Let $A$ and $B$ be two square matrices of the same size where $A$ is invertible and the kernel of $B$ is $1$-dimensional. Consider the function $f:\mathbb C \to \mathbb M$ defined by $$f(z) = z^2I + Az+B.$$ I use $\mathbb M$ to denote the space of square matrices in which $A$ and $B$ live and write $I\in\mathbb M$ for the identity matrix. I am interested in the behavior of $f^{-1}(z) = f(z)^{-1}$ around $z=0$. I suspect that $f^{-1}$ around $z=0$ has the form $$\sum_{n=-1}^\infty a_n z^n.$$ Could someone help me with calculating such expansion?
I would like the coefficients as explicitly as possible. The standard trick of partial fraction decomposition and using $$(I-X)^{-1} = \sum_{n=0}^\infty X^n$$ does not work as it is not clear that we are able to factor $f(z)$.
Example: (As requested by Rodrigo.) Suppose that $\mathbb M$ consist of $2\times 2$ matrices. Then take $A = 2I$ and $B=\text{diag}(1,0)$. I want to look at the inverse of $$ f(z) = \begin{pmatrix} z^2 + 2z+1 & 0 \\ 0 & z^2 +2z \end{pmatrix}, $$ which is given by $$ \frac{1}{\det(f(z))} \begin{pmatrix} z^2+2z & 0 \\ 0 & z^2+2z+1 \end{pmatrix}, $$ where $\det(f(z)) = z(z^2+2z+1)(z+2)$. We then clearly see that $f(z)$ splits up in a part that is analytic at $z=0$ and a part that is not $$ f(z)^{-1} = \frac{1}{(z^2+2z+1)(z+2)}\begin{pmatrix} z+2 & 0 \\ 0 & z+2 \end{pmatrix} + \frac 1 {\det(f(z))}\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}. $$ Now I see that if I want to know the behavior of $f(z)^{-1}$ around $z=0$, I just need to consider the Laurent expansion of a `normal' rational function $\mathbb C \to \mathbb C$ that I can calculate.