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.


1 Answer 1


$\newcommand{\la}{\lambda}$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.

  • $\begingroup$ 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? $\endgroup$
    – Hans
    May 21, 2021 at 5:20
  • $\begingroup$ @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 $\lambda$. $\endgroup$ May 21, 2021 at 12:53
  • $\begingroup$ 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. $\endgroup$
    – Hans
    May 21, 2021 at 15:10
  • $\begingroup$ @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. $\endgroup$ May 21, 2021 at 16:14
  • 1
    $\begingroup$ @Hans : I think you computed the entry incorrectly. Here is the link to Mathematica calculations: u.pcloud.link/publink/… $\endgroup$ May 24, 2021 at 15:00

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