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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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Posts: 12,067
Registered: 7/15/05
Re: Matrices of rank at least k
Posted: Nov 28, 2012 6:01 PM
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On Wed, 28 Nov 2012 17:51:37 -0500, quasi <quasi@null.set> wrote:

>Kaba wrote:
>>quasi wrote:
>>> Let m,n be positive integers, and let k be an integer with
>>> 0 <= k <= min(m,n). The set T_k of m x n matrices of
>>> rank <= k is easily seen to be closed since, for each k,
>>> there is a polynomial P_k in m*n variables with real
>>> coefficients such that an m x n matrix A with real entries
>>> satisfies the condition rank(A) <= k iff the coefficients
>>> of A satisfy P_k = 0. Regarding P_k as a function from
>>> R^(mxn) to R, P_k is continuous, hence ((P_k)^(-1))(0)
>>> is closed. It follows that T_k is closed for all k. In
>>> particular, for each k, T_(k-1) is closed, and thus,
>>> the set of matrices with rank >= k is open.

>>Sounds good. But how to prove the existence of the
>>polynomials P_k?

>An m x n matrix A has rank <= k
> iff some k x k submatrix of A has determinant 0,
> iff the product of the determinants of all k x k
> submatrices of A is equal to 0.

No, that's wrong -- sorry.

It's something vaguely like that, but not what I said.

>But the determinant of a square matrix is a polynomial in the

I don't have time now to fix my it, but there surely is a
polynomial P_k -- just not the one I described above.


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