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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!

1 Answers 1

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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 $a_2$.

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    Cool !! So for any analytic function the coefficients $$\eqalign{a_0&=f(0)\cr a_1&=\frac{f(i\theta)-f(-i\theta)}{i2\theta}\cr a_2&=\frac{2f(0)-f(i\theta)-f(-i\theta)}{2\theta^2}\cr}$$ can be used to evaluate the matrix function $$f(W)=a_2W^2+a_1W+a_0I$$ – 2017-01-07
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    Hi hans, thanks for your explanation. I am wondering where does $f(z)=\frac{z}{e^z-1}$ come from? – 2017-01-07
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    Probably because of my limited math background, I have a hard time understand your solution. Is that say according to Sylvester interpolation, I know that $F^{-1}$ will have the form $F^{-1}=a_2W^2+a_1W+a_0I$ and the unknowns are $a_0,a_1,a_2$. Since I know that $F = \beta(\theta)W^2+\alpha(\theta)W+I$ we have that $f(x)=\beta(\theta)x^2+\alpha(\theta)x +1$, where $x\in\mathbb{R}$ is a scalar. – 2017-01-07
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    okay try again, let's use the mysterious $f(z)=\frac{z}{e^z-1}$ I do get $a_1=-1/2$ and in the same way, I can verify $a_2$, but again where does $f(z)$ come from? – 2017-01-07
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    @ShudaLi I recognized the coefficients of $F$ in your problem statement as the coefficients of the function $\phi(z)=\frac{e^z-1}{z}$ commonly used in ODE solvers. Therefore if you're trying to calculate its reciprocal, it would be the function $f(z)$ in my answer. It seems strange that you don't even know the function that you're trying to evaluate. – 2017-01-07
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    Wow, thanks a lot for pointing that out to me! I've been reading the notes by Ethan about Lie group for more than 10 times, but didn't realise Eqn(65) and (7) have such an amazing connection until you told me. Thank you so much I learn a lot today! – 2017-01-07
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    Okay, I was wrong in my first time because $f(x)=\beta(\theta)x^2+\alpha(\theta)x +1$ is the original analytic function of $W$, I need to find the inverse of $f$, luckily, $f(x)=\frac{e^z-1}{z}$ and its inverse is $f^{-1}(x)=\frac{z}{e^z-1}$ which is a lot simpler than $f^{-1}(x)=\frac{1}{\beta(\theta)x^2+\alpha(\theta)x +1}$, right? sorry for my naive questions, now everything makes perfect sense to me! this is so cool! Thanks again hans and greg! – 2017-01-07
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    Sorry hans, I just realise that $W$ as an odd order skey-symmetric matrix is not invertible, therefore, we can't say $f(W)=\frac{e^W-I}{W}=(e^W-1)W^{-1}$. We might not be able to apply your method in this case, but it is pretty amazing that it still gives the correct answer. I wonder if there is an explanation for it. – 2017-01-08
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    @ShudaLi That is definitely NOT the way to calculate the phi function. Note that $$\phi(0) = 1$$ If you take the Taylor expansion for $\exp(x)$ and subtract $1$ then divide by $x$ you are left with another Taylor expansion with an infinite radius of convergence. You should apply the phi function on the *scalar* eigenvalues of $W$ in order to calculate the *scalar* coefficients of the Sylvester Interpolant. – 2017-01-09
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    Since the Taylor series for $\phi(x)$ has an infinite radius of convergence, it can be evaluated for ALL matrices, whether or not they are singular. And if you divide the Taylor series by the characteristic polynomial, the remainder will be the Sylvester interpolant. – 2017-01-09
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    @hans, Thanks for the update! in the last sentence, which Taylor series do you mean? those of $f(z) = \frac{e^z-1}{z}$ or $f(z)=e^z$, the characteristic polynomial of $W$ which is $t^3 + t\omega_1^2+t\omega_2^2+t\omega_3^2$? how do you get the Sylvester interpolant? The Sylvester interpolant are $a_2$, $a_1$ and $a_0$? – 2017-01-09
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    The way polynomial division works is like this $$f(x) = p(x)\,q(x)+r(x)$$where the "function" $f$ is of degree $M$, the characteristic polynomial $p$ (of $A$) is of degree $n$, the "quotient" $q$ is of degree $(M/n)$, and the "remainder" $r$ is of degree $\le n$. This equality holds when the matrix $A$ is substituted for the scalar $x$. The clever bit is that, according to Cayley-Hamiltion, $p(A)=0$ so $$f(A)=r(A)$$and $r$ is what we have been calling the Sylvester interpolant in our discussions. – 2017-01-10
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    @ShudaLi Just realized that I forgot to mention your user name in my previous comment. – 2017-01-11