On Tuesday, 3 July 2012 10:36:37 UTC+8, Dima Pasechnik wrote: > > > > 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 :–) >
by the way, for generating random elements it might be better to use the Bruhat decomposition, which is unique. See http://en.wikipedia.org/wiki/Bruhat_decomposition > > >> 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
