Consider the following code :
R.<x,y,z> = QQ[];
p=(x-y)(y-z)(x-z);
J = matrix(R,[[x-y],[y]]);
J.rank();
The answer displayed is 1. This is the rank over the polynomia ring I suppose. I would like to compute rank over the field of rationals. That is if there are rational numbers a,b and c such that a diff(p,x) + b diff(p,y) + c diff(p,z) =0 but a,b,c non zero then rank should be 3.
What is the way to do it ?
Yes it is the rank over the polynomial ring. Houw could it be the rank over the rationals ? Note that x and y do not belong to the rational field, but the rank depends on the rational values you associate to them (the zero vector has rank 0, the other have rank one):
sage: J.substitute({x:0,y:0}).rank()
0
sage: J.substitute({x:0,y:1}).rank()
1
This makes no sense. The Jacobian matrix can be computed easily (in sage):
sage: var( 'x,y,z' );
sage: J = jacobian( p, [x,y,z] )
sage: J
[ (x - y)*(y - z) + (x - z)*(y - z) (x - y)*(x - z) - (x - z)*(y - z) -(x - y)*(x - z) - (x - y)*(y - z)]
and this is a 3x1 matrix. Its rank is at most one! The rank is one, if $x,y,z$ are seen as (transcendental) independent variables. If we want to see for which numbers $x,y,z$ the rank is not one (which is generically the case), but zero (not three, this makes no sense), and this is a different question starting from a different question, then the following finds the point $(x,y,z)$:
sage: solve( [ component==0 for component in J[0] ], [x,y,z] )
[[x == r1, y == r1, z == r1]]
$p'(x,y,z)$ is zero iff $x=y=z$.
Asked: 2017-04-20 19:08:36 +0100
Seen: 900 times
Last updated: Apr 21 '17
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The matrix is defined over a "bigger" ring, $R$. So it is computed over the ring of definition....
Using
J.rank?orJ.rank??gives more info. In our case, the work is delegated toJ.pivots(). (The length of this list of pivots is the rank.)