Let's consider an equation:
\begin{equation}
\frac{d\vec{r}}{dt}=\vec{\omega}\times\vec{r}.
\end{equation}
This equation might be expressed via "angular velocity" matrix:
\begin{equation}\label{eq:3}\tag{1}
\frac{d\vec{r}}{dt}= \hat{\Omega} \vec{r},
\end{equation}
where the "angular velocity" matrix is
\begin{equation}
\Omega =
\begin{pmatrix}
0 & \omega_z & -\omega_y \\
-\omega_z & 0 & \omega_x \\
\omega_y & -\omega_x & 0 \\
\end{pmatrix}.
\end{equation}
Also let's introduce matrixes $\hat{\vec{S}}_{x,y,z}$ ($SO(3)$ generators):
\begin{multline}
\Omega =
\omega_x
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0 \\
\end{pmatrix}
+
\omega_y
\begin{pmatrix}
0 & 0 & -1 \\
0 & 0 & 0 \\
1 & 0 & 0 \\
\end{pmatrix}
+
\omega_z
\begin{pmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
= \\ =
i \omega_x
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0 \\
\end{pmatrix}
+
i \omega_y
\begin{pmatrix}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0 \\
\end{pmatrix}
+
i\omega_z
\begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}.
\end{multline}
\begin{multline}
\hat{\vec{S}} =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0 \\
\end{pmatrix}
\vec{e}_x
+
\begin{pmatrix}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0 \\
\end{pmatrix}
\vec{e}_y
+
\begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\vec{e}_z =
\hat{S}_x \vec{e}_x +
\hat{S}_y \vec{e}_y +
\hat{S}_z \vec{e}_z
.
\end{multline}
So, we can write a formal solution of~\eqref{eq:3} as:
\begin{equation}\label{eq:solmatrix}\tag{2}
\vec{r} = e^{i (\vec{\omega} t\cdot\vec{\hat{S}})} \vec{r}_0.
\end{equation}
Now, if I use some convinient axes, I can write \eqref{eq:solmatrix} as:
\begin{equation}\label{eq:solmatrix1}
\vec{r} = e^{i (\omega t\cdot\hat{S}_{z'})} \vec{r}_0,
\end{equation}
and in explicit form (using Taylor series expansion of $e^M$) it looks like:
\begin{equation}\label{eq:so2}\tag{3}
\vec{r}(t)=\begin{pmatrix} \cos(\omega t) & -\sin(\omega t) & 0\\ \sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1 \end{pmatrix} \vec{r}_0 = \hat{R}_{z'}(\omega t) \vec{v}_0 ,
\end{equation}
where $\hat{R}_{z'}(\omega t)$ is a rotation matrix about $z'$.
Now a question. How can I write the solution \eqref{eq:solmatrix} in explicit form (via rotation matrix like \eqref{eq:so2})? I tried, but there are powers of ($\sum_m i \omega_m \hat{S}_m)^n$ in Taylor series, which is not clear how to simplify. I can't see what they should collapse into.
Best Answer
I'll focus my answer on computing the matrix exponential. Consider components of the unit vector $\hat{\omega}^a = \omega^a /|\omega| $. Splitting the exponential into odd/ even terms
$$ E=\sum_{n=0}^\infty \frac{\left( i |\omega|\hat{\omega}^a S^a \right)^n}{n!} = 1+\sum_{n=1}^\infty \frac{(-1)^n |\omega|^{2n}(\hat{\omega}^aS^a)^{2n}}{(2n)!}+i \sum_{n=0}^\infty \frac{(-1)^n |\omega|^{2n+1}(\hat{\omega}^aS^a)^{2n+1}}{(2n+1)!} $$
The components $jk$ of the $\operatorname{so}(3)$ rotation generator $a$ are $S^{ajk}=i\epsilon^{ajk} $. Consider
$$\left[ (\hat{\omega}^aS^a)^2\right]^{jl}=\hat{\omega}^a \hat{\omega}^b S^{ajk}S^{bkl}=\delta^{jl}-\hat{\omega}^j\hat{\omega}^l$$
The last equality follows$^\dagger $ by using the 'contracted epsilon identity' as well as the fact that $\hat{\omega}^a \hat{\omega}^a =1 $. We also note that RHS is a projection operator$^\ddagger$, hence
$$\left[ (\hat{\omega}^aS^a)^{2n}\right]^{jl}=\delta^{jl}-\hat{\omega}^j\hat{\omega}^l$$
$$\left[ (\hat{\omega}^aS^a)^{2n+1}\right]^{jl}=\left[ \hat{\omega}^aS^a \right]^{jl}$$ Substitute back into the sums
$$ E^{jl}=\delta^{jl}+\sum_{n=1}^\infty \frac{(-1)^n |\omega|^{2n}}{(2n)!}(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l) +i \sum_{n=0}^\infty \frac{(-1)^n |\omega|^{2n+1}}{(2n+1)!}\left[ \hat{\omega}^aS^a \right]^{jl} $$
$$E^{jl}=\delta^{jl}+(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l)\left(\cos(|\omega|)-1\right) +i\left[ \hat{\omega}^aS^a \right]^{jl}\sin(|\omega|)$$
$$E^{jl}=\hat{\omega}^j\hat{\omega}^l+(\delta^{jl}-\hat{\omega}^j\hat{\omega}^l)\cos(|\omega|) +i\left[ \hat{\omega}^aS^a \right]^{jl}\sin(|\omega|)$$
Those are the components of the matrix. In matrix form it's probably not going to be very pretty, so I'll leave it there.
$\dagger$ Like this
$$ \hat{\omega}^a \hat{\omega}^b S^{ajk}S^{bkl}=i^2\hat{\omega}^a \hat{\omega}^b \epsilon^{ajk}\epsilon^{bkl}=\hat{\omega}^a \hat{\omega}^b \epsilon^{ajk}\epsilon^{blk}=\hat{\omega}^a \hat{\omega}^b \left( \delta^{ab}\delta^{jl}-\delta^{al}\delta^{bj} \right) \\= \hat{\omega}^a \hat{\omega}^a \delta^{jl}-\hat{\omega}^j \hat{\omega}^l = \delta^{jl}-\hat{\omega}^j \hat{\omega}^l $$
$\ddagger$ By direct computation $$ (\delta^{jl}-\hat{\omega}^j \hat{\omega}^l)(\delta^{lm}-\hat{\omega}^l \hat{\omega}^m)= \delta^{jl}\delta^{lm}+\hat{\omega}^j \hat{\omega}^l \hat{\omega}^l \hat{\omega}^m-\delta^{jl}\hat{\omega}^l \hat{\omega}^m-\delta^{lm}\hat{\omega}^j \hat{\omega}^l \\ =\delta^{jm}-\hat{\omega}^j \hat{\omega}^m$$