# Quaternion kinematics transition matrix closed form

For rigid body rotation discrete time kinematics equation is as follows

$$R_{k+1} = T_r \cdot R_k$$

where:

$$R$$ is a rotation matrix

$$T_r = e^{-[a\times]}$$ is a transition matrix

$$[a\times]$$ means a skew symmetric matrix from vector $$a$$

$$a=\omega \cdot \delta t$$ is a rotation vector

$$\omega$$ is angular velocity vector in body frame

$$\delta t$$ is a time step.

Using Rodrigue's formula one can have a closed form solution for matrix exponent so no infinite series calculation required.

$$T_r=I+sin(\theta)\left[e\times \right] + (1-cos(\theta))\left[e\times \right]^2$$

where

$$e = \frac{a}{||a||}$$ is a unit vector of rotation

$$\theta = ||a||$$ is angle of rotation around that unit vector

When dealing with quaternions similar expression is

$$q_{k+1} = T_q \cdot q_k$$

where

$$T_q = e^{\frac{1}{2}\begin{bmatrix}-\left[a\times\right] && a\\ -a^T&& 0 \end{bmatrix}}$$

Hope I have put not many mistakes in the expressions above. But general idea should be clear... Is there a closed form solution for quaternion discrete kinematics transition matrix $$T_q$$?

$$\begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}^2=\|a\|^2\begin{bmatrix}-I_3 & 0 \\ 0 & -1\end{bmatrix}$$
$$\exp\Big(\frac{1}{2}\begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}\Big)\,=\,\cos\!\Big(\frac{\theta}{2}\Big)I_4+\sin\!\Big(\frac{\theta}{2}\Big)\begin{bmatrix}-[e\times] & e \\ -e^T & 0\end{bmatrix}.$$
• @aliko It is textbook how to exponentiate diagonalizable matrices, and this is block-diagonal with $2\times2$ blocks which we also know how to exponentiate. Indeed, the matrix with the $a$s in it is none other than the matrix representing right-multiplication-by-$a$ on the quaternions with respect to the ordered basis $\{i,j,k,1\}$, so we can expect the matrix exponential $\exp(\tfrac{1}{2}[\cdots])$ to be the matrix representing right-multiplication by the quaternion exponential $\exp(\frac{1}{2}a)=\cos(\theta/2)+\sin(\theta/2)e$. Commented Aug 5, 2023 at 8:51
• To block-diagonalize it, BTW, extend $e=\hat{a}$ to an orthonormal basis $\{\hat{a},b,c\}$ of $\Bbb R^3$, then extend further with $d=e_4$, so then the matrix with $a$s in it acts block-diagonally with two $2\times2$ blocks both $\theta=\|a\|$ times $90^\circ$ rotation matrices with respect to the basis $\{b,c,\hat{a},d\}$. Commented Aug 5, 2023 at 9:00