We can compute a rotation matrix \( R \in SO(3) \) from an angle \(\theta\) and axis \(l\) (unit vector)

\[
R = I + \sin(\theta)C + (1-\cos \theta)C^2
\]

where \(C\) is the antisymmetric matrix:

\[
C = \begin{bmatrix}
0 & -l_z & l_y \\
l_z & 0 & -l_x \\
-l_y & l_x & 0
\end{bmatrix}
\]

Our proof will use the Taylor's formula to define the exponential of a matrix \(M\)

\[
e^M = \sum_{k=0}^{\infty}\frac{M^k}{k!}
\]

where \(M^0\) is defined to be the identity matrix \(I\). Because \(C^3 = -C\) and the Taylor expansions:

\begin{align}
\sin \theta &= \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots + \frac{(-1)^n}{(2n+1)!}\theta^{2n+1} + \cdots \\
\cos \theta &= 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots + \frac{(-1)^n}{(2n)!}\theta^{2n} + \cdots
\end{align}

then we obtain

\begin{align}
R &= e^{\theta C}\\
&= \sum_{k=0}^{\infty}\frac{(\theta C)^k}{k!} \\
&= I + \frac{1}{1!}\theta C + \frac{1}{2!}(\theta C)^2 + \frac{1}{3!}(\theta C)^3 + \frac{1}{4!}(\theta C)^4 + \frac{1}{5!}(\theta C)^5 + \frac{1}{6!}(\theta C)^6 + \cdots \\
&= I + (\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots)C + (\frac{\theta^2}{2!} - \frac{\theta^4}{4!} + \frac{\theta^6}{6!} - \cdots)C^2 \\
&= I + \sin\theta C + (1-\cos \theta)C^2
\end{align}

2016-07-08 Comments

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