proof of the determinant condition for a sequence of vectors


Let x1,x2, be a sequence of d dimensional vectors. Assume that there is C:NdR{0} such that

n1++nd=n0<n1<<ndC(n1,,nd)det[xn1,xn2,,xnd]=0 (1)

for every nN. Then det[xn1,xn2,,xnd]=0 for all (n1,,nd)Nd.


Introduce a linear order over the set of ordered tuples: (n1,n2,,nd)(n^1,n^2,,n^d) if (i=1dni,n^d,n^d-1,,n^1) precedes (i=1dn^i,nd,nd-1,,n1) lexicographically. Let (n1,n2,,nd) be the minimal (according to the above order) ordered tuple for which

det[xn1,xn2,,xnd]0. (2)

Take another ordered tuple, (n^1,n^2,,n^d), such that i=1dni=i=1dn^i. By minimality, if (nd,nd-1,,n1) precedes (n^d,n^d-1,,n^1) lexicographically then det[xn^1,xn^2,,xn^d]=0. Otherwise, let i{0,1,,d-1} be the first index such that nd-in^d-i (more specifically, nd-i>n^d-i). Then, n^d-j=nd-j for j=0,,i-1 and n^d-j<nd-i for j=i,,d-1. Therefore,


for all m=1,2,,d (some because of repeated columns and the others because j=1dnj-nd-i+n^m<j=1dnj). Since the vectors xn1,xn2,,xnd are linearly independentMathworldPlanetmath, we get that


In particular det[xn^1,xn^2,,xn^d]=0. Therefore, (1) reduces to det[xn1,xn2,,xnd]=0 which contradicts (2).

Title proof of the determinantMathworldPlanetmath condition for a sequence of vectors
Canonical name ProofOfTheDeterminantConditionForASequenceOfVectors
Date of creation 2013-03-22 14:33:46
Last modified on 2013-03-22 14:33:46
Owner GeraW (6138)
Last modified by GeraW (6138)
Numerical id 5
Author GeraW (6138)
Entry type Proof
Classification msc 15A15