The matrix ( V ) in question is a Vandermonde-like matrix constructed from bivariate monomials ( x_i^b y_i^a ) for ( 0 \leq a, b \leq m-1 ), with ( n = m^2 + 1 ) columns (points ( (x_i, y_i) )) and ( m^2 ) rows. To determine its kernel, we analyze the linear independence of the evaluated monomials.

Key Analysis:

• The kernel of ( V ) consists of vectors ( \mathbf{c} = (c_1, \ld...

Think of your matrix as the evaluation matrix of the bivariate polynomial space Π = span{x^b y^a : 0 ≤ a ≤ m−1, 0 ≤ b ≤ m−1}, which has dimension m^2. Your matrix V has one column per point (x_i, y_i) and one row per monomial in Π, so V ∈ R^{m^2 × (m^2+1)}.

1. Dimension of the kernel

• For “generic” points (x_i, y_i) (i.e., away from an algebraic exceptional set), rank(V) = m^2, hence dim ker(V) ...

Looking at this matrix, I notice it has a very specific structure related to bivariate polynomial interpolation. Let me analyze its kernel.

The matrix has dimensions , so it has a one-dimensional kernel by the rank-nullity theorem (assuming it has full rank ).

To find the kernel, we need to find a non-zero vector such that $...

<think> We are given a matrix V with dimensions (n-1) x n, where n = m^2 + 1. The matrix has the following structure:

• The first m rows are the standard Vandermonde rows: [1, 1, ..., 1], [x_1, x_2, ..., x_n], [x_1^2, x_2^2, ..., x_n^2], ..., [x_1^{m-1}, x_2^{m-1}, ..., x_n^{m-1}].
• The next m rows are multiplied by y_i: [y_1, y_2, ..., y_n], [y_1 x_1, y_2 x_2, ..., y_n x_n], ..., [y_1 x_1^{m-1}...

The short answer is that nothing essentially new is hiding here – one still gets a one–dimensional null‐space and it is (up to scale) given by the vector of cofactors (minors) of the ((n!-!1)\times n) matrix. Equivalently, the null–vector is just the list of coefficients of the unique bivariate polynomial of bi‐degree (\le(m-1,m-1)) vanishing on your (m^2+1) points.

1. Cofactor‐formula. ...

Let

 Pₘ := { p(x ,y)=∑{0≤a,b≤m−1} c{ab} x^{b}y^{a} } (dim Pₘ = m²)

be the space of bivariate polynomials whose separate degree in x and y is ≤ m−1. For the m² monomials that form the standard basis of Pₘ we put them one under another; after a harmless re–ordering the (m²)-by-n matrix that evaluates these monomials at the n points (x₁ ,y₁), … ,(x_n ,y_n) is exactly the matrix ...

Kernel of a Vandermonde-like Matrix

The problem asks about finding the kernel of a matrix with entries formed from powers of two variables , where . This is an interesting generalization of the standard Vandermonde matrix case.

Analysis of the Matrix Structure

First, let's understand what we're working with:

• The matrix has r...

An excellent and insightful question. Yes, a similar analytical form for the kernel exists, and it's a beautiful generalization of the 1D Vandermonde case. The key is to re-interpret the problem in terms of polynomial interpolation in two variables.

Summary of the Result

Let's first clarify the structure of your matrix V. Based on your description, the matrix is (m^2) x (m^2 + 1). Let `n...

Let

P = {(x1 ,y1 ), … , (xn ,yn )}, n = m^2+1 , m ≥ 1
V = ( vαβ (j) ) ∈ ℝm^2×n , vαβ (j) = xjβ yjα , 0 ≤ α,β ≤ m−1.

(So the rows are all monomials yα xβ with α,β < m; there are m^2 rows.)

Goal
Describe Ker V and decide when it is one–dimensional.

1 Row rank and the geometric condition ——————————————————————————— Let

Πm = { bivariate polynomials p(x,y)=∑ α,β< m cαβ xβ yα...