Derivative of rotation matrix produces angular velocity vector I am trying to understand how and why the angular velocity vector is related to the derivative of a rotation matrix.
I understand that for a rotation matrix $R(t)$ one has $R(t)R(t)^T = I $, so $\dot{R}R^T + R\dot{R}^T = 0$ which shows that the matrix $R(t)\dot{R}^T = \hat{\Omega}$  is skew symmetric. I understand that then we can write $\hat{\Omega}$ in vector form $\omega$ such that we get for all vectors $v$, $\hat{\Omega}v = \omega \times  v$.
But what else do we need to show that that $\omega$ is the angular velocity vector? Can we assert that $\omega$ will be unique?
 A: Take any basis vector $\hat{u}$ that is riding on a rotating coordinate frame and find as far as the components as measured by the inertial frame you have
$$ \frac{\rm d}{{\rm d}t} \hat{u} = \vec{\omega}  \times \hat{u}  \tag{1}$$
Now recognize that the rotation matrix $\mathbf{R}$ just has the three basis vectors of the body frame in its columns
$$ \mathbf{R} = \left[ \begin{array}{c|c|c} \hat{i} & \hat{j} & \hat{k} \end{array} \right] \tag{2}$$
and the relationship
$$ \frac{\rm d}{{\rm d}t} \mathbf{R} = \vec{\omega}  \times \mathbf{R} \tag{3}$$
is just a shortcut for
$$  \begin{aligned}
  \frac{\rm d}{{\rm d}t} \hat{i} & = \vec{\omega}  \times \hat{i} \\
  \frac{\rm d}{{\rm d}t} \hat{j} & = \vec{\omega}  \times \hat{j} \\
  \frac{\rm d}{{\rm d}t} \hat{k} & = \vec{\omega}  \times \hat{k} \\
\end{aligned} $$
So your question is really how do you prove (1)?
My favorite method is stating that under rotations the length of the basis vector must remain one $\| \hat{u} \| = 1$, or in derivative form
$$ \frac{\rm d}{{\rm d}t} \sqrt{ \hat{u} \cdot \hat{u} } = 0 $$ which quickly simplifies to
$$ \hat{u} \cdot \frac{\rm d}{{\rm d}t} \hat{u} = 0 $$
which is interpreted as the derivative must be perpendicular to the direction (pretty intuitive) and that one (or only way) to ensure this is to use a cross product to define the derivative as the cross product is guaranteed to be perpendicular to both arguments.
$$ \hat{u} \cdot ( \vec{\omega} \times \hat{u}) = 0 $$
The above leads to
$$ \frac{\rm d}{{\rm d}t} \vec{r} = \vec{v} = \vec{\omega} \times \vec{r} $$ for $\vec{r}$ riding on the body and $\vec{\omega}$ is the angular velocity vector.
A: The transformation matrix $~\mathbf R~$ transformed  the  components of a vector
from body system (Index B) to inertial system (Index I)
$$(\mathbf u)_I=\mathbf R\,(\mathbf u)_B$$
take the time derivative you obtain the velocity components $~\mathbf v~$ in inertial system
$$(\mathbf v)_I=\mathbf{\dot{R}}\,(\mathbf u)_B$$
with
$$\mathbf{\dot{R}}=\,\mathbf R\,\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\  \omega_{{z}}&0&-\omega_{{x}}\\  
-\omega_{{y}}&\omega_{{x}}&0\end {array} \right]$$
$\Rightarrow$
$$(\mathbf v)_I=\mathbf R\,\left(\mathbf\omega\times\mathbf u\right)_B
$$
or
$$\underbrace{\mathbf R^T\,(\mathbf v)_I}_{(\mathbf v)_B}=\left(\mathbf\omega\times\mathbf u\right)_B$$
thus you obtain
$$\mathbf v=\mathbf\omega\times\mathbf u$$

with :
\begin{align*}
&\mathbf{\dot{R}}=\mathbf{R}\,\mathbf{\tilde{\omega}}\\
&\Rightarrow\\  &\mathbf{\tilde{\omega}}=\mathbf{R}^T\,\mathbf{\dot{R}}\quad
  \text{and}\quad
  \mathbf{\tilde{\omega}}+\mathbf{\tilde{\omega}}^T=\mathbf{0}\\
  &\Rightarrow\\
  &\mathbf{R}^T\,\mathbf{\dot{R}}+\,\mathbf{\dot{R}}^T\,\mathbf{R}=\mathbf{0}\\
  &\text{or}\\  
  &\mathbf{R}^T\,\mathbf{R}=\mathbf{I}
\end{align*}
