proof that Sylvester’s matrix equals the resultant

In the derivationPlanetmathPlanetmath of Sylvester’s matrix for the resultantMathworldPlanetmath, it was seen that if two polynomialsMathworldPlanetmathPlanetmathPlanetmath have a common root, then Sylvester’s determinant will equal zero. Since two polynomials have a common root if and only if their resultant is zero, it follows that if the resultant is zero, then Sylvester’s determinant equals zero. In this entry, we shall use this fact to show that Sylvester’s determinant equals the resultant.

The secret is to view both Sylvester’s determinant and the resultant as functions of the roots. A more precise way of saying what this means is that we will study polnomials in the indeterminates a0,r1,r2,,rm,b0,s1,s2,,sn. We will regard a1,a2,,am,b1,b2,,bm as polynomials in these variables using the expression of coefficients of a polynomial as symmetric functions of its roots, e.g.


Note that ak is a kth order polynomial in the ri’s and bk is a kth order polynomial in the si’s.

Let R be the polynomial


and let D be the polynomial which is gotten by replacing occurrences of a1,a2,,am,b1,b2,,bm in Sylvester’s matrix by their expressions in of a0,r1,r2,,rm,b0,s1,s2,,sn. We want to show that R=D.

First, note that, in each row of Sylvester’s matrix, every entry is multiplied by either an a0 or a b0. By a fundamental property of determinants, this means that we may pull all those factors of a0 and b0 outside the determinant. Since there are n rows containing a0 and m rows containing b0, this means that D=a0mb0nD(r1,,rm,s1,,sn). Note that these factors correspond to the powers of a0 and b0 in the definition of R. Hence, to show that D=R it only remains to show that D=R, where


Second, note that the degree of D is not greater than the degree of R. From the definition, it is obvious that R is a polynomial of degree mn. By examining Sylvester’s determinant and keeping in mind that ak and bk are of degree k, it is not hard to see that the degree of D cannot exceed mn.

Third, we will show that R divides D. In the derivation of the Sylvester determinant, we saw that if ri=sj for some choice of i and j, then D=0, and hence D=0. The only way for a non-zero polynomial to to equal zero when ri=sj is for ri-si to be a factor of the polynomial. It is easy to see that D is not the zero polynomialMathworldPlanetmath, and hence, si-sj must be a factor of D. This means that every factor of R is also a factor of D. Since all the factors of R occur with multiplicityMathworldPlanetmath one, it follows that D is a multipleMathworldPlanetmath of R.

Combining the of the last two paragraphs, we come to the conclusionMathworldPlanetmath that D must be a constant multiple of R. To determine the constant of proportionality, all one needs to do is to compare the values of the two polynomials for a set of value of the variables for which they to not vanish. For instance, one could try r1=r2==rm=1 and s1=s2==sn=0. Both R and D equal 1 for this special set of values, and hence R=D.

Title proof that Sylvester’s matrix equals the resultant
Canonical name ProofThatSylvestersMatrixEqualsTheResultant
Date of creation 2013-03-22 14:36:50
Last modified on 2013-03-22 14:36:50
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 8
Author rspuzio (6075)
Entry type Definition
Classification msc 13P10