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. 



Inverting (via Taylor expansion) a sum of (rank-deficient) skew-symmetric matrix and (rank-deficient) Diagonal matrix
Just an idea.


I think the best way to start is to expand on the structure of the matrices $D, C$. For example, $D$ can be readily seen to be expressible in the form $$\alpha M\_1\otimes I\_2=\alpha\begin{bmatrix}
0 & \mathbf{0}^T\\
\mathbf{0} & J
\end{bmatrix}\otimes I\_2$$ where $J=diag(1,2,,\cdots)$. I think the structure of matrix $A$ can be expressed as $\beta M\_2\otimes B$ where $$M\_2=\begin{bmatrix}0 & 1 & 0 & 0 & \cdots\\
1 & 0 & \sqrt{2} & 0 & \cdots\\
0 & \sqrt{2} & 0 & \sqrt{3} & \cdots\\
0& 0 & \sqrt{3} & 0 & \cdots\\
\vdots & \vdots & \vdots & \vdots & \ddots\end{bmatrix},\\B=\begin{bmatrix}
0 & 1\\
-1 & 0\end{bmatrix}$$ Now the problem becomes finding $(\epsilon\beta M\_2\otimes B+\alpha M\_1\otimes I\_2 )^{-1}.$


