quasi
Posts:
12,067
Registered:
7/15/05


Re: Matrices of rank at least k
Posted:
Nov 28, 2012 6:01 PM


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_(k1) 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

