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Aug 17, 2018 at 14:06 vote accept Alpha001
Aug 17, 2018 at 14:06 vote accept Alpha001
Aug 17, 2018 at 14:06
S Jul 4, 2018 at 17:28 history suggested Rodrigo de Azevedo CC BY-SA 4.0
Minor edits
Jul 4, 2018 at 17:27 review Suggested edits
S Jul 4, 2018 at 17:28
Jul 3, 2018 at 23:53 answer added Igor Rivin timeline score: 5
Jul 3, 2018 at 23:17 comment added Alpha001 @RobertIsrael Is it clear how it would change the dimension?
Jul 3, 2018 at 22:35 comment added Peter Shor @RobertIsrael: Oops. I completely missed the Hermitian requirement.
Jul 3, 2018 at 22:31 comment added Robert Israel Given a Hermitian matrix with $r$ positive and $r$ negative eigenvalues, any sufficiently close Hermitian matrix of rank $2r$ will also have $r$ positive and $r$ negative eigenvalues. On the other hand, the restriction to Hermitian matrices does change the dimension.
Jul 3, 2018 at 20:50 review Close votes
Jul 4, 2018 at 11:00
Jul 3, 2018 at 20:48 comment added Alpha001 I thougth that the restriction of $r$ postivie and $r$ negative eigenvalues reduces the dimension. I don't got your point in total. Could you explain it with a few more words?
Jul 3, 2018 at 19:08 comment added Peter Shor Why should the dimension be any different? This partitions the manifold of dimension $2n(2r) - (2r)^2$ into a finite number (15) of submanifolds.
S Jul 3, 2018 at 19:03 history suggested Glorfindel CC BY-SA 4.0
typos corrected, formatting
Jul 3, 2018 at 19:00 review Suggested edits
S Jul 3, 2018 at 19:03
Jul 3, 2018 at 18:52 review First posts
Jul 3, 2018 at 19:00
Jul 3, 2018 at 18:48 history asked Alpha001 CC BY-SA 4.0