In a system modelling context, I am facing some problems with computing the matrix exponential of the following $N \times N$ ($N$ can be arbitrarily large) block matrix:
$$ \mathcal{M} = \left( \begin{array}{c|c} \mathcal{A} & \mathcal{B} \\ \hline \mathcal{C} & \mathcal{D} \end{array} \right) \in \mathbb{R}^{N \times N}, $$ where $\mathcal{A} \in \mathbb{R}^{k \times k}$, $\mathcal{B} \in \mathbb{R}^{k \times (N-k)}$, $\mathcal{C} \in \mathbb{R}^{(N-k) \times k}$ and $\mathcal{D} \in \mathbb{R}^{(N-k) \times (N-k)}$ and $N > k$.
I found the following general properties of the blocks and their elements:
- $\mathcal{A}$ is a diagonal matrix with all elements $<0$
- $\mathcal{D}$ is a diagonal matrix with all elements $>0$
- The elements of $\mathcal{B}$ and $\mathcal{C}$ are related by $(\mathcal{C})_{j,i} = -a_{i,j+k} (\mathcal{B})_{i,j} $, where $a_{i,j+k} \in \mathbb{R}^+$ denote some parameters and $(\mathcal{B})_{i,j} \in \mathbb{R}^+$
- $\sum_j (\mathcal{A})_{i,j} + \sum_j (\mathcal{B})_{i,j} = 0$ and $\sum_j (\mathcal{C})_{i,j} + \sum_j (\mathcal{D})_{i,j} = 0$, i.e. rows add up to zero
I found that I can define new matrices such that $\mathcal{M} = \mathcal{S} + \mathcal{T}$, with
$$ \begin{align} \mathcal{S} &= \frac{\mathcal{M}-\mathcal{M}^T}{2}, \quad\text{skew-symmetric ($\text{rank}(\mathcal{S})\neq 1$ in general) and} \\ \mathcal{T} &= \frac{\mathcal{M}+\mathcal{M}^T}{2}, \quad\text{symmetric.} \end{align} $$ Unfortunately $[\mathcal{S},\mathcal{T}] \neq \mathcal{0}$, $[\mathcal{S},[\mathcal{S},\mathcal{T}]] \neq \mathcal{0}$ and $[\mathcal{T},[\mathcal{T},\mathcal{S}]] \neq \mathcal{0}$, meaning that this discovery is no help in computing $e^{\mathcal{M}} = e^{\mathcal{S} + \mathcal{T}}$.
Given the properties (1. - 4.), I want to prove the existence of $\color{blue}{(\mathcal{D}^\prime)^{-1}} \in \mathbb{R}^{(N-k)\times(N-K)}$, belonging to the matrix exponential
$$e^{\mathcal{M}}= \left( \begin{array}{c|c} \mathcal{A}^\prime_{k \times k} & \mathcal{B}^\prime_{k \times (N-k)} \\ \hline \mathcal{C}^\prime_{(N-k) \times k} & \mathcal{D}^\prime_{(N-k) \times (N-k)} \end{array} \right) \in \mathbb{R}^{N \times N},$$
My approach so far was to compute an explicit formula for $e^\mathcal{M}$ and to deduce the invertibility of $\mathcal{D}^\prime$ from that. Long story short, I didn't succeed in finding such a general expression for the $N \times N$ case.
Do you have any ideas about how to prove the existence of the inverse of $\mathcal{D}^\prime$? Is it possible to show a relation of the form $(\mathcal{C}^\prime)_{j,i} = - A_{i,j+k} (\mathcal{B}^\prime)_{i,j}$, with $ A_{i,j+k} \in \mathbb{R}^+$?
(Example) For $k = 2$ and $N = 5$, $\mathcal{M}$ looks like this:
$$ \mathcal{M} = \left( \begin{array}{ccccc} -b_{1,1}-b_{1,2}-b_{1,3} & 0 & b_{1,1} & b_{1,2} & b_{1,3} \\ 0 & -b_{2,1}-b_{2,2}-b_{2,3} & b_{2,1} & b_{2,2} & b_{2,3} \\ -a_{1,3} b_{1,1} & -a_{2,3} b_{2,1} & a_{1,3} b_{1,1}+a_{2,3} b_{2,1} & 0 & 0 \\ -a_{1,4} b_{1,2} & -a_{2,4} b_{2,2} & 0 & a_{1,4} b_{1,2}+a_{2,4} b_{2,2} & 0 \\ -a_{1,5} b_{1,3} & -a_{2,5} b_{2,3} & 0 & 0 & a_{1,5} b_{1,3}+a_{2,5} b_{2,3} \\ \end{array} \right) $$