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Topic:
Matrices of rank at least k
Replies:
12
Last Post:
Nov 29, 2012 1:15 PM




Re: Matrices of rank at least k
Posted:
Nov 29, 2012 10:07 AM


Am 29.11.12 09:08, schrieb Robin Chapman: > 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).
Using the determinant is probably the most efficient argument, but I find it interesting, that it an argument which is closer to the definition of rank is also possible.
A matrix is of rank less than k if for all sets of k columns there is a linear dependency between these columns. A linear dependency is given by an equality involving a nonzero vector in R^k, and we can assume length 1, i.e. a point in the (k1)sphere. The matrices of rank less than k together with the linear dependencies proving that they are of rank less than k are therefore a closed subset of the product of R^{m x n} and n choose k (k1)spheres. Since these spheres are compact, the projection of this set to R^{m x n} is also closed.



