[sage-support] Re: Constructing subsapces of vector spaces
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]]]).kernel() 125 loops, best of 3: 2.15 ms per loop sage: %timeit V.submodule_with_basis([V.linear_combination_of_basis(b.list()) for b in mat.kernel().basis()]) 625 loops, best of 3: 1.39 ms per loop so it does seem to be slower, although one example is hardly definitive. Andrew -- You received this message because you are subscribed to the Google Groups sage-support group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
[sage-support] Re: Constructing subsapces of vector spaces
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 homomorphism. Indeed, %timeit V.hom([(ZZ^3)(v) for v in [[1,2,3],[2,1,4],[3,3,7]]]).kernel() 125 loops, best of 3: 2.15 ms per loop sage: %timeit V.submodule_with_basis([V.linear_combination_of_basis(b.list()) for b in mat.kernel().basis()]) 625 loops, best of 3: 1.39 ms per loop so it does seem to be slower, although one example is hardly definitive. I wouldn't draw any conclusions from that. In one timeit you are still constructing the matrix/hom whereas in the other you already have it. Anyway, indeed, the whole module (homomorphism) code is incredibly inefficient because special cases such a ZZ-modules haven't been special-cased at all yet (except for computing the kernel code). If you want speed, write it out in matrices: sage: %timeit matrix([[1,2,3],[2,1,4],[3,3,7]]).transpose().right_kernel_matrix()*matrix([[0,1,2,3],[2,3,1,4],[1,3,2,1]]) 625 loops, best of 3: 668 µs per loop As long as you only need submodules it's not so bad. Once you really need quotients (especially non-free ones) you probably are better off using modules. -- You received this message because you are subscribed to the Google Groups sage-support group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
[sage-support] Re: Constructing subsapces of vector spaces
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 where I want to find a submodule of a module V which is annihilated by a bunch of maps. What I want to do something like the following: V= some free submodule of rank v inside a space of rank d\ge v. mat=[[0 for col in range(w)] for row in range(v)] # fairly sparse matrix which describes the map f on bases of V and W ...compute non-zero entries of the matrix ``mat`` V = V.intersection( matrix(mat).kernel() ) where v=dim V, w=dim W where f is a map from V to W. Of course, this doesn't quite work because matrix(mat).kernel() returns a submodule of ZZ^v with respect to its standard basis whereas the space V has a very different basis in general. To get around this I found myself writing kernel=V.submodule_with_basis([V.linear_combination_of_basis(b.list()) for b in matrix(pmat).kernel().basis()]) V=V.intersection(kernel) This strikes me as being a very convoluted way of writing something is presumably quite common: writing the standard basis elements as linear combinations of another basis. My question is: is there a better way to do this? For a little while, I tried defining W=FreeModule(ZZ,w) and then constructing the map f explicitly as a homomorphism from V to W. Doing it this way does make f.kernel() into a submodule of V, but this approach just seems to create a lot of extra overhead with very little benefit as I essentially still had to construct the matrix ``mat`` above and used it to define ``f`` with respect to the bases of ``V`` and ``W``. In fact, all this really did was shift my problem above of writing the basis of the kernel in terms of the basis of V to the problem of writing the images of f in terms of the standard basis of W. (Even though I imagine that this second approach is much slower I really should have checked by doing some profiling before rewriting as above, but I didn't...) Any thoughts? Cheers, Andrew -- You received this message because you are subscribed to the Google Groups sage-support group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
[sage-support] Re: Constructing subsapces of vector spaces
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 basis matrix: [0 1 2 3] [2 3 1 4] [1 3 2 1] sage: mat=matrix([[1,2,3],[2,1,4],[3,3,7]]); mat.kernel() Free module of degree 3 and rank 1 over Integer Ring Echelon basis matrix: [ 1 1 -1] The problem that is V is isomorphic to Z^3, but it is represented as a subspace of Z^4, whereas the kernel is a subspace of Z^3. As I mentioned, sage: V.submodule_with_basis([V.linear_combination_of_basis(b.list()) for b in mat.kernel().basis()]) Free module of degree 4 and rank 1 over Integer Ring User basis matrix: [1 1 1 6] does give the kernel as a subspace of V. I was just wondering if there was a better way of doing this. Cheers, Andrew -- You received this message because you are subscribed to the Google Groups sage-support group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
[sage-support] Re: Constructing subsapces of vector spaces
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 could have: sage: V Free module of degree 4 and rank 3 over Integer Ring User basis matrix: [0 1 2 3] [2 3 1 4] [1 3 2 1] sage: mat=matrix([[1,2,3],[2,1,4],[3,3,7]]); mat.kernel() Free module of degree 3 and rank 1 over Integer Ring Echelon basis matrix: [ 1 1 -1] The problem that is V is isomorphic to Z^3, but it is represented as a subspace of Z^4, whereas the kernel is a subspace of Z^3. As I mentioned, So, when you're computing the kernel, you should tell the system that you want a kernel of a homomorphism on V; not on abstract Z^3: sage: A=ZZ^4 sage: V=A.submodule_with_basis([[0,1,2,3],[2,3,1,4],[1,3,2,1]]) sage: phi=V.hom([(ZZ^3)(v) for v in [[1,2,3],[2,1,4],[3,3,7]]]) sage: phi.kernel() Free module of degree 4 and rank 1 over Integer Ring Echelon basis matrix: [1 1 1 6] -- You received this message because you are subscribed to the Google Groups sage-support group. To post to this group, send email to sage-support@googlegroups.com. To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com. Visit this group at http://groups.google.com/group/sage-support?hl=en.
