## The annihilator of a subset of a dual vector space

Let $V$ be a vector space over a field $F$ and let $\widehat{V} = \mathsf{Hom}_F(V,F)$ denote the dual vector space of $V$. Given $S \subseteq \widehat{V}$, define $\mathsf{Ann}(S) = \{v \in V \ |\ f(v) = 0\ \mathrm{for\ all}\ f \in S \}$. (This set is called the annihilator of $S$ in $V$.)

1. Prove that $\mathsf{Ann}(\widehat{S})$ is a subspace of $V$ for all $\widehat{S} \subseteq \widehat{V}$.
2. Suppose $\widehat{W}_1$ and $\widehat{W}_2$ are subspaces of $\widehat{V}$. Prove that $\mathsf{Ann}(\widehat{W}_1 + \widehat{W}_2) = \mathsf{Ann}(\widehat{W}_1) \cap \mathsf{Ann}(\widehat{W}_2)$ and $\mathsf{Ann}(\widehat{W}_1 \cap \widehat{W}_2) = \mathsf{Ann}(\widehat{W}_1) + \mathsf{Ann}(\widehat{W}_2)$.
3. Let $\widehat{W}_1, \widehat{W}_2 \subseteq \widehat{V}$ be subspaces. Prove that $\mathsf{Ann}(\widehat{W}_1) = \mathsf{Ann}(\widehat{W}_2)$ if and only if $\widehat{W}_1 = \widehat{W}_2$.
4. Prove that, for all $\widehat{S} \subseteq \widehat{V}$, $\mathsf{Ann}(\widehat{S}) = \mathsf{Ann}(\mathsf{span}\ \widehat{S})$.
5. Assume $V$ is finite dimensional with basis $B = \{v_i\}_{i=1}^n$, and let $\widehat{B} = \{\widehat{v}_i\}_{i=1}^n$ denote the basis dual to $B$. Prove that if $\widehat{S} = \{\widehat{v}_i\}_{i=1}^k$ for some $1 \leq k \leq n$, then $\mathsf{Ann}(\widehat{S}) = \mathsf{span} \{v_i\}_{i=k+1}^n$.
6. Assume $V$ is finite dimensional. Prove that if $\widehat{W} \subseteq \widehat{V}$ is a subspace, then $\mathsf{dim}\ \mathsf{Ann}(\widehat{W}) = \mathsf{dim}\ V - \mathsf{dim}\ \widehat{W}$.

[This needs to be cleaned up.]

Recall that a bounded lattice is a tuple $(L, \wedge, \vee, \top, \bot)$, where $\wedge$ and $\vee$ are binary operators on $L$ and $\top$ and $\bot$ are elements of $L$ satisfying the following:

1. $\wedge$ and $\vee$ are associative and commutative,
2. $\top$ and $\bot$ are identity elements with respect to $\wedge$ and $\vee$, respectively, and
3. $a \wedge (a \vee b) = a$ and $a \vee (a \wedge b) = a$ for all $a,b \in L$. (Called the “absorption laws”.)

If $L_1$ and $L_2$ are bounded lattices, a bounded lattice homomorphism is a mapping $\varphi : L_1 \rightarrow L_2$ that preserves the operators- $\varphi(a \wedge b) = \varphi(a) \wedge \varphi(b)$, $\varphi(a \vee b) = \varphi(a) \vee \varphi(b)$, $\varphi(\bot) = \bot$, and $\varphi(\top) = \top$. As usual, a lattice homomorphism which is also bijective is called a lattice isomorphism.

The interchangability of $\wedge$ and $\vee$ (and of $\bot$ and $\top$) immediately suggests the following definition. Given a bounded lattice $L$, we define a new lattice $\widehat{L}$ having the same base set as $L$ but with the roles of $\wedge$ and $\vee$ (and of $\bot$ and $\top$) interchanged. This $\widehat{L}$ is called the dual lattice of $L$.

Let $V$ be a vector space (of arbitrary dimension) over a field $F$. We let $\mathcal{S}_F(V)$ denote the set of all $F$-subspaces of $V$. We claim that $(\mathcal{S}_F(V), \cap, +, V, 0)$ is a bounded lattice. The least obvious of the axioms to check are the absorption laws. Indeed, note that for all subspaces $U,W \subseteq V$, we have $U \cap (U + W) = U$ and $U + (U \cap W) = U$.

Now let $V$ be a vector space (again of arbitrary dimension) over a field $F$, and let $\widehat{V} = \mathsf{Hom}_F(V,F)$ denote its dual space. If $S \subseteq \widehat{V}$ is an arbitrary subset and $\mathsf{Ann}(S)$ is defined as above, note that $f(0) = 0$ for all $f \in S$, and that if $x,y \in \mathsf{Ann}(S)$ and $r \in F$, we have $f(x+ry) = f(x)+rf(y) = 0$ for all $f \in S$. By the submodule criterion, $\mathsf{Ann}(S) \subseteq V$ is a subspace.

Now define $A : \mathcal{S}_F(\widehat{V}) \rightarrow \widehat{\mathcal{S}_F(V)}$ by $A(\widehat{W}) = \mathsf{Ann}(\widehat{W})$. We claim that if $V$ is finite dimensional, then $A$ is a bounded lattice homomorphism.

1. ($A(\widehat{0}) = V$) Note that for all $v \in V$, we have $\widehat{0}(v) = 0$. Thus $V = \mathsf{Ann}(\widehat{0}) = A(\widehat{0})$. ($\widehat{0}$ is the zero function $V \rightarrow F$.)
2. ($A(\widehat{V}) = 0$) Suppose there exists a nonzero element $v \in \mathsf{Ann}(\widehat{V})$. Then there exists a basis $E$ of $V$ containing $v$, and we may construct a homomorphism $\varphi : V \rightarrow F$ such that $\varphi(v) \neq 0$. In particular, $v \notin A(\widehat{V})$. On the other hand, it is certainly the case that $0 \in A(\widehat{V})$. Thus we have $A(\widehat{V}) = 0$.
3. ($A(\widehat{W}_1 + \widehat{W}_2) = A(\widehat{W}_1) \cap A(\widehat{W}_2)$) $(\subseteq)$ Let $v \in A(\widehat{W}_1 + \widehat{W}_2)$. Then for all $f + g \in \widehat{W}_1 + \widehat{W}_2$, we have $(f+g)(v) = f(v) + g(v) = 0$. In particular, if $f \in \widehat{W}_1$, then $f(v) = (f+0)(v) = 0$, so that $v \in A(\widehat{W}_1)$. Similarly, $v \in A(\widehat{W}_2)$, and thus $v \in A(\widehat{W}_1) \cap A(\widehat{W}_2)$. $(\supseteq)$ Suppose $v \in A(\widehat{W}_1) \cap A(\widehat{W}_2)$. Then for all $f+g \in \widehat{W}_1 + \widehat{W}_2$, we have $(f+g)(v) = f(v)+g(v) = 0$; thus $v \in A(\widehat{W}_1+\widehat{W}_2)$. Thus $A(\widehat{W}_1 + \widehat{W}_2) = A(\widehat{W}_1) \cap A(\widehat{W}_2)$.
4. ($A(\widehat{W}_1 \cap \widehat{W}_2) = A(\widehat{W}_1) + A(\widehat{W}_2)$) $(\supseteq)$ Suppose $v \in A(\widehat{W}_1)$. Then for all $f \in \widehat{W}_1$, $f(v) = 0$. In particular, for all $f \in \widehat{W}_1 \cap \widehat{W}_2$. Thus $v \in A(\widehat{W}_1 \cap \widehat{W}_2)$. Similarly we have $A(\widehat{W}_2) \subseteq A(\widehat{W}_1 \cap \widehat{W}_2)$; thus $A(\widehat{W}_1) + A(\widehat{W}_2) \subseteq A(\widehat{W}_1 \cap \widehat{W}_2)$. $(\subseteq)$ First, we claim that this inclusion holds for all pairs of one dimensional subspaces. If $\widehat{W}_1$ and $\widehat{W}_2$ intersect in a dimension 1 subspace (that is, $\widehat{W}_1 = \widehat{W}_2$), then certianly $A(\widehat{W}_1 \cap \widehat{W}_2) \subseteq A(\widehat{W}_1) + A(\widehat{W}_2)$. If they intersect trivially, then we have $\widehat{W}_1 = F\widehat{t}_1$ and $\widehat{W}_2 = F\widehat{t}_2$, and $A(\widehat{W}_1) = \mathsf{ker}\ \widehat{t}_1$ and $A(\widehat{W}_2) = \mathsf{ker}\ \widehat{t}_2$. Now $\widehat{t}_1$ and $\widehat{t}_2$ are nonzero linear transformations $V \rightarrow F$, and so by the first isomorphism theorem for modules their kernels have dimension $(\mathsf{dim}\ V) - 1$. Note that linear transformations $V \rightarrow F$ are realized (after fixing a basis) by matrices of dimension $1 \times \mathsf{dim}\ V$; in particular, if $\widehat{t}_1$ and $\widehat{t}_2$ have the same kernel, then they are row equivalent, and so are $F$-multiples of each other. Thus we have $A(\widehat{W}_1) + A(\widehat{W}_2) = V$. Now suppose $\widehat{W}_1 = \sum \widehat{W}_{1,i}$ and $\widehat{W}_2 = \sum \widehat{W}_{2,i}$ are sums of one dimensional subspaces. We have $A(\widehat{W}_1 \cap \widehat{W}_2) = A((\sum \widehat{W}_{1,i}) \cap (\sum \widehat{W}_{2,j}))$ $= A(\sum (\widehat{W}_{1,i} \cap \widehat{W}_{2,j}))$ $= \bigcap A(\widehat{W}_{1,i} \cap \widehat{W}_{2,j})$. From the one-dimensional case, this is equal to $\bigcap (A(\widehat{W}_{i,1} + A(\widehat{W}_{2,j}) = (\bigcap A(\widehat{W}_{1,i})) + (\bigcap A(\widehat{W}_{2,i}))$ $= A(\widehat{W}_1) + A(\widehat{W}_2)$. (Note that our proof depends on $V$ being finite dimensional.)

Thus $A$ is a bounded lattice homomorphism. We claim also that $A$ is bijective. To see surjectivity, let $W \subseteq V$ be a subspace. Define $\widehat{W} = \{ f \in \widehat{V} \ |\ \mathsf{ker}\ f \supseteq W \}$. We claim that $A(\widehat{W}) = W$. To see this, it is clear that $W \subseteq A(\widehat{A})$. Moreover, there is a mapping $f \in \widehat{W}$ whose kernel is exactly $W$, and thus $A(\widehat{W}) = W$. Before we show injectivity, we give a lemma.

Lemma: Let $\widehat{W} \subseteq \widehat{V}$ be a subspace with basis $\{\widehat{v}_i\}_{i=1}^k$, and extend to a basis $\{\widehat{v}_i\}_{i=1}^n$. Let $\{v_i\}_{i=1}^n$ be the dual basis to $\{\widehat{v}_i\}_{i=1}^n$, obtained using the natural isomorphism $V \cong \widehat{\widehat{V}}$. Then $A(\widehat{W}) = \mathsf{span}\ \{v_i\}_{i=k+1}^n$. Proof: Let $\sum \alpha_i v_i \in A(\widehat{W})$. In particular, we have $\widehat{v}_j(\sum \alpha_i v_i) = \alpha_j = 0$ for all $1 \leq j \leq k$. Thus $\sum \alpha_iv_i \in \mathsf{span}\ \{v_i\}_{i=k+1}^n$. Conversely, note that $\widehat{v}_j(v_i) = 0$ for all $k+1 \leq i \leq n$, so that $\mathsf{span}\ \{v_i\}_{i=k+1}^n \subseteq A(\widehat{W})$. $\square$

In particular, we have $\mathsf{dim}\ \widehat{W} + \mathsf{dim}\ A(\widehat{W}) = \mathsf{dim}\ V$. Now suppose $A(\widehat{W}) = \mathsf{span}\ \{v_i\}_{i=1}^k$, and extend to a basis $\{v_i\}_{i=1}^n$ of $V$. Let $\{\widehat{v}_i\}_{i=1}^n$ denote the dual basis. Note that for all $f \in \widehat{W}$, writing $f = \sum \alpha_i \widehat{v}_i$, we have $\alpha_j = f(v_j) = 0$ whenever $1 \leq j \leq k$. In particular, $\widehat{W} \subseteq \mathsf{span}\ \{\widehat{v}_i\}_{i=k+1}^n$. Condiering dimension, we have equality. Now to see injectivity for $A$, note that if $A(\widehat{W}_1) = A(\widehat{W}_2)$, then $\widehat{W}_1$ and $\widehat{W}_2$ share a basis- hence $\widehat{W}_1 = \widehat{W}_2$, and so $A$ is injective.

Thus, as lattices, we have $\mathcal{S}_F(\widehat{V}) \cong \widehat{\mathcal{S}_F(V)}$.

Finally, note that it is clear we have $\mathsf{Ann}(\mathsf{span}\ S) \subseteq \mathsf{Ann}(S)$. Conversely, if $v \in \mathsf{Ann}(S)$ and $f = \sum \alpha_i s_i \in \mathsf{span}\ S$, then $f(v) = 0$. Thus $\mathsf{Ann}(S) = \mathsf{Ann}(\mathsf{span}\ S)$.