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Linear functional equivalence in vsp and subsp

  1. Mar 3, 2007 #1
    I am writing a solution for the following problem, I hope someone can correct it, because I am not sure what I am missing.

    Q. V is a finite dim. vsp over K, and W is a subspace of V. Let f be a linear functional on W. Show that there exists a linear functional g on V s. t. g(w)=f(w).

    Ans. So far:
    For all w in W, w = c_1*v_1 + c_2*v_2 + ... + c_n*v_n (where v_i for all i is the basis of V and c_i for all i are scalars part of K).

    f(p*w1 + q*w1) = p*f(w1) + q*f(w2)

    But since w for any w can be represented by a linear comnbination of the basis for V, there exists g, s.t.

    f(w) = c_1*g(v_1)+c_2*g(v_2)+....

    How can I complete this? Can I make the above assertion?
  2. jcsd
  3. Mar 4, 2007 #2

    matt grime

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    You're going from the wrong direction.

    You need to start from W and extend to V. Picking a basis for V won't help you. Try picking a basis of W, then using the standard results.

    You don't need to pick a basis explicitly either, if it helps. You can just think in terms of complementary subspaces.
  4. Mar 4, 2007 #3


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    Given any basis, BW, for W, there exist a basis, BV for V that contains BW. There are an infinite number of ways to define g so that it reduces to f on W.
  5. Mar 4, 2007 #4
    That is of course the basis extension principle (theorem?).

    So here is my revised take on the Ans (please tell me if I am reasoning it):

    Take some [tex]B_W = { w_1, w_2, w_3, ..., w_n} [/tex], n = dim W <= dim V

    [tex]B_W \subset B_V[/tex] for some Basis of V.

    We know [tex]f(p*w1 + q*w2) = p*f(w1) + q*f(w2)[/tex] over all [tex]w1, w2 \subset W[/tex].

    But what about all the elements r of V not in W (for [tex] r \in V and \notin W [/tex]? Those can be anything we want. Therefore there exists linear functional g on V s.t. when restricted to W, g(w) = f(w).

    Is this complete?
  6. Mar 4, 2007 #5

    matt grime

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    No. Saying 'they can be anything we want' doesn't prove anything at all. How do you know that there is a choice of g that makes it linear? Prove there is by writing down such a choice (by picking a basis of V if necessary). You were told, and understood the basis extension principle, so why didn't you use it?

    Forget about w1, w2 in W. You have nothing to prove for them, since f is already a linear functional on W. Just give me one example of a linear functional on V that agrees with f on W. Just one. One of the simplest ones, for instance.
  7. Mar 4, 2007 #6
    Ok here is one:

    If there exists linear functional g on V such that [tex] g(\{ w_1, w_2, w_3,..., w_n, v_n_+_1, v_n_+_2, ... , v_r \}) = k_1*w_1 + k_2*w_2 + .... + k_r*v_r [/tex] where r = dim V and for i=1 to r, [tex] k_i, w_i, v_i \in K field [/tex]

    This g is I think called a linear form and when it is restricted to W, g(w) = f(w)
    Last edited: Mar 4, 2007
  8. Mar 4, 2007 #7

    matt grime

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    Why is that equal to f? (Since you've not specifed what f is this is not an answer.) I see what you're getting at, but that doesn't mean that you get the benefit of the doubt. Here is the answer as I would write it:

    Pick any W' complementary to W. Any element v has a unique representation v=w+w' with w in W and w' in W'. Define g(w+w')=f(w).

    You should think if you understand it, and try to write out the basis dependent version.
  9. Mar 4, 2007 #8
    Is it possible to just define g(w') to be 0? It seems too simple.
  10. Mar 4, 2007 #9

    matt grime

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    Why? Look at your attempt. It's what you were attempting to do.
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