Dual Annihilators (11.3.3)

Dummit and Foote Abstract Algebra, section 11.3, exercise 3:

Let $S \subseteq V$ for $V$ some finite dimensional space. Define $\text{Ann}(S)=\{v∈V~|~f(v)=0\text{ for all }f∈S\}$.

(a) Show $\text{Ann}(S)$ is a subspace of $V$.
(b) Let $W_1,W_2 \subseteq V^*$ be subspaces. Show $\text{Ann}(W_1+W_2)=\text{Ann}(W_1)∩\text{Ann}(W_2)$ and $\text{Ann}(W_1∩W_2)=\text{Ann}(W_1)+\text{Ann}(W_2)$.
(c) Prove $W_1=W_2 ⇔ \text{Ann}(W_1)=\text{Ann}(W_2)$.
(d) Prove $\text{Ann}(S)=\text{Ann}(\text{span }S)$.
(e) Assume $V$ has for basis $\{v_1,...,v_n\}$. Prove that if $S=\{v_1^*,...,v_k^*\}$ then $\text{Ann}(S)$ has basis $\{v_{k+1},...,v_n\}$.
(f) Assume $V$ is finite dimensional. Prove that if $W^* \subseteq V^*$ is a subspace then $\text{dim }\text{Ann}(W^*) = \text{dim }V - \text{dim }W^*$.

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 transformation$$F : V → F^k$$$$F(v)=(v_1^*(v),...,v_k^*(v))$$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$$\text{Ann}(W_1∩W_2)=\text{Ann}(W_1)+\text{Ann}(W_2)⇔$$$$\text{Ann}(\text{Ann}(W_1∩W_2))=\text{Ann}(\text{Ann}(W_1)+\text{Ann}(W_2))⇔$$$$W_1∩W_2=\text{Ann}(\text{Ann}(W_1))∩\text{Ann}(\text{Ann}(W_2))=W_1∩W_2~\square$$