a proof of (rv)^ = rv^r’

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

$$% <![CDATA[ begin{align} & ({bf R v})^{land} = {bf Rv^{land}R}^{T} \ Leftrightarrowquad & ({bf R v})^{land}{bf R} = {bf Rv^{land}} \ Leftrightarrowquad &forall {bf u}in mathbb{R}^3, ({bf Rv})^{land}{bf Ru}={bf Rv}^{land}{bf u} \ Leftrightarrowquad &forall {bf u}in mathbb{R}^3, ({bf Rv})times({bf Ru})={bf R(vtimes u)} end{align} %]]>$$

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

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

It can be intuitively comprehended in a geometric perspective:
imagine that ${bf v,u}$ are two general 3D vectors, and $({bf vtimes u})$ 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 ${bf v,u,vtimes u}$, both of their norms and relative orientations would not change,
hence the cross product of $({bf Rv})$ and $({bf Ru})$ is still ${bf R(vtimes u)}$.