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The resolvent of a matrix $\mathbf{A}$ is defined as

\begin{equation} \mathbf{G}_{\mathbf{A}}(z) = \left(\mathbf{A} - z \mathbf{1}_n\right)^{-1}, \quad z \in \mathbb{C} \setminus \sigma(\mathbf{A}), \end{equation}

where $\mathbf{1}_n$ is the $n \times n$ identity matrix and $\sigma(\mathbf{A})$ is the spectrum of $\mathbf{A}$. In the infinite-size limit $n \to \infty$, the spectral density $\rho(\lambda)$ is related to the trace of the resolvent by

\begin{equation} \rho(\lambda) = \lim_{n \to \infty} -\left. \frac{1}{\pi n} \frac{\partial}{\partial \bar{z}} \operatorname{Tr} \mathbf{G}_{\mathbf{A}}(z) \right|_{z = \lambda}, \end{equation}

where $\frac{\partial}{\partial \bar{z}} = \frac{\partial}{\partial x} + \mathrm{i} \frac{\partial}{\partial y}$. This can be intuitively understood with an analogy to electrostatics (see e.g. [1]).

Question: The resolvent $\mathbf{G}_{\mathbf{A}}(z)$ is undefined for $z \in \sigma(\mathbf{A})$, yet the derivative is taken precisely at $z = \lambda \in \sigma(\mathbf{A})$. This appears even more problematic for non-Hermitian matrices, as adding a small regularization $z \to z + \mathrm{i}\varepsilon$ does not resolve the issue in the limit $\varepsilon \to 0$. Why is it valid to compute this derivative despite $\mathbf{G}_{\mathbf{A}}(z)$ being undefined at $z = \lambda$?

Additionally, for non-Hermitian matrices, $\mathbf{G}_{\mathbf{A}}(z)$ is not analytic in regions where $\rho(\lambda) \neq 0$. However, many papers (e.g. [2]) use formal series expansions of $\mathbf{G}_{\mathbf{A}}(z)$ in these regions to derive results. How is this justified when analyticity is typically required for series convergence? The whole point in [1] was that, for non-H matrices, while the resolvent still carries information about the spectrum it is not possible to get it as a series expansion because $\mathbf{G}$ is not analytic.

Any pedagogical references related to these questions is also warmly welcomed... Thank you!

[1] Sommers, H. J., et al. "Spectrum of large random asymmetric matrices." Phys. Rev. Lett. 60.19 (1988): 1895.
[2] Brézin, E., & Zee, A. "Non-Hermitian delocalization: multiple scattering and bounds." Nucl. Phys. B 509.3 (1998): 599–614.

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  • $\begingroup$ To be honest, the displayed formula makes zero sense to me -- what even is $A$? You have a limit $n\to\infty$, so $A$ is supposedly an $n\times n$ matrix for all $n$ simultaneously? $\endgroup$ Commented Nov 23, 2024 at 11:43
  • $\begingroup$ $A$ is some given matrix of size $n\times n$. A simple example is when all entries of the matrix are I.I.D Gaussian random variables with 0 mean and variance $1/n$. In the limit $n\to\infty$ the distribution of eigenvalues is a uniform distribution on the unit disk on the complex plane. $\endgroup$ Commented Nov 24, 2024 at 5:18

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Indeed, if $A$ is non-Hermitian you cannot use the inverse $(A-z)^{-1}$ to study eigenvalues near a complex number $z$. To apply resolvent techniques to a non-Hermitian matrix $A$ you need to first symmetrize it, $$H(z)=\begin{pmatrix}0&A-z\\ (A-z)^\ast&0 \end{pmatrix}.$$ Then Girko's formula$^\ast$ relates the eigenvalues $\lambda$ of $A$ to the resolvent of the Hermitian matrix $H(z)$ on the imaginary axis. A review of this technique, and how it works in the limit that the dimension $n$ of $A$ goes to infinity, is Fluctuations in the spectrum of non-Hermitian i.i.d. matrices (2022).

$^\ast$ V.L. Girko, Theory of Probability and its Applications 29, 694 (1985). $$\sum_{\lambda}f(\lambda)=-\frac{1}{4\pi}\int\Delta f(z)\int_0^\infty\operatorname{Im}\operatorname{Tr}[H(z)-i\eta]^{-1}d\eta\, d^2 z.$$
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  • $\begingroup$ Thanks for the reference, and I never realized it was Girko that introduced this Hermitization method! I always thought it was Feinberg and Zee... However this doesn't entirely answer my question: While I cannot use the inverse $(A-z)^{-1}$, it seems like the relation between $\rho(z)$ and $(A-z)^{-1}$ still holds. Even when $(A-z)^{-1}$ is not defined. Am I mistaken? Second, in the paper by Brézin and Zee I referred, they express $(A-z)^{-1}$ as a formal series, which should not be possible since $(A-z)^{-1}$ is not analytic? $\endgroup$ Commented Nov 24, 2024 at 5:24
  • $\begingroup$ the relation between the spectral density $\rho(z)=\sum_{\lambda\in\sigma(A)}\delta(z-\lambda)$ and the resolvent $G(z)=(A-z)^{-1}$ is $$\rho(z)=-\frac{1}{\pi}\frac{\partial}{\partial z^\ast}\operatorname{Tr}G(z).$$ This equation is formally correct, but not a useful relation because the right-hand-side is singular [it vanishes when $z\notin\sigma(A)$ and diverges when $z\in\sigma(A)$]. If you average the right-hand-side, the divergence is regularized, this is how Feinberg and Zee proceed. $\endgroup$ Commented Nov 24, 2024 at 11:34
  • $\begingroup$ I see. Still not sure why we can express the resolvent as a formal series (ref Brézin and Zee) despite being singular but I get the idea $\endgroup$ Commented Nov 26, 2024 at 10:01

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