This post is about power series of the form:

$$ \alpha_0 I + \sum_{n=0}^\infty \alpha_n [v]_\times^n. $$

Assuming $v \neq 0$ we can rewrite any such power series as: $$ \beta_0 I + \sum_{n=1}^\infty \beta_n [u]_\times^n, \qquad u = \frac{v}{\|v\|}, \qquad \beta_n = \alpha_n \|v\|^n. $$

Based on the power relationships discussed in a previous post, this can be further simplified to: $$ \beta_0 I + \sum_{n=1}^\infty \beta_n [u]_\times^n = \beta_0 I + \left(\sum_{n=0}^\infty (-1)^n \beta_{2n+1}\right) [u]_\times + \left(\sum_{n=1}^\infty (-1)^{n+1}\beta_{2n}\right) [u]_\times^2 $$

This makes it clear that any power series in $[u]_x$ can always be rewritten as at most three terms:

$$ \begin{equation} \alpha I + \beta [u]_\times + \gamma [u]_\times^2, \qquad \|u\| = 1 \label{pow} \end{equation} $$

Applying $[u]_\times^2 = uu^T - I$, we can find one last simplification: $$ \begin{equation} a I + b [u]_\times + c uu^T, \qquad \|u\| = 1\label{proj} \end{equation} $$ with $a = \alpha - \gamma$, $b = \beta$ and $c = \gamma$.

The Rodrigues rotation formula for $\exp(\theta [u]_\times)$ is a special case of such a power series worth writing out explicitly: $$ \begin{align} \exp([\theta u]_\times) &= I + \sin(\theta) [u]_\times + (1-\cos(\theta)) [u]_\times^2 \\
&= \cos(\theta) I + \sin(\theta) [u]_\times + (1-\cos(\theta)) uu^T \\
\end{align} $$

Eigen-vector Analysis

A previous post reviewed how the eigenvector-eigenvalue pairs for $[u]_\times$. When $u \neq 0$ there pairs were:

• $(u, 0)$
• $(v + jw, -\|u\|j)$
• $(v - jw, \|u\|j)$

where $v$ was any vector orthogonal to $u$ and $w := \frac{1}{\|u\|}[u]_\times v$.

Its easy to see that matrices of the form $\eqref{proj}$ have the same eigenvectors as $[u]_\times$ owing to $u^Tv = u^Tw = 0$, so that the eigen-vector / eigen-value pairs are:

• $(u, a + c)$
• $(v + jw, a-bj)$
• $(v - jw, a+bj)$

Inverse

There are several ways to show that the matrices of the form $\eqref{proj}$ have an inverse of the form: $$ (a I + b [u]_\times + c uu^T)^{-1} = d I + e [u]_\times + f uu^T, $$ with: $$ \begin{equation} \begin{bmatrix}d \\ e \\ f\end{bmatrix} = \frac{1}{a^2+b^2}\begin{bmatrix}a \\ -b \\ \frac{b^2-ca}{a+c}\end{bmatrix}. \label{eq:def} \end{equation} $$ if and only if $a + c \neq 0$ and $a^2 + b^2 > 0$. One approach is to apply the eigen-analysis of the previous section, but I prefer the following direct proof.

Proof: If $a + c= 0$: $$ (a I + b [u]_\times + c uu^T) u = (a + c) u = 0 $$ so the matrix is not invertible. Similarly, if $a^2 + b^2 = 0$ then: $$ (a I + b [u]_\times + c uu^T) = c uu^T $$ which is clearly not invertible.

Otherwise, we explicitly expand the product of these two matrices: $$ \begin{gather} (a I + b [u]_\times + cuu^T) (d I + e [u]_\times + f uu^T) \\
=\\
ad I + ae [u]_\times + af uu^T + bd [u]_\times + be uu^T - be I + c(d+f) uu^T\\
=\\
(ad-be) I + (bd+ae) [u]_\times + (cd + be + (a+c)f) uu^T. \end{gather} $$ For the second matrix to be the inverse of the first, requires that: $$ \begin{equation} \begin{bmatrix} a & -b & 0 \\
b & a& 0 \\
c & b & a+c \end{bmatrix} \begin{bmatrix} d \\ e \\ f \end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} \label{eq:inv_lin_sys} \end{equation} $$ and $\eqref{eq:def}$ solves this equation.

The inverse of the matrix in Equation $\eqref{eq:inv_lin_sys}$ is easy to find by applying the identity: $$ \begin{bmatrix} A & 0 \\
B & C \end{bmatrix}^{-1} = \begin{bmatrix} A^{-1} & 0 \\
-C^{-1}BA^{-1} & C^{-1} \end{bmatrix}. $$