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


Re: Matrices of rank at least k
Posted:
Nov 28, 2012 5:51 PM


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.
But the determinant of a square matrix is a polynomial in the entries.
quasi

