```Date: Nov 29, 2012 9:44 AM
Author: Kaba
Subject: Re: Matrices of rank at least k

29.11.2012 10:08, Robin Chapman wrote: > On 28/11/2012 20:56, Kaba wrote: >> Hi, >> >> An exercise in a book on smooth manifolds asks me to prove that >> (m x n)-matrices (over R) of rank at least k is an open subset of >> R^{m x n} (and thus an open submanifold). It is intuitively clear to me >> why that is true: an arbitrary small perturbation can add one or more to >> the rank of a matrix, but if a matrix is of rank k, then there is a >> small open neighborhood in which the rank stays the same. So I should be >> able to find a small open neighborhood around each at-least-k rank >> matrix which still stays in the set, therefore proving the claim. How do >> I find such a neighborhood? > > A matrix has rank at least k iff it has a nonsingular k by k submatrix > The set of matrices where that particular submatrix is nonsingular > serves as the required open neighbourhood. (It is defined by > the nonvanishing of a determinant).Yep. After finding an open neighborhood for the submatrix by the continuity of the determinant, the other elements do not contribute to this sub-determinant. Therefore one can pick any open neighborhood for them, and then the product neighborhood gives the required open neighborhood.-- http://kaba.hilvi.org
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