Dear Martin,
Thank you so much. It works!
Can we make it faster?
It took 17 seconds for my problem but
M1.LLL() took only 3 seconds. Of course I understand we
are calculating extra matrix U.
Thanks again for your help.
Regards,
Santanu
On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
sage-devel@googlegroups.com> wrote:
> Hi there,
>
> This should do the trick:
>
> sage: from fpylll import *
> sage: A = random_matrix(ZZ, 6, 90)
> sage: U = IntegerMatrix.identity(6)
> sage: B = LLL.reduction(IntegerMatrix.from_matrix(A),
> U).to_matrix(matrix(ZZ, 6,
> 90))
> sage: U = U.to_matrix(matrix(ZZ, 6,6))
> sage: B == U*A
> True
> sage: abs(U.det())
> 1
>
> Cheers,
> Martin
>
>
> Santanu Sarkar <sarkar.santanu....@gmail.com> writes:
> > Dear all,
> > I have a matrix M1 with integer entries with 90 rows and 6 columns.
> > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
> > corresponding unimodular transformation matrix T such that
> > T*M1=M2. We can find T by
> > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
> > becomes 0 i.e., T.det()=0.
> > I want T.det()=1.
> >
> > Best regards,
> > Santanu
>
>
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