I have a Markov chain, whose transition intensity matrix (or 'generator matrix') is:
I know that, to get from this to the transition probability matrix $\textbf{P(t)}$ (where $t$ is time), I need to use the Kolmogorov backwards equation, and that:
$\textbf{P(t)}=e^{\textbf{Q}t}$
Entries in the $\textbf{P(t)}$ matrix are denoted $p_{ij}$. The first entry in the probability transition matrix should then be
$p_{11}=e^{-(\lambda_{12}+\lambda_{17})t}$
Diagonal entries such as that are easy to work out (assuming that I have that correctly). But I am stuck on the other elements. I suspect this is because there are so many states involved that it is just intractable to find the probability matrix. Is this correct, or is there a simple solution?
This is not a complete answer, but it's too long for a comment. One way to do this, and I don't know that it's the best way, is to use the fact that $e^{At+Bt}=e^{At}e^{Bt}.$ Now write $Q=D+N$ where $Q$ is a diagonal matrix and $N$ is nilpotent. That is, $D$ is the matrix with the given diagonal elements, and $N$ has all the off-diagonal elements. Since $N$ is strictly upper triangular, $N$ is nilpotent.
The exponential of a diagonal matrix is trivial to compute, and for the nilpotent matrix you just use the power series definition of the exponential, which turns out to be a polynomial, since $N$ is nilpotent.
Still, this will be a mess to do by hand. If you want a symbolic solution, I'd certainly suggest using a computer algebra system. For that matter, if you want a numerical solution, I'd certainly suggest using numerical software.
I've been playing around with this a bit, and I see that it's a lot easier to write down an explicit formula than I thought it would be, even symbolically. I don't know exactly what you're trying to do, but since I've worked it out (at least part way,) I'll outline the idea.
Here $D$ and $N$ are as above. We will have $N^7 = 0,$ so $$e^{Nt}= \sum_{n=0}^6{\frac{N^n}{n!}t^n}$$ $N$ has nonzero elements only on the first superdiagonal and in the last column. It turns out that $N^2$ has nonzero entries only on the second superdiagonal and in the last column. Furthermore, they're easy to specify. For the superdiagonal, $$n_{k,k+2}=\begin{align}\begin{cases}\lambda_{k,k+1}\lambda_{k+1,k+2}, &1 \le k \le 4\\ 0, &k \ge 5 \end{cases}\end{align}$$ For the last column, we have $$n_{k,7}=\begin{align}\begin{cases}\lambda_{k,k+1}\lambda_{k+1,7}, &1 \le k \le 4\\ 0, &k \ge 5 \end{cases}\end{align}$$ Here of course $N = (n_{ij})_{1\le i,j \le 7}$. For $N^3$ there will only be nonzero entries on the third superdiagonal and in the last column, and it's easy to write down specific formulas as above. Each nonzero entry will be a product of 3 of the lambdas. The same pattern extends to the higher powers of $N$. When we add up all the powers, the only terms that aren't monomials occur in the last column.
Finally, $e^{Dt} = diag(e^{d_{11}t}, e^{d_{22}t}, ...,1, 1)$ and multplying by a diagonal matrix is trivial, of course.