# A question about the Vandermonde determinant

We know that the Vandermonde determinant of order $n$ is the determinant defined as follows:

$$\begin{vmatrix} 1&x_1&x_1^2&\dots&x_1^{n-1}\\ 1&x_2&x_2^2&\dots&x_2^{n-1}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&x_n&x_n^2&\dots&x_n^{n-1}\\ \end{vmatrix}=\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$

For any or some special nonempty subset $S\subseteq \{(i,j)\mid1\leq i<j\leq n\}$，does there exist some matrix $A_S$ which is similar to the Vandermonde matrix (for example, every element of $A_S$ is something just like ${x_i}^j$) such that $$\begin{vmatrix} A_S \end{vmatrix}=\prod\limits_{(i,j)\in S}(x_j-x_i)$$ $$\text{or}$$$$\begin{vmatrix} A_S \end{vmatrix}\neq 0 \quad\text{if and only if}\prod\limits_{(i,j)\in S}(x_j-x_i)\neq 0\quad?$$

• How is $A_S$ derived from $A$ and $S$? Are the values for indices not in $S$ supposed to be zero? – Pushpendre Jul 2 '15 at 5:24

It is not what you are looking for, I'm sure, but there is a matrix which is $N \times N$ for $N=2|S|.$ It is almost entirely zeros except that on the main diagonal there are $2 \times 2$ blocks $$\begin{vmatrix} 1&x_j\\ 1&x_i \end{vmatrix},$$ one for each desired term $x_i-x_j.$
There are certainly more compact solutions for some choices of $S$, such as the Vandermonde case of all pairs.
• Thank you very much for your help!You said:"There are certainly more compact solutions for some choices of S." Would you please give some examples for such $S$. – user173856 Jul 4 '15 at 16:09
• If you want the same result as the Vandermonde determinant, this method uses a much bigger determinant. $n^2-n$ instead of $n.$ – Aaron Meyerowitz Jul 9 '15 at 0:29