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.

About these ads

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

12 Comments »

  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

  4. [...] sum must be orthogonal. Incidentally, this shows that the direct sum between a subspace and its orthogonal complement is also a direct sum of inner product [...]

    Pingback by The Category of Inner Product Spaces « The Unapologetic Mathematician | May 6, 2009 | Reply

  5. [...] 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 empty, since the inner product is [...]

    Pingback by Orthogonal Complements and the Lattice of Subspaces « The Unapologetic Mathematician | May 7, 2009 | Reply

  6. [...] Complementation is a Galois Connection We now know how to take orthogonal complements of subspaces in an inner product space. It turns out that this process (and itself again) forms an [...]

    Pingback by Orthogonal Complementation is a Galois Connection « The Unapologetic Mathematician | May 19, 2009 | Reply

  7. [...] (and, in particular, eigenspaces) of self-adjoint transformations. Specifically, the fact that the orthogonal complement of an invariant subspace is also [...]

    Pingback by Invariant Subspaces of Self-Adjoint Transformations « The Unapologetic Mathematician | August 11, 2009 | Reply

  8. [...] orthonormal basis of all of . Just set to be the span of all the new basis vectors, which is the orthogonal complement of the image of , and let be the inclusion of into . We can then combine to get a unitary [...]

    Pingback by The Singular Value Decomposition « The Unapologetic Mathematician | August 17, 2009 | Reply

  9. [...] vector in the subspace defined by the new one. That is, we want the new parallelepiped to span the orthogonal complement to the subspace we start [...]

    Pingback by The Hodge Star « The Unapologetic Mathematician | November 9, 2009 | Reply

  10. [...] kernel of consists of all vectors orthogonal to the gradient vector , and the line it spans is the orthogonal complement to the kernel. Similarly, the kernel of consists of all vectors orthogonal to each of the gradient [...]

    Pingback by Extrema with Constraints I « The Unapologetic Mathematician | November 25, 2009 | Reply

  11. [...] nonzero vector spans a line , and the orthogonal complement — all the vectors perpendicular to — forms an -dimensional subspace , which we can use [...]

    Pingback by Reflections « The Unapologetic Mathematician | January 18, 2010 | Reply

  12. [...] we just consider them as vector spaces, we already know this: the orthogonal complement is exactly the subspace we need, for . I say that if is a -invariant subspace of , then is as [...]

    Pingback by Invariant Forms « The Unapologetic Mathematician | September 27, 2010 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: