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I am trying to prove that the composition of two reflections in non-parallel lines (i.e. lines that intersect) is a rotation.

From observation I can see that using $L_1$ as the $x$-axis and $L_2$ as the $y$-axis (so the angle $\theta = \frac{\pi}{2}$) that the composition $r_{L_1L_2}$ is a rotation around the origin (the point of intersect) by $\pi$ ($= 2\theta$).

How can I prove that the composition of any reflections (not just my example) is a rotation? Any pointers are appreciated.

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1 Answer 1

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Note $L_i^t=-L_i$. Moreover, $L_i^2=I$, since $L_i$ is a reflection. Therefore $L_1L_2(L_1L_2)^t=L_1L_2[(-L_2)(-L_1)]=L_1L_2L_2L_1=L_1IL_1=L_1^2=I$, and similary $(L_1L_2)^tL_2L_1=I$. Thus $(L_1L_2)^{-1}=(L_1L_2)^t$ and $\mathrm{det}(L_1L_2)=1$. Hence $L_1L_2$ is a rotation.

Note that in the case of the lines to be parallel, then the rotation is simply the identity matrix (trivial rotation).

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  • $\begingroup$ How do you know $L^t_i = -L_i$? $\endgroup$ Commented Mar 16, 2016 at 20:33
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    $\begingroup$ Write explicitely the matrix $L_i$: $L_i=\left(\begin{array}{cc}\cos2\theta_i&\sin2\theta_i\\\sin2\theta_i&-\cos2\theta_i\end{array}\right)$, where $\theta_i$ is the counterclockwise angle with respect to the x-axis of the line $\ell_i$ $\endgroup$ Commented Mar 16, 2016 at 20:39
  • $\begingroup$ Thanks! How come $(L_1L_2)^t = (-L_2)(-L_1)$ and doesn't just equal $L_2^tL_1^t$? $\endgroup$ Commented Mar 17, 2016 at 7:19
  • $\begingroup$ I only wrote the intermediate step $\endgroup$ Commented Mar 17, 2016 at 7:27

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