## 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

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

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 ${\bf R(v\times u)}$.