# proof that Sylvester’s matrix 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 $a_{0},r_{1},r_{2},\ldots,r_{m},b_{0},s_{1},s_{2},\ldots,s_{n}$. We will regard $a_{1},a_{2},\ldots,a_{m},b_{1},b_{2},\ldots,b_{m}$ as polynomials in these variables using the expression of coefficients of a polynomial as symmetric functions of its roots, e.g.

 $a_{1}=a_{0}(r_{1}+r_{2}+\cdots)$
 $a_{2}=a_{0}(r_{1}r_{2}+r_{1}r_{3}+\cdots)$

Note that $a_{k}$ is a $k^{\hbox{th}}$ order polynomial in the $r_{i}$’s and $b_{k}$ is a $k^{\hbox{th}}$ order polynomial in the $s_{i}$’s.

Let $R$ be the polynomial

 $R=a_{0}^{n}b_{0}^{m}\prod_{i=1}^{m}\prod_{j=1}^{n}(r_{i}-s_{j})$

and let $D$ be the polynomial which is gotten by replacing occurrences of $a_{1},a_{2},\ldots,a_{m},b_{1},b_{2},\ldots,b_{m}$ in Sylvester’s matrix by their expressions in of $a_{0},r_{1},r_{2},\ldots,r_{m},b_{0},s_{1},s_{2},\ldots,s_{n}$. We want to show that $R=D$.

First, note that, in each row of Sylvester’s matrix, every entry is multiplied by either an $a_{0}$ or a $b_{0}$. By a fundamental property of determinants, this means that we may pull all those factors of $a_{0}$ and $b_{0}$ outside the determinant. Since there are $n$ rows containing $a_{0}$ and $m$ rows containing $b_{0}$, this means that $D=a_{0}^{m}b_{0}^{n}D^{\prime}(r_{1},\ldots,r_{m},s_{1},\ldots,s_{n})$. Note that these factors correspond to the powers of $a_{0}$ and $b_{0}$ in the definition of $R$. Hence, to show that $D=R$ it only remains to show that $D^{\prime}=R^{\prime}$, where

 $R^{\prime}=\prod_{i=1}^{m}\prod_{j=1}^{n}(r_{i}-s_{j})$

Second, note that the degree of $D^{\prime}$ is not greater than the degree of $R^{\prime}$. From the definition, it is obvious that $R^{\prime}$ is a polynomial of degree $mn$. By examining Sylvester’s determinant and keeping in mind that $a_{k}$ and $b_{k}$ are of degree $k$, it is not hard to see that the degree of $D^{\prime}$ cannot exceed $mn$.

Third, we will show that $R^{\prime}$ divides $D^{\prime}$. In the derivation of the Sylvester determinant, we saw that if $r_{i}=s_{j}$ for some choice of $i$ and $j$, then $D=0$, and hence $D^{\prime}=0$. The only way for a non-zero polynomial to to equal zero when $r_{i}=s_{j}$ is for $r_{i}-s_{i}$ to be a factor of the polynomial. It is easy to see that $D^{\prime}$ is not the zero polynomial  , and hence, $s_{i}-s_{j}$ must be a factor of $D^{\prime}$. This means that every factor of $R^{\prime}$ is also a factor of $D^{\prime}$. Since all the factors of $R^{\prime}$ occur with multiplicity  one, it follows that $D^{\prime}$ is a multiple  of $R^{\prime}$.

Combining the of the last two paragraphs, we come to the conclusion  that $D^{\prime}$ must be a constant multiple of $R^{\prime}$. 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 $r_{1}=r_{2}=\cdots=r_{m}=1$ and $s_{1}=s_{2}=\cdots=s_{n}=0$. Both $R^{\prime}$ and $D^{\prime}$ equal $1$ for this special set of values, and hence $R^{\prime}=D^{\prime}$.

Title proof that Sylvester’s matrix equals the resultant ProofThatSylvestersMatrixEqualsTheResultant 2013-03-22 14:36:50 2013-03-22 14:36:50 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Definition msc 13P10