How to calculate the inverse of $F= I + \alpha(\theta)W+\beta(\theta)W^2$, where $W=[\omega]_\times=-\varepsilon\cdot\omega$, $\theta=||\omega||$, $\omega\in\mathbb{R}^3$ is a 3-D vector, $\alpha(\theta) =\frac{1-\cos\theta}{\theta^2}$, and$\beta(\theta)=\frac{\theta-\sin\theta}{\theta^3}$. $I$ is an identity matrix of 3 by 3.
I found the solution from here (equation 78), which is :
$$ F^{-1}=I-\frac{1}{2}W+\frac{2-2\cos\theta-\theta\sin\theta}{\theta^2(2-2\cos\theta)}W^2 $$
But how is it calculated?
I tried the following $$ FF^{-1}=(I + \frac{1-\cos\theta}{\theta^2}W+\frac{\theta-\sin\theta}{\theta^3}W)(I-\frac{1}{2}W+\frac{2-2\cos\theta-\theta\sin\theta}{\theta^2(2-2\cos\theta)}W^2)\\ =I +\frac{1-\cos\theta}{\theta^2}W+\frac{\theta-\sin\theta}{\theta^3}W^2-\frac{1}{2}W-\frac{1-\cos\theta}{2\theta^2}W^2-\frac{\theta-\sin\theta}{2\theta^3}W^3+\frac{2-2\cos\theta-\theta\sin\theta}{\theta^2(2-2\cos\theta)}W^2) +\frac{1-\cos\theta}{\theta^2}\frac{2-2\cos\theta-\theta\sin\theta}{\theta^2(2-2\cos\theta)}W^3+\frac{\theta-\sin\theta}{\theta^3}\frac{2-2\cos\theta-\theta\sin\theta}{\theta^2(2-2\cos\theta)}W^3 $$ I couldn't see any chance to make the long formula into an identity matrix. Can anyone help to explain how to derive the inverse of $F$?
Many thanks!
The skew matrix $W$ has 3 distinct eigenvalues: $\lambda = \{-i\theta, 0, +i\theta\}$
Knowing this we can calculate $f(z)$ via Sylvester interpolation.
Using Cayley-Hamilton, we can express any analytic function of $W$ as $$\eqalign{ f(W) &= a_2W^2 + a_1W + a_0I \cr\cr }$$ Now write down the corresponding scalar equation for each eigenvalue $$\eqalign{ f(\lambda_k) &= a_2\lambda_k^2 + a_1\lambda_k + a_0 \cr }$$ and solve these 3 scalar equations to determine the {$a_k$} coefficients.
In this case, the function is $$f(z)=\frac{z}{e^z-1}$$ To get started, the equation for the zero eigenvalue yields $$\eqalign{ f(0) &= a_2*0 + a_1*0 + a_0 \cr 1 &= a_0 \cr }$$ The remaining two equations are conjugates $$\eqalign{ f(+i\theta) &= -\theta^2a_2 + i\theta a_1 + 1 \cr f(-i\theta) &= -\theta^2a_2 - i\theta a_1+ 1 \cr }$$ Subtracting them yields $$\eqalign{ f(+i\theta)-f(-i\theta) &= i2\theta a_1 \cr a_1 &= \frac{f(+i\theta)-f(-i\theta)}{i2\theta} \cr }$$ While adding them yields $$\eqalign{ f(+i\theta)+f(-i\theta) &= 2-2\theta^2\,a_2 \cr a_2 &= \frac{2-f(+i\theta)-f(-i\theta)}{2\theta^2} \cr }$$ Now you just need to evaluate the functions and use some trigonometric identities to find nicer expressions for the coefficients.
Indeed, $a_1$ does reduce to $-\frac{1}{2}$
I'm still working on $a_2$.