Of $$\prod_{1 \leq i < j \leq n} (x_i - x_j)$$

Taking into account the matrix $$\begin{pmatrix}1 & x_1 & x_1^2 & \cdots & x_1^n \\ 1& x_2 & x_2^2 & \cdots & x_2^n\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & x_n^2 & \cdots & x_n^n\end{pmatrix}$$

I have found that, for $n=3$ the multiplication gives $x_1 x_2 (x_1 - x_2) + x_2x_3(x_2-x_3)+x_1x_3(x_1-x_3)$ which uses all the combinations for $n=3$. Also, I've found that the order of the result will be $n-1$ and that I will get $n!$ monomials. I don't know how to find the answer though.

  • 1
    $\begingroup$ See math.stackexchange.com/questions/1247286/… $\endgroup$
    – reuns
    Commented Apr 10, 2017 at 9:13
  • $\begingroup$ I didn't even know there was a name for this! Thank you so much! $\endgroup$
    – The Bosco
    Commented Apr 10, 2017 at 9:26
  • $\begingroup$ If $x_i$ are the roots of some polynomial, it is $\delta = \sqrt{\Delta}$ the discriminant, important in Galois theory (field extensions) $\endgroup$
    – reuns
    Commented Apr 10, 2017 at 9:31

1 Answer 1


Call the Vandermonde determinant:

$$V_n(x_1,x_2,\ldots,x_n)= \left\lvert \begin{matrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1& x_2 & x_2^2 & \cdots & x_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1}\end{matrix} \right\rvert$$

Then if we express this via the cofactor expansion of the bottom row it is a polynomial in $x_n$ with $V_{n-1}(x_1,x_2,\ldots,x_{n-1})$ being the coefficient of $x_n^{n-1}$ .

We can also see that replacing $x_n$ with any of $x_1,x_2,\ldots,x_{n-1}$ produces a determinant with two equal rows so they must be the $n$ roots of our polynomial.

Hence we can write the polynomial in terms of its roots


then solving the reccurrence by iterating $V_{n-1}(x_1,x_2,\ldots,x_{n-1})$ gives the desired product in the question.

This determinant comes in handy for a proof of the hook length formula for Standard Young's Tableaux.


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