3
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.