Properties of matrix exponential I know that the solution to system $x' = Ax$ is $e^{At}$, and I'm aware of various methods to calculate the exponential numerically. However I wonder if there are some analytical results.
Namely, I'm interested in matrices of type $A_{i,i} = -A_{i+1,i}$; $A_{n,n}=0$. These matrices are the infinitesimal generators of a Markov chain, where you transit from state 1 to state n through various steps. Would it be possible to calculate $[e^{At}]_{n,1}$, i.e. transition probability from state $1$ to state $n$ analytically as a function of time $t$?
 A: This seems to transform the original problem into a more complicated one... A direct approach is as follows.
For every $1\leqslant i\leqslant n-1$, let $a_i=-A_{i,i}$. The hitting time $T$ of $n$ starting from $1$ is the sum of $n-1$ independent random variables, each exponential with parameter $a_i$. In particular,
$$
E[\mathrm e^{-sT}]=\prod_i\frac{a_i}{a_i+s}.
$$
Assume first that the parameters $a_i$ are distinct. Then the RHS is
$$
\sum_ib_i\frac{a_i}{a_i+s},\qquad b_i=\prod_{j\ne i}\frac{a_j}{a_j-a_i},
$$
hence the distribution of $T$ has density $f$, where, for every $t\geqslant0$,
$$
f(t)=\sum_ib_ia_i\mathrm e^{-a_is}.
$$
In particular,
$$
[\mathrm e^{At}]_{n1}=P[T\leqslant t]=\int_0^tf=\sum_ib_i(1-\mathrm e^{-a_it})=1-\sum_ib_i\mathrm e^{-a_it}.
$$
If some parameters $a_i$ coincide, consider the limit of the expression above when $a_i-a_j\to0$ for some $i\ne j$. This limit is finite and coincides with $[\mathrm e^{At}]_{n1}$. Note that $1-[\mathrm e^{At}]_{n1}$ is always a linear combination of the functions $t\mapsto t^{k-1}\mathrm e^{-at}$, where $a=a_i$ for at least some $1\leqslant i\leqslant n-1$, and $1\leqslant k\leqslant\#\{1\leqslant i\leqslant n-1\mid a_i=a\}$.
A: So all you are saying is that it's in fact a mixture of exponential distributions!? Quite amazning, if true. Can you give some references?
Igor: they're zero. The last replier has the correct interpretation of the problem, but I'm not certain yet, if the answer is right.
