Let us consider the most general motion of a rigid body. Two arbitrary points of the body, $i$ and $j$ must not change their distance $d_{ij}$ during motion. Therefore,$$(\vec{r}_j - \vec{r}_i)^2 = d_{ij}^2 = \text{const.}$$ Differentiating, we have $$(\vec{v}_j - \vec{v}_i) \cdot (\vec{r}_j - \vec{r}_i)=0.$$From here, we can conclude that the relative velocity can be written in the form $$(\vec{v}_j - \vec{v}_i) = \vec{\omega}_{ij} \times (\vec{r}_j - \vec{r}_i),$$ for some vector $\vec{\omega}_{ij}$, which, in general, depends on the pair of points $(i,j)$ in consideration.
Is there an easy way to show that $\vec{\omega}_{ij} = \vec{\omega}$ is, in fact, the same for all pairs of particles? It seems to me that it should be possible to prove this just by using linear algebra, without any physical considerations.
UPDATE #1: My attempt - consider three particles $i,j,k$ and write $$(\vec{v}_j - \vec{v}_i) = \vec{\omega}_{ij} \times (\vec{r}_j - \vec{r}_i)\\(\vec{v}_k - \vec{v}_j) = \vec{\omega}_{jk} \times (\vec{r}_k - \vec{r}_j)\\(\vec{v}_i - \vec{v}_k) = \vec{\omega}_{ki} \times (\vec{r}_i - \vec{r}_k)$$Adding these three equations and rearranging we have $$(\vec{\omega}_{ij}-\vec{\omega}_{ki}) \times \vec{r}_i + (\vec{\omega}_{jk}-\vec{\omega}_{ij}) \times \vec{r}_j + (\vec{\omega}_{ki}-\vec{\omega}_{jk}) \times \vec{r}_k = \vec{0}.$$ Now, in general, my position vectors $\vec{r}$ are linearly independent. Does this imply that the brackets must vanish?