proof that Sylvester's matrix equals the resultant

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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.