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Recall the following:
Matrix exponential of a skew symmetric matrix

The conclusion is:

$$e^C = I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$$

  1. $x = \sqrt{a_1^2+a_2^2+a_3^2}$
  2. $C=\left( \begin{array}{ccc} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{array} \right).$

And by the fact that $$e^C\in SO(3),$$ i.e., the exponential of skew-symmetric matrix is an element of $SO(3)$.
(Please see the reference (p.4): http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=5766283)


Also

The differential equation for the rotation matrix: $$\dot{C} = -\omega^{\times}C$$ where $\omega^{\times} = \left( \begin{array}{ccc} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \\ \end{array} \right)$

And we can think that the rotation axis with angular velocity in this dynamic is

$\hat{\omega} = \left( \begin{array}{ccc} \omega_1 \\ \omega_2 \\ \omega_3\end{array} \right)$


My question is suppose (rotation about the $z$-axis)

$\hat{\omega} = \left( \begin{array}{ccc} 0 \\ 0 \\ \omega_3\end{array} \right)$ i.e.,

$\omega^{\times} = \left( \begin{array}{ccc} 0 & -\omega_3 & 0 \\ \omega_3 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$

How to derive the corresponding rotation matrix is:

$R=\left( \begin{array}{ccc} \cos\omega t & -\sin\omega t & 0 \\ \sin\omega t & \cos\omega t & 0 \\ 0 & 0 & 1 \\ \end{array} \right).$

There is an identity matrix $I$ in the equation; how to deal with that?

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    $\begingroup$ Have you computed all the components of the Rodriguez formula, esp. $C^2$ for your special case? $\endgroup$ Feb 18, 2017 at 16:42

1 Answer 1

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For rotation about the $e_3$, or z-axis, let $$\eqalign{ B &= \omega^\times\,t \cr b &= \omega_3\,t \cr G_3 &= B/b \cr }$$ $G_3$ is a skew-symmetric (0-1) matrix, whose square is $$\eqalign{ G_3^2 &= e_3e_3^T-I \cr\cr }$$

Evaluating the exponential formula $$\eqalign{ R = e^B &= \frac{\sin b}{b}\,B &+\,I + \frac{1-\cos b}{b^2}\,B^2 \cr &= (\sin b)\,G &+\,I + (1-\cos b)\,G^2 \cr &= (\sin b)\,G &+\,(\cos b)\,I + (1-\cos b)\,e_3e_3^T \cr \cr}$$ The first 2 terms yield the familiar $(2\times 2)$ cosine-sine rotation matrix.

The third term sets the lower right element by subtracting the $\cos(b)$ factor coming from the $(3\times 3)$ identity matrix in the second term, and replacing it with a value of 1.

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