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Let be $L_1$ and $L_2$ two lines $$ L_1 = \{ x \in \mathbb{R}^2 | (n_1,x)=0 \} $$ $$L_2 =\{ x \in \mathbb{R}^2 | (n_2,x)=0 \} $$

and $S_1$ and $S_2$ reflections on those lines.

I want to prove that $S_1 \circ S_2 = S_2 \circ S_1 \leftrightarrow L_1=L_2 $ or $(n_1,n_2)=0 $

The statement is quite clear if I draw it out..I need help formulating the proof. Any help appreciated

wondering1123
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  • Using polar coordinates. $L_1:\phi=\phi_1$, $L_2:\phi=\phi_2$. Then $S_1:\phi\to 2\phi_1-\phi$, $S_2:\phi\to 2\phi_2-\phi$. I believe $(S_2 \circ S_1) (A) = S_2(S_1(A))$. Then $S_2\circ S_1: \phi\to 2\phi_2-2\phi_1+\phi$ , $S_1\circ S_2: \phi\to 2\phi_1-2\phi_2+\phi$. And $S_1\circ S_2=S_2\circ S_1 \Leftrightarrow 4(\phi_1-\phi_2)=2k\pi, k\in\mathbb{Z}$. If lines $L_1$ and $L_2$ are distinct, then $\phi_1-\phi_2 \neq k\pi, k\in \mathbb{Z}$, then $\phi_1-\phi_2=\frac{\pi}2+k\pi,k\in\mathbb{Z}$. Then $(n_1,n_2)=0$. – Ivan Kaznacheyeu Apr 26 '22 at 11:54

1 Answers1

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Here's one approach: by the way the lines are defined they intersect at the origin. By rotating the whole set-up you can assume that $n_1=(k,0)$ for some non-zero $k$, so $L_1$ is the 'y-axis'. Hence as a matrix transformation $$S_1=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.$$ Let $\theta$ be the angle from $L_1$ to $L_2$ measured anti-clockwise, and let $R_\theta$ be the rotation matrix. Then you can express $$S_2=R_\theta S_1R_\theta^{-1}.$$ Compute this explicitly as a matrix in the variable $\theta$ and then compare the matrices $S_1S_2$ and $S_2S_1$.

David Sheard
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