Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.