Operations on Linear Transformations in 2D

Lessons

Senior 4 · Mathematics

Operations on Linear Transformations in 2D

5 lessons

Lesson 1 of 5

Reflection in a line through the origin

25 min

The lake as a mirror

On a still morning the surface of Lake Kivu works as a mirror. Every point of a cloud or a hill appears below the waterline, on the vertical through the point and exactly as far below the water as the point is above it. The waterline is the mirror line.

In S2 you reflected shapes in the axes and in $y = x$ by counting squares. In the previous unit you wrote those reflections as matrices. This lesson finds the matrix of the reflection in any line through the origin, and with it the properties every reflection shares.

Definition

Reflection in a line: let $L$ be a line through $O$ . The reflection in $L$ sends each point $P$ to the point $P^{'}$ such that $L$ is the perpendicular bisector of $P P^{'}$ ; points of $L$ stay where they are. $L$ is the axis (mirror line) of the reflection.

Example: in the line $y = x$ , $P (5, 1)$ goes to $P^{'} (1, 5)$ : the midpoint $(3, 3)$ is on $y = x$ , and $\overrightarrow{PP'} = (-4, 4)$ is perpendicular to the direction $(1, 1)$ of the line.

A formula from the projection

Let $\hat{u}$ be a unit vector along $L$ . The foot of the perpendicular from the tip of $\vec{v}$ to $L$ is the tip of the projection $\vec{p} = (\vec{v} \cdot \hat{u})\,\hat{u}$ (the dot product gives the signed length along $L$ ). This foot is the midpoint of $\vec{v}$ and its image $S(\vec{v})$ , so $\vec{p} = \frac{1}{2}\big(\vec{v} + S(\vec{v})\big)$ and

S(\vec{v}) = 2(\vec{v} \cdot \hat{u})\,\hat{u} - \vec{v}.

In the diagram $L$ is $y = \frac{1}{2}x$ , with $\hat{u} = \frac{1}{\sqrt{5}}(2, 1)$ , and $\vec{v} = (5, 0)$ . Then $\vec{v} \cdot \hat{u} = \frac{10}{\sqrt{5}}$ , $\vec{p} = \frac{10}{5}(2, 1) = (4, 2)$ and $S(\vec{v}) = (8, 4) - (5, 0) = (3, 4)$ . Because the dot product is linear in $\vec{v}$ , so is $S$ : every reflection in a line through $O$ is a linear transformation.

Important

Theorem (matrix of a reflection). The reflection in the line through $O$ that makes an angle $\alpha$ with the positive $x$ -axis has matrix

S_\alpha = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}.

Proof. Take $\hat{u} = (\cos\alpha, \sin\alpha)$ in the formula. For $\vec{e}_1$ : $\vec{e}_1 \cdot \hat{u} = \cos\alpha$ , so $S(\vec{e}_1) = (2\cos^2\alpha - 1,\ 2\sin\alpha\cos\alpha) = (\cos 2\alpha,\ \sin 2\alpha)$ . For $\vec{e}_2$ : $\vec{e}_2 \cdot \hat{u} = \sin\alpha$ , so $S(\vec{e}_2) = (2\sin\alpha\cos\alpha,\ 2\sin^2\alpha - 1) = (\sin 2\alpha,\ -\cos 2\alpha)$ . These are the columns. $\blacksquare$

For the line $y = m x$ , with $m = \tan\alpha$ , the double-angle formulas $\cos 2\alpha = \frac{1 - m^2}{1 + m^2}$ and $\sin 2\alpha = \frac{2m}{1 + m^2}$ give

S = \frac{1}{1 + m^2}\begin{pmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{pmatrix}.

Special cases and shared properties

Axis	$\alpha$	$2\alpha$	Matrix
$x$ -axis	$0^\circ$	$0^\circ$	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
$y = x$	$45^\circ$	$90^\circ$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
$y$ -axis	$90^\circ$	$180^\circ$	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
$y = - x$	$135^\circ$	$270^\circ$	$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$

The general matrix shows what every reflection $S$ in a line through $O$ has in common:

$\det S = -\cos^2 2\alpha - \sin^2 2\alpha = -1$ : areas are kept, orientation is reversed.
$S^{2} = I$ (multiply out and use $\cos^2 + \sin^2 = 1$ ): reflecting twice brings every point back, so $S^{-1} = S$ .
$S$ is symmetric ( $S^{\mathsf{T}} = S$ ) and the sum of its diagonal entries is 0.
Vectors along the axis are fixed, $S(\vec{v}) = \vec{v}$ ; vectors perpendicular to it are reversed, $S(\vec{v}) = -\vec{v}$ .

Worked example 1: the reflection in $y = 2 x$

Find the matrix of the reflection in $y = 2 x$ and the image of $P (5, 0)$ . Check the answer geometrically.

Solution. With $m = 2$ : $S = \frac{1}{5}\begin{pmatrix} 1 - 4 & 4 \\ 4 & 4 - 1 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} -3 & 4 \\ 4 & 3 \end{pmatrix}$ . Then $S\begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$ , so $P^{'} = (- 3, 4)$ .

Check: the midpoint of $P P^{'}$ is $(1, 2)$ , which is on $y = 2 x$ ; $\overrightarrow{PP'} = (-8, 4)$ and $(-8, 4) \cdot (1, 2) = 0$ , so $P P^{'}$ is perpendicular to the axis. Also $O P = O P^{'} = 5$ . ✓

Worked example 2: an axis given by its angle

Find the image of $Q (2, 0)$ under the reflection in the line through $O$ at $30^\circ$ to the $x$ -axis.

Solution. $2\alpha = 60^\circ$ , so $S = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}$ and $Q' = S\begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}$ . Geometrically, $Q$ is at $0^\circ$ , the axis at $30^\circ$ , so the image is at $60^\circ$ , still 2 units from $O$ : $(2\cos 60^\circ,\ 2\sin 60^\circ) = (1, \sqrt{3})$ . ✓

Worked example 3: recognising a reflection

Show that $M = \begin{pmatrix} 0.6 & 0.8 \\ 0.8 & -0.6 \end{pmatrix}$ is a reflection and find its axis.

Solution. $M$ has the form $\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$ with $\cos 2\alpha = 0.6$ and $\sin 2\alpha = 0.8$ (and ${0.6}^{2} + {0.8}^{2} = 1$ ), so it is a reflection; as a check, $\det M = -0.36 - 0.64 = -1$ . Its axis is the set of fixed vectors: $M\vec{v} = \vec{v}$ gives $0.6 x + 0.8 y = x$ , so $0.8 y = 0.4 x$ and the axis is $y = \frac{1}{2}x$ . (This is the line of the diagram: $M$ sends $(5, 0)$ to $(3, 4)$ .)

Warning

Only lines through the origin give linear reflections. The reflection in $y = 1$ sends $(0, 0)$ to $(0, 2)$ , so it is not linear and has no $2 \times 2$ matrix. Also, do not use $\alpha$ where the matrix needs $2\alpha$ : for the axis at $30^\circ$ the entries are $\cos 60^\circ$ and $\sin 60^\circ$ .

Tip

Three quick tests catch most wrong reflection matrices: the determinant must be $- 1$ , the diagonal entries must add up to $0$ , and the two off-diagonal entries must be equal.

Joel Speranza Math: Reflecting in the line y = mx with matrix transformations (about 7 min). Watch how the images of the two basis vectors give the matrix of the reflection.

✦

Practice

Work each one, then open the answers.

Write the matrix of the reflection in (a) $y = - x$ (b) the line through $O$ at $60^\circ$ to the $x$ -axis (c) $y = 3 x$ .
Find the image of $(4, 2)$ under the reflection in $y = 3 x$ , and check that its distance from $O$ is unchanged.
Show that $\begin{pmatrix} -0.8 & 0.6 \\ 0.6 & 0.8 \end{pmatrix}$ is the matrix of a reflection and find its axis.
A reflection in a line $y = m x$ sends $(5, 0)$ to $(3, 4)$ . Find $m$ .
Prove that $S_\alpha^2 = I$ by multiplying the matrices.
Use $S(\vec{v}) = 2(\vec{v} \cdot \hat{u})\hat{u} - \vec{v}$ to reflect $\vec{v} = (1, 7)$ in the line through $O$ in the direction $(3, 4)$ .
Find the image of the $x$ -axis under the reflection in $y = 2 x$ , as an equation.

Show answersHide answers

(a) $\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ (b) $2\alpha = 120^\circ$ : $\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$ (c) $\frac{1}{10}\begin{pmatrix} -8 & 6 \\ 6 & 8 \end{pmatrix}$ .
$\frac{1}{10}(-32 + 12,\ 24 + 16) = (-2, 4)$ . $\|(4, 2)\| = \|(-2, 4)\| = \sqrt{20}$ . ✓ (Midpoint $(1, 3)$ is on $y = 3 x$ .)
$\cos 2\alpha = -0.8$ , $\sin 2\alpha = 0.6$ , $0.64 + 0.36 = 1$ ; $\det = -0.64 - 0.36 = -1$ . Fixed vectors: $- 0.8 x + 0.6 y = x$ , so $y = 3 x$ .
The midpoint $(4, 2)$ of $(5, 0)$ and $(3, 4)$ lies on the axis, so $m = \frac{2}{4} = \frac{1}{2}$ .
The top-left entry of $S_\alpha^2$ is $\cos^2 2\alpha + \sin^2 2\alpha = 1$ ; the top-right is $\cos 2\alpha \sin 2\alpha - \sin 2\alpha \cos 2\alpha = 0$ ; similarly the others give $0$ and $1$ .
$\hat{u} = (0.6, 0.8)$ , $\vec{v} \cdot \hat{u} = 6.2$ : $S(\vec{v}) = (7.44, 9.92) - (1, 7) = (6.44, 2.92)$ . Check: ${6.44}^{2} + {2.92}^{2} = 50 = 1^{2} + 7^{2}$ .
$\vec{e}_1$ goes to $(- 0.6, 0.8)$ , so the $x$ -axis goes to the line through $O$ and $(- 3, 4)$ : $y = -\frac{4}{3}x$ .

Quick check

What is the determinant of the matrix of any reflection in a line through the origin?

Key points

The reflection in a line through $O$ at angle $\alpha$ has matrix $\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$ ; for $y = m x$ it is $\frac{1}{1 + m^2}\begin{pmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{pmatrix}$ .
Vector form: $S(\vec{v}) = 2(\vec{v} \cdot \hat{u})\hat{u} - \vec{v}$ .
Every such reflection has $\det S = -1$ , $S^{2} = I$ , $S^{\mathsf{T}} = S$ ; it fixes its axis and reverses the perpendicular direction.
Reflections in lines not through $O$ are not linear.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Photo: View of the sky and its reflection on Lake Kivu, MONUSCO / Myriam Asmani, Wikimedia Commons, CC BY-SA 2.0.
Video: Reflecting in the line y = mx with Matrix Transformations, Joel Speranza Math, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1The lake as a mirror

2A formula from the projection

3Special cases and shared properties

The lake as a mirror

A formula from the projection

Special cases and shared properties