Translation as a product of two reflections.

Question

We have $f(x,y)=(x-3,y+4)$,then how can we express its translation as a product of two reflection.

MY TRY: This is not a linear transformation so, i can not use the concept of affine transformation.Thank you.

Answer 1

Let us consider straight lines with equations:

$$\tag{1}\begin{cases}L_1 \ \text{(in blue):} \ &y&=&\frac34x\\L_2 \ \text{(in red):} \ &y&=&\frac34x+\frac{25}{8}\end{cases}$$

as shown on the figure below. Let $S_i$ be the (orthogonal) symmetry with respect to $(L_i)$.

Being given an initial point, $M_1$, let $M_2=S_1(M_1)$ and

$$M_3=S_2(M_2)=S_2 \circ S_1(M_1)=T_{\vec{V}}(M_1) \ \ \ \iff \ \ \ \vec{M_1M_3}=\vec{V}$$

where $\vec{V}=\binom{-3}{ \ 4}$.

How theses straight lines have been chosen ? Their common normal vector is $\vec{V}$, and they have been taken in such a way that $(L_2)$ is the image of $(L_1)$ by a $\frac12\vec{V}$ translation. Any other group of lines of the form

$$\begin{cases}L_1 \ \text{(in blue):} \ &y&=&\frac34x+k\\L_2 \ \text{(in red):} \ &y&=&\frac34x+\frac{25}{8}+k\end{cases}$$

($k \in \mathbb{R}$) would answer the question.

In a reciprocal way, there are no other group of lines that are solutions to this problem.

Remark 1 : $S_1 \circ S_2$ (in that order) is the reciprocal translation by $-\vec{V}$.

Remark 2 : the fact that $S_2 \circ S_1=T_{\vec{V}}$ is possible if and only if $(L_2)$ is the image of $(L_1)$ by a $\frac12\vec{V}$ translation, is strikingly analogous to the fact that the composition of 2 symmetries wrt to intersecting lines $(L_1)$ and $(L_2)$ is a rotation $S_2 \circ S_1=R_{\theta}$ if and only if the angle between $(L_1)$ and $(L_2)$ is $\theta/2$.