Part 2 · Tensors

2.6 Orthogonal tensors

Source: Favata’s notes, appendix A.2, p. 15–16 (orthogonal tensors, Orth, Rot)

Objectives

Understand orthogonal tensors as the maps that keep lengths and angles.
Tell rotations from reflections by the determinant.
Recipe: check $Q^{⊤} Q = I$ ; write a rotation about a given axis.
Pictures: the unit circle mapped onto itself; the letter turned, or mirrored.
Theorem: $Q$ is orthogonal iff $Q^{⊤} = Q^{- 1}$ .
Vocabulary: orthogonal tensor, $Orth$ , rotation, $Rot$ , reflection.

Keeping the scalar product

Definition 2.6.1Orthogonal tensor

$Q \in Lin$ is orthogonal if it preserves the scalar product: $Qu \cdot Qv = u \cdot v \forall u, v \in V .$

Lengths and angles are made of scalar products, so both are kept:

with $v = u$ : $∣ Qu ∣^{2} = u \cdot u$ , so $∣ Qu ∣ = ∣ u ∣$ ;
$cos ϑ = \frac{u \cdot v}{∣ u ∣∣ v ∣} = \frac{Qu \cdot Qv}{∣ Qu ∣∣ Qv ∣}$ .

A body moved by an orthogonal tensor keeps its shape: nothing is stretched, nothing is sheared. That is why they reappear as the gradients of rigid deformations.

Loading the demo…

Figure 2.6.1 The unit circle (dotted) and the square. For the rotations and the reflection, the circle lands on itself and the square stays a unit square: lengths and right angles are kept. The reflection mirrors the F, and its determinant says so. Compare with the shear and the stretch, where the circle becomes an ellipse.

Characterisation

Theorem 2.6.2Orthogonal means transpose = inverse

$Q$ is orthogonal iff $Q^{⊤} Q = Q Q^{⊤} = I, that is, Q^{⊤} = Q^{- 1} .$

Proof

⇒ (in the notes): move $Q$ across the scalar product, $u \cdot v = Qu \cdot Qv = Q^{⊤} Qu \cdot v$ . So $(u - Q^{⊤} Qu) \cdot v = 0$ for every $v$ ; take $v$ equal to the bracket to get $u = Q^{⊤} Qu$ for every $u$ .

The other order (beyond the notes): $(det Q)^{2} = det (Q^{⊤} Q) = 1$ , so $Q$ is invertible (Theorem 2.5.2); multiply $Q^{⊤} Q = I$ by $Q^{- 1}$ on the right.

⇐ (beyond the notes): $Qu \cdot Qv = Q^{⊤} Qu \cdot v = u \cdot v$ .

Recipe 2.6.3Orthogonal from the columnsbeyond the notes

$(Q^{⊤} Q)_{ij} = Q_{ki} Q_{kj} = Q e_{i} \cdot Q e_{j}$ . So $Q$ is orthogonal iff its columns are orthonormal: it sends the basis ${e_{i}}$ to another orthonormal basis. To check a matrix, check that each column has length $1$ and that different columns are orthogonal.

Rotations and reflections

From $Q^{⊤} Q = I$ : $(det Q)^{2} = det Q^{⊤} det Q = 1$ , so $det Q = \pm 1$ .

Definition 2.6.4Orth and Rot

$Orth := {Q \in Lin : Q^{⊤} Q = Q Q^{⊤} = I}, Rot := {R \in Orth : det R = 1} .$ An orthogonal $Q$ with $det Q = 1$ is a rotation; with $det Q = - 1$ , a reflection: it keeps lengths and angles but flips orientation (Key observation 2.5.6).

beyond the notes

Products and inverses of orthogonal tensors are orthogonal: $(Q_{1} Q_{2})^{⊤} Q_{1} Q_{2} = Q_{2}^{⊤} Q_{1}^{⊤} Q_{1} Q_{2} = I$ . Determinants multiply, so the same holds inside $Rot$ . Two reflections make a rotation.

Examples (beyond the notes):

Example 2.6.5Rotation about e₃beyond the notes

$[R] = cos θ sin θ 0 - sin θ cos θ 0 001, det R = cos^{2} θ + sin^{2} θ = 1.$ Column 1 is where $e_{1}$ goes: turned by $θ$ in the 1–2 plane. This is the matrix of Exercise 3.5.1.

Example 2.6.6Rotation about any axisbeyond the notes

For a unit vector $e$ and an angle $θ$ , $R = e \otimes e + cos θ (I - e \otimes e) + sin θ E, Eu = e \times u .$ It keeps the component along the axis (Proposition 2.1.7), and turns the component in the normal plane by $θ$ , using the skew tensor of the axis (Theorem 2.3.3). With $e = e_{3}$ it is the matrix above.

Example 2.6.7Reflectionbeyond the notes

$I - 2 e \otimes e$ reflects across the plane normal to the unit vector $e$ : the plane is kept, $e \mapsto - e$ . It is symmetric, and $(I - 2 e \otimes e)^{2} = I - 4 e \otimes e + 4 e \otimes e = I$ , so it is orthogonal with determinant $- 1$ .

Warning

$- I$ is orthogonal, and in 3D $det (- I) = (- 1)^{3} = - 1$ : the point inversion is a reflection, not a rotation. In the plane $det (- I) = + 1$ , and it is the rotation by $18 0^{\circ}$ . Statements about orientation depend on the dimension.

Non-example 2.6.8Determinant 1 is not enoughbeyond the notes

The simple shear has $det = 1$ and keeps areas, but it is not orthogonal: $∣ F e_{2} ∣ = 1 + γ^{2} \neq = 1$ . Keeping volumes is much weaker than keeping lengths. Pick the shear in the demo above and watch the circle.

beyond the notes

A real eigenvalue of an orthogonal tensor is $\pm 1$ : from $Qv = λ v$ and $∣ Qv ∣ = ∣ v ∣$ , $∣ λ ∣ = 1$ . A rotation of 3D space always has the eigenvalue $1$ : its axis.

Exercises

Exercise 2.6.1beyond the notespractice (A.5)

Let $Q \in Orth$ , $A \in Lin$ . Show $∣ QA ∣ = ∣ A ∣$ and $tr (Q^{⊤} AQ) = tr A$ .

Solution

$∣ QA ∣^{2} = tr ((QA)^{⊤} QA) = tr (A^{⊤} Q^{⊤} QA) = tr (A^{⊤} A) = ∣ A ∣^{2}$ . $tr (Q^{⊤} AQ) = tr (AQ Q^{⊤}) = tr A$ , by the cyclic property.

Exercise 2.6.2beyond the notes

Is $[Q] = \frac{1}{3} 122 21 - 2 2 - 2 1$ orthogonal? A rotation or a reflection?

Solution

Each column has length $\frac{1}{3} 1 + 4 + 4 = 1$ , and $(1, 2, 2) \cdot (2, 1, - 2) = (1, 2, 2) \cdot (2, - 2, 1) = (2, 1, - 2) \cdot (2, - 2, 1) = 0$ : the columns are orthonormal, so $Q$ is orthogonal. $det [Q] = \frac{1}{27} (- 3 - 12 - 12) = - 1$ : a reflection. In fact $[Q] = I - 2 e \otimes e$ with $e = (1, - 1, - 1) / 3$ , the reflection across the plane normal to $e$ .

Exercise 2.6.3beyond the notes

Let $Q \in Orth$ . Show that $Qu \times Qv = (det Q) Q (u \times v)$ : the vector product follows a rotation, and flips sign under a reflection.

Solution

For every $w$ , the determinant gives $Qw \cdot (Qu \times Qv) = det Q w \cdot (u \times v)$ (both sides vanish when $w, u, v$ are dependent). And $w \cdot (u \times v) = Qw \cdot Q (u \times v)$ , since $Q$ is orthogonal. So $Qw \cdot (Qu \times Qv - det Q Q (u \times v)) = 0$ for every $w$ ; $Qw$ ranges over all of $V$ , so the bracket is $0$ . The vector product depends on the handedness (Definition 1.3.5).