Skip to main content
8 events
when toggle format what by license comment
Oct 27, 2016 at 18:56 vote accept solver6
Oct 24, 2016 at 21:13 answer added YCor timeline score: 2
Oct 24, 2016 at 20:30 comment added YCor It can be shown that the $\mathbb{C}$-subalgebra $A$ generated by $R$ also satisfies the property that any 2 elements have a common eigenvector (because $R$ is Zariski-dense in $A$ and this property is Zariski-closed). Hence it is no restriction to assume that $R$ is a subalgebra.
Oct 24, 2016 at 19:01 history edited YCor CC BY-SA 3.0
Reformulated question so that it has broader interest
Oct 23, 2016 at 12:54 history edited solver6 CC BY-SA 3.0
deleted 30 characters in body
Oct 23, 2016 at 12:41 comment added abx No, take for $R$ the subalgebra of matrices preserving $U$.
Oct 23, 2016 at 12:12 review First posts
Oct 23, 2016 at 12:15
Oct 23, 2016 at 12:11 history asked solver6 CC BY-SA 3.0