I have the following problem: A matrix $C\in \mathbb{R}^{2N}$, where $C=\epsilon A+D$

$\epsilon A=(C-C')/2$ is skew symmetric with "block" anti-diagonal structure of size 4.

$ D=(C+C')/2$ (Diagonal matrix) with "block" diagonal structure of size 2.

$D=\begin{bmatrix} 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ 0 & 0 & \alpha & \dots \\ 0 & 0 & 0& \alpha & \dots \\ 0 & 0 & 0& 0 & 2\alpha \dots \\ 0 & 0 & 0& 0 & 0& 2\alpha \dots\\ \vdots\\ 0 & 0 & 0& 0 & 0& \dots &(N-1)\alpha &0\\ 0 & 0 & 0& 0 & 0& 0& \dots &(N-1)\alpha \end{bmatrix}$

And

$A=\begin{bmatrix} 0 & 0 & 0 & -\beta \dots \\ 0 & 0 & \beta & \dots \\ 0 & -\beta & 0 & 0&0&-\sqrt{2}\beta\dots \\ \beta & 0 & 0& 0 & \sqrt{2}\beta&\dots \\ 0 & 0 & 0& -\sqrt{2}\beta & 0 \dots \\ 0 & 0 & \sqrt{2}\beta& 0 & 0& 0 \dots\\ \vdots\\ 0 & 0 & 0& 0 &\dots&-\sqrt{N-1}\beta&0 &0\\ 0 & 0 & 0&\dots \sqrt{N-1}\beta& 0& 0& &0 \end{bmatrix}$

Here $\alpha,\beta$ are constants of order 1.

I want to expand the inverse of $C$ in terms of $\epsilon$<<1, by writing

$C^{-1}=(D+\epsilon A)^{-1}$.

Note that both $D$ and $A$ are of rank $2N-2$.

Due to rank-deficiency of $D$, I cannot simply invert $D$ and do the obvious Taylor series.

Does any one have any ideas on how to proceed in this ?

The difficulty seems to be due to differently sized "block"-wise structures of $A$ and $D$.

For example, if instead the matrix $A$ was "block"-anti-diagonal of size 2 (same as block size of $D$), it seems like the Taylor series gives the right result where I replace $D^{-1}$ with pseudo-inverse and only invert in the non-singular subspace.