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proof that Sylvester's matrix equals the resultant

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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

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.

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