Operations on Linear Transformations in 2D

Lessons

Senior 4 · Mathematics

Operations on Linear Transformations in 2D

5 lessons

Lesson 4 of 5

Operations on linear transformations and composites

25 min

Two mirrors make a turn

Look into a kaleidoscope and you see one small scene repeated around a centre. Inside are mirrors set at fixed angles. A copy seen after one reflection is back to front; a copy seen after two reflections is the right way round again, only turned. Two reflections have combined into a rotation.

This lesson combines linear transformations in three ways (adding them, multiplying them by scalars and doing one after another) and uses matrices to predict the result. You will prove the kaleidoscope's rule exactly: the angle of the rotation is twice the angle between the mirrors.

Definition

Operations on linear transformations: let $f$ and $g$ be linear transformations of $\mathbb{R}^2$ with matrices $A$ and $B$ , and let $k$ be real.

Operation	Definition	Matrix
Sum	$(f + g)(\vec{v}) = f(\vec{v}) + g(\vec{v})$	$A + B$
Scalar multiple	$(kf)(\vec{v}) = k\,f(\vec{v})$	$k A$
Composite (" $g$ then $f$ ")	$(f \circ g)(\vec{v}) = f(g(\vec{v}))$	$A B$
Power	$f^n = f \circ f \circ \cdots \circ f$ ( $n$ times)	$A^{n}$
Inverse (when $\det A \ne 0$ )	$f^{-1}(f(\vec{v})) = \vec{v}$	$A^{- 1}$

Each result is again linear. For the composite: $f(g(\vec{u} + \vec{v})) = f(g(\vec{u}) + g(\vec{v})) = f(g(\vec{u})) + f(g(\vec{v}))$ , and $f(g(k\vec{u})) = f(kg(\vec{u})) = kf(g(\vec{u}))$ .

Example: for $f(x, y) = (x + y,\ 2y)$ and $g(x, y) = (x - y,\ x)$ , $(2f - g)(x, y) = (2x + 2y - x + y,\ 4y - x) = (x + 3y,\ -x + 4y)$ , with matrix $2\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} - \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ -1 & 4 \end{pmatrix}$ .

Worked example 1: sums and composites

$f(x, y) = (2x - y,\ x + 3y)$ and $g(x, y) = (y,\ -x)$ . Find the matrices of $f + g$ , $f \circ g$ and $g \circ f$ , and the image of $(1, 1)$ under $f \circ g$ .

Solution. $A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ ( $g$ is the rotation through $-90^\circ$ ).

A + B = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}, \quad AB = \begin{pmatrix} 1 & 2 \\ -3 & 1 \end{pmatrix}, \quad BA = \begin{pmatrix} 1 & 3 \\ -2 & 1 \end{pmatrix}.

$(f \circ g)(1, 1) = AB\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ . Check directly: $g (1, 1) = (1, - 1)$ , then $f(1, -1) = (2 + 1,\ 1 - 3) = (3, -2)$ . ✓ Since $AB \ne BA$ , the order of composition matters.

Composites of rotations and reflections

Write $R_\theta$ for the rotation about $O$ through $\theta$ , and $S_\alpha$ for the reflection in the line through $O$ at angle $\alpha$ to the $x$ -axis (matrix $\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$ ). Multiplying the matrices and using the addition formulas gives three rules:

Composite	Result	Determinants
$R_\alpha R_\beta$	$R_{\alpha + \beta}$	$1 \times 1 = 1$
$S_\beta S_\alpha$ (first $S_\alpha$ , then $S_\beta$ )	$R_{2(\beta - \alpha)}$	$(- 1) (- 1) = 1$
$R_\theta S_\alpha$ and $S_\alpha R_\theta$	$S_{\alpha + \theta/2}$ and $S_{\alpha - \theta/2}$	$1 \times (-1) = -1$

Proof of the middle rule. The top-left entry of $S_\beta S_\alpha$ is $\cos 2\beta\cos 2\alpha + \sin 2\beta\sin 2\alpha = \cos(2\beta - 2\alpha)$ ; the bottom-left is $\sin 2\beta\cos 2\alpha - \cos 2\beta\sin 2\alpha = \sin(2\beta - 2\alpha)$ . The other two entries work out to $-\sin(2\beta - 2\alpha)$ and $\cos(2\beta - 2\alpha)$ , which is $R_{2(\beta - \alpha)}$ . $\blacksquare$

In the diagram the mirrors are the $x$ -axis ( $\alpha = 0^\circ$ ) and the line at $\beta = 30^\circ$ . Reflecting in the first and then the second turns the triangle through $2(30^\circ - 0^\circ) = 60^\circ$ . Reflect in the other order and the turn is $-60^\circ$ .

Worked example 2: naming a composite

Describe the single transformation equivalent to (a) the reflection in $y = x$ followed by the reflection in the $y$ -axis (b) the rotation through $90^\circ$ followed by the reflection in the $x$ -axis.

Solution. (a) $y = x$ is at $\alpha = 45^\circ$ , the $y$ -axis at $\beta = 90^\circ$ : $S_{90^\circ}S_{45^\circ} = R_{2(90^\circ - 45^\circ)} = R_{90^\circ}$ . Check: $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = R_{90^\circ}$ . ✓

(b) $S_{0^\circ}R_{90^\circ} = S_{0^\circ - 45^\circ} = S_{-45^\circ}$ , the reflection in $y = - x$ . Check: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$ . ✓

Worked example 3: the inverse of a composite

$f$ and $g$ have matrices $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$ . Find the matrix of $(f \circ g)^{-1}$ .

Solution. $AB = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$ , $\det(AB) = 2$ , so $(AB)^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 0.5 & 0 \\ -0.5 & 1 \end{pmatrix}$ .

Second way: $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ , with matrix $B^{-1}A^{-1} = \begin{pmatrix} 1 & 0.5 \\ 0 & 0.5 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 0.5 & 0 \\ -0.5 & 1 \end{pmatrix}$ . ✓ To undo " $g$ , then $f$ ", undo $f$ first.

Warning

Read composites from right to left. $f \circ g$ means "first $g$ ", and its matrix $A B$ has $B$ next to the vector: $AB\vec{v} = A(B\vec{v})$ . For two reflections the order changes the sign of the angle: $S_\beta S_\alpha = R_{2(\beta - \alpha)}$ but $S_\alpha S_\beta = R_{2(\alpha - \beta)}$ . And $f + g$ is not " $f$ then $g$ ": it adds the two outputs.

Tip

Before multiplying, predict the type from the determinants: rotation × rotation and reflection × reflection give rotations ( $\det = 1$ ); a rotation with a reflection gives a reflection ( $\det = -1$ ). Then the angle rule gives the details.

Dr. Trefor Bazett: Visualizing composition of linear transformations, aka matrix multiplication (about 14 min). Watch how doing one transformation after another is read from the matrix product.

✦

Practice

Work each one, then open the answers.

$f(x, y) = (2x - y,\ x + 3y)$ and $g(x, y) = (y,\ -x)$ . Find the matrices of $3 f$ and $f - 2 g$ .
Describe the single transformation: the reflection in the $x$ -axis followed by the reflection in the $y$ -axis.
Describe the single transformation: the reflection in $y = x$ followed by the reflection in the $x$ -axis.
Describe the single transformation: the rotation through $90^\circ$ followed by the reflection in $y = x$ .
Mirrors through $O$ are at $10^\circ$ and $55^\circ$ to the $x$ -axis. Find the image of $(2, 0)$ after reflecting in the first and then the second.
Find the matrix of $R_{30^\circ}$ followed by $R_{60^\circ}$ , and check that it is $R_{90^\circ}$ .
$A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$ are the matrices of $f$ and $g$ . Find the matrices of $f \circ g$ , $g \circ f$ and $(g \circ f)^{-1}$ .
Prove that the composite of a rotation about $O$ and a reflection in a line through $O$ , in either order, is a reflection in a line through $O$ .

Show answersHide answers

$3A = \begin{pmatrix} 6 & -3 \\ 3 & 9 \end{pmatrix}$ ; $A - 2B = \begin{pmatrix} 2 & -3 \\ 3 & 3 \end{pmatrix}$ .
$S_{90^\circ}S_{0^\circ} = R_{180^\circ}$ : the central symmetry about $O$ .
$S_{0^\circ}S_{45^\circ} = R_{-90^\circ}$ : the rotation through $90^\circ$ clockwise. Matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ .
$S_{45^\circ}R_{90^\circ} = S_{45^\circ - 45^\circ} = S_{0^\circ}$ : the reflection in the $x$ -axis.
$S_{55^\circ}S_{10^\circ} = R_{90^\circ}$ , so $(2, 0) \mapsto (0, 2)$ .
$R_{60^\circ}R_{30^\circ} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = R_{90^\circ}$ .
$f \circ g$ : $AB = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}$ . $g \circ f$ : $BA = \begin{pmatrix} 3 & 6 \\ 0 & 1 \end{pmatrix}$ . $(BA)^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -6 \\ 0 & 3 \end{pmatrix}$ .
Both factors are orthogonal, so the product is orthogonal; its determinant is $1 \times (-1) = -1$ ; an orthogonal matrix with determinant $- 1$ is a reflection in a line through $O$ .

Quick check

f has matrix A and g has matrix B. What is the matrix of g ∘ f, that is, first f and then g?

Key points

Matrices of $f + g$ , $k f$ , $f \circ g$ , $f^{n}$ , $f^{- 1}$ : $A + B$ , $k A$ , $A B$ , $A^{n}$ , $A^{- 1}$ .
$R_\alpha R_\beta = R_{\alpha + \beta}$ ; $S_\beta S_\alpha = R_{2(\beta - \alpha)}$ ; a rotation combined with a reflection is a reflection.
Composites are read from right to left, and order matters.
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ , matrix $B^{-1}A^{-1}$ .

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.
Image: Kaleidoscope-abc, Koperczak, Wikimedia Commons, CC BY 3.0.
Video: Visualizing Composition of Linear Transformations (aka Matrix Multiplication), Dr. Trefor Bazett, embedded from YouTube.

1Two mirrors make a turn

2Composites of rotations and reflections

Two mirrors make a turn

Composites of rotations and reflections