# Orthogonal Transformations _ Benson and Grove on Finite Reflection Groups

1. Mar 2, 2012

### Math Amateur

I am reading Grove and Benson's book on Finite Reflection Groups and am struggling with some of the basic linear algebra.

Some terminology from Grove and Benson:

V is a real Euclidean vector space

A transformation of V is understood to be a linear transformation

The group of all orthogonal transformations of V will be denoted O(V)

Then in chapter 2, Grove and Benson write the following:

If T $\in$ O(V), then T is completely determined by its action on the basis vectors $e_1$ = (1,0) and $e_2$ = (0,1).

If T$e_1 = ( \mu , \nu )$, then ${\mu}^2 + {\nu}^2 = 1$ and $T e_2 = \pm ( - \nu , \mu)$

Peter

2. Mar 2, 2012

### Fredrik

Staff Emeritus
Since your basis vectors only have two components, I assume that we're now dealing with the case V=ℝ2.

$$\begin{pmatrix}\mu\\ \nu\end{pmatrix}=Te_1=\begin{pmatrix}T_{11} & T_{12}\\ T_{21} & T_{22}\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}T_{11}\\ T_{21}\end{pmatrix}$$ This is the first column of the matrix T. The columns of an orthogonal matrix are orthonormal. This is easy to see from the condition $T^T T=1$. So the orthogonality of T implies that $\mu^2+\nu^2=1$, and also that the second column has norm 1 and is orthogonal to the first.

3. Mar 2, 2012

### Math Amateur

Thanks Fredrik

Definitely dealihng with case $\mathbb{R^2}$

Will just reflect on what you wrote!

Still puzzling a bit about showing that $T e_2 = \pm ( - \nu , \mu)$

Peter

4. Mar 2, 2012

### Fredrik

Staff Emeritus
You know that Te2 is the second column of T, that $(\mu,\nu)$ is the first column, and that the columns are orthonormal. So you only have to prove that if $(\alpha,\beta)\cdot (\mu,\nu)=0$ and $|(\alpha,\beta)|=1$, then $(\alpha,\beta)=\pm(-\nu,\mu)$.

5. Mar 2, 2012

### Math Amateur

Thanks for the clarification Fredrik