# Linear functional equivalence in vsp and subsp

JamesTheBond
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?

## Answers and Replies

Homework Helper
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.

Homework Helper
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.

JamesTheBond
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 $$B_W = { w_1, w_2, w_3, ..., w_n}$$, n = dim W <= dim V

$$B_W \subset B_V$$ for some Basis of V.

We know $$f(p*w1 + q*w2) = p*f(w1) + q*f(w2)$$ over all $$w1, w2 \subset W$$.

But what about all the elements r of V not in W (for $$r \in V and \notin W$$? 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?

Homework Helper
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.

JamesTheBond
Ok here is one:

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

This g is I think called a linear form and when it is restricted to W, g(w) = f(w)

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Homework Helper
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.

JamesTheBond
Is it possible to just define g(w') to be 0? It seems too simple.