I tried to create a script for Markov reliability modelling according to this page. I have a tough time to understand this particular line from the website:
The average probabilities for the interval beginning at time nT is given by
$$\overline{P(t)_n} = \frac{1}{T} \int _{t=0}^T (A e^{\textbf{M}T})^n P(0) \, {\rm d} t = \frac{e^{\textbf{M}T} - \textbf{I}}{\textbf{M}T} (A e^{\textbf{M}T})^n P(0)$$ where I is the identity matrix. Notice that the numerator on the right hand side is divisible by MT, so it isn’t necessary to invert the M matrix.
I tried to find the inverse, but the matrix is non-invertible. My question is, how should I approach this problem?
Let us consider only the case where the matrix of interest is diagonalizable. Suppose we have diagonalizable matrix $\rm A$, whose eigendecomposition is
$$\rm A = V \Lambda V^{-1} = V \begin{bmatrix} \Lambda_1 & \rm{O}\\ \rm{O} & \rm{O}\end{bmatrix} V^{-1}$$
where diagonal matrix $\Lambda_1$ contains the non-zero eigenvalues of $\rm A$ on its main diagonal. After some work, it can be shown that
$$e^{t \rm{A}} = \rm V \begin{bmatrix} e^{t \Lambda_1} & \rm{O}\\ \rm{O} & \rm{I}\end{bmatrix} V^{-1}$$
Hence, integrating over interval $[t_i, t_f]$, we obtain
$$\int_{t_i}^{t_f} e^{t \rm{A}} \,\rm{d}t = \rm V \begin{bmatrix} \displaystyle\int_{t_i}^{t_f} e^{t \Lambda_1} \,\rm{d}t& \rm{O}\\ \rm{O} & (t_f - t_i) \, \rm{I}\end{bmatrix} V^{-1}$$
and, since $\Lambda_1$ is invertible, the integral should be straightforward to compute.