## Orthogonal Complements and the Lattice of Subspaces

We know that the poset of subspaces of a vector space is a lattice. Now we can define complementary subspaces in a way that doesn’t depend on any choice of basis at all. So what does this look like in terms of the lattice?

First off, remember that the “meet” of two subspaces is their intersection, which is again a subspace. On the other hand their “join” is their sum as subspaces. But now we have a new operation called the “complement”. In general lattice-theory terms, a complement of an element in a bounded lattice (one that has a top element and a bottom element ) is an element so that and .

In particular, since the top subspace is itself, and the bottom subspace is we can see that the orthogonal complement satisfies these properties. The intersection is trivial, since the inner product is positive-definite as a bilinear form, and the sum is all of , as we’ve seen.

Even more is true. The orthogonal complement is involutive (when is finite-dimensional), and order-reversing, which makes it an “orthocomplement”. In lattice-theory terms, this means that , and that if then .

First, let’s say we’ve got two subspaces of . I say that . Indeed, if is a vector in then it for all . But since any is also a vector in , we can see that , and so as well. Thus orthogonal complementation is

Now let’s take a single subspace of , and let be a vector in . If is any vector in , then by the (conjugate) symmetry of the inner product and the definition of . Thus is a vector in , and so . Note that this much holds whether is finite-dimensional or not.

On the other hand, if is finite-dimensional we can take an orthonormal basis of and expand it into an orthonormal basis of all of . Then the new vectors form a basis of , so that . A vector in is orthogonal to every vector in exactly when it can be written using only the first basis vectors, and thus lies in . That is, when is finite-dimensional.