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Homeproof that Sylvester's matrix equals the resultant

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# proof that Sylvester’s matrix equals the resultant

In the derivation of Sylvester’s matrix for the resultant, it was seen that if two polynomials 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 $a_{0},r_{1},r_{2},\ldots,r_{m},b_{0},s_{1},s_{2},\ldots,s_{n}$. 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 terms 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 observations 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 $R^{{\prime}}=D^{{\prime}}$.

## Mathematics Subject Classification

13P10*no label found*

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## Comments

## degree mn

I was wondering if it was possible to explain why the degree of D cannot be greater than mn. I am considering the matrix and thinking of Leibniz's formula for working out the determinant (permutations of entries) but I cannot work out why in the general case that the largest possible permutation of entries has degree mn. Thanks

## Re: degree mn

You can indeed use the expansion of the determinant in terms of permutations of entries in order to show that the degree of D is precisely $mn$.

To be clear, let us work with the matrix that starts with the row

\[a_0,\ldots,a_m,0,\ldots,0\]

and progressively shifting the row to the right, $n$ times, and then the same for the $b_i$'s.

This matrix has size $(m+n)\times (m+n)$. Consider a permutation $p_1,\ldots,p_m,p_{m+1},\ldots,p_{m+n}$ of $\{1,\ldots,m+n\}$. The corresponding entry of the determinant is

\[\prod_{i=1}^{m} a_{p_{i}-i} \prod_{j=1}^n b_{p_{m+j}-j}. \]

Since $a_k$ and $b_k$ have degree $k$ as symmetric function of the roots, this term has total degree

\begin{eqnarray*}

\sum_{i=1}^m p_{i}-i +\sum_{j=1}^n p_{m+j}-j &=& \left(\sum p_i \right) - m(m+1)/2 - n(n+1)/2 \\

&=& (m+n)(m+n+1)/2 - m(m+1)/2 - n(n+1)/2 \\

&=& mn

\end{eqnarray*}

Thus every nonzero term in the determinant has degree $mn$.

Hope this helps.