Matrix exponential of the sum of two skew-symmetric matrices

360 Views Asked by At

This is my first message in this site. I'm a mechanical engineer with, amongst others, an interest in inertial navigation. I'm currently reading the book "Principles of GNSS, Inertial and Multisensor Integrated Navigation Systems" from Paul D. Groves. On the development of one equation (eq. 5.26 in the second edition of the book) I came across this step:

$$C_b^e (t) \exp⁡[(Ω_{ib}^b-Ω_{ie}^b ) τ_i ]=C_b^e (t) \exp⁡(Ω_{ib}^b τ_i )-C_b^e (t)[\exp⁡(Ω_{ie}^b τ_i )-I_3 ]$$

The $C$s are rotation matrices. The $\Omega$s are the skew-symmetric matrices associated to rotation velocities between two reference systems. The sub- and superindexes $i$, $e$ and $b$ refer to $inertial$, $earth$ and $body$ reference systems. $t$ is time, $\tau_i$ is the time lapse between two sample/integration steps.

What I can't understand is how the matrix exponential of the difference of two matrices can convert in the difference of the matrix exponentials of those two matrices (plus the identity matrix). I've seen this operation in several places on the book, so I doubt it's an erratum. I guess one might prove it by means of the series expansion of the matrix exponential function and some property of skew-symmetric matrices, but I didn't get very far this way.

Can somebody please cast some light on this step? Thanks in advance!

Greetings

Guillermo

1

There are 1 best solutions below

0
On

If we proceed by orders of $\tau_i$, the zeroth and first orders come out right, and the second order yields

$${\Omega^b_{ib}}^2-\Omega^b_{ib}\Omega^b_{ie}-\Omega^b_{ie}\Omega^b_{ib}+{\Omega^b_{ie}}^2={\Omega^b_{ib}}^2-{\Omega^b_{ie}}^2\;,$$

or

$$\Omega^b_{ib}\Omega^b_{ie}+\Omega^b_{ie}\Omega^b_{ib}=2{\Omega^b_{ie}}^2\;.$$

This does not hold for skew-symmetric matrices in general (e.g. it doesn't hold for the generators of rotations about the $x$ and $y$ axes), so the equation is either wrong or depends on specific properties of the matrices.