Let be a vector space over a field and let denote the dual vector space of . Given , define . (This set is called the *annihilator* of in .)

- Prove that is a subspace of for all .
- Suppose and are subspaces of . Prove that and .
- Let be subspaces. Prove that if and only if .
- Prove that, for all , .
- Assume is finite dimensional with basis , and let denote the basis dual to . Prove that if for some , then .
- Assume is finite dimensional. Prove that if is a subspace, then .

[This needs to be cleaned up.]

Recall that a *bounded lattice* is a tuple , where and are binary operators on and and are elements of satisfying the following:

- and are associative and commutative,
- and are identity elements with respect to and , respectively, and
- and for all . (Called the “absorption laws”.)

If and are bounded lattices, a bounded lattice homomorphism is a mapping that preserves the operators- , , , and . As usual, a lattice homomorphism which is also bijective is called a lattice isomorphism.

The interchangability of and (and of and ) immediately suggests the following definition. Given a bounded lattice , we define a new lattice having the same base set as but with the roles of and (and of and ) interchanged. This is called the *dual lattice* of .

Let be a vector space (of arbitrary dimension) over a field . We let denote the set of all -subspaces of . We claim that is a bounded lattice. The least obvious of the axioms to check are the absorption laws. Indeed, note that for all subspaces , we have and .

Now let be a vector space (again of arbitrary dimension) over a field , and let denote its dual space. If is an arbitrary subset and is defined as above, note that for all , and that if and , we have for all . By the submodule criterion, is a subspace.

Now define by . We claim that if is finite dimensional, then is a bounded lattice homomorphism.

- () Note that for all , we have . Thus . ( is the zero function .)
- () Suppose there exists a nonzero element . Then there exists a basis of containing , and we may construct a homomorphism such that . In particular, . On the other hand, it is certainly the case that . Thus we have .
- () Let . Then for all , we have . In particular, if , then , so that . Similarly, , and thus . Suppose . Then for all , we have ; thus . Thus .
- () Suppose . Then for all , . In particular, for all . Thus . Similarly we have ; thus . First, we claim that this inclusion holds for all pairs of one dimensional subspaces. If and intersect in a dimension 1 subspace (that is, ), then certianly . If they intersect trivially, then we have and , and and . Now and are nonzero linear transformations , and so by the first isomorphism theorem for modules their kernels have dimension . Note that linear transformations are realized (after fixing a basis) by matrices of dimension ; in particular, if and have the same kernel, then they are row equivalent, and so are -multiples of each other. Thus we have . Now suppose and are sums of one dimensional subspaces. We have . From the one-dimensional case, this is equal to . (Note that our proof depends on being finite dimensional.)

Thus is a bounded lattice homomorphism. We claim also that is bijective. To see surjectivity, let be a subspace. Define . We claim that . To see this, it is clear that . Moreover, there is a mapping whose kernel is exactly , and thus . Before we show injectivity, we give a lemma.

Lemma: Let be a subspace with basis , and extend to a basis . Let be the dual basis to , obtained using the natural isomorphism . Then . Proof: Let . In particular, we have for all . Thus . Conversely, note that for all , so that .

In particular, we have . Now suppose , and extend to a basis of . Let denote the dual basis. Note that for all , writing , we have whenever . In particular, . Condiering dimension, we have equality. Now to see injectivity for , note that if , then and share a basis- hence , and so is injective.

Thus, as lattices, we have .

Finally, note that it is clear we have . Conversely, if and , then . Thus .