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.

--

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