Matrix morphism addition ignores bases

[rbeezer]

When two module or vector space morphisms are added, the matrix_morphism module simply adds their matrix representations, even if the representations have wildly different bases. To wit,

S, T, R are all versions of the identity linear transformation, so S+T, S+R should just be 2I.

sage: u = vector([1,2])
sage: S = linear_transformation(QQ^2, QQ^2, [[1,0],[0,1]])
sage: T = linear_transformation(QQ^2, QQ^2, [[1,0],[0,1]])
sage: good = ((S+T)(u) == S(u) + T(u))
sage: B  = [vector([0,1]), vector([1,0])]
sage: V = (QQ^2).subspace_with_basis(B)
sage: R = T.restrict_domain(V)
sage: bad = ((S+R)(u) == S(u) + R(u))
sage: ugly = (S+R).matrix()
sage: good, bad, ugly
(
             [1 1]
True, False, [1 1]
)

CC: @toadrush @roed314

Component: linear algebra

Keywords: matrix morphism sum addition

Issue created by migration from https://trac.sagemath.org/ticket/18520

[rbeezer]

comment:1

Packaging up a fix now. The result (sum) will have the bases from self, meaning that the matrix representation of right will need to change.

[mkoeppe]

comment:3

Moving to 9.4, as 9.3 has been released.

[roed314]

comment:5

I think the issue is that S and R don't have the same domain. We should either raise an error in this case, or compose with the coercion map V -> QQ^2.

[mjungmath]

comment:8

Replying to @roed314:

I think the issue is that S and R don't have the same domain. We should either raise an error in this case, or compose with the coercion map V -> QQ^2.

The problem I see here is that there is really no preferred coercion:

sage: S = linear_transformation(QQ^2, QQ^2, [[1,0],[0,1]])
sage: T = linear_transformation(QQ^2, QQ^2, [[1,0],[0,1]])
sage: B  = [vector([0,1]), vector([1,0])]
sage: V = (QQ^2).subspace_with_basis(B)
sage: R = T.restrict_domain(V)
sage: S+R
Free module morphism defined by the matrix
[1 1]
[1 1]
Domain: Vector space of dimension 2 over Rational Field
Codomain: Vector space of dimension 2 over Rational Field
sage: R+S
Free module morphism defined by the matrix
[1 1]
[1 1]
Domain: Vector space of degree 2 and dimension 2 over Rational Field
User basis matrix:
[0 1]
[1 0]

I'd opt for an error message.

[mjungmath]

comment:9

But in any case, the coercion needs to be fixed. In fact, nothing happens to R or S when coerced.