(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):v_1,v_2∈\text{Ann}(S)⇒f(v_1+αv_2)=f(v_1)+αf(v_2)=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 → F^kF(v)=(v_1^*(v),...,v_k^*(v))Under this terminology, we can see \text{Ann}(S)=\text{ker }F. Letting A be the k × n reduced row echelon matrix of F from basis e_1,...,e_n (of V) to e_1,...,e_k (of F^k) we observe the linear functionals manifest as linearly independent rows, so that the kernel is of dimension n-k=\text{dim }V-\text{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 \text{Ann}(S)=\{v^*∈V^*~|~v^*(s)=0~\text{for all }s∈S\} to be the dual notion of the \text{Ann} defined above. A parallel argument by evaluations at v_i gives \text{dim Ann}(S)=\text{dim }V^*-\text{dim span }S. When S is a subspace of V or of V^* we show \text{Ann}(\text{Ann}(S))=S since clearly ⊇ holds, and dimension restrictions show ⊆, which now easily gives (c).
We now show \text{Ann}(W_1+W_2)=\text{Ann}(W_1)∩\text{Ann}(W_2) when W_1,W_2 are subspaces of V^* (or V). (⊇) Let v∈\text{Ann}(W_1)∩\text{Ann}(W_2). We have (w_1+w_2)(v)=w_1(v)+w_2(v)=0 (respectively, v(w_1+w_2)=v(w_1)+v(w_2)=0). (⊆) Let v∈\text{Ann}(W_1+W_2). We have w_1(v)=(w_1+0)(v)=0 and similarly w_2(v)=0 (respectively, v(w_1)=v(w_1+0)=0 and v(w_2)=0). Now, to finish, we have\text{Ann}(W_1∩W_2)=\text{Ann}(W_1)+\text{Ann}(W_2)⇔\text{Ann}(\text{Ann}(W_1∩W_2))=\text{Ann}(\text{Ann}(W_1)+\text{Ann}(W_2))⇔W_1∩W_2=\text{Ann}(\text{Ann}(W_1))∩\text{Ann}(\text{Ann}(W_2))=W_1∩W_2~\square
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