Orthogonal Complements and the Lattice of Subspaces

We know that the poset of subspaces of a vector space $V$ 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 $x$ in a bounded lattice $L$ (one that has a top element ${1}$ and a bottom element ${0}$ ) is an element $\neg x\in L$ so that $x\vee\neg x=1$ and $x\wedge\neg x=0$ .

In particular, since the top subspace is $V$ itself, and the bottom subspace is $\mathbf{0}$ we can see that the orthogonal complement $U^\perp$ satisfies these properties. The intersection $U\cap U^\perp$ is trivial, since the inner product is positive-definite as a bilinear form, and the sum $U+U^\perp$ is all of $V$ , as we’ve seen.

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

First, let’s say we’ve got two subspaces $U\subseteq W$ of $V$ . I say that $W^\perp\subseteq U^\perp$ . Indeed, if $p$ is a vector in $W^\perp$ then it $\langle w,p\rangle=0$ for all $w\in W$ . But since any $u\in U$ is also a vector in $W$ , we can see that $\langle u,p\rangle=0$ , and so $p\in U^\perp$ as well. Thus orthogonal complementation is

Now let’s take a single subspace $U$ of $V$ , and let $u$ be a vector in $U$ . If $v$ is any vector in $U^\perp$ , then $\langle v,u\rangle=\overline{\langle u,v\rangle}=0$ by the (conjugate) symmetry of the inner product and the definition of $U^\perp$ . Thus $u$ is a vector in $\left(U^\perp\right)^\perp$ , and so $U\subseteq U^{\perp\perp}$ . Note that this much holds whether $V$ is finite-dimensional or not.

On the other hand, if $V$ is finite-dimensional we can take an orthonormal basis $\left\{e_i\right\}_{i=1}^n$ of $U$ and expand it into an orthonormal basis $\left\{e_i\right\}_{i=1}^d$ of all of $V$ . Then the new vectors $\left\{e_i\right\}_{i=n+1}^d$ form a basis of $U^\perp$ , so that $V=U\oplus U^\perp$ . A vector in $V$ is orthogonal to every vector in $U^\perp$ exactly when it can be written using only the first $n$ basis vectors, and thus lies in $U$ . That is, $U^{\perp\perp}=U$ when $V$ is finite-dimensional.

7 Comments »

[…] I want to prove two equations that hold in any orthocomplemented lattice. They are the famous DeMorgan’s […]

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[…] . Clearly the symmetry of the situation shows us that we only need to check one direction. So if , we know that , and also that . And thus we see that […]

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[…] subspace they span. The kernel of is contained in the kernel of , and orthogonal complements are order-reversing, which means that must lie within the span of the . That is, there must be real numbers so […]

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[…] can push further and make this into an orthocomplemented lattice. We define the orthocomplement of an idempotent […]

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[…] We’ve seen before that the power set — the set of all the subsets of — is an orthocomplemented lattice. That is, we can take meets (intersections) , joins (unions) and complements of subsets […]

Pingback by Algebras of Sets « The Unapologetic Mathematician | March 15, 2010 | Reply
Two minor remarks:

“The intersection $U cap U^\perp$ is empty”
0 is in it, so it’s not really “empty”.

In the 2nd line from below, I think $\{e_i\}_{i=n}^d$ should read $\{e_i\}_{i=n+1}^d$ .

Comment by wildildildlife | August 29, 2010 | Reply
Yes, “trivial” would be better, and the $n+1$ thing was an oversight. Thanks.

Comment by John Armstrong | August 29, 2010 | Reply

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