- #1

- 222

- 0

Let V be a vector space over the field R and S be a subspace of V. Now let f:S→R be a linear transformation. Prove that T can be extended to all the vector space V, i.e., there exists f':V→R such that f' restricted to S equals f.

3. The Attempt at a Solution .

In this exercise I am supposed to use Zorn's lemma to prove the existence of f'. I define a partially ordered set as follows: P={(U,T) where U is a subspace of V which contains the subspace S and T is a linear transformation T:U→R that satisfies T restricted to S equals f}. Then, I define the relation (U,T)≤(U',T') if and only if UcU' and T' restricted to U equals T. I need to verify that the relation I defined is indeed an order relation on P, but I didn't have problems with that; it's very easy to check for reflexivity, antisymmetry and transitivity.

I want to show that every totally ordered set of P has an upper bound, so let P' be a totally ordered set of P. If I choose the set X to be the union of all elements (U',T') of P', then (U',T')≤X for all (U',T') in P'. Now I can apply Zorn's lemma: P has a maximal element which I will denote M=(W,T'') where W is a subspace of V and T'' is a linear transformation, T'':W→R. By definition of maximal element, W is a subspace which includes all subspaces which contain the subspace S and T'' is a linear transformation that satisfies T'' restricted to S equals f. To complete the proof, I need to show that the domain of T'' is all the vector space V. Suppose V is not the domain of T''. So the domain W has to be a proper subspace of V. Then there is an element v in V which is not in W and Wu<v> is another subspace of V which includes W. I define h to be the linear transformation on Wu{v} as: h restricted to W equals T'' and h(v)=0. Then (W,T'')≤(Wu<v>,h). This is absurd since (W,T'') was the maximal element of P. Then V must be the domain of T'' and T'' restricted to S equals f so T'' is the linear transformation I was looking for.

Well, I know my proof is verbose, but is it correct?