Timeline for Sums of Unitary Matrices

8 events

when toggle format	what		by	license	comment
Jan 19, 2012 at 1:21	vote	accept	dick lipton
Jan 18, 2012 at 17:18	answer	added	Denis Serre		timeline score: 2
Jan 18, 2012 at 16:10	answer	added	Mikael de la Salle		timeline score: 7
Jan 18, 2012 at 15:58	comment	added	Matthew Daws		@Federico: Yes; but that's why I wrote "Indeed..." To be more explicit, there is $\lambda>0$ with $\lambda J=h$ a self-adjoint contraction. Then $2h=u+u^$ for the $u$ I gave, and so $J = (1/2\lambda)(u+u^)$.
Jan 18, 2012 at 15:30	comment	added	Federico Poloni		Am I wrong, or the 4-unitaries theorem does not apply here? The OP asks for a representation of the form $J=\alpha \sum Q_i$, not $J=\sum \alpha_i Q_i$.
Jan 18, 2012 at 14:51	comment	added	Koen S		A way to picture it geometrically: consider the line $x_1 = x_2 = \dots x_n$ (the $x_i$ being the coordinates of the vector space) and the rotation of 180 degrees around this line. The sum of a vector and his image rotation then lies on this line, from which one then can derive that the sum of the identity matrix and the matrix corresponding with the rotation is of the form $\lambda J$.
Jan 18, 2012 at 14:41	comment	added	Matthew Daws		In a C$^$-algebra any operator is the linear combination of 4 unitaries... Indeed, every self-adjoint $h$ with $\\|h\\|\leq 1$ can be written as $(u+u^)/2$ where $u=h+i(1-h^2)^{1/2}$. As your $J$ is the scalar-multiple of a projection, we conclude that $f(n)=2$ for all $n$.
Jan 18, 2012 at 14:22	history	asked	dick lipton	CC BY-SA 3.0