From 2bbb3199a4774f2880954ad66faa187dbb020339 Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 7 Dec 2020 16:15:17 +0100
Subject: More matrix talk.

---
 bezout.tex | 13 ++++++++++++-
 1 file changed, 12 insertions(+), 1 deletion(-)

(limited to 'bezout.tex')

diff --git a/bezout.tex b/bezout.tex
index 783d547..6873e58 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -116,7 +116,18 @@ $\langle(\partial_i\partial_j H_0)^2\rangle=p(p-1)\kappa/2N$, $\rho_0(\lambda)$
   \left(\frac{\mathop{\mathrm{Im}}(\lambda e^{i\theta})}{a^{p-2}-|\kappa|}\right)^2
   <\frac{p(p-1)}{2a^{p-2}}
 \end{equation}
-where $\theta=\frac12\arg\kappa$ \cite{Nguyen_2014_The}.
+where $\theta=\frac12\arg\kappa$ \cite{Nguyen_2014_The}. The eigenvalue
+spectrum of $\partial\partial H$ therefore is than of an ellipse whose center
+is shifted by $p\epsilon$.
+
+The eigenvalue spectrum of the Hessian of the real part, or equivalently the
+eigenvalue spectrum of $(\partial\partial H)^\dagger\partial\partial H$, is the
+singular value spectrum of $\partial\partial H$. This is a more difficult
+problem and to our knowledge a closed form for arbitrary $\kappa$ is not known.
+We have worked out an implicit form for this spectrum using the saddle point of
+a replica calculation for the Green function. blah blah blah\dots
+
+The transition from a one-cut to two-cut singular value spectrum naturally corresponds to the origin leaving the support of the eigenvalue spectrum. Weyl's theorem requires that the product over the norm of all eigenvalues must not be greater than the product over all singular values. Therefore, the absence of zero eigenvalues implies the absence of zero singular values.
 
 \bibliographystyle{apsrev4-2}
 \bibliography{bezout}
-- 
cgit v1.2.3-54-g00ecf

