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Reflections - Kane's Book

I am reading Kane - Reflection Groups and Invariant Theory and need help with property A-4 of the properties of reflections stated on page 7

(see attachment - Kane _ Reflection Groups and Invariant Theory - pages 6-7)

On page 6 Kane mentions he is working in http://latex.codecogs.com/png.latex? \ell dimensional Euclidean space ie http://latex.codecogs.com/png.latex? E = R^{\ell} where http://latex.codecogs.com/png.latex? R^{\ell} has the usual inner product (x,y).

In defining reflections with respect to vectors Kane writes:

" Given http://latex.codecogs.com/png.latex? 0 \ne \alpha \in E let http://latex.codecogs.com/png.latex?...lpha}\subset E be the hyperplane

http://latex.codecogs.com/png.latex?...alpha ) = 0 \}

We then define the reflection http://latex.codecogs.com/png.latex?...ngrightarrow E by the rules

http://latex.codecogs.com/png.latex?...a} \cdot x = x if http://latex.codecogs.com/png.latex? x \in H_{\alpha}

http://latex.codecogs.com/png.latex?...pha = - \alpha "

Then Kane states that the following property follows:

(A-4) If is an orthogonal automorphism of E then

**Can someone explicitly and formally demonstrate A-4?**

I am not really sure of the meaning of but I assume

And I assume that

Can anyone help me formulate an explicit and detailed proof of A-4

Peter

Peter

Re: Reflections - Kane's Book

first of all, realize that is an equality of two sets.

so suppose . then for some .

so and because is orthogonal, , that is

so , so .

now suppose we have , so that .

since is an orthogonal automorphism, it is bijective, and its inverse is also orthogonal.

so . but .

this means that so ,

which shows that , thus the two sets are equal.

the second equality, is an equality of functions, so we must show that the two functions have the same value at every point of .

so let . since is surjective, we can write for some ,

where . so then we have:

for all .

this shows the two functions are indeed equal on all of .