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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.

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

where $n = m^2+1$ and $(x_i, y_i) \neq (x_j, y_j)$ for $i \neq j$

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 $i \neq j$?

(Cross)

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  • $\begingroup$ I assume that $V$ is over a field $F$. Let $v\in\mathbb F^n$ be any vector. Then $Vv=0$ iff $aVv=0$ for each $a\in F^{n-1}$, that is if $v$ is orthogonal to any row of the matrix $aV$. The latter holds iff $v$ is orthogonal to any row $(P(x_1,y_1),P(x_2,y_2),\dots,P(x_n,y_n))$, where $P(x,y)=\sum_{i,j}^{m-1} a_{i,j} x^iy^j$, and all $a_{i,j}$ belong to $F$. $\endgroup$ Commented Sep 24, 2023 at 11:06

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