quasi
Posts:
9,076
Registered:
7/15/05
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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
>entries.
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.
quasi
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