Given $R\in SO(3)$ and $w\in \mathbb R^3$, prove the following identity

Question

Given $R\in SO(3)$ and $w\in \mathbb R^3$, and $a \times b= \hat a b$ prove the following identity: $$R {\hat w} R^T=\widehat{(Rw)}$$

It's from the textbook A Mathematical Introduction to Robotic Manipulation and the property is listed as a lemma. It is of course possible to prove this by direct calculation, but I'm wondering if someone can help me prove this without explicitly doing the matrix multiplications.

Answer 1

Let $v\in \Bbb R^3$. By the definition, $$R\hat{w}R^Tv=R\left(w\times R^Tv\right).$$ We claim that this equals $$\widehat{Rw}v=(Rw)\times v.$$ Because $R$ is an invertible operator, $v=Ru$ for some $u\in\Bbb R^3$. So the first equation becomes $$R\hat{w}R^Tv=R(w\times R^TRu)=R(w\times u),$$ noting that $R^T=R^{-1}$, while the second equation is $$\widehat{Rw}v=(Rw)\times(Ru).$$

In other words, we have to show that $R(w\times u)=(Rw)\times (Ru)$ for every $w,u\in\Bbb R^3$. Let $x,y\in \Bbb R^3$. Because $R\in SO(3)\subseteq O(3)$, $R^TR=I$ so that $$(Rx)\cdot (Ry)=x^T R^TRy=x^Ty=x\cdot y.$$ In particular when $x=(w\times u)$, we have $$R(w\times u)\cdot Ry=Rx\cdot Ry=x\cdot y=(w\times u)\cdot y.$$ On the other hand, the oriented volumes of the parallelepiped given by vectors $w,u,y$ and the one given by vectors $Rw,Ru,Ry$ are related by $$(Rw\times Ru)\cdot Ry=(\det R) (w\times u)\cdot y$$ due to the change-of-coordinates formula (think about volume integration in three dimensions, $\det R$ is the Jacobian determinant). But $\det R=1$, so $R(w\times u)\cdot Ry= (Rw\times Ru)\cdot Ry$ for all $y$. Since the dot product is a non-degenerate bilinear form on $\Bbb R^3$ (and $R$ is an automorphism of $\Bbb R^3$), $$Rw\times Ru=R(w\times u).$$

Interestingly if $R\in O(3)$ but $R\notin SO(3)$ (that is $\det R=-1$), then we have $R\hat{w}R^T=-\widehat{Rw}$.

Answer 2

You can think of this as an instance of the change of basis formula for matrices. More concretely, you need to start by observing that $R(a\times b) = Ra\times Rb$ (the easiest way I know to check this is to use the fact that cross product of vectors is more naturally interpreted as wedge product, and it is standard that for any linear map $T$, we have $T(a\wedge b) = Ta\wedge Tb$). Then $$(Ra)\times b = R(a\times R^{-1}b) = R\hat a (R^{-1}b) = (R \hat a R^{-1})b.$$ Since $R$ is orthogonal, $R^{-1} = R^\top$.