The Unapologetic Mathematician

Mathematics for the interested outsider

Orthogonal Complements

An important fact about the category of vector spaces is that all exact sequences split. That is, if we have a short exact sequence

\displaystyle\mathbf{0}\rightarrow U\rightarrow V\rightarrow W\rightarrow\mathbf{0}

we can find a linear map from W to V which lets us view it as a subspace of V, and we can write V\cong U\oplus W. When we have an inner product around and V is finite-dimensional, we can do this canonically.

What we’ll do is define the orthogonal complement of U\subseteq V to be the vector space

\displaystyle U^\perp=\left\{v\in V\vert\forall u\in U,\langle u,v\rangle=0\right\}

That is, U^\perp consists of all vectors in V perpendicular to every vector in U.

First, we should check that this is indeed a subspace. If we have vectors v,w\in U^\perp, scalars a,b, and a vector u\in U, then we can check

\displaystyle\langle u,av+bw\rangle=a\langle u,v\rangle+b\langle u,w\rangle=0

and thus the linear combination av+bw is also in U^\perp.

Now to see that U\oplus U^\perp\cong V, take an orthonormal basis \left\{e_i\right\}_{i=1}^n for U\subseteq V. Then we can expand it to an orthonormal basis \left\{e_i\right\}_{i=1}^d of V. But now I say that \left\{e_i\right\}_{i=n+1}^d is a basis for U^\perp. Clearly they’re linearly independent, so we just have to verify that their span is exactly U^\perp.

First, we can check that e_k\in U^\perp for any k between n+1 and d, and so their span is contained in U^\perp. Indeed, if u=u^ie_i is a vector in U, then we can calculate the inner product

\displaystyle\langle u^ie_i,e_k\rangle=\bar{u^i}\langle e_i,e_k\rangle=\bar{u^i}\delta_{ik}=0

since i\leq n and k\geq n+1. Of course, we omit the conjugation when working over \mathbb{R}.

Now, let’s say we have a vector v\in U^\perp\subseteq V. We can write it in terms of the full basis \left\{e_k\right\}_{k=1}^d as v^ke_k. Then we can calculate its inner product with each of the basis vectors of U as

\displaystyle\langle e_i,v^ke_k\rangle=v^k\langle e_i,e_k\rangle=v^k\delta_{ik}=v^i

Since this must be zero, we find that the coefficient v^i of e_i must be zero for all i between {1} and n. That is, U^\perp is contained within the span of \left\{e_i\right\}_{i=n+1}^d

So between a basis for U and a basis for U^\perp we have a basis for V with no overlap, we can write any vector v\in V uniquely as the sum of one vector from U and one from U^\perp, and so we have a direct sum decomposition as desired.

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May 4, 2009 - Posted by | Algebra, Linear Algebra


  1. The fact that every exact sequence splits is that every module is projective. Isn’t this the same as saying the ring in question (here a field) is semisimple?

    Comment by Zygmund | May 5, 2009 | Reply

  2. That sounds right, but I’m not really digging into ring theory like that.

    Comment by John Armstrong | May 5, 2009 | Reply

  3. Yeah, I was trying to remember something I read a while back from Cartan and Eilenberg. Anyway, so the property then is fairly unique since semisimple algebras are basically products of matrix algebras (over division rings though).

    Comment by Zygmund | May 5, 2009 | Reply

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