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Kaba
Posts:
289
Registered:
5/23/11
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Re: Matrices of rank at least k
Posted:
Nov 29, 2012 9:44 AM
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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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