Try mat.kernel_on(V).
Also, if v is a vector in the ambient space which happens to lie in V then
V.coordinates(v) will give its coordinates w.r.t. the basis of V.
John Cremona
On Monday, December 3, 2012 1:38:38 AM UTC, Andrew Mathas wrote:
>
> Hi All,
>
> I have been playing with some code
Hi John,
Thanks for the reply, but you have my problem "upside down" as I don't need
to restrict from the ambient space to the subspace but rather to extend
from the subspace to the ambient space.
For example, I could have:
sage: V
Free module of degree 4 and rank 3 over Integer Ring
User bas
On Monday, December 3, 2012 5:09:40 PM UTC-8, Andrew Mathas wrote:
>
> Hi John,
>
> Thanks for the reply, but you have my problem "upside down" as I don't
> need to restrict from the ambient space to the subspace but rather to
> extend from the subspace to the ambient space.
>
> For example, I
Thanks Nils. I this is similar to, but more elegant than, what I tried
earlier. I went back to the solution above, however, I thought that there
was probably a lot of overhead in creating the space ZZ^3 and the
homomorphism. Indeed,
%timeit V.hom([(ZZ^3)(v) for v in [[1,2,3],[2,1,4],[3,3,7]]]).
On Tuesday, December 4, 2012 3:14:27 AM UTC-8, Andrew Mathas wrote:
>
> Thanks Nils. I this is similar to, but more elegant than, what I tried
> earlier. I went back to the solution above, however, I thought that there
> was probably a lot of overhead in creating the space ZZ^3 and the
> homomo
