Preamble
In angular dynamics, it is customary to describe the rotation of a rigid object with the rotation of a reference frame attached to it, say having basis $\vec{a},\vec{b},\vec{c}$ as unit vector of the coordinate axes. The angular velocity is the rate at which the unit vectors rotate in space. I restrict myself to orthonormal, 'right-handed' coordinate systems.
I came across an approach that I found elegant in which the angular velocity of this frame $\vec{\omega}$ is defined as $$ \omega_a = \frac{d\vec{b}}{dt} \cdot \vec{c},\quad \omega_b = \frac{d\vec{c}}{dt} \cdot \vec{a}, \quad \omega_c = \frac{d\vec{a}}{dt} \cdot \vec{b}$$ Given a vector with constant lenght but variable direction, the following is always a true construction $$ \frac{d\vec{a}}{dt} = \underbrace{\left(\frac{d\vec{a}}{dt} \cdot \vec{a} \right)}_{\textrm{perpendicular, hence 0}} \vec{a} + \left(\frac{d\vec{a}}{dt} \cdot \vec{b} \right) \vec{b} + \left(\frac{d\vec{a}}{dt} \cdot \vec{c} \right) \vec{c} $$ Since the condition of perpendicularity $\vec{a} \cdot \vec{b} = 0$ also leads to $$ \frac{d\,}{dt} \vec{a} \cdot \vec{b} = 0 = \frac{d\vec{a}}{dt} \cdot \vec{b} + \frac{d\vec{b}}{dt} \cdot \vec{a} \quad \Rightarrow \quad \omega_c = - \frac{d\vec{b}}{dt} \cdot \vec{a} $$ and since the like holds for $\frac{d\vec{b}}{dt}$, $\frac{d\vec{c}}{dt}$, $\vec{b} \cdot \vec{c}$, $\vec{c} \cdot \vec{a}$, after a little playing around one gets \begin{equation} \frac{d\vec{a}}{dt} = \vec{\omega} \times \vec{a} \\ \frac{d\vec{b}}{dt} = \vec{\omega} \times \vec{b} \\ \frac{d\vec{c}}{dt} = \vec{\omega} \times \vec{c} \end{equation}
This demonstration partly uses the fact that in the basis $\vec{a},\vec{b},\vec{c}$, where the reasoning is being developed, it obviously holds $\vec{a}=(1,0,0)$, $\vec{b}=(0,1,0)$, $\vec{c}=(0,0,1)$.
Intuition
It seems clear to me that the expressions of the time derivatives as cross products must hold when the operands are expressed in any arbitrary basis. Note that I am still referring to the rotating motion of that same reference frame $\vec{a},\vec{b},\vec{c}$. However, the vectors $\vec{\omega},\vec{a},\ldots$ may well be expressed in any other reference frame than $\vec{a},\vec{b},\vec{c}$. The two concepts --- a reference frame that moves and eventually describes motion of objects in that frame, and a reference frame that describes the vectors --- are conceptually quite different, I believe, although they might coincide.
If this intuition is right, there is the matrix $R$ that rotates the vectors in the old basis $\vec{a},\vec{b},\vec{c}$ into the new basis, say, $\vec{i},\vec{j},\vec{k}$. Therefore, the following relationships should also hold: \begin{equation} \frac{dR\vec{a}}{dt} = R\vec{\omega} \times R\vec{a} \\ \frac{dR\vec{b}}{dt} = R\vec{\omega} \times R\vec{b} \\ \frac{dR\vec{c}}{dt} = R\vec{\omega} \times R\vec{c} \end{equation} but I kind of get stuck in a brute-force approach that turns out to be unwieldy all too soon.
Question
Is there an elegant way in linear algebra that shows that expressing the rate of change of the unit vector by means of a cross product with the angular velocity of the frame it belongs to, holds for any reference frame you describe the chosen vectors with?
Correction to faults and flaws welcome.