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 $A\_1 = \operatorname{diag}(\lambda\_1,\lambda\_2)$ for $\lambda\_i \in \mathbb C$ and $A\_2: V^2 \to V^2.$


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


Question: Can we express the resolvent in the form


$$K = \begin{pmatrix} T\_1(T\_2-\lambda)^{-1} & \* \\ \* & T\_3 (T\_4-\lambda)^{-1}\end{pmatrix}$$


where $T\_1,..,T\_4$ are some matrices and $\*$ elements I do not really care about.



Inverse of block matrix
$\newcommand{\la}{\lambda}$The answer is no. Indeed, let
\begin{equation}
 A=\left(
\begin{array}{cccc}
 1 & 0 & 2 & -1 \\
 0 & -2 & 1 & 3 \\
 2 & 1 & -1 & 0 \\
 -1 & 3 & 0 & 2 \\
\end{array}
\right). 
\end{equation}
Suppose that the desired result holds for some matrices $T\_1,\dots,T\_4$.


Then, letting $L$ denote the upper-left $2\times2$ block of the matrix $K=(A-\la)^{-1}$, we will have $L=T\_1(T\_2-\la)^{-1}$. So,
$L(T\_2-\la)=T\_1$.
So, the upper-left entry, say $p(\la)$, of the $2\times2$ matrix $L(T\_2-\la)$ cannot depend on $\la$, and hence $p'(\la)=0$ for all $\la$.


However, writing $T\_2=\left(
\begin{array}{cc}
 u & v \\
 x & y \\
\end{array}
\right)$, we see that
$p'(\la)(74 - 20 \la^2 + \la^4)^2$ is a polynomial in $\la$ of degree $\le6$, with respective coefficients $u-1$ and $2(u-6)$ of $\la^6$ and $\la^5$, and these two coefficients cannot simultaneously vanish. So, it is not true that $p'(\la)=0$ for all $\la$, and thus
we do get a contradiction.


