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.