Reflection of Plane in Line through Origin is Linear Operator

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.

$\blacksquare$