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Vector space basis proof

  1. Dec 1, 2011 #1
    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. jcsd
  3. Dec 1, 2011 #2

    jgens

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    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}
     
  4. Dec 1, 2011 #3
    It's not specified, so i guess it's f(x) = 0 for all x in S.
     
  5. Dec 1, 2011 #4

    jgens

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    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?
     
  6. Dec 1, 2011 #5
    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.
     
  7. Dec 1, 2011 #6

    jgens

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    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?
     
  8. Dec 1, 2011 #7
    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?
     
  9. Dec 1, 2011 #8

    jgens

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    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.
     
  10. Dec 1, 2011 #9
    Ok, thank you, i'll take one more ride :).
     
  11. Dec 2, 2011 #10
    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.
     
  12. Dec 2, 2011 #11

    jgens

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    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
  13. Dec 2, 2011 #12
    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.
     
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