Timeline for Finding a particular matrix factor

10 events

when toggle format	what		by	license	comment
Jun 26, 2018 at 19:20	vote	accept	Ludwig
Jun 26, 2018 at 12:32	answer	added	Alex Gavrilov		timeline score: 1
Jun 26, 2018 at 8:51	comment	added	Achim Krause		Doesn't it follow immediately from Liouville? $f(x) = 1/f(x^{-1})$ together with the fact that $f$ has no zero at $0$ shows that it is bounded.
Jun 25, 2018 at 20:43	history	edited	Ludwig	CC BY-SA 4.0	deleted 10 characters in body
Jun 25, 2018 at 16:43	comment	added	Jochen Glueck		Yes. Now, if we could prove that every holomorphic solution of the functional equation which has no poles and no zeros in the unit disk is, say, constant (i.e. identically $1$ or $-1$) this would impose a severe restriction on the possible choices of $C(x)$. But unfortunately I don't see at the moment whether each such solution of the functional equation is constant.
Jun 25, 2018 at 14:42	history	edited	Ludwig	CC BY-SA 4.0	added 6 characters in body
Jun 25, 2018 at 14:41	comment	added	Ludwig		@JochenGlueck: If you solve your determinant equation then a solution satisfying the desired requirements is given by any $C(x)$ such that $\det(C(x))=1$. But perhaps I didn't understand you question.
Jun 25, 2018 at 4:45	comment	added	Jochen Glueck		Have you tried to first solve the functional equation $f(x)f(x^{-1}) = 1$ for the determinant $f(x) := \det(C(x))$ of $C(x)$?
Jun 24, 2018 at 14:54	history	edited	Ludwig	CC BY-SA 4.0	added 7 characters in body
Jun 24, 2018 at 4:12	history	asked	Ludwig	CC BY-SA 4.0