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Kaba
Posts:
289
Registered:
5/23/11


Re: Matrices of rank at least k
Posted:
Nov 29, 2012 9:44 AM


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 atleastk 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 subdeterminant. Therefore one can pick any open neighborhood for them, and then the product neighborhood gives the required open neighborhood.
 http://kaba.hilvi.org



