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Jun 26, 2020 at 14:07 history edited Federico Poloni CC BY-SA 4.0
renamed U -> M so that people won't think it is unitary
Jun 26, 2020 at 14:06 comment added Federico Poloni @Abeginnermathmatician no, $X$ is not unitary in general, unfortunately. Thompson's setting uses matrix pencils, i.e., linear expressions $\lambda A + \mu B$ in the variables $\lambda, \mu$, so he has to specify that $X$ does not depend on $\mu$ nor $\lambda$, i.e., it is "constant". However, this setting is equivalent to considering matrix pairs $(A,B)$, and in that case $X$ is just a general invertible complex matrix.
Jun 26, 2020 at 13:31 comment added A beginner mathmatician Thank you again. For the similarity can we take $X$ in the paper to be unitary? The matrix $X$ is referred to as "nonsingular constant matrix".
Jun 26, 2020 at 6:12 comment added Federico Poloni @Abeginnermathmatician You are correct that in my paper the theorem is phrased for real symmetric matrices only, but it holds without changes for complex Hermitian ones. I have now edited my answer and references to point to the original paper by Thompson that we cite; that reference treats directly the Hermitian-Hermitian case.
Jun 26, 2020 at 6:11 history edited Federico Poloni CC BY-SA 4.0
moved to Thompson paper
Jun 26, 2020 at 6:02 comment added Federico Poloni @Abeginnermathmatician Oops, I wrote this answer looking at the Arxiv preprint, where we had a variant with a fourth block type E4 (EKCF vs. EWCF). I have fixed it, as well as the theorem and table number according to the published version.
Jun 26, 2020 at 6:01 history edited Federico Poloni CC BY-SA 4.0
fixed references
Jun 26, 2020 at 3:17 comment added A beginner mathmatician Thank you very much. Actually that is what was looking for. I think in your paper A and B are real matrices . Right? Also could not see E_4. I can see E_1,E_2 and E_3 conditions.
Jun 26, 2020 at 3:14 vote accept A beginner mathmatician
Jun 25, 2020 at 17:42 history answered Federico Poloni CC BY-SA 4.0