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I am working on moving least square approximation. To well-define of this method, the following matrix of size $n \times m$ must be of rank $m$. For $m=3$ $${\bf A }=\begin{pmatrix} 1&x_1&y_1\\ 1&x_2&y_2\\ \vdots & \vdots & \vdots\\ 1&x_n&y_n \end{pmatrix},$$ where $x_i,y_i$ are point's coordinates lie in the supprt and $n$ is the number of them. Also, for $m=6$ $${\bf A }=\begin{pmatrix} 1&x_1&y_1&x_1^2&x_1y_1&y_1^2\\ 1&x_2&y_2&x_2^2&x_2y_2&y_2^2\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1&x_n&y_n&x_n^2&x_n y_n&y_n^2 \end{pmatrix}.$$ If for the first one, there are only $3$ points in the support and these points are the vertices of a triangle, can we sure that $rank(A)=3?.$ Does anyone what the best strategy for finding support size is? For uniform distributed points usually we can take $r=1.1*\delta$ for $m=3$ and $r=2.1*\delta$ for $m=6$, where $\delta$ is the distance between points. But for a mesh obtained by the adaptive technique or non-uniform this is an important thing that should be choose carefully not too large and not too small.

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    $\begingroup$ These matrices are called Vandermonde matrices. Their rank depends on the geometric configuration of the points; this is a standard issue in constructing finite elements (and you might wish to look at, e.g., Ciarlet's book). (For $m=3$, the vertices of a -- nondegenerate -- triangle indeed lead to full rank.) I'm really not sure what the support has to do with it, though -- the support of what? $\endgroup$ – Christian Clason Jun 29 '16 at 21:25
  • $\begingroup$ Thank you for your comment, the support of the weight function in the MLS approximation should be contained at least $m$ points. $\endgroup$ – Rosa Jun 30 '16 at 6:03
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    $\begingroup$ But the rank of the Vandermonde matrix only depends on the geometric configuration, not the actual distance (although the condition number does). So you just pick your points, say, inside the unit simplex, and then use an affine transformation to scale them down to fit inside the given support. That's exactly how it's done in finite element methods. $\endgroup$ – Christian Clason Jun 30 '16 at 6:57

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