Isometries of the plane

Question

Let $m$ be an orientation-reversing isometry. Prove algebraically the $m^2$ is a translation.

What I attempted: We know that $m$ is an orientation-reversing isometry i.e. it is either a reflection or a glide-reflection.

I started by consider the case when $m$ is a reflection.

The matrix representing this transformation is given by $\begin{pmatrix} \cos \phi & \sin \phi\\ \sin \phi & -\cos \phi \end{pmatrix}$

I calculated the square of the above matrix and obtained the identity matrix, which means that $m^2$ is a translation of $v=\begin{pmatrix}0 \\ 0 \end{pmatrix}$

I am not sure if this is correct..

With the glide-reflection, I am not sure how to proceed, since I don't know how to express neither the endomorphism "glide-reflection" in terms of the endomorphism rotation, reflection, translation and neither the matrix representing this transformation.

user35603 · Accepted Answer · 2013-10-16T12:39:48.670

4

Hint: Consider an orientation-reversing isometry as $m:x \mapsto Ax+b$, where $A$ is a reflaction matrix (so, $A^2=I$), $b$ is a translation vector. And try to find $m^2$.

edited Oct 16 '13 at 12:39

answered Oct 16 '13 at 12:28

user35603

3,002

Oh so that is the glide reflection transformation right ? – amir Oct 16 '13 at 12:32
@user43418: Yes, if $b \neq \vec{0}$. – user35603 Oct 16 '13 at 12:33
And is what I did for the reflection correct ? – amir Oct 16 '13 at 12:35
@user43418: yes. – user35603 Oct 16 '13 at 12:37
Ok. Thank you for your help – amir Oct 16 '13 at 12:39

Isometries of the plane

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