(a) Show Ann(S) is a subspace of V.
(b) Let W1,W2⊆V∗ be subspaces. Show Ann(W1+W2)=Ann(W1)∩Ann(W2) and Ann(W1∩W2)=Ann(W1)+Ann(W2).
(c) Prove W1=W2⇔Ann(W1)=Ann(W2).
(d) Prove Ann(S)=Ann(span S).
(e) Assume V has for basis {v1,...,vn}. Prove that if S={v∗1,...,v∗k} then Ann(S) has basis {vk+1,...,vn}.
(f) Assume V is finite dimensional. Prove that if W∗⊆V∗ is a subspace then dim Ann(W∗)=dim V−dim W∗.
Proof: We shall approach this problem from another angle to obtain the final result first, and then recover the rest as implications. But first, (a):v1,v2∈Ann(S)⇒f(v1+αv2)=f(v1)+αf(v2)=0 ∀f∈S Let S be a finite set of linearly independent linear functionals v∗1,...,v∗k, perhaps a basis for an arbitrary subspace of V∗. Define a linear transformationF:V→FkF(v)=(v∗1(v),...,v∗k(v))Under this terminology, we can see Ann(S)=ker F. Letting A be the k×n reduced row echelon matrix of F from basis e1,...,en (of V) to e1,...,ek (of Fk) we observe the linear functionals manifest as linearly independent rows, so that the kernel is of dimension n−k=dim V−dim span S. This is (f), since (d) is immediately evident. (e) then follows easily, as the ⊇ argument for the basis is clear and dimensional restrictions show it spans.
When S⊆V, define the subspace Ann(S)={v∗∈V∗ | v∗(s)=0 for all s∈S} to be the dual notion of the Ann defined above. A parallel argument by evaluations at vi gives dim Ann(S)=dim V∗−dim span S. When S is a subspace of V or of V∗ we show Ann(Ann(S))=S since clearly ⊇ holds, and dimension restrictions show ⊆, which now easily gives (c).
We now show Ann(W1+W2)=Ann(W1)∩Ann(W2) when W1,W2 are subspaces of V∗ (or V). (⊇) Let v∈Ann(W1)∩Ann(W2). We have (w1+w2)(v)=w1(v)+w2(v)=0 (respectively, v(w1+w2)=v(w1)+v(w2)=0). (⊆) Let v∈Ann(W1+W2). We have w1(v)=(w1+0)(v)=0 and similarly w2(v)=0 (respectively, v(w1)=v(w1+0)=0 and v(w2)=0). Now, to finish, we haveAnn(W1∩W2)=Ann(W1)+Ann(W2)⇔Ann(Ann(W1∩W2))=Ann(Ann(W1)+Ann(W2))⇔W1∩W2=Ann(Ann(W1))∩Ann(Ann(W2))=W1∩W2 ◻
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