Orthogonal Transformations _ Benson and Grove on Finite Reflection Groups

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SUMMARY

The discussion focuses on the concepts presented in Grove and Benson's book on Finite Reflection Groups, specifically regarding orthogonal transformations in a real Euclidean vector space V, denoted as O(V). It is established that for a transformation T in O(V), the action on basis vectors e_1 and e_2 determines T completely. The participants confirm that if Te_1 = (μ, ν), then it follows that μ² + ν² = 1, and they seek to prove that Te_2 = ±(-ν, μ) based on the orthonormality of the transformation matrix.

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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 Te_1 = ( \mu , \nu ), then {\mu}^2 + {\nu}^2 = 1 and T e_2 = \pm ( - \nu , \mu)

Can someone please help by proving why the last statement is true?

Peter
 
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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

Appreciate your help

Peter, Math Hobbyist
 
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