Dear Volker Braun,
Thanks so much that is what I need
Thanks
Doaa

On 25 April 2012 14:05, Volker Braun <[email protected]> wrote:
> Note that this is a question about convex polyhedra since it involves
> inequalities! You can't solve it with linear algebra alone. A simple way is
> to intersect your null space with the positive orthant, which gives you a
> cone in your null space whose elements have all positive entries:
>
> sage: kernel = Polyhedron(lines=[
> ....:   [ 1,  0,  1,  0, -1,  0,  0],
> ....:   [ 0,  1,  1,  0,  0,  0,  0],
> ....:   [ 0,  0,  0,  1,  1,  0,  0],
> ....:   [ 0,  0,  0,  0,  0,  1,  1]])
> sage: positive = Polyhedron(rays=identity_matrix(7).columns())
> sage: pos_ker = kernel.intersection(positive)
> sage: pos_ker
> A 4-dimensional polyhedron in ZZ^7 defined as the convex hull of 1 vertex
> and 4 rays
> sage: pos_ker.rays()
> (A ray in the direction (0, 0, 0, 0, 0, 1, 1), A ray in the direction (0, 0,
> 0, 1, 1, 0, 0), A ray in the direction (0, 1, 1, 0, 0, 0, 0), A ray in the
> direction (1, 0, 1, 1, 0, 0, 0))
>
>
>
>
> On Tuesday, April 24, 2012 9:36:04 AM UTC-4, Doaa El-Sakout wrote:
>>
>> Hi
>>
>> I am using right_kernel to find a kernel, for example the result is
>> [ 1  0  1  0 -1  0  0]
>> [ 0  1  1  0  0  0  0]
>> [ 0  0  0  1  1  0  0]
>> [ 0  0  0  0  0  1  1]
>>
>> How can I (for an arbitrary kernel) get positive entries only,
>> by making a linear combination of rows?
>>
>> Regards,
>> Doaa
>
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