This post is a tribute to the fact that the matrix logarithm of a $3\times 3$ rotational matrix is

\[\log{R} = \frac{\theta}{2\sin{\theta}}(R^T - R) \label{eq:logm} \tag{1}\]

where $\theta$ satisfies $1 + 2\cos(\theta) = \text{trace}(R)$.

I think this is beautiful and surprising: The matrix logarithm is proportional to the simplest skew-symmetric matrix one can create from $R$: $R^T - R$.

We will argue that a matrix logarithm of a rotational matrix will output a skew-symmetric matrix, but that a priori does not have to mean the relationship is as simple as it is!

A more thorough treatment of this topic can be found in this paper.

Matrix exponentials

Matrix exponentials are neat. They appear in the (analytical) solutions to linear systems of ordinary differential equations (ODEs):

\[\mathbf{\dot{x}} = A\mathbf{x} \quad \Rightarrow \quad \mathbf{x}(t) = e^{At}\mathbf{x}_0 \label{eq:ode} \tag{2}\]

in which

\[e^A = 1 + A + \frac{1}{2} A^2 + \frac{1}{3!}A^3 + \ldots + \frac{1}{n!}A^n + \ldots\]

There is a famous paper by MATLAB founder Cleve Moler on how there are many ways to compute it, but all of them are dubious. Luckily, we will only be dealing with a specific type of small systems in this post: $3 \times 3$ skew-symmetric systems. A skew-symmetric system is one that satisfies

\[S^T = - S\]

It follows rather directly that

\[(e^S)^Te^S = e^{S^T}e^S = e^{-S}e^S = e^{-S + S} = e^0 = I\]

The matrix exponential of a skew-symmetric matrix is orthogonal. And conversely the matrix logarithm of an orthogonal matrix will be skew-symmetric.

We have not actually used the $3\times 3$ size of the matrices yet, but we’ll need that to also understand the $\theta$ which appears in equation \eqref{eq:logm}

We will take a detour through some application: 3D curves

3D curves

For sufficiently smooth curves in 3D, one can define an orthogonal coordinate frame at every point along the curve consisting of the 3 vectors:

$T$: the tangent
$N$: the normal
$B$: the binormal

A picture says a thousand words:

Frenet-Serret frame

This frame is known as the Frenet-Serret frame. Associated are the Frenet-Serret equations which establish the relationship between the change of these frames when one moves along the curve, and the curvature $\kappa$ and torsion $\tau$ of the 3D curve:

\[\begin{bmatrix} \mathbf{T'} \\ \mathbf{N'} \\ \mathbf{B'} \end{bmatrix} = \begin{bmatrix} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix} \begin{bmatrix} \mathbf{T} \\ \mathbf{N} \\ \mathbf{B} \end{bmatrix}\]

Et voila, we have our skew-symmetric matrix! Solving these differential equations is generally not as simple as using equation \eqref{eq:ode}, as $\kappa$ and $\tau$ are usually functions of time/arc length. But matrix exponentials can be useful when designing dedicated integrators for these types of systems, which will be a topic for a future blog post.

Note that these are not general skew-symmetric matrices, as the $(1, 3)$ and $(3, 1)$ entries are zero. Parallel transport frames and Darboux frames are alternative frames for which these entries might not be zero.

Why would one be interested in using all of these equations? Well, the applications are plenty: rollercoasters are real life examples of freeform curves, flight paths of planes and drones as well, freeform metal tubes are used in many design applications, etc…

3D rotation formalisms

Wikipedia has a very extensive page on rotation formalisms in 3D, which lists are relates formalisms like

Euler angles
Rotation matrices
Quaternions
Axis-angle representations

Matrix exponentials are abundant. Even in 2D this is actually the case when we consider the relationship between angles, complex numbers and exponentials through:

\[e^{i\theta} = \cos(\theta) + i \sin(\theta)\]

In 3D, the key connection between the matrix exponentials and logarithms and the rotation formalisms is that the skew-symmetric matrix is very closely related to the axis-angle representation. A $3\times 3$ skew-symmetric matrix only has 3 ‘degrees-of-freedom’ $\mathbf{u} = (u_1, u_2, u_3)$:

\[S_u = \begin{bmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{bmatrix}\]

The ‘axis’ in axis-angle is only a direction, so a normalized $\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|}$. The angle $\theta$ from equation \eqref{eq:logm} is precisely the length of $\mathbf{u}$:

\[\theta = \|\mathbf{u}\|\]

Finally, we can remark the relationship between the cross product and matrix vector multiplication with $S$

Scipy experiments

import numpy as np
import scipy.linalg