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

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Carsten Schultz

Posts: 3
Registered: 11/29/12
Re: Matrices of rank at least k
Posted: Nov 29, 2012 10:07 AM
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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 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).


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 non-zero vector in R^k, and we can assume
length 1, i.e. a point in the (k-1)-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 (k-1)-spheres. Since these spheres are compact, the
projection of this set to R^{m x n} is also closed.





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