Reflection of Plane in Line through Origin is Linear Operator

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Theorem

Let $M$ be a straight line in the plane $\R^2$ passing through the origin.

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


Then $s_M$ is a linear operator for every straight line $M$ through the origin.


Proof

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


To prove that $s_M$ is a linear operator it is sufficient to demonstrate that:

$(1): \quad \forall P_1, P_2 \in \R^2: \map {s_M} {P_1 + P_2} = \map {s_M} {P_1} + \map {s_M} {P_2}$
$(2): \quad \forall \lambda \in \R: \map {s_M} {\lambda P_1} = \lambda \map {s_M} {P_1}$


So, let $P_1 = \tuple {x_1, y_1}$ and $P_2 = \tuple {x_2, y_2}$ be arbitrary points in the plane.

\(\ds \map {s_M} {P_1 + P_2}\) \(=\) \(\ds \tuple {\paren {x_1 + x_2} \cos 2 \alpha + \paren {y_1 + y_2} \sin 2 \alpha, \paren {x_1 + x_2} \sin 2 \alpha - \paren {y_1 + y_2} \cos 2 \alpha}\) Equations defining Plane Reflection: Cartesian
\(\ds \) \(=\) \(\ds \tuple {x_1 \cos 2 \alpha + y_1 \sin 2 \alpha, x_1 \sin 2 \alpha - y_1 \cos 2 \alpha} + \tuple {x_2 \cos 2 \alpha + y_2 \sin 2 \alpha, x_2 \sin 2 \alpha - y_2 \cos 2 \alpha}\)
\(\ds \) \(=\) \(\ds \map {s_M} {P_1} + \map {s_M} {P_2}\) Equations defining Plane Reflection: Cartesian


and:

\(\ds \forall \lambda \in \R: \, \) \(\ds \map {s_M} {\lambda P_1}\) \(=\) \(\ds \map {s_M} {\lambda \tuple {x_1, y_1} }\) Definition of $P_1$
\(\ds \) \(=\) \(\ds \tuple {\lambda x_1 \cos 2 \alpha + \lambda y_1 \sin 2 \alpha, \lambda x_1 \sin 2 \alpha - \lambda y_1 \cos 2 \alpha}\) Equations defining Plane Reflection: Cartesian
\(\ds \) \(=\) \(\ds \lambda \tuple {x_1 \cos 2 \alpha + y_1 \sin 2 \alpha, x_1 \sin 2 \alpha - y_1 \cos 2 \alpha}\)
\(\ds \) \(=\) \(\ds \lambda \map {s_M} {P_1}\) Equations defining Plane Reflection: Cartesian

Hence the result.

$\blacksquare$


Sources

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