Assume we are given for a transition between two time points $t\_0 = 0$ and $t\_1$ a matrix relationship, eventually describing the solution of a system of linear with non-constant coefficients,
$$Y(t\_1) = \exp(\Omega(t\_1,0))Y(0),$$
or, in a more general setting,
$$Y(t\_1) = \exp(\Omega(t\_1,0))Y(0)+c(t\_1,0),$$
where the $\Omega(t\_1,0)$ is the Magnus expansion just as in <https://en.wikipedia.org/wiki/Magnus_expansion> and <http://arxiv.org/pdf/0810.5488.pdf> .


If I assume that the system of linear ODEs is given by,
$$\frac{dY}{dt} = A(t)Y(t)$$
or, more generally,
$$\frac{dY}{dt} = A(t)Y(t)+b(t),$$
is there any way to get out some information about $A(t)$ or $\int\_0^{t\_1} A(t) dt$ or $b(t)$ by just knowing $\Omega(t\_1,0))$ (and $c(t\_1,0)$).



Getting out a system of linear ODEs by knowing the Magnus expansion
You can directly get the trace of $A$ from the identity


$${\rm det}\,\left[\exp\bigl(\Omega(t,0)\bigr)\right]=\exp\left[\int\_0^t {\rm tr}\,A(s)ds\right]$$


The full matrix $A$ is determined by $\Omega$ via the inverse Magnus expansion [see equation 4.2 from this [thesis](http://www.dms.uaf.edu/~bueler/tcarlsonMS.pdf)]:


$$A=\sum\_{k=0}^{\infty}\frac{(-1)^k}{(k+1)!}\left[\sum\_{q=0}^{k}(-1)^q{k\choose q}\Omega^{q}\frac{d\Omega}{dt}\Omega^{k-q}\right]$$


This is useful if $\Omega$ is proportional to some small parameter $\epsilon$, so that the expansion gives you $A$ as a power series in $\epsilon$.


