# Inverse of block matrix II

This is a follow-up question on a previous question of mine that had a negative answer. I tried some examples and believe the following has a chance to be true.

Let $$V$$ be a finite-dimensional vector space and consider the space $$X=V\times V\times V\times V.$$

Consider the block matrix

$$A = \begin{pmatrix} A_1 & A_2 \\ A_2^* & -A_1\end{pmatrix}$$

where $$A_1:V^2 \to V^2$$ is a diagonal matrix $$A_1 = \operatorname{diag}(0,\lambda)$$ for $$\lambda \in \mathbb C$$ and $$A_2: V^2 \to V^2$$ a matrix that shall not be further specified.

We then consider $$K=(A-\lambda)^{-1}.$$

Question: Can we express the resolvent in the form

$$K = \begin{pmatrix} T_1(\lambda)(T_2-\lambda)^{-1}+T_3(\lambda)(T_4-\lambda)^{-1} & * \\ * & T_5(\lambda)(T_6-\lambda)^{-1}+T_7(\lambda)(T_8-\lambda)^{-1}\end{pmatrix}$$

where $$T_1,T_3,T_5,T_7$$ are some matrices that are entire functions of $$\lambda$$ and $$T_2,T_4,T_6,T_8$$ are independent of $$\lambda$$. In addition, $$*$$ are elements I do not really care about, so they are allowed to be anything.