Date: Nov 28, 2012 6:18 PM
Author: Kaba
Subject: Re: Matrices of rank at least k

29.11.2012 1:05, quasi wrote:
> An m x n matrix A has rank <= k
>
> iff every (k+1) x (k+1) submatrix of A has determinant 0,
>
> iff the sum of the squares of the determinants of all
> (k+1) x (k+1) submatrices of A has determinant 0,


That sounds correct, and finishes the proof. Thanks:)

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