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Orthogonal Transformations _ Benson and Grove on Finite Reflection Groups

  1. Mar 2, 2012 #1
    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 [itex]\in[/itex] O(V), then T is completely determined by its action on the basis vectors [itex]e_1[/itex] = (1,0) and [itex]e_2[/itex] = (0,1).

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

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

    Peter
     
  2. jcsd
  3. Mar 2, 2012 #2

    Fredrik

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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.
     
  4. Mar 2, 2012 #3
    Thanks Fredrik

    Definitely dealihng with case [itex] \mathbb{R^2}[/itex]

    Will just reflect on what you wrote!

    Still puzzling a bit about showing that [itex] T e_2 = \pm ( - \nu , \mu) [/itex]

    Peter
     
  5. Mar 2, 2012 #4

    Fredrik

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    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)##.
     
  6. Mar 2, 2012 #5
    Thanks for the clarification Fredrik

    Appreciate your help

    Peter, Math Hobbyist
     
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