Operations on Linear Transformations in 2D

Lessons

Senior 4 · Mathematics

Operations on Linear Transformations in 2D

5 lessons

Lesson 2 of 5

Central symmetry and rotation about the origin

25 min

Turning without stretching

When a big wheel turns, every seat moves along a circle around the hub, and every spoke turns through the same angle. Nothing gets longer or shorter, and the seats keep their order around the rim. This is a rotation. Turning through exactly half a revolution is special: every point ends up on the opposite side of the centre at the same distance, which is called the central symmetry.

Both transformations, with the centre at the origin, are linear. This lesson derives their matrices and shows that multiplying rotation matrices gives you the addition formulas of trigonometry.

Definition

Central symmetry (point reflection, half-turn) with centre $C$ : each point $P$ is sent to the point $P^{'}$ such that $C$ is the midpoint of $P P^{'}$ . With centre $O$ ,

(x, y) \mapsto (-x, -y), \qquad \text{matrix } \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I .

With centre $C (a, b)$ , the midpoint condition gives $P' = (2a - x,\ 2b - y)$ ; this moves the origin unless $C = O$ , so only the central symmetry about $O$ is linear.

Example: about $O$ , $(3, -5) \mapsto (-3, 5)$ ; about $C (1, 2)$ , $(4, -1) \mapsto (2 - 4,\ 4 + 1) = (-2, 5)$ .

Definition

Rotation about $O$ through $\theta$ : the transformation that sends each point $P$ to $P^{'}$ with $O P^{'} = O P$ and the angle from $\overrightarrow{OP}$ to $\overrightarrow{OP'}$ equal to $\theta$ , measured anticlockwise (a negative $\theta$ turns clockwise). The origin stays fixed. We write $R_\theta$ for its matrix.

Example: the rotation through $90^\circ$ sends $(4, 2)$ to $(- 2, 4)$ .

The rotation matrix

Turn $\vec{e}_1 = (1, 0)$ through $\theta$ : it stays on the unit circle and lands at $(\cos\theta, \sin\theta)$ . Turn $\vec{e}_2 = (0, 1)$ , which starts at $90^\circ$ : it lands at angle $90^\circ + \theta$ , the point $(\cos(90^\circ + \theta), \sin(90^\circ + \theta)) = (-\sin\theta, \cos\theta)$ . These are the columns:

R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, \qquad R_\theta\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x\cos\theta - y\sin\theta \\ x\sin\theta + y\cos\theta \end{pmatrix}.

(A rotation is linear because it carries the parallelogram built on $\vec{u}$ and $\vec{v}$ to the parallelogram built on their images, and stretches nothing.) The central symmetry about $O$ is $R_{180^\circ} = -I$ : in the plane, a half-turn and a point reflection are the same transformation.

$\theta$	$90^\circ$	$180^\circ$	$270^\circ$	$60^\circ$	$45^\circ$
$R_\theta$	$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$	$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$	$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$	$\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$	$\frac{\sqrt{2}}{2}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$

Worked example 1: rotating a point

Find the image of $P (4, 2)$ under the rotation about $O$ through (a) $90^\circ$ (b) $60^\circ$ .

Solution. (a) $R_{90^\circ}\begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 - 2 \\ 4 + 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ .

(b) $R_{60^\circ}\begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \cdot \frac{1}{2} - 2 \cdot \frac{\sqrt{3}}{2} \\ 4 \cdot \frac{\sqrt{3}}{2} + 2 \cdot \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 2 - \sqrt{3} \\ 2\sqrt{3} + 1 \end{pmatrix} \approx \begin{pmatrix} 0.268 \\ 4.464 \end{pmatrix}$ .

Check the length: $(2 - \sqrt{3})^2 + (2\sqrt{3} + 1)^2 = (7 - 4\sqrt{3}) + (13 + 4\sqrt{3}) = 20 = 4^2 + 2^2$ . ✓

Important

Properties of rotations about $O$

$\det R_\theta = \cos^2\theta + \sin^2\theta = 1$ : areas and orientation are both kept.
$R_\alpha R_\beta = R_{\alpha + \beta}$ : turning through $\beta$ and then through $\alpha$ is turning through $\alpha + \beta$ . In particular rotations about $O$ commute, and $R_\theta^n = R_{n\theta}$ .
$R_\theta^{-1} = R_{-\theta} = R_\theta^{\mathsf{T}}$ : to undo a turn, turn back, and the inverse is simply the transpose.

A bonus: the addition formulas. Multiply out the left-hand side of $R_\alpha R_\beta = R_{\alpha + \beta}$ :

\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}\begin{pmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{pmatrix} = \begin{pmatrix} \cos\alpha\cos\beta - \sin\alpha\sin\beta & \cdots \\ \sin\alpha\cos\beta + \cos\alpha\sin\beta & \cdots \end{pmatrix}.

Comparing the first column with that of $R_{\alpha + \beta}$ gives $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ .

Worked example 2: recognising a rotation

Show that $M = \begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$ is a rotation about $O$ , find its angle, and find $M^{- 1}$ .

Solution. $M$ has the form $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ with $\cos\theta = 0.6$ , $\sin\theta = 0.8$ , and ${0.6}^{2} + {0.8}^{2} = 1$ . Both are positive, so $\theta$ is acute: $\theta = \cos^{-1} 0.6 \approx 53.1^\circ$ . The inverse is the rotation through $-53.1^\circ$ : $M^{-1} = M^{\mathsf{T}} = \begin{pmatrix} 0.6 & 0.8 \\ -0.8 & 0.6 \end{pmatrix}$ .

Worked example 3: the rotation that takes one point to another

A rotation about $O$ sends $A (3, 4)$ to $B (0, 5)$ . Find its matrix.

Solution. Both points are 5 units from $O$ , as they must be. Write $R_\theta\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3\cos\theta - 4\sin\theta \\ 3\sin\theta + 4\cos\theta \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}$ . Solving $3 c - 4 s = 0$ , $4 c + 3 s = 5$ (Cramer: $D = 9 + 16 = 25$ ) gives $c = \frac{20}{25} = 0.8$ and $s = \frac{15}{25} = 0.6$ . So $R = \begin{pmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{pmatrix}$ , a turn of about $36.9^\circ$ . Check: $(2.4 - 2.4,\ 1.8 + 3.2) = (0, 5)$ . ✓

Warning

Rotation or reflection? Look at the signs. A rotation matrix has equal diagonal entries and opposite off-diagonal entries, $\begin{pmatrix} c & -s \\ s & c \end{pmatrix}$ , and determinant $+ 1$ . A reflection matrix has opposite diagonal entries and equal off-diagonal entries, $\begin{pmatrix} c & s \\ s & -c \end{pmatrix}$ , and determinant $- 1$ . Also, the minus sign belongs in the top-right corner for an anticlockwise turn; putting it bottom-left turns the other way.

Tip

Remember the columns, not the matrix: " $\vec{e}_1$ goes to $(\cos\theta, \sin\theta)$ , and $\vec{e}_2$ is always a quarter-turn ahead of it, at $(-\sin\theta, \cos\theta)$ ."

Khan Academy: Linear transformation examples, rotations in R2 (about 18 min). Watch how the rotation matrix is built from where the two basis vectors land.

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Practice

Give exact values where you can.

Write the matrix of the rotation about $O$ through (a) $270^\circ$ (b) $30^\circ$ (c) $-45^\circ$ .
Find the image of $(6, - 2)$ under the central symmetry with centre (a) $O$ (b) $C (2, 1)$ .
Find the image of $(2, 4)$ under the rotation about $O$ through $30^\circ$ .
Show that $\begin{pmatrix} -0.8 & -0.6 \\ 0.6 & -0.8 \end{pmatrix}$ is a rotation and find its angle to one decimal place.
Find $R_{60^\circ}^3$ and describe it.
Use the addition formulas to prove that $R_\theta R_{-\theta} = I$ .
Find the rotation about $O$ that sends $(5, 0)$ to $(3, 4)$ , as a matrix.
Explain why the central symmetry about $O$ cannot be a reflection in a line.

Show answersHide answers

(a) $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ (b) $\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}$ (c) $\frac{\sqrt{2}}{2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ .
(a) $(- 6, 2)$ (b) $(4 - 6,\ 2 + 2) = (-2, 4)$ .
$\left(2 \cdot \frac{\sqrt{3}}{2} - 4 \cdot \frac{1}{2},\ 2 \cdot \frac{1}{2} + 4 \cdot \frac{\sqrt{3}}{2}\right) = (\sqrt{3} - 2,\ 1 + 2\sqrt{3}) \approx (-0.268, 4.464)$ .
Form $\begin{pmatrix} c & -s \\ s & c \end{pmatrix}$ with $c = - 0.8$ , $s = 0.6$ , $c^{2} + s^{2} = 1$ . $\sin\theta \gt 0$ and $\cos\theta \lt 0$ , so $\theta$ is in the second quadrant: $\theta = 180^\circ - 36.9^\circ = 143.1^\circ$ .
$R_{60^\circ}^3 = R_{180^\circ} = -I$ : the central symmetry about $O$ .
$R_\theta R_{-\theta} = R_{\theta + (-\theta)} = R_0 = I$ ; entry by entry, the top-left is $\cos\theta\cos\theta + \sin\theta\sin\theta = 1$ and the top-right is $\cos\theta\sin\theta - \sin\theta\cos\theta = 0$ , and so on.
The angle has $\cos\theta = \frac{3}{5}$ , $\sin\theta = \frac{4}{5}$ : $\begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$ .
Its determinant is $(- 1) (- 1) - 0 = 1$ , but every reflection in a line has determinant $- 1$ . (It also fixes no line pointwise: only $O$ is fixed.)

Quick check

What is the image of (4, 2) under the rotation of 90° anticlockwise about O?

Key points

The central symmetry about $O$ is $(x, y) \mapsto (-x, -y)$ , matrix $- I$ : the half-turn $R_{180^\circ}$ .
$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ : its columns are the images of $\vec{e}_1$ and $\vec{e}_2$ .
$\det R_\theta = 1$ , $R_\alpha R_\beta = R_{\alpha + \beta}$ , $R_\theta^{-1} = R_{-\theta} = R_\theta^{\mathsf{T}}$ .
Centres other than $O$ give transformations that 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: Ferris wheel, Nice, December 2008, Cayambe, Wikimedia Commons, CC BY-SA 3.0.
Video: Linear transformation examples: Rotations in R2, Khan Academy, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Turning without stretching

2The rotation matrix

Turning without stretching

The rotation matrix