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Mar 26, 2014 at 17:45 comment added mohi I can now actually do o(nlog^2 n) by truncation I think. I would like o(n) but I would actually be happy with an argument that gives o(nlog n)
Mar 25, 2014 at 8:41 comment added zouzias I see. Your bound O(n (log n)^3) is tight up to a (logn)^2 factor for general rank-one samples. What are you shooting for? O(n) number of samples?
Mar 24, 2014 at 17:25 comment added mohi Thanks, I am familiar with all of these references. In some sense all of them require that the matrices are bounded. Hence, one has to apply truncation first. This is where I loose the log factors.
Mar 23, 2014 at 16:32 history answered zouzias CC BY-SA 3.0