Plane Reflection is Involution

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Theorem

Let $M$ be a straight line in the plane passing through the origin.

Let $s_M$ be the reflection of $\R^2$ in $M$.


Then $s_M$ is an involution in the sense that:

$s_M \circ s_M = I_{\R^2}$

where $I_{\R^2}$ is the identity mapping on $\R_2$.

That is:

$s_M = {s_M}^{-1}$


Proof

Let the angle between $M$ and the $x$-axis be $\alpha$.


Let $P = \tuple {x, y}$ be an arbitrary point in the plane.

Then from Equations defining Plane Reflection:

$\map {s_M} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$


Thus:

\(\ds \map {s_M \circ s_M} P\) \(=\) \(\ds \map {s_M} {\map {s_M} P}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {s_M} {\tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha} }\)
\(\ds \) \(=\) \(\ds {\tuple {\paren {x \cos 2 \alpha + y \sin 2 \alpha} \cos 2 \alpha + \paren {x \sin 2 \alpha - y \cos 2 \alpha} \sin 2 \alpha, \paren {x \cos 2 \alpha + y \sin 2 \alpha} \sin 2 \alpha - \paren {x \sin 2 \alpha - y \cos 2 \alpha} \cos 2 \alpha} }\)
\(\ds \) \(=\) \(\ds \tuple {x \cos^2 2 \alpha + y \sin 2 \alpha \cos 2 \alpha + x \sin^2 2 \alpha - y \sin 2 \alpha \cos 2 \alpha, x \sin 2 \alpha \cos 2 \alpha + y \sin^2 2 \alpha - x \sin 2 \alpha \cos 2 \alpha + y \cos^2 2 \alpha}\)
\(\ds \) \(=\) \(\ds \tuple {x \paren {\cos^2 2 + \sin^2 2 \alpha}, y \paren {\sin^2 2 \alpha + \cos^2 2 \alpha} }\) simplifying
\(\ds \) \(=\) \(\ds \tuple {x, y}\) Sum of Squares of Sine and Cosine

Hence the result.

$\blacksquare$


Sources

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