> I have tried with linalg.solve(a, b), where I put a as the > Matrix A - eigen value* unit matrix, and b as the zero matrix. But the > solution returned is a zero matrix, which I really find disappointing.
So if you're trying to solve (A - \lambda I) x = b, try appending an extra row to your (A - \lambda I) matrix with all ones; the output of this will be the some of the elements of your eigenvector. Append a 1 to the end of your b. This will exclude the case where x = 0 as valid solution, as you'll then require that \sum_i x_i = 1. You'll probably want to renormalize properly afterwards. Quick and dirty, but should work. --Hoyt -- ++++++++++++++++++++++++++++++++++++++++++++++++ + Hoyt Koepke + University of Washington Department of Statistics + http://www.stat.washington.edu/~hoytak/ + hoy...@gmail.com ++++++++++++++++++++++++++++++++++++++++++ _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
