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).

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