## A Proof of (Rv)^ = Rv^R'

Feb 2, 2018

There is a common property about the skew-symmetric mapping $()^\land$ of a 3D vector:

For any rotation matrix $\bf R$ and 3D vector $\bf v$, there is always $({\bf R v})^{\land} = {\bf Rv^{\land}R}^{T}$

This property is of wide use in robotic dynamics and state estimation. This post provides a simple and intuitive proof of it.

We utilize the relationship between the skew-symmetric mapping and the cross product. It is known that for any ${\bf v,u}\in \mathbb{R}^3$, there is always ${\bf v}^{\land}{\bf u}={\bf v \times u}$. Therefore, we have

\begin{aligned} & ({\bf R v})^{\land} = {\bf Rv^{\land}R}^{T} \\ \Leftrightarrow\quad & ({\bf R v})^{\land}{\bf R} = {\bf Rv^{\land}} \\ \Leftrightarrow\quad &\forall {\bf u}\in \mathbb{R}^3, ({\bf Rv})^{\land}{\bf Ru}={\bf Rv}^{\land}{\bf u} \\ \Leftrightarrow\quad &\forall {\bf u}\in \mathbb{R}^3, ({\bf Rv})\times({\bf Ru})={\bf R(v\times u)} \end{aligned}

The last line in the equation above is true base on the invariance property of the cross product under rotation (check Wikipedia), i.e. for any ${\bf v,u}\in \mathbb{R}^3$, there is always

$({\bf Rv})\times({\bf Ru})={\bf R(v\times u)}$

It can be intuitively comprehended in a geometric perspective: imagine that ${\bf v,u}$ are two general 3D vectors, and $({\bf v\times u})$ would be a 3D vector perpendicular to both of ${\bf v,u}$, with a norm of ${\bf |v| |u|} \sin ({\bf u, v})$. If applying an identical rotation to every one of ${\bf v,u,v\times u}$, both of their norms and relative orientations would not change, hence the cross product of $({\bf Rv})$ and $({\bf Ru})$ is still ${\bf R(v\times u)}$.

Tags: robotics

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