In a physics related problem I am given a matrix $\mathcal{M} \in \mathbb{R}^{N \times N}$, defined by
$$\mathcal{M} = e^{\mathcal{R}} = \left(\begin{array}{c|c} \mathcal{A} & \mathcal{B} \\ \hline\mathcal{C} & \mathcal{D} \end{array} \right),$$
where
$$ \mathcal{A} \in \mathbb{R}^{k \times k} \quad \mathcal{B} \in \mathbb{R}^{k \times (N-k)} \quad \mathcal{C} \in \mathbb{R}^{(N-k) \times k} \quad \mathcal{D} \in \mathbb{R}^{(N-k) \times (N-k)} $$
whose elements are all different from zero but not possible (at least not easy) to specify in general. The only matrix I can specify in general is $\mathcal{R}$ whose elements add up to $0$ in each row and thus $\sum_j (\mathcal{M})_{ij} = 1$.
Instead of using the knowledge about the elements $(\mathcal{R})_{ij}$, I want to use that due to the application of the matrix exponential, I know that $\det(\mathcal{M}) \neq 0$ since $\mathcal{M}^{-1}$ always exists.
My first idea to proof that $\mathcal{D}^{-1}$ exists if $\mathcal{M}^{-1}$ exists was a circular proof, where I assumed that the inverse of $\mathcal{D}$ exists to show that it exists using the formula
$$ \det(\mathcal{M}) = \det(\mathcal{A}-\mathcal{B}\mathcal{D}^{-1}\mathcal{C}) \times \det(\mathcal{D}) \neq 0.$$
It can thus be omitted - My second idea was to use the general definition of the inverse of a block matrix to conclude that since the inverse of a matrix exponential always exists, $\mathcal{D}^{-1}$ has to exist as well (using a formula of the kind shown in the section "Block matrix inversion" on wikipedia https://en.wikipedia.org/wiki/Block_matrix and the knowledge that $\mathbf{P} = \mathcal{M}^{-1}$ exists, where $\mathbf{P}$ is in the notation of the wikipedia article).
Do you think it is possible to show invertibility of $\mathcal{D}$ by the given properties of $\mathcal{M}$?
I'm not sure if I understand the question correctly; however, the exponential of $$ \begin{pmatrix} 0 & -x\\ x & 0 \end{pmatrix} $$ has positive lower-right coordinate for $x=1$ and negative for $x=2$. So it is zero for some $x$.