25 Nov
2006
25 Nov
'06
3:35 p.m.
There's a construction using Vandermonde determinants: look at the Vandermonde curve t -> ( t, t^2, t^3, ..., t^n) in R^n, or for symmetry of computation, put it in a hyperplane in R^(n+1), t -> ( 1, t, t^2, t^3, ..., t^n). For any n+1 distinct points, the determinant of the associated (n+1)X (n+1) matrix is non-zero. This determinant is the volume of the simplex spanned by the points times (n+1)! Therefore, for k <= n, any k points on the Vandermonde curve are affinely independent. Bill On Nov 25, 2006, at 11:29 AM, Fred lunnon wrote:
Alternatively, we might ask for an explicit d x oo matrix such that every subset of d rows has maximum rank. I can't see how to do this off- hand.