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.
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Two minor remarks:
“The intersection
is empty”
0 is in it, so it’s not really “empty”.
In the 2nd line from below, I think
should read
.
Yes, “trivial” would be better, and the
thing was an oversight. Thanks.