# Orthogonal Transformations _ Benson and Grove on Finite Reflection Groups

• Math Amateur
In summary: Looks like you may be struggling with the concept of determinants. A determinant is a matrix that satisfies the following equation:$$\det(A)=\alpha_{ij}^{\mathrm{diag}}+\beta_{ijk}^{\mathrm{diag}}$$In the equation, the $\alpha_{ij}$s and $\beta_{ijk}$s are the determinant coefficients of A. The $\alpha_{ij}$s and $\beta_{ijk}$s are always non-zero, and the determinant of a matrix is always positive.To see that Te2 is the second column of T, you need to show that the equation ##(\alpha
Math Amateur
Gold Member
MHB
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

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.

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

Math Amateur said:
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
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)##.

Thanks for the clarification Fredrik

Peter, Math Hobbyist

## 1. What are orthogonal transformations?

Orthogonal transformations are transformations that preserve the length and angle of vectors. This means that the distance between points and the angles between lines remain unchanged after the transformation is applied.

## 2. Who are Benson and Grove?

Benson and Grove are two mathematicians who have extensively studied and researched finite reflection groups, which are groups of orthogonal transformations that can be generated by a finite set of reflections.

## 3. What are finite reflection groups?

Finite reflection groups are groups of orthogonal transformations that can be generated by a finite set of reflections. These groups have important applications in various areas of mathematics, such as geometry, group theory, and representation theory.

## 4. What is the significance of studying finite reflection groups?

Studying finite reflection groups allows us to better understand the properties and structures of these groups, which have important applications in various areas of mathematics. It also helps us to classify and characterize these groups, which can lead to new insights and discoveries in mathematics.

## 5. How are orthogonal transformations related to finite reflection groups?

Finite reflection groups are groups of orthogonal transformations, meaning that they preserve the length and angle of vectors. These groups are generated by a finite set of reflections, which are specific types of orthogonal transformations. Therefore, studying orthogonal transformations is crucial in understanding finite reflection groups.

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