Asymptotic expansion of the inverse of a matrix valued function 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.
 A: 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.
