3.3 Orthogonal matrices, rotations and reflections

The matrices that never stretch anything are exactly those whose columns are orthonormal. In the plane they are rotations and reflections; in space every rotation turns about a fixed axis. Their inverse is free, which is why computers love them.

A general matrix stretches some directions and squashes others. This chapter studies the matrices that do neither: they move vectors around without changing any length or any angle. In the plane those are rotations and reflections. In space they include every way of turning a rigid object. They have one more property that makes them precious in computing: their inverse is their transpose, so undoing them costs nothing and loses no accuracy.

Before you start

Chapter 2.9 (orthonormal sets, norms) and 3.1 (where $Q$ in $A = Q\Lambda Q^{\mathsf T}$ already had orthonormal columns).

Quick checkFor orthonormal vectors

q_1, q_2

, what is

q_1 \cdot q_2

and

q_1 \cdot q_1

Quick checkWhat is

\det(AB)

The picture

Watch the orange unit square: under “Rotate 90°” and “Reflect” it stays a unit square, only turned or flipped. The grid stays square too. Now try “Shear” or “Stretch”: squares become parallelograms or rectangles, so lengths or angles change. The maps that keep the grid square are this chapter’s subject.

Where this came from· Rotations, from Euler to the Apollo guidance computer

Leonhard Euler presented a paper to the St. Petersburg Academy in 1775 (printed in 1776) showing that any turning of a rigid body about a fixed point is a rotation about a single axis: a ball turned any way keeps two opposite points where they were. He wanted a way to describe the motion of rigid bodies with a few numbers. In 1843 William Rowan Hamilton found the quaternions, which describe 3D rotations compactly. In 1846 Arthur Cayley published a formula that builds an orthogonal matrix from an antisymmetric one. The idea outlived them. The Apollo guidance computer stored the orientation of its inertial platform as a 3 by 3 matrix of this kind (called REFSMMAT), and Apollo documents describe an orthogonalization step when computing it. Any program that multiplies many rotation matrices must restore orthogonality now and then, because rounding errors slowly turn a rotation into a slight stretch.

Definitions and theorems

Definition A

Orthogonal matrix

A real $n \times n$ matrix $Q$ is orthogonal if its columns form an orthonormal basis of $\mathbb R^n$ . Equivalently, $Q^{\mathsf T}Q = I$ , because entry $(i, j)$ of $Q^{\mathsf T}Q$ is the dot product of column $i$ with column $j$ . Then $Q^{-1} = Q^{\mathsf T}$ , and so also $QQ^{\mathsf T} = I$ : the rows are orthonormal too.

In one lineA square real matrix whose columns are orthonormal; equivalently $Q^{\mathsf T}Q = I$ , so $Q^{-1} = Q^{\mathsf T}$ .

Why it existsThese are exactly the matrices that preserve length, and their inverse costs nothing to compute.

Watch out· The name is misleading

An “orthogonal matrix” has orthonormal columns, not only perpendicular ones. $\begin{bmatrix}2&0\\0&3\end{bmatrix}$ has perpendicular columns but is not orthogonal. The name is historical and every textbook uses it.

Definition B

Isometry

A linear map $Q: \mathbb R^n \to \mathbb R^n$ is an isometry if $\lVert Qx \rVert = \lVert x \rVert$ for every $x$ .

In one lineA linear map that preserves length: $\lVert Qx \rVert = \lVert x \rVert$ for every $x$ .

Why it existsIt is the precise meaning of 'moves things without distorting them'.

Theorem A Apostol 5.10; Axler 7.42

Orthogonal means length-preserving

For a real $n \times n$ matrix $Q$ , the following are equivalent:

$Q^{\mathsf T}Q = I$ .
$(Qx) \cdot (Qy) = x \cdot y$ for all $x, y$ (dot products, hence angles, are preserved).
$\lVert Qx \rVert = \lVert x \rVert$ for all $x$ (lengths are preserved).

In one lineFor a square real matrix Q, these are equivalent: QᵀQ = I; Q preserves dot products; Q preserves lengths.

Why it is trueDot products and lengths determine each other (polarization), and QᵀQ = I is exactly 'dot products survive'.

ProofAxler 7.42

The idea: 1 ⇒ 2 by moving Qᵀ across. 2 ⇒ 3 by taking x = y. 3 ⇒ 2 by the polarization identity, which recovers dot products from lengths. 2 ⇒ 1 by testing on standard basis vectors.

(1 ⇒ 2) If

Q^{\mathsf T}Q = I

then

(Qx) \cdot (Qy) = x^{\mathsf T}Iy = x \cdot y

(Qx) \cdot (Qy) = (Qx)^{\mathsf T}Qy = x^{\mathsf T}Q^{\mathsf T}Qy

(2 ⇒ 3) Take

y = x

\lVert Qx \rVert^2 = \lVert x \rVert^2

A length is the square root of a dot product with itself.

(3 ⇒ 2)

x \cdot y = \frac12\big(\lVert x + y \rVert^2 - \lVert x \rVert^2 - \lVert y \rVert^2\big)

. Apply it to

Qx, Qy

, use

Q(x + y) = Qx + Qy

, and each length on the right is unchanged by

Q

Expand

\lVert x + y \rVert^2 = \lVert x \rVert^2 + 2x \cdot y + \lVert y \rVert^2

and solve for

x \cdot y

: the polarization identity.

(2 ⇒ 1)

(Q^{\mathsf T}Q)_{ij} = (Qe_i)\cdot(Qe_j) = e_i \cdot e_j

, which is 1 if

i = j

and 0 otherwise. So

Q^{\mathsf T}Q = I

\blacksquare

Entry

(i, j)

Q^{\mathsf T}Q

(Qe_i) \cdot (Qe_j)

, which by 2 equals

e_i \cdot e_j

Theorem A standard

Determinant of an orthogonal matrix

If $Q$ is orthogonal, $\det Q = \pm 1$ . Orthogonal matrices with $\det Q = 1$ are called rotations (or proper orthogonal); those with $\det Q = -1$ reverse orientation.

In one lineAn orthogonal matrix has determinant +1 (a rotation) or −1 (it includes a reflection).

Why it is truedet(QᵀQ) = det(Q)² = det I = 1.

Proofstandard

The idea: Take determinants of QᵀQ = I.

1 = \det I = \det(Q^{\mathsf T}Q) = \det Q^{\mathsf T}\det Q = (\det Q)^2

Product rule (2.7), and

\det Q^{\mathsf T} = \det Q

(2.7).

\det Q = \pm 1

\blacksquare

In the picture: area is unchanged (

|\det| = 1

), and the sign says whether the square was flipped.

A real number whose square is 1.

Worked example

Prove: the product

PQ

of two orthogonal matrices is orthogonal, and so is

Q^{-1}

Solution.

State what must be shown. We need $(PQ)^{\mathsf T}(PQ) = I$ . That is the definition of orthogonal.
Reverse the transpose of a product. $(PQ)^{\mathsf T}(PQ) = Q^{\mathsf T}P^{\mathsf T}PQ$ . The transpose of a product reverses the order (2.5).
Cancel from the middle. $P^{\mathsf T}P = I$ , so this is $Q^{\mathsf T}Q = I$ . Both $P$ and $Q$ are orthogonal. The inner pair collapses first, then the outer pair.
Inverse. $Q^{-1} = Q^{\mathsf T}$ , and $(Q^{\mathsf T})^{\mathsf T}Q^{\mathsf T} = QQ^{\mathsf T} = I$ . We showed $QQ^{\mathsf T} = I$ in the definition (rows are orthonormal too).

Answer. Both are orthogonal. $\blacksquare$ Geometric meaning: two length-preserving moves in a row preserve length, and so does undoing one.

Worked example

Prove: an orthogonal

Q

sends an orthonormal basis

u_1, \dots, u_n

to an orthonormal basis.

Solution.

Compute the dot products of the images. $(Qu_i)\cdot(Qu_j) = u_i \cdot u_j$ . Orthogonal matrices preserve dot products (the theorem above).
Read off. That equals $1$ when $i = j$ and $0$ when $i \ne j$ . The $u_i$ are orthonormal.
Count. The $n$ vectors $Qu_i$ are orthonormal, so independent (2.9), and $n$ independent vectors in $\mathbb R^n$ form a basis (2.2). Orthonormal sets are independent, and a basis needs exactly $n$ vectors.

Answer. $Qu_1, \dots, Qu_n$ is an orthonormal basis. $\blacksquare$ Example: the standard basis goes to the columns of $Q$ .

The plane: every orthogonal matrix is a rotation or a reflection

Definition A

Rotation matrix

The rotation matrix by angle $\theta$ is $R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$ . Its first column is $\hat\imath$ turned by $\theta$ , and its second is $\hat\jmath$ turned by $\theta$ .

In one line $R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$ turns the plane counterclockwise by θ.

Why it existsIts columns are where the two axis arrows land after turning by θ.

Theorem A standard

Orthogonal 2 by 2 matrices

Every $2 \times 2$ orthogonal matrix is either $R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$ (determinant 1) or $\begin{bmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{bmatrix}$ (determinant $-1$ ), which reflects across the line at angle $\theta/2$ .

In one lineEvery 2×2 orthogonal matrix is a rotation R_θ (det 1) or a reflection across a line (det −1).

Why it is trueThe first column is a unit vector at some angle θ; the second must be a unit vector perpendicular to it, and there are only two choices.

Proofstandard

The idea: Column 1 is a unit vector, so it is (cos θ, sin θ) for some θ. Column 2 is a unit vector perpendicular to it: only ±(−sin θ, cos θ).

The first column is

(\cos\theta, \sin\theta)

for some

\theta

A unit vector in the plane lies on the unit circle, so it has this form for some angle

\theta

The second column is a unit vector perpendicular to it, so it is

(-\sin\theta, \cos\theta)

(\sin\theta, -\cos\theta)

The vectors perpendicular to

(a, b)

are the multiples of

(-b, a)

; length 1 leaves two choices.

The first choice gives

R_\theta

with determinant

\cos^2\theta + \sin^2\theta = 1

; the second gives determinant

-1

\blacksquare

Compute the two determinants.

Worked example

Q = \frac1{13}\begin{bmatrix}5&12\\12&-5\end{bmatrix}

orthogonal? If so, is it a rotation or a reflection, and across which line?

Solution.

Test the columns. Column 1 is $\frac1{13}(5, 12)$ with squared length $\frac{25 + 144}{169} = 1$ . Column 2 is $\frac1{13}(12, -5)$ with squared length $\frac{144 + 25}{169} = 1$ . Their dot product is $\frac{60 - 60}{169} = 0$ . Orthonormal columns are the definition. Unit length and perpendicular are two separate checks, and both are needed.
Find the determinant. $\det Q = \frac{-25 - 144}{169} = -1$ . $+1$ means a rotation, $-1$ means a reflection (the plane theorem).
Match the reflection form. $\cos\theta = \frac5{13}$ and $\sin\theta = \frac{12}{13}$ , so $\theta \approx 67.4°$ . Compare with $\begin{bmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{bmatrix}$ .
Find the mirror. It is the line at angle $\frac\theta2 \approx 33.7°$ , which has slope $\tan\frac\theta2 = \frac{\sin\theta}{1 + \cos\theta} = \frac{12/13}{18/13} = \frac23$ , so direction $(3, 2)$ . Check: $Q(3, 2) = \frac1{13}(15 + 24,\ 36 - 10) = (3, 2)$ , so the mirror line is fixed.

Answer. Orthogonal, determinant $-1$ : a reflection across the line through $(3, 2)$ .

Worked example

Show

R_\alpha R_\beta = R_{\alpha + \beta}

, and explain what that says about trigonometry.

Solution.

Multiply. $R_\alpha R_\beta = \begin{bmatrix}\cos\alpha\cos\beta - \sin\alpha\sin\beta & -(\cos\alpha\sin\beta + \sin\alpha\cos\beta)\\ \sin\alpha\cos\beta + \cos\alpha\sin\beta & \cos\alpha\cos\beta - \sin\alpha\sin\beta\end{bmatrix}$ . Row-times-column multiplication (2.5).
Compare with $R_{\alpha+\beta}$ . Turning by $\beta$ then by $\alpha$ is turning by $\alpha + \beta$ , so the product must equal $\begin{bmatrix}\cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) & \cos(\alpha+\beta)\end{bmatrix}$ . Composition of maps is multiplication of their matrices (2.5).
Read off the entries. $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ . Equal matrices have equal entries.

Answer. The addition formulas you memorised in school are the statement “two rotations make a rotation”, read entry by entry. Rotations of the plane also commute: $R_\alpha R_\beta = R_\beta R_\alpha$ .

Your answerRotate the vector

(3, 4)

90°

counterclockwise. What is the first coordinate of the result?

Worked example

Find the inverse of

R_\theta

, and compute

R_{30°}^{-1}

Solution.

Use the transpose. $R_\theta$ is orthogonal, so $R_\theta^{-1} = R_\theta^{\mathsf T} = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$ . For an orthogonal matrix the inverse is the transpose, with no elimination.
Recognize it. Since $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$ , this is $R_{-\theta}$ . Turning back by $\theta$ undoes turning forward by $\theta$ .
Put in $30°$ . $\cos 30° = \frac{\sqrt3}2$ and $\sin 30° = \frac12$ , so $R_{30°}^{-1} = \begin{bmatrix}\frac{\sqrt3}2 & \frac12\\ -\frac12 & \frac{\sqrt3}2\end{bmatrix}$ . Substitute in the formula of step 1.

Answer. $R_\theta^{-1} = R_{-\theta} = R_\theta^{\mathsf T}$ .

Worked example

Write the matrix of the reflection across the line through the origin at angle

\varphi

. Then find the reflection across

y = x

Solution.

Set up. Let $u = (\cos\varphi, \sin\varphi)$ be the unit vector along the mirror. A vector $x$ splits as the part along the mirror, $(u \cdot x)u$ , plus the part across it, $x - (u \cdot x)u$ . The first part is the projection onto the line (2.9).
Flip the cross part. The reflection keeps the first part and negates the second: $Fx = (u \cdot x)u - \big(x - (u \cdot x)u\big) = 2(u \cdot x)u - x$ . A mirror leaves points on it fixed and sends the perpendicular offset to its negative.
Read off the matrix. Since $(u \cdot x)u = uu^{\mathsf T}x$ , we get $F = 2uu^{\mathsf T} - I = \begin{bmatrix}2\cos^2\varphi - 1 & 2\sin\varphi\cos\varphi\\ 2\sin\varphi\cos\varphi & 2\sin^2\varphi - 1\end{bmatrix} = \begin{bmatrix}\cos 2\varphi & \sin 2\varphi\\ \sin 2\varphi & -\cos 2\varphi\end{bmatrix}$ . Double-angle formulas. This is the reflection form of the plane theorem with $\theta = 2\varphi$ .
Take $y = x$ . Here $\varphi = 45°$ , so $2\varphi = 90°$ and $F = \begin{bmatrix}0&1\\1&0\end{bmatrix}$ . $\cos 90° = 0$ , $\sin 90° = 1$ . Check: it swaps $x$ and $y$ , which is what reflecting across $y = x$ does.

Answer. $F = \begin{bmatrix}\cos 2\varphi & \sin 2\varphi\\ \sin 2\varphi & -\cos 2\varphi\end{bmatrix}$ ; for $y = x$ it is $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ with determinant $-1$ .

Your answerReflect the point

(2, 5)

across the line

y = x

. What is the first coordinate of the result?

Quick checkIs

\begin{bmatrix}0.6 & 0.8\\ 0.8 & -0.6\end{bmatrix}

a rotation or a reflection?

Space: every rotation has an axis

Theorem B Euler 1775; standard

Euler's rotation theorem

If $Q$ is a $3 \times 3$ orthogonal matrix with $\det Q = 1$ , then $Qv = v$ for some $v \ne O$ . So $Q$ is a rotation about the axis through $v$ .

In one lineEvery 3×3 rotation (orthogonal, det 1) fixes a line through the origin: it is a turn about an axis.

Why it is truedet(Q − I) = 0 for a 3×3 rotation, so 1 is an eigenvalue; its eigenvector is the axis.

Proofstandard

The idea: Show det(Q − I) = 0 using QᵀQ = I, det Q = 1 and the fact that for 3×3 matrices det(−A) = −det A.

Q - I = Q - QQ^{\mathsf T} = Q(I - Q^{\mathsf T})

Write

I = QQ^{\mathsf T}

(rows of an orthogonal matrix are orthonormal too), then factor

Q

out on the left.

\det(Q - I) = \det Q \cdot \det(I - Q^{\mathsf T}) = \det(I - Q)

Product rule,

\det Q = 1

, and

\det(I - Q^{\mathsf T}) = \det\big((I - Q)^{\mathsf T}\big) = \det(I - Q)

\det(I - Q) = \det\big(-(Q - I)\big) = -\det(Q - I)

For a

3 \times 3

matrix, multiplying every row by

-1

multiplies the determinant by

(-1)^3 = -1

\det(Q - I) = -\det(Q - I)

, hence

\det(Q - I) = 0

A number equal to its own negative is 0.

There is

v \ne O

with

(Q - I)v = O

, that is,

Qv = v

\blacksquare

Q - I

is not invertible, so it has a non-zero null vector (2.6).

Proof workshopRebuild the proof of Euler's rotation theorem

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Beyond the course: for understanding· Why only odd dimensions

The proof used $(-1)^3 = -1$ . In even dimensions it fails: $R_{90°}$ in the plane fixes no direction. In $\mathbb R^4$ a rotation can turn two perpendicular planes at once and fix nothing.

Worked example Spagnolie Lecture 8

Find the axis and angle of the rotation

Q

that sends

(x, y, z)

(z, x, y)

Solution. Step through it below. The stages are: confirm it is a rotation, find the axis, find the angle from the trace, solve.

Answer. A rotation by $120°$ about the line through $(1, 1, 1)$ .

Your answerA

3 \times 3

rotation matrix has trace 2. By what angle in degrees (between 0 and 180) does it turn?

Worked example classic

Let

P = \begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}

(swap

x

and

y

). It has

\det P = -1

. Write it as a rotation followed by

x \mapsto -x

, and find an eigenvalue

-1

eigenvector.

Solution.

Check orthogonal. The columns are the standard basis vectors in a different order, so $P^{\mathsf T}P = I$ . Orthonormal columns.
Negate it. $-P$ is orthogonal too, and $\det(-P) = (-1)^3\det P = (-1)(-1) = 1$ . Negating all three columns multiplies the determinant by $(-1)^3$ . Now $-P$ is a rotation.
Apply Euler. There is $v \ne O$ with $(-P)v = v$ . Solving, $Pv = -v$ gives $v = (1, -1, 0)$ : $P(1, -1, 0) = (-1, 1, 0) = -v$ . Euler’s theorem gives an axis for the rotation $-P$ , and that axis is a $-1$ -eigenvector of $P$ .
Find the angle. $\operatorname{tr}(-P) = -1 = 1 + 2\cos\theta$ , so $\theta = 180°$ . Trace formula for rotations.

Answer. $P = -R$ where $R = -P$ is a half-turn about $(1, -1, 0)$ , and $-1$ is an eigenvalue of $P$ with eigenvector $(1, -1, 0)$ . $\blacksquare$

Complex version and eigenvalues

Definition A

Unitary matrix

A complex $n \times n$ matrix $U$ is unitary if $U^*U = I$ , where $U^* = \overline{U}^{\mathsf T}$ . Real unitary matrices are exactly the orthogonal ones.

In one lineA complex square matrix with $U^*U = I$ (conjugate transpose); the complex version of orthogonal.

Why it existsIt preserves complex lengths, and it is how quantum states evolve.

Theorem A Apostol 5.10

Eigenvalues of an orthogonal matrix

If $Q$ is unitary (in particular real orthogonal) and $Qv = \lambda v$ with $v \ne O$ , then $|\lambda| = 1$ .

In one lineEvery eigenvalue of an orthogonal (or unitary) matrix has absolute value 1.

Why it is trueQ preserves length, so Qv = λv forces |λ| = 1.

ProofApostol 5.10

The idea: Length is preserved: ‖v‖ = ‖Qv‖ = ‖λv‖ = |λ|‖v‖.

\lVert v \rVert = \lVert Qv \rVert

Unitary matrices preserve length: with the complex inner product of 3.1,

\lVert Uv \rVert^2 = (Uv, Uv) = (v, U^*Uv) = (v, v)

by the adjoint property and

U^*U = I

\lVert Qv \rVert = \lVert \lambda v \rVert = |\lambda|\,\lVert v \rVert

Qv = \lambda v

, and a scalar comes out of a length with its absolute value:

\lVert \lambda v \rVert^2 = \lambda\bar\lambda\,(v, v) = |\lambda|^2 \lVert v \rVert^2

|\lambda| = 1

\blacksquare

For

R_\theta

the eigenvalues are

e^{\pm i\theta}

, on the unit circle (0.6).

Combining the two lines gives

\lVert v \rVert = |\lambda|\,\lVert v \rVert

, and

\lVert v \rVert \ne 0

, so divide.

Worked example

Find the eigenvalues of

R_\theta

and check they have absolute value 1. For which

\theta

is an eigenvalue real?

Solution.

Characteristic polynomial. $\det(R_\theta - \lambda I) = (\cos\theta - \lambda)^2 + \sin^2\theta = \lambda^2 - 2\cos\theta\,\lambda + 1$ . Determinant of $\begin{bmatrix}\cos\theta - \lambda & -\sin\theta\\ \sin\theta & \cos\theta - \lambda\end{bmatrix}$ , using $\cos^2\theta + \sin^2\theta = 1$ .
Solve. $\lambda = \cos\theta \pm \sqrt{\cos^2\theta - 1} = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$ . Quadratic formula, and $\cos^2\theta - 1 = -\sin^2\theta$ . Euler’s formula from 0.6.
Absolute value. $|\cos\theta \pm i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$ . Matches the theorem.
Real cases. The eigenvalues are real only when $\sin\theta = 0$ : $\theta = 0$ gives $1, 1$ and $\theta = \pi$ gives $-1, -1$ . Otherwise no direction of the plane stays on its own line, as the picture at the top shows for $90°$ .

Answer. $\lambda = e^{\pm i\theta}$ , on the unit circle.

Quick checkA

3 \times 3

matrix is orthogonal with determinant

-1

. Which is guaranteed?

Quick checkWhich of these could be an eigenvalue of a real orthogonal matrix?

Your answerWhat is the determinant of the reflection

\begin{bmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta\end{bmatrix}

Proof workshopRebuild: an orthogonal matrix preserves lengthsAxler 7.42

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Where it is used

Used inside maths· Inside maths: the SVD and QR

Chapter 3.4 writes every matrix as $U\Sigma V^{\mathsf T}$ : rotate, stretch along axes, rotate, with $U, V$ orthogonal. The QR factorisation $A = QR$ (2.10) is Gram–Schmidt in matrix form, with $Q$ orthogonal.

Used in the real world· Numerical computing: why orthogonal matrices are numerically safe

Multiplying by an orthogonal matrix cannot magnify errors, because it preserves lengths: an error vector of size $10^{-16}$ stays size $10^{-16}$ . That is why reliable algorithms (QR for least squares, the eigenvalue algorithms inside NumPy and MATLAB) are built from sequences of rotations and reflections (Givens rotations and Householder reflections) rather than general row operations.

Used in the real world· Graphics and robotics: orientation

A robot arm’s pose is a chain of joint rotations, each a rotation matrix about the joint’s axis; multiplying them gives the hand’s orientation (forward kinematics). Game engines store orientations as quaternions and convert to rotation matrices for drawing; interpolating between two orientations smoothly (“slerp”) follows the shortest arc on the space of rotations. Because rounding slowly destroys orthogonality, a program that multiplies many rotations re-orthonormalises (for example with a Gram–Schmidt step) from time to time.

Traps

Watch out· Orthogonal columns are not enough

The columns must also have length 1. Check $Q^{\mathsf T}Q = I$ , not only that the columns are perpendicular.

Watch out· 3D rotations do not commute

$R_\alpha R_\beta = R_\beta R_\alpha$ in the plane, but in space two quarter-turns about different axes give different results in different orders (Spagnolie’s Lecture 8, chapter 2.4).

Problems

check

Is $\frac15\begin{bmatrix}3&-4\\4&3\end{bmatrix}$ orthogonal? A rotation or a reflection? By what angle?

Solution

Columns $(\frac35, \frac45)$ and $(-\frac45, \frac35)$ are unit and perpendicular, so yes. Determinant $\frac{9 + 16}{25} = 1$ : a rotation, by $\theta$ with $\cos\theta = \frac35$ , $\sin\theta = \frac45$ , about $53.13°$ .

check

What is the inverse of $R_\theta$ ? Check with $R_\theta^{\mathsf T}$ .

Solution

$R_{-\theta}$ , turning back. $R_\theta^{\mathsf T} = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix} = R_{-\theta}$ , using $\cos(-\theta) = \cos\theta$ , $\sin(-\theta) = -\sin\theta$ .

use

Find the axis and angle of the rotation $Q = \begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}$ (Spagnolie’s $T$ from Lecture 8).

Solution

$Q(1,0,0) = (1,0,0)$ , so the axis is the $x$ -axis. $\operatorname{tr} Q = 1 = 1 + 2\cos\theta$ gives $\cos\theta = 0$ , $\theta = 90°$ .

use

Write the matrix of reflection across the line $y = \sqrt3\,x$ and check it fits the theorem on $2 \times 2$ orthogonal matrices.

Solution

The line is at angle $\varphi = 60°$ , so $\theta = 2\varphi = 120°$ and the matrix is $\begin{bmatrix}\cos 120° & \sin 120°\\ \sin 120° & -\cos 120°\end{bmatrix} = \begin{bmatrix}-\frac12 & \frac{\sqrt3}2\\ \frac{\sqrt3}2 & \frac12\end{bmatrix}$ . Determinant $-\frac14 - \frac34 = -1$ . Check: it fixes $(1, \sqrt3)$ , since $-\frac12 + \frac32 = 1$ and $\frac{\sqrt3}2 + \frac{\sqrt3}2 = \sqrt3$ .

prove

Prove that the product of two orthogonal matrices is orthogonal, and the inverse of an orthogonal matrix is orthogonal.

Solution

$(PQ)^{\mathsf T}(PQ) = Q^{\mathsf T}P^{\mathsf T}PQ = Q^{\mathsf T}Q = I$ . For the inverse: $Q^{-1} = Q^{\mathsf T}$ and $(Q^{\mathsf T})^{\mathsf T}Q^{\mathsf T} = QQ^{\mathsf T} = I$ . (So the orthogonal matrices are closed under products and inverses, and contain $I$ : they form a group.)

prove

Prove that an orthogonal matrix maps an orthonormal basis to an orthonormal basis.

Solution

If $(u_i)$ is orthonormal, $(Qu_i)\cdot(Qu_j) = u_i \cdot u_j$ (dot products preserved), so the images are orthonormal; $n$ orthonormal vectors in $\mathbb R^n$ form a basis.

stretchclassic

Show that every $3 \times 3$ orthogonal matrix with determinant $-1$ is a rotation followed by the reflection $x \mapsto -x$ . Conclude that $-1$ is always one of its eigenvalues.

Solution

$-Q$ is orthogonal with $\det(-Q) = (-1)^3\det Q = 1$ , so $-Q = R$ is a rotation and $Q = -R$ . By Euler’s theorem $Rv = v$ for some $v \ne O$ , so $Qv = -v$ : $-1$ is an eigenvalue.

Summary

Orthogonal $Q$ : orthonormal columns, $Q^{\mathsf T}Q = I$ , $Q^{-1} = Q^{\mathsf T}$ .
Equivalent: preserves dot products, preserves lengths (an isometry).
$\det Q = \pm 1$ : $+1$ rotations, $-1$ orientation-reversing.
In the plane: rotations $R_\theta$ and reflections, nothing else. $R_\alpha R_\beta = R_{\alpha+\beta}$ is the addition formula.
In space: every rotation has an axis (Euler), and its angle comes from $\operatorname{tr} Q = 1 + 2\cos\theta$ .
Eigenvalues have $|\lambda| = 1$ . Unitary is the complex version.
Orthogonal matrices never magnify errors, which is why numerical algorithms are built from them.

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3.3 Orthogonal matrices, rotations and reflections

The picture

Definitions and theorems

Orthogonal matrix

Isometry

Orthogonal means length-preserving

Determinant of an orthogonal matrix

The plane: every orthogonal matrix is a rotation or a reflection

Rotation matrix

Orthogonal 2 by 2 matrices

Space: every rotation has an axis

Euler's rotation theorem

Complex version and eigenvalues

Unitary matrix

Eigenvalues of an orthogonal matrix

Where it is used

Traps

Problems

Summary

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Where this sits

Comes from

Leads to

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