On Tuesday, 3 July 2012 09:56:04 UTC+8, Charles Bouillaguet wrote: > > > Mhh, why not? If A = LUP we just write AP^-1 = LU, hence for each LU we > > construct there are as many As as there are permutation matrices, or am > I > > missing something (again :))? > > I am not sure that the LUP decomposition is unique (I understand that > the LU is).
An invertible matrix need not have an LU decomposition. E.g. A=[[0,1],[1,1]] does not have it. It's evident over F_2: L can only take 2 values, and U can only take 2 values, so you can't have more than 4 different matrices of the form LU :–) > If A has more distinct LUP factorizations than B, then A > is more likely to be produced by this process than B.... > > Charles > -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
