Cayley-Klein Parameters I have a very simple question(I guess )to ask
$$\frac{d\mathbf{m}}{dt}= \mathbf{C} \times \mathbf{m}$$
where $\mathbf{m}$ and $\mathbf{C}$ are vectors. Assume that $\mathbf{C}$ is constant over a certain period of time $[0,T]$.
Then would someone please explain me how  can we find a rotation matrix using Cayley-Klein parameter so that, for $t\in [0,T]$, we can express $\mathbf{m}(t)=R(t) \mathbf{m}_0$? Here $R(t)$ is a rotation matrix and $\mathbf{m}_0$ is the initial vector.
I know that, in $[0,T]$, it can be solved analytically as $\mathbf{m}(t)=exp(At) \mathbf{m}_0$. Moreover would anyone please explain the relation between this two solution?
 A: I do not know anything about Cayley-Klein parameters, sorry, but the rest of you question is easy. 
We start noticing that it is always possible to suppose that ${\bf C}$ is directed along $z$, just fixing our reference frame suitably.  So we can write ${\bf C}= c{\bf e}_z$ whit $c>0$ (otherwise everything is trivial and the solution is ${\bf m}(t)= {\bf m}_0$ for $t\in [0,T]$, so that $R(t)=I$ and $A=0$).
With this choice:
$${\bf e}_z \times {\bf m}= S {\bf e}_z$$
where 
$$ S= 
\begin{bmatrix}
0& 1 & 0\\
-1& 0 & 0\\
0 & 0 &0
\end{bmatrix}
$$
So your equation is: 
$$\frac{d\bf m}{dt} = c S {\bf m}(t)\:$$
Using this equation and taking the derivative of both sides we obtain:
$$\frac{d^{2}\bf m}{dt^{2}} = c S \frac{d{\bf m}}{dt} = c^2 S^2 {\bf m}(t)\:$$
and so on with all orders of derivatives, eventually getting:
$$\frac{d^{n}\bf m}{dt^{n}} = c^nS^n {\bf m}(t)\:.$$
For $t=0$ we have in particular:
$$\frac{d^{n}\bf m}{dt^{n}}|_{t=0} = c^nS^n {\bf m}_0\:.$$
We have found all coefficients of Taylor expansion of the solution, that therefore reads:
$${\bf m}(t)= \sum_{n=0}^{+\infty} \frac{t^n}{n!}\frac{d^{n}\bf m}{dt^{n}}|_{t=0} = \sum_{n=0}^{+\infty} \frac{(t c S)^n}{n!} {\bf m}_0
= e^{tcS} {\bf m}_0\:.$$
The fact that this series converges absolutely can be proved exactly as for the series of the exponential for complex numbers, just  replacing the magnitude of complex numbers with the norm of matrices. The fact that the series with all its derivatives converges uniformly allows one to swap the symbol of derivative and the symbol of series, proving that the series solves the initial differential equation. Uniqueness of the solution is a known general result especially valid for linear equations as the considered one.  
The matrix $A$ in your question is therefore given by $cS$. 
Finally, let us establish that $e^{tcS} = R(t) \in SO(3)$, namely it is a rotation in $\mathbb R^3$. $3D$ Rotations are real $3\times 3$ matrices, $B$, verifying $BB^t=I$. So, it would be enough proving that 
$$e^{tcS} (e^{tcS})^t = I\:.$$
By direct inspection you see that $S^t=-S$ and:
$$(e^{tcS})^t= \left( \sum_{n=0}^{+\infty} \frac{(t c S)^n}{n!}\right)^t=
  \sum_{n=0}^{+\infty} \frac{(t c S^t)^{n}}{n!}= \sum_{n=0}^{+\infty} \frac{(-t c S)^{n}}{n!} = e^{-tcS}\:. $$
Thus, as wanted: 
$$e^{tcS} (e^{tcS})^t = e^{tcS} e^{-tcS} = e^{tcS-tcS} = e^{0S} = I\:,$$
where I used the fact that, if $A$ and $B$ commute, then, exploiting exactly the same proof as for numbers $e^Ae^B=e^{A+B}$.
