A matrix valued function is of the form $\psi:\mathbb{R}_+\to\mathbb{R}^{n\times n}$ and it is known that $\psi(\lambda)$ is always a positive definite matrix. The asymptotic exapnsion of $\psi(\lambda)$ is given as $$\psi(\lambda) = A + \frac{B_1}{\lambda}+ \frac{B_2}{\lambda^2}+ \frac{B_3}{\lambda^3}+...$$ where $A$ is an $n\times n$ all ones matrix and $B_1,B_2,B_3...$ are nonsingular $n\times n$ matrices.

I want to get an asymptotic expansion for $\psi(\lambda)^{-1}$. I don't want to compute the exact coefficient matrices but want to show that it is of the form $$\psi(\lambda)^{-1} = \lambda D_{-1} + D_0 + \frac{D_1}{\lambda} + \frac{D_2}{\lambda^2} + \frac{D_3}{\lambda^3} +...$$ where $D_{-1},D_0,D_1,D_2,..$ are $n\times n$ matrices and $D_{-1}$ is a non zero matrix.

What I know

Neumann's series and this post.

PS : if this isn't a research level problem, please let me know.


If we disregard the positivity constraint, this is not true in general, the leading order term can be of order $n-1$ rather than of order 1.

The problem is treated in Laurent expansion of the inverse of perturbed, singular matrices. The leading order term is of the order of the singularity in $A$, which for $A$ an $n\times n$ all-1 matrix is of order $n-1$.

Here is an example for $n=3$: $$\psi=\left( \begin{array}{ccc} 1 & 1+\frac{1}{\lambda} & 1+\frac{4}{\lambda} \\ 1+\frac{1}{\lambda} & 1 & 1+\frac{1}{\lambda} \\ 1+\frac{4}{\lambda} & 1+\frac{1}{\lambda}& 1 \\ \end{array} \right)$$ has inverse of order $\lambda^2$: $$\psi^{-1}=\left( \begin{array}{ccc} -\frac{1}{8} {\lambda} (2 {\lambda}+1) & \frac{1}{2} {\lambda} ({\lambda}+1) & -\frac{1}{8} {\lambda} (2 {\lambda}-1) \\ \frac{1}{2} {\lambda} ({\lambda}+1) & -{\lambda} ({\lambda}+2) & \frac{1}{2} {\lambda} ({\lambda}+1) \\ -\frac{1}{8} {\lambda} (2 {\lambda}-1) & \frac{1}{2} {\lambda} ({\lambda}+1) & -\frac{1}{8} {\lambda} (2 {\lambda}+1) \\ \end{array} \right).$$

I do not know how/if the positivity constraint modifies the order.

  • $\begingroup$ I notice I did not use your additional condition that $\psi$ is positive definite for all $\lambda$. $\endgroup$ – Carlo Beenakker Aug 7 '20 at 13:44
  • $\begingroup$ Yes, that's something need to be resolved. To see if it can make the leading $n-2$ matrices zero. $\endgroup$ – Rajesh D Aug 7 '20 at 13:45
  • $\begingroup$ Oh the paper clearly says $H_{-s}$ is not zero. So thats about it? $\endgroup$ – Rajesh D Aug 7 '20 at 13:48
  • $\begingroup$ I am confused by that statement; it does not hold, for example, for the matrix $A+I/\lambda$ for $n=3$, then the leading order term is of order $\lambda$, not $\lambda^2$. $\endgroup$ – Carlo Beenakker Aug 7 '20 at 13:53
  • $\begingroup$ In some of my computations too I observed the same. I get $\lambda$ increase and not any higher powers. $\endgroup$ – Rajesh D Aug 7 '20 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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