 2, 2011

flyerpower

You're right. I was confused because i thought f and f[sub k] are not the same functions. Now it makes sense, thank you.