Angular velocity as eigenvector of rotation map In physics the angular velocity $\omega$ usually is defined as follows:

For a map $B: \mathbb{R} \to SO(3)$ define $\Omega(t):=\dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $\omega(t) \in \mathbb{R}^3$ satisfying $\Omega(t) x=\omega(t) \times x$ for $x\in\mathbb{R}^3$. 

Does $\omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $\omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $\omega_j(t) = \frac{1}{2}\sum_{k,l=1}^3\epsilon_{jkl}\Omega_{kl}(t)$ (were $\epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)\omega(t))_j=-\frac{1}{2}\sum_{k,l,m=1}^3\epsilon_{ikl}B_{ji}(t)\epsilon_{ikl}\Omega_{kl}(t)=-\frac{1}{2}\sum_{k,l,m,n=1}^3\epsilon_{ikl}B_{ji}(t)\dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?

No this is not true in general. Take for example
$$ B = \begin{pmatrix} 0 & -\sin\phi & -\cos\phi\\ 0 & \cos\phi & -\sin\phi\\ 1 & 0 &0 \end{pmatrix}$$
Then
$$\dot B = \dot\phi\begin{pmatrix} 0 & -\cos\phi & \sin\phi \\ 0 & -\sin\phi & -\cos\phi \\ 1 & 0 & 0\end{pmatrix} 
\quad\text{and}\quad \Omega = \dot\phi\begin{pmatrix} 0 & -1& 0 \\
 1 & 0 & 0 \\
 0 & 0 & 0  \end{pmatrix}, \omega = \begin{pmatrix}0\\0\\-\dot\phi\end{pmatrix} $$
However, if $\dot\phi(0) \neq 0$ then $\omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $\exp(\Omega)$.
And indeed, since $\Omega \omega = \omega \times \omega = 0$ we have $\exp(\Omega)\omega = \omega$.
