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authorJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-07 16:56:48 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2020-12-07 16:56:48 +0100
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Added silly citation.
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@@ -144,7 +144,12 @@ 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.
+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 \cite{Weyl_1912_Das}.
+Therefore, the absence of zero eigenvalues implies the absence of zero singular
+values.
\bibliographystyle{apsrev4-2}
\bibliography{bezout}