The Unapologetic Mathematician

Mathematics for the interested outsider

A Lemma on Reflections

Here’s a fact we’ll find useful soon enough as we talk about reflections. Hopefully it will also help get back into thinking about linear transformations and inner product spaces. However, if the linear algebra gets a little hairy (or if you’re just joining us) you can just take this fact as given. Remember that we’re looking at a real vector space V equipped with an inner product \langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle.

Now, let’s say \Phi is some finite collection of vectors which span V (it doesn’t matter if they’re linearly independent or not). Let \sigma be a linear transformation which leaves \Phi invariant. That is, if we pick any vector \phi\in\Phi then the image \sigma(\phi) will be another vector in \Phi. Let’s also assume that there is some n-1-dimensional subspace P which \sigma leaves completely untouched. That is, \sigma(v)=v for every v\in P. Finally, say that there’s some \alpha\in\Phi so that \sigma(\alpha)=-\alpha (clearly \alpha\notin P) and also that \Phi is invariant under \sigma_\alpha. Then I say that \sigma=\sigma_\alpha and P=P_\alpha.

We’ll proceed by actually considering the transformation \tau=\sigma\sigma_\alpha, and showing that this is the identity. First off, \tau definitely fixes \alpha, since


so \tau acts as the identity on the line \mathbb{R}\alpha. In fact, I assert that \tau also acts as the identity on the quotient space V/\mathbb{R}\alpha. Indeed, \sigma_\alpha acts trivially on P_\alpha, and every vector in V/\mathbb{R}\alpha has a unique representative in P_\alpha. And then \sigma acts trivially on P, and every vector in V/\mathbb{R}\alpha has a unique representative in P.

This does not, however, mean that \tau acts trivially on any given complement of \mathbb{R}\alpha. All we really know at this point is that for every v\in V the difference between v and \tau(v) is some scalar multiple of \alpha. On the other hand, remember how we found upper-triangular matrices before. This time we peeled off one vector and the remaining transformation was the identity on the remaining n-1-dimensional space. This tells us that all of our eigenvalues are {1}, and the characteristic polynomial is (T-1)^n, where n=\dim(V). We can evaluate this on the transformation \tau to find that (\tau-1)^n=0

Now let’s try to use the collection of vectors \Phi. We assumed that both \sigma and \sigma_\alpha send vectors in \Phi back to other vectors in \Phi, and so the same must be true of \tau. But there are only finitely many vectors (say k of them) in \Phi to begin with, so \tau must act as some sort of permutation of the k vectors in \Phi. But every permutation in S_k has an order that divides k!. That is, applying \tau k! times must send every vector in \Phi back to itself. But since \Phi is a spanning set for V, this means that \tau^{k!}=1, or that \tau^{k!}-1=0

So we have two polynomial relations satisfied by \tau, and \tau will clearly satisfy any linear combination of these relations. But Euclid’s algorithm shows us that we can write the greatest common divisor of these relations as a linear combination, and so \tau must satisfy the greatest common divisor of T^{k!}-1 and (T-1)^n. It’s not hard to show that this greatest common divisor is T-1, which means that we must have \tau-1=0 or \tau=1.

It’s sort of convoluted, but there are some neat tricks along the way, and we’ll be able to put this result to good use soon.

January 19, 2010 - Posted by | Algebra, Geometry, Linear Algebra


  1. […] is all the data we need to invoke our lemma, and conclude that is actually equal to . Specifying the action on the generators of is enough to […]

    Pingback by The Weyl Group of a Root System « The Unapologetic Mathematician | January 21, 2010 | Reply

  2. You’ve a “$lateh_\alpha$” which ought to be fixed.

    Comment by Blake Stacey | February 21, 2010 | Reply

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