11.2.13  Sylvester matrix of two polynomials and resultant

Given two polynomials A(x)=∑_i=0ⁱ⁼ⁿ a_ixⁱ and B(x)=∑_i=0^i=mb_ixⁱ, their Sylvester matrix is a square matrix of size m+n where m=degree(B) and n=degree(A). The m first lines are made with the coefficients of A, so that:

and the n further lines are made with the coefficients of B, so that:

The determinant of a Sylvester polynomial is the resultant of the two polynomials. If A and B have integer coefficients with non-zero resultant r, then the polynomials equation

has a unique solution U,V such that degree(U)<degree(B) and degree(V)<degree(A), and this solution has integer coefficients.

Remark.

The discriminant of a polynomial is the resultant of the polynomial and its derivative.

The sylvester command computes Sylvester matrices.

• sylvester takes two arguments: P and Q, two polynomials.
• sylvester(P,Q) returns the Sylvester matrix of P and Q.

The resultant command computes the resultant of two polynomials.

• resultant takes three arguments:
  • P and Q, two polynomials.
  • x, a variable.
• resultant(P,Q,x) returns the resultant of P and Q.

Example

Examples using the resultant.

1. Let F₁ and F₂ be two fixed points in the plane and A be a variable point on the circle with center F₁ and radius 2a. Find the cartesian equation of the set of points M, intersection of the line F₁A and of the perpendicular bisector of F₂A.

   Geometric solution.

   Since |MF₁|+|MF₂|=|MF₁|+|MA|=|F₁A|=2a, the point M is on an ellipse with focus F₁,F₂ and major axis 2a.

   Analytic solution.

   In the Cartesian coordinate system with center F₁ and x-axis having the same direction as the vector F₁F₂, the coordinates of A are:

   A=(2acos(θ),2asin(θ))

   where θ is the (Ox,OA) angle. Now choose t=tan(θ/2) as parameter, so that the coordinates of A are rational functions with respect to t. More precisely:

   A=(ax,ay)=

   ⎛ ⎜ ⎜ ⎝

   2a

   1−t2 1+t2

   ,2a

   2t 1+t2

   ⎞ ⎟ ⎟ ⎠

   .

   If |F₁F₂|=2c and if I is the midpoint of AF₂, then since the coordinates of F₂ are F₂=(2c,0), the coordinates of I are

   I=(c+ax/2;ay/2)=

   ⎛ ⎜ ⎜ ⎝

   c+a

   1−t2 1+t2

   ;a

   2t 1+t2

   ⎞ ⎟ ⎟ ⎠

   .

   IM is orthogonal to AF₂, hence M=(x;y) satisfies the equation eq₁=0 where

   eq1:=(x−ix)(ax−2c)+(y−iy)ay.

   But M=(x,y) is also on F₁A, hence M satisfies the equation eq₂=0 where

   eq2:=y/x−ay/ax.

   The resultant of both equations with respect to t, resultant(eq1,eq2,t), is a polynomial eq₃ depending on the variables x,y, independent of t which is the cartesian equation of the set of points M when t varies.

   To obtain the resultant:

   ax:=2*a*(1-t^2)/(1+t^2);ay:=2*a*2*t/(1+t^2); ix:=(ax+2*c)/2; iy:=(ay/2); eq1:=(x-ix)*(ax-2*c)+(y-iy)*ay; eq2:=y/x-ay/ax; factor(resultant(eq1,eq2,t))

   −64(x2+y2)(x2 a2−x2 c2+−2 x a2 c+2 x c3−a4+2 a2 c2+a2 y2−c4)

   The factor −64 (x²+y²) is always different from zero, hence the locus equation of M:

   x2a2−x2c2+−2xa2c+2xc3−a4+2a2c2+a2y2−c4=0.

   If the frame origin is O, the middle point of F₁F₂, then this is the cartesian equation of an ellipse. To make the change of origin F₁M=F₁O+OM:

   normal(subst(x^2*a^2-x^2*c^2+-2*x*a^2*c+2*x*c^3-a^4+2*a^2*c^2+a^2*y^2-c^4, [x,y]=[c+X,Y]))

   X2 a2−X2 c2+Y2 a2−a4+a2 c2

   or if b²=a²−c²:

   normal(subst(-c^2*X^2+c^2*a^2+X^2*a^2-a^4+a^2*Y^2,c^2=a^2-b^2))

   X2 b2+Y2 a2−a2 b2

   that is to say, after division by a²b², M satisfies the equation

   X2 a2

   +

   Y2 b2

   =1.

2. Let F₁ and F₂ be fixed points and A a variable point on the circle with center F₁ and radius 2a. Find the cartesian equation of the hull of D, the segment bisector of F₂A.

   The segment bisector of F₂A is tangent to the ellipse of focus F₁,F₂ and major axis 2a.

   In the Cartesian coordinate system with center F₁ and x-axis having the same direction as the vector F₁F₂, the coordinates of A are:

   A=(2acos(θ);2asin(θ)),

   where θ is the (Ox,OA) angle. Choose t=tan(θ/2) as parameter such that the coordinates of A are rational functions with respect to t. More precisely:

   A=(ax;ay)=

   ⎛ ⎜ ⎜ ⎝

   2a

   1−t2 1+t2

   ;2a

   2t 1+t2

   ⎞ ⎟ ⎟ ⎠

   .

   If |F₁F₂|=2c and I is the midpoint of AF₂:

   F2=(2c,0),   I=(c+ax/2;ay/2)=

   ⎛ ⎜ ⎜ ⎝

   c+a

   1−t2 1+t2

   ;a

   2t 1+t2

   ⎞ ⎟ ⎟ ⎠

   .

   Since D is orthogonal to AF₂, the equation of D is eq₁=0 where

   eq1:=(x−ix)(ax−2c)+(y−iy)ay.

   So, the hull of D is the locus of M, the intersection point of D and D′ where D′ has equation eq₂:=d/dteq₁=0.

   To obtain the resultant:

   ax:=2*a*(1-t^2)/(1+t^2);ay:=2*a*2*t/(1+t^2); ix:=(ax+2*c)/2; iy:=(ay/2); eq1:=normal((x-ix)*(ax-2*c)+(y-iy)*ay); eq2:=normal(diff(eq1,t)); factor(resultant(eq1,eq2,t))

   −64a2(x2+y2)(x2 a2−x2 c2+−2x a2c+ 2x c3−a4+2a2c2+a2y2−c4)

   The factor −64 a²(x²+y²) is always different from zero, therefore the locus equation is:

   x2a2−x2c2−2xa2c+2xc3−a4+2a2c2+a2y2−c4=0

   If O, the midpoint of F₁F₂, is chosen as origin, you find again the cartesian equation of the ellipse:

   X2 a2

   +

   Y2 b2

   =1.