Hi Xiangrui, I used lambda = 0.1...It is possible that 2 users ranked in movies in a very similar way...
I agree that increasing lambda will solve the problem but you agree this is not a solution...lambda should be tuned based on sparsity / other criteria and not to make a linearly dependent hessian matrix linearly independent... Thanks. Deb On Thu, Mar 6, 2014 at 7:20 PM, Xiangrui Meng <men...@gmail.com> wrote: > If the matrix is very ill-conditioned, then A^T A becomes numerically > rank deficient. However, if you use a reasonably large positive > regularization constant (lambda), "A^T A + lambda I" should be still > positive definite. What was the regularization constant (lambda) you > set? Could you test whether the error still happens when you use a > large lambda? > > Best, > Xiangrui >
