# Inverse of block matrix

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.

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

• Would you mind I ask some simple questions which confuse me? If $p'(\lambda)=0$, should $p'(\lambda)(74 - 20 \lambda^2 + \lambda^4)^2$ not equal to $0$? Where does $(74 - 20 \lambda^2 + \lambda^4)^2$ come from?
– Hans
May 21 '21 at 5:20
• @Hans : $p'(\lambda)$ is a certain rational function of $\lambda$ with denominator $(74 - 20 \lambda^2 + \lambda^4)^2$, so that, as noted in the answer, $p'(\lambda)(74 - 20 \lambda^2 + \lambda^4)^2$ is a polynomial in $\lambda$. May 21 '21 at 12:53
• I am confused. Your second paragraph says $L=T_1(T_2-\lambda)^{-1}$. $p(\lambda)$ being the upper-left entry of $T_1=L(T_2-\lambda)$ which is a constant matrix is a constant. How is it a rational function of $\lambda$? There must be a typo or something.
– Hans
May 21 '21 at 15:10
• @Hans : $p(\lambda)$ must be constant if we assume that the desired result holds. As shown in the answer, this assumption leads to a contradiction, which implies that the assumption was false. So, yes, we do have a (useful) contradiction here. May 21 '21 at 16:14
• @Hans : I think you computed the entry incorrectly. Here is the link to Mathematica calculations: u.pcloud.link/publink/… May 24 '21 at 15:00