Inverse of Euler parameter rotation matrix formulation [Barfoot 6.6.2

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Reading through Barfoot's State Estimation for Robotics and I am curious regarding one of the problems:

Show that $C^{-1}=C^{T}$ starting from $C = \cos\theta I > (1-\cos\theta)aa^{T} + \sin\theta a^{\wedge}$

I've worked out that $C^T = \cos\theta I + (1-\cos\theta) aa^T - \sin\theta a^\wedge$ but am not sure how to proceed with the inverse, I've tried multiplying $CC^T$ but this didn't seem easily reducable.