Recall the following:
Matrix exponential of a skew symmetric matrix
The conclusion is:
$$e^C = I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$$
- $x = \sqrt{a_1^2+a_2^2+a_3^2}$
- $C=\left( \begin{array}{ccc} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{array} \right).$
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?