# The Unapologetic Mathematician

## 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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Posted by John Armstrong | 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 | 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 | 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 | 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 | 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 | 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 | 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 | 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 | 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 | 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 | September 27, 2010 | Reply

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