On Jun 20, 2009, at 6:07 PM, rickhg12hs wrote: > > On Jun 19, 6:50 am, Pierre <pierre.guil...@gmail.com> wrote: >> is there an easy way of creating a subdiagonal matrix, that is with 1 >> just under the diagonal and 0's elsewhere ? > > This is a hack (especially since I'm no mathematician and no expert in > Sage or any of its components), but it seems to work. A formal proof > would be an interesting read. > > Here's an example: > > n = 10 > m = matrix(ZZ,n) > m[range(n-1),range(n-1)] = identity_matrix(n-1) > m_subd = m.LLL() > > Works with some large n I played with.
Another option sage: n = 10 sage: m = block_matrix([0, zero_matrix(1,1), identity_matrix(n), 0]) sage: m.subdivide() # get rid of the block divisions sage: m [0 0 0 0 0 0 0 0 0 0 0] [1 0 0 0 0 0 0 0 0 0 0] [0 1 0 0 0 0 0 0 0 0 0] [0 0 1 0 0 0 0 0 0 0 0] [0 0 0 1 0 0 0 0 0 0 0] [0 0 0 0 1 0 0 0 0 0 0] [0 0 0 0 0 1 0 0 0 0 0] [0 0 0 0 0 0 1 0 0 0 0] [0 0 0 0 0 0 0 1 0 0 0] [0 0 0 0 0 0 0 0 1 0 0] [0 0 0 0 0 0 0 0 0 1 0] You can also use this to get anything you want on the sub-diagonal (or pretty much anywhere). sage: m = block_matrix([0, zero_matrix(1,1), diagonal_matrix([10, 20, 30, 40]), 0]); m.subdivide(); m [ 0 0 0 0 0] [10 0 0 0 0] [ 0 20 0 0 0] [ 0 0 30 0 0] [ 0 0 0 40 0] - Robert --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
