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Jul 18, 2016 at 7:58 vote accept Michael
Jul 12, 2016 at 11:50 comment added Dirk Comment for ii): You can solve $A^TAx=A^Tb$ by CG without even forming $A^TA$ (side note: contrary to the widespread believe this is not such a bad idea, even for ill-conditioned matrices, since the convergence of CG is more influenced by the clustering of the singular values and not only by it condition number).
Jul 11, 2016 at 21:25 answer added Cristóbal Guzmán timeline score: 4
Jul 7, 2016 at 21:10 history asked Michael CC BY-SA 3.0