On Tuesday, 3 July 2012 11:26:54 UTC+8, Dima Pasechnik wrote: > > > > 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 > > I take the last part back, sorry, it makes little sense. What certainly is doable is the following: 1) uniformly select a random maximal flag 2) uniformly select a random element U in the subgroup stabilizing the "canonical" maximal flag, i.e. the one stabilized by the upper triangular matrices. Then an element MU, for M being a matrix mapping the "canonical" flag to the flag selected at step 1, will be uniformly distributed over the whole group G (as step 1 selects a coset of U in G).
>> >> >>> 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 [email protected] To unsubscribe from this group, send an email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
