Bill Allombert on Thu, 29 Oct 2015 11:03:47 +0100 |
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Re: qfgaussred() for singular matrices |
On Wed, Oct 28, 2015 at 12:22:18PM +0100, Jeroen Demeyer wrote: > Hello, > > it is not clear to me if qfgaussred() is officially supported for > singular matrices. > > I am asking because of this unexpected output: > > gp> qfgaussred([1,1;1,1]) > %1 = > [1 1] > > [1 0] > > The problem is that the new basis vectors are (1,1) and (1,1) which > are not linearly independent. Note that this does not seem to contradict the documentation. > It would be useful if the vectors > > x_i + sum_{i != j} a_ij x_j > > (where x_i is a given basis) would always form a new basis, even in > the singular case. You can extract the linear forms associated to non-zero diagonal terms and use matsupplement to get a basis. Cheers, Bill.