# Derivative of rotation matrix in a form skew-symmetric matrix

I am working on an application of CV, in which a way to calculate the derivative of rotation matrix is involved.

$$R$$ is the rotation matrix and $$R \in SO(3)$$ Also, $$R$$ is changing with $$t$$ giving $$R(t)$$.

$$R(t)R(t)^\mathrm{T} = I$$ is known. We calculate the derivative of $$R(t)R(t)^\mathrm{T}$$ which gives us a skew symmetric matrix $$\dot{R}(t)R(t)^\mathrm{T} = -R(t)\dot{R}{}^\mathrm{T}(t) =: \phi(t)$$, where $$\phi (t) = \left[ \begin{matrix} 0 & -\phi_3 & \phi_2 \\ \phi_3 & 0 & -\phi_1\\ -\phi_2 & \phi_1 & 0 \end{matrix} \right].$$

The derivative of $$R(t)$$ is then given by $$\dot{R}(t) = \phi (t) R(t).$$

To derive the Lie algebra, later, it gives the 1st order expansion of $$R(t)$$ $$R(t) \approx R(t_0) + \dot{R}(t_0)\ (t-t_0) = I + t\ \phi (t_0) ,$$ where $$t_0 = 0$$ and $$R(t_0) = I$$. By given an assumption that $$\phi (t_0) = \phi_0$$ around $$t_0$$, it gets $$\dot{R}(t) = \phi (t_0) R(t) = \phi_0 R(t).$$ This is the one I cannot get, i.e. the last equation is my question. Any suggestion would be appreciated.

• Related : Velocity in a turning reference frame. – Frobenius Jun 16 at 17:32
• with Euler-Rodrigues-Formel is the rotation matrix $R=\exp \left( \left[ \overrightarrow{\varphi }\right] _{x}\right)$ where $\vec \varphi$ is the rotation axis and $\vec\varphi\cdot\vec\varphi=1$ – Eli Jun 16 at 20:40

$$\phi$$ can be thought of as the tangent vector $$\text{d}/\text{d}t$$ of $$R(t)$$. For any $$R(t)$$, we can use the exponential map to obtain $$R(t_0 + t) = R(t_0) + t\frac{\text{d}}{\text{d}t} R(t_0) + \frac{1}{2!}\left(t \frac{\text{d}}{\text{d}t}\right)^2 R(t_0) + \frac{1}{3!}\left(t \frac{\text{d}}{\text{d}t}\right)^3 R(t_0) + \ldots \\ = \sum^\infty_{n = 0} \frac{1}{n!} \left(t \frac{\text{d}}{\text{d}t}\right)^n = e^{t \phi } R |_{t_0}$$

By definition $$R(t_0 = 0) = I$$, we have $$R(t) = e^{t\phi}$$

and we can differentiate the matrix exponential to obtain $$\dot{R}(t) = \phi e^{t\phi} = \phi_0 R(t)$$

• Thanks, Vincent. I forget this full expansion. – MakeItGreatAgain Jun 17 at 6:33

Use your $$t_0 = 0$$ so $$R(0) = I$$, $$\dot{R}(t)R(t)^\mathrm{T} =: \phi(t)= \phi(0)+ O(t)\leadsto \\ \dot{R}(t)= (\phi_0+ O(t) ) R(t)= (\phi_0+ O(t) )( I+ O(t))= \phi_0+ O(t) .$$
The order t terms vanish at 0, so $$\dot{R}(0)=\phi_0$$, antisymmetric, alright.

• Thanks as well, Cosmas. – MakeItGreatAgain Jun 17 at 6:34
• Cosmas, I have a little more to ask. It is simply about the symbol in my question you edited. Is that OK to use $$\phi(t) = \left[ \begin{matrix} x & x & x \\ x & x & x \\ x & x& x \end{matrix} \right]$$ instead of $\phi(t)^{\wedge}$ representing a skew symmetric matrix of vector $\phi(t)$. I want them in my assignment report. – MakeItGreatAgain Jun 17 at 6:42
• Cosmas, I have a little more to ask. It is simply about the symbol in my question you edited. Is that OK to use \phi(t) = \left [ – MakeItGreatAgain Jun 17 at 6:42
• No, the object you defined is manifestly antisymmetric, as you demonstrate. – Cosmas Zachos Jun 17 at 10:28