Orthogonal transformations – Operations on Linear Transformations in 2D (Senior 4 Mathematics) | Ganzaa

Operations on Linear Transformations in 2D

Lessons

Senior 4 · Mathematics

Operations on Linear Transformations in 2D

5 lessons

Lesson 3 of 5

Orthogonal transformations

25 min

Moving a shape without bending it

When a carpenter turns a door on its hinges or a machine turns a metal part, the object is rigid: every length and every angle in it stays the same. Of the transformations you know, rotations and reflections behave like this. A shear does not: in the diagram it turns the unit square into a parallelogram in which one side has grown to length $\sqrt{2}$ and the angle at $O$ has shrunk to $45^\circ$ . An enlargement keeps angles but changes lengths.

Which matrices give rigid motions? The answer is a simple test on the matrix, and it shows that in the plane there are only two kinds: rotations and reflections.

Definition

Orthogonal transformation: a linear transformation $f$ of $\mathbb{R}^2$ is orthogonal when it preserves the dot product:

f(\vec{u}) \cdot f(\vec{v}) = \vec{u} \cdot \vec{v} \quad \text{for all } \vec{u}, \vec{v} \in \mathbb{R}^2 .

Taking $\vec{u} = \vec{v}$ gives $\|f(\vec{v})\| = \|\vec{v}\|$ : an orthogonal transformation preserves lengths, hence distances, and (through $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|\,\|\vec{v}\|}$ ) angles. Conversely, a linear map that preserves every length preserves the dot product, because $\vec{u} \cdot \vec{v} = \frac{1}{4}\big(\|\vec{u} + \vec{v}\|^2 - \|\vec{u} - \vec{v}\|^2\big)$ .

Example: every rotation $R_\theta$ and every reflection $S_\alpha$ is orthogonal; the shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is not, since it sends $(0, 1)$ , of length 1, to $(1, 1)$ , of length $\sqrt{2}$ .

Definition

Orthonormal basis: a basis $\{ \vec{u}_1, \vec{u}_2 \}$ of $\mathbb{R}^2$ in which both vectors are unit vectors and they are orthogonal: $\vec{u}_1 \cdot \vec{u}_1 = \vec{u}_2 \cdot \vec{u}_2 = 1$ and $\vec{u}_1 \cdot \vec{u}_2 = 0$ .

Orthogonal matrix: a square matrix $A$ with $A^{\mathsf{T}}A = I$ , that is, $A^{-1} = A^{\mathsf{T}}$ . Because the entries of $A^{\mathsf{T}}A$ are the dot products of the columns of $A$ , this says exactly that the columns of $A$ form an orthonormal basis. For this reason some books call such matrices, and the transformations they represent, orthonormal.

Example: $\{ (0.6, 0.8), (-0.8, 0.6) \}$ is an orthonormal basis: $0.36 + 0.64 = 1$ twice, and $- 0.48 + 0.48 = 0$ . So $\begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$ is an orthogonal matrix.

Important

Theorem 1. A linear transformation with matrix $A$ is orthogonal if and only if $A^{\mathsf{T}}A = I$ .

Proof. Write the dot product as a matrix product, $\vec{u} \cdot \vec{v} = \vec{u}^{\mathsf{T}}\vec{v}$ (a row times a column). Then $A\vec{u} \cdot A\vec{v} = (A\vec{u})^{\mathsf{T}}(A\vec{v}) = \vec{u}^{\mathsf{T}}(A^{\mathsf{T}}A)\vec{v}$ . If $A^{\mathsf{T}}A = I$ , this is $\vec{u}^{\mathsf{T}}\vec{v} = \vec{u} \cdot \vec{v}$ . Conversely, if the dot product is preserved, apply it to $\vec{e}_1, \vec{e}_2$ : the columns $A\vec{e}_1$ , $A\vec{e}_2$ have the same dot products as $\vec{e}_1, \vec{e}_2$ , so they are orthonormal and $A^{\mathsf{T}}A = I$ . $\blacksquare$

Theorem 2 (classification). A $2 \times 2$ orthogonal matrix has $\det A = \pm 1$ , and it is either a rotation $R_\theta$ ( $\det A = 1$ ) or a reflection $S_\alpha$ in a line through $O$ ( $\det A = -1$ ).

Proof. $(\det A)^2 = \det A^{\mathsf{T}} \det A = \det(A^{\mathsf{T}}A) = \det I = 1$ . The first column is a unit vector, so it is $(\cos\theta, \sin\theta)$ for some $\theta$ . The second is a unit vector perpendicular to it, so it is $(-\sin\theta, \cos\theta)$ or $(\sin\theta, -\cos\theta)$ . The first choice gives $R_\theta$ , with determinant 1; the second gives $\begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}$ , the reflection in the line at angle $\frac{\theta}{2}$ , with determinant $- 1$ . $\blacksquare$

Worked example 1: testing and classifying

Decide whether each matrix is orthogonal; if it is, say whether it is a rotation or a reflection. (a) $A = \frac{1}{5}\begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}$ (b) $B = \frac{1}{13}\begin{pmatrix} 5 & 12 \\ 12 & -5 \end{pmatrix}$ (c) $C = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$

Solution.

(a) Columns $\frac{1}{5}(3, 4)$ and $\frac{1}{5}(-4, 3)$ : lengths $\frac{5}{5} = 1$ , dot product $\frac{-12 + 12}{25} = 0$ . Orthogonal, $\det A = \frac{9 + 16}{25} = 1$ : the rotation with $\cos\theta = 0.6$ , $\sin\theta = 0.8$ , $\theta \approx 53.1^\circ$ .

(b) Columns $\frac{1}{13}(5, 12)$ and $\frac{1}{13}(12, -5)$ : both of length 1, dot product $\frac{60 - 60}{169} = 0$ . Orthogonal, $\det B = \frac{-25 - 144}{169} = -1$ : a reflection. Its axis is the line of fixed vectors: $\frac{5x + 12y}{13} = x$ gives $12 y = 8 x$ , the line $y = \frac{2}{3}x$ .

Worked example 2: completing an orthogonal matrix

Find every orthogonal matrix whose first column is $(0.28, 0.96)$ , and say what each one does.

Solution. ${0.28}^{2} + {0.96}^{2} = 0.0784 + 0.9216 = 1$ , so the first column is a unit vector. The second column must be a unit vector perpendicular to it: $(- 0.96, 0.28)$ or $(0.96, - 0.28)$ . So

\begin{pmatrix} 0.28 & -0.96 \\ 0.96 & 0.28 \end{pmatrix} \ (\det = 1,\ \text{a rotation}), \qquad \begin{pmatrix} 0.28 & 0.96 \\ 0.96 & -0.28 \end{pmatrix} \ (\det = -1,\ \text{a reflection}).

Worked example 3: coordinates in an orthonormal basis

Find the coordinates of $\vec{v} = (5, 10)$ in the orthonormal basis $\vec{u}_1 = (0.6, 0.8)$ , $\vec{u}_2 = (-0.8, 0.6)$ .

Solution. If $\vec{v} = a\vec{u}_1 + b\vec{u}_2$ , dotting with $\vec{u}_1$ kills the second term: $a = \vec{v} \cdot \vec{u}_1 = 3 + 8 = 11$ . Likewise $b = \vec{v} \cdot \vec{u}_2 = -4 + 6 = 2$ . Check: $11(0.6, 0.8) + 2(-0.8, 0.6) = (6.6 - 1.6,\ 8.8 + 1.2) = (5, 10)$ . ✓ With an orthonormal basis, no system of equations is needed: each coordinate is one dot product.

Warning

"Orthogonal matrix" needs more than perpendicular columns. $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ has perpendicular columns but is not orthogonal: the columns must also be unit vectors. And $\det A = \pm 1$ is necessary but not enough: the shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has determinant 1 and is not orthogonal.

Tip

To test a $2 \times 2$ matrix, check three numbers: the two column lengths (both 1) and the dot product of the columns (0). Then the sign of the determinant tells you rotation ( $+ 1$ ) or reflection ( $- 1$ ).

Khan Academy: Orthogonal matrices preserve angles and lengths (about 11 min). Watch for the step that writes a dot product as a row times a column and uses the transpose.

✦

Practice

Work each one, then open the answers.

Which are orthogonal? (a) $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (b) $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ (c) $\frac{1}{5}\begin{pmatrix} 4 & -3 \\ 3 & 4 \end{pmatrix}$ (d) $\begin{pmatrix} 0.5 & 0.5 \\ -0.5 & 0.5 \end{pmatrix}$
For each orthogonal matrix in question 1, say whether it is a rotation or a reflection and give its angle or axis.
Find the two orthogonal matrices whose first column is $(0.8, - 0.6)$ , and classify them.
For $A = \frac{1}{13}\begin{pmatrix} 5 & -12 \\ 12 & 5 \end{pmatrix}$ and $\vec{v} = (2, 3)$ , compute $A\vec{v}$ and check that $\|A\vec{v}\| = \|\vec{v}\|$ .
Find the coordinates of $\vec{v} = (7, 1)$ in the orthonormal basis $\vec{u}_1 = (0.6, 0.8)$ , $\vec{u}_2 = (0.8, -0.6)$ .
Prove that the product of two orthogonal matrices is orthogonal.
Find $a$ and $b$ so that $\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix}$ is orthogonal.
Show that the enlargement $3 I$ keeps right angles but is not an orthogonal transformation.

Show answersHide answers

(a) yes (b) no: column $(2, 1)$ has length $\sqrt{5}$ (c) yes: columns of length 1, dot product $\frac{-12 + 12}{25} = 0$ (d) no: columns have length $\sqrt{0.5}$ .
(a) $\det = -1$ : reflection in $y = x$ . (c) $\det = 1$ : rotation with $\cos\theta = 0.8$ , $\sin\theta = 0.6$ , $\theta \approx 36.9^\circ$ .
$\begin{pmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{pmatrix}$ : $\det = 1$ , rotation through about $-36.9^\circ$ . $\begin{pmatrix} 0.8 & -0.6 \\ -0.6 & -0.8 \end{pmatrix}$ : $\det = -1$ , a reflection.
$A\vec{v} = \frac{1}{13}(10 - 36,\ 24 + 15) = (-2, 3)$ ; both lengths are $\sqrt{13}$ .
$a = \vec{v} \cdot \vec{u}_1 = 4.2 + 0.8 = 5$ , $b = \vec{v} \cdot \vec{u}_2 = 5.6 - 0.6 = 5$ : coordinates $(5, 5)$ .
$(AB)^{\mathsf{T}}(AB) = B^{\mathsf{T}}A^{\mathsf{T}}AB = B^{\mathsf{T}}IB = B^{\mathsf{T}}B = I$ .
Second column unit: $\frac{a^2 + b^2}{2} = 1$ ; perpendicular to the first: $\frac{a + b}{2} = 0$ . So $b = - a$ , $a^{2} = 1$ : $(a, b) = (1, - 1)$ or $(- 1, 1)$ .
$3\vec{u} \cdot 3\vec{v} = 9\,\vec{u} \cdot \vec{v}$ , which is 0 whenever $\vec{u} \cdot \vec{v} = 0$ , so right angles stay right angles. But $\|3\vec{e}_1\| = 3 \ne 1$ , so lengths change.

Quick check

The shear with matrix rows (1, 1) and (0, 1) is an orthogonal transformation, because its determinant is 1.

Key points

An orthogonal transformation preserves the dot product, hence lengths, distances and angles.
Its matrix satisfies $A^{\mathsf{T}}A = I$ : the columns form an orthonormal basis, and $A^{-1} = A^{\mathsf{T}}$ .
In the plane, orthogonal means rotation ( $\det = 1$ ) or reflection in a line through $O$ ( $\det = -1$ ).
In an orthonormal basis, each coordinate of $\vec{v}$ is a dot product $\vec{v} \cdot \vec{u}_i$ .

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.
Video: Orthogonal matrices preserve angles and lengths, Khan Academy, embedded from YouTube.

1Moving a shape without bending it

Moving a shape without bending it