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The kernel of a Vandermonde matrix can be determined using this formula.

The following type of matrix has a similar structure, and should also have a one-dimensional kernel.

$$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 \\ x_1 & x_2 & x_3 & \ldots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \ldots & x_n^{m-1}\\ y_1 & y_2 & y_3 & \ldots & y_n \\ y_1x_1 & y_2x_2 & y_3x_3 & \ldots & y_nx_n \\ y_1x_1^2 & y_2x_2^2 & y_3x_3^2 & \ldots & y_nx_n^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y_1x_1^{m-1} & y_2x_2^{m-1} & y_3x_3^{m-1} & \ldots & y_nx_n^{m-1}\\ y_1^2x_1 & y_2^2x_2 & y_3^2x_3 & \ldots & y_n^2x_n\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y_1^{m-1}x_1^{m-1} & y_2^{m-1}x_2^{m-1} & y_3^{m-1}x_3^{m-1} & \ldots & y_n^{m-1}x_n^{m-1}\\ \end{bmatrix} \in \mathbb{R}^{(n-1)\times n}$$

where $n = m^2+1$ and $(x_i, y_i) \neq (x_j, y_j)$ for $i \neq j$; i.e. there are $m$ groups of $m$ rows with all possible combinations of powers $y^ax^b$ and one more column than rows.

Does a similar analytical form exist for it? Or, would additional constraints be required, like $x_i^ay_i^b \neq x_i^cy_i^d$ for $i \neq j$?

(Crossposted to MO)

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  • $\begingroup$ Shouldn't $n=2m+1$ rather than $n=m^2+1$? $\endgroup$ – Abel Jun 2 '13 at 13:26
  • $\begingroup$ There are $m$ groups of $m$ rows (All possible combinations of powers $x^iy^j$); the matrix should have one more column than rows to give a 1-dimensional kernel, therefore $n=m^2+1$. $\endgroup$ – trion Jun 2 '13 at 14:22
  • $\begingroup$ Ah, I see now. Nevermind. $\endgroup$ – Abel Jun 2 '13 at 14:24
  • $\begingroup$ Edited the question to make it clearer. $\endgroup$ – trion Jun 2 '13 at 14:27
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    $\begingroup$ Questions: have you tried using Lagrange interpolation, as in the regular proof of the kernel of the van der monde matrix? Also, have you tried getting people on mathematica.stackexchange to experiment with different $m$? $\endgroup$ – Brian Rushton Jun 8 '13 at 14:32

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