# Inverting (via Taylor expansion) a sum of (rank-deficient) skew-symmetric matrix and (rank-deficient) Diagonal matrix

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.

## 1 Answer

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 $(\epsilon\beta M_2\otimes B+\alpha M_1\otimes I_2 )^{-1}.$