Property on glide reflections

Question

I need to prove that the conjugate of a glide reflection is a glide reflection.

What I have tried: Let $m: X= \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & -\cos \phi \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\begin{pmatrix} a_1\\ a_2 \end{pmatrix}$

and $n:X= \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} \cos \psi & \sin \psi \\ \sin \psi & -\cos \psi \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\begin{pmatrix} b_1\\ b_2 \end{pmatrix}$

I wanted to calculate $n \circ m \circ n^{-1}$ and show that it has the form of a glide reflection but what is $n^{-1}$ ?

Answer 1

Hint: The involved matrix, let's call her $M_{\psi}$, is invertible so simply solve $y=M_{\psi}x+b$ for $x$.

Answer 2

If $${\rm Ref}\ (\phi) =\begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & -\cos \phi \end{pmatrix} $$ then $m(X) = {\rm Ref}\ (\phi)X + A$ and $ n(X)={\rm Ref}\ (\psi)X+B$.

Note that ${\rm Ref}\ (\phi)$ is reflection wrt a line which contains $(\cos\ \phi/2, \sin\ \phi/2)$ and ${\rm Ref}\ (\theta_1){\rm Ref}\ (\theta_2$ is a $(\theta_1-\theta_2)$-rotation matrix.

And $$ {\rm Ref}\ (\theta)^2=I,\ n^{-1}(X)={\rm Ref}\ (\psi)(X-B)$$

Then $$ nmn^{-1}(X) ={\rm Ref}\ (\psi)[{\rm Ref}\ (\phi){\rm Ref}\ (\psi)(X-B) + A ] + B$$ $$={\rm Ref}\ (\psi){\rm Ref}\ (\phi){\rm Ref}\ (\psi)(X) + [ -{\rm Ref}\ (\psi){\rm Ref}\ (\phi){\rm Ref}\ (\psi)(B) + {\rm Ref}\ (\psi) A + B] $$

So here we have a claim : ${\rm Ref}\ (\psi){\rm Ref}\ (\phi){\rm Ref}\ (\psi)$ is a reflection. In general $${\rm Ref}\ (\psi){\rm Ref}\ (\phi){\rm Ref}\ (\theta) = {\rm Ref}\ (\psi - \phi + \theta)$$