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Jun 28, 2018 at 14:17 comment added Gericault Post edited, the matrix is not hermitian ! Hence the spectrum is indeed complex
Jun 28, 2018 at 8:26 comment added Carlo Beenakker I have asked the OP for clarification, I was under the impression the question referred to a real spectrum.
Jun 27, 2018 at 21:31 comment added Tobias Fritz This answer addresses the symmetric/hermitian case, right? Does this imply anything about the unsymmetric case, which the original question seems to be asking about?
Jun 27, 2018 at 20:50 history answered Carlo Beenakker CC BY-SA 4.0