Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Every nonzero vector space can be viewed as a space of functions

  1. Jul 22, 2010 #1
    The problem statement, all variables and given/known data

    Let V be a nonzero vector space over a field F, and suppose that S is a bases for V. Let C(S,F) denote the vector space of all functions f ∈ Ω(S,F) (i.e. the set of functions from S to a field F) such that f(s) = 0 for all but a finite number of vectors in S. Let Ψ: C(S,F) → V be defined by Ψ(f) = 0 if f is the zero function, and Ψ(f) = Σ {s ∈ S, f(s) ≠ 0} f(s)s, otherwise. Prove that Ψ is an isomorphism.

    The attempt at a solution

    Okay, I've already proved that Ψ is linear, that it is 1-1, but I'm having troubles proving that it is onto. Here's what I've done:

    I'd like to be able to show that for any v ∈ V there is a f ∈ Ω(S,F) such that Ψ(f) = v. So v = (a1)s1 + (a2)s2 + ... where S = {s1, s2, ...} is a basis for V (I haven't been told whether V is finite or infinite dimensional). However Ψ(f), when f is nonzero, is a linear combination of FINITE number of elements of the basis. I do realize we could write Ψ(f) = f(s1)s1 + f(s2)s2 + ... where some of these will be zero as f is only nonzero at a finite number of them. See if v = (a1)s1 + a2(s2) + ... + (an)sn then I could define f to be f(si) = ai and I would be done. But the problem is that I can't see and can't show that Ψ(f) could ever equal a vector in v which is a linear combination of an infinite number of elements of the basis.

    Any help? This isn't homework, I'm taking a look at linear algebra on my own this summer. Thanks a lot!

    EDIT: I've figured it out!! :)

    The vectors in V will be written as a linear combination of FINITE number of vectors of S - this is by definition! I hadn't read a section on Maximal linearly independent sets (the course I'll be taking in the fall skips it) so I hadn't seen this result.
    Last edited: Jul 22, 2010
  2. jcsd
  3. Jul 23, 2010 #2
    Well, S here is the basis for both vector spaces, i.e., the free vector space on
    S has S as its basis, so the two should be isomorphic.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook