On Fri, Apr 5, 2013 at 2:40 AM, Koobas <[email protected]> wrote: > Anyways, I saw no particular reason for the method to fail with k > approaching or exceeding m and n. > It does if there is no regularization. > But with regularization in place, k can be pretty much anything. >
Ahh... this is an important point and it should handle all of the issues of poor conditioning. The regularizer takes the rank deficient A and makes it reasonably well conditioned. How well conditioned depends on the choice of lambda, the regularizing scale constant.
