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Subspaces of Functions- Linear Algebra

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Which of the following are subspaces of F[R] = {f |f:R-->R}?
    a) U = {f e F[R]|f(-1)f(1)=0
    b) V = " |f(1)+f(2)=0
    c) S = " |f(x)=f(-x)
    d) T = " |f(1)<= 0


    2. Relevant equations



    3. The attempt at a solution

    I got S and V or c) and b), is that correct?

    I thought U or a) would be too, but just found out that it is not closed under addition:
    Lets say if: f(1)=2 f(-1)=0 whose product is 0, however another possibility is f(1)=0 f(-1)=a number (non-zero), so if you add them it would not be in V (2+0, 0 + a number) whose product would not be zero. Am I correct?

    I somehow know whether it is in the subspace or not, but no idea how to write a proof. How do I show it? Do I do as I typed above, like with actual number examples? or do I have to do it with symbols? ex. (f+g)(x) = f(x) + g(x) there closed under addition...like that?

    Thanks
     
  2. jcsd
  3. Sep 30, 2012 #2
    To show its NOT a vector space, what you've done suffices (although I'm not following your notations of those 4 sets). You only need one counterexample: so showing its not closed under addition suffices.

    To show that something IS a vector space (subspace in this case) however, is much harder. You must verify the 10 (if my memory serves me) axioms for vector space. But since you're trying to show its a subspace, many of these follow directly from U,V,S, and T inheriting their operations from F[r].
    As you mentioned, closure does not follow trivially (like associativity does). So go over the axioms of vector fields to see which ones don't follow trivially, and you must write a proof for each of these.

    For instance, if you think something IS a subspace, one of the things you must show is closure (as you've mentioned). So let 2 arbitrary elements be in your (supposed) subspace, and show that their 'sum' is in your subspace. Then take another arbitrary element in your subspace and show that its 'scalar product' is in your subspace. Then conclude that since the elements were arbitrarily chosen, closure holds for ALL elements of the subspace.

    You must do this for all of the axioms!

    Good luck! Hope this helps!!
     
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