Does R' = R*R'*R^(-1) hold?

Question

Given two 3D rotation matrices ($3\times 3$) $R$ and $R'$, does this equivalence hold?:

$R' = R*R'*R^{-1}$

My intuition tells me so, but I can't find a formal proof for it.

Thanks.

Answer 1

If $R' = RR'R^{-1}$ holds, then $R'R = RR'$. So you are basically asking if 3D rotation matrices commute. The answer would have been yes for 2D, but for 3D the answer is no.

Consider $R = R_x(\tfrac{\pi}{2}) = \begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}$ and $R' = R_z(\tfrac{\pi}{2}) = \begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}$.

We have $R_x(\tfrac{\pi}{2})R_z(\tfrac{\pi}{2}) = \begin{bmatrix}0&-1&0\\0&0&-1\\1&0&0\end{bmatrix}$ but $R_z(\tfrac{\pi}{2})R_x(\tfrac{\pi}{2}) = \begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$.

Answer 2

The equation you wrote is equivalent to $R' R = R R'$. Do all 3-D rotations commute?

Answer 3

What does $R'$ denote? If, as other answerers have supposed, it is just some other rotation matrix, then the answer is, as they have noted, no.

If you mean "$R$ transpose", then, as before, this is

$$ RR^T \stackrel{?}{=} R^TR $$

which is asking whether it is normal or not. In this case, rotations are normal, so the answer is yes.