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](https://en.wikipedia.org/wiki/Neumann_series) and this [post](https://math.stackexchange.com/a/1268242/2987).


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



Asymptotic expansion of the inverse of a matrix valued function
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](https://e-archivo.uc3m.es/bitstream/handle/10016/23560/laurent_JCP_2015_ps.pdf). 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.


