quasi
Posts:
10,967
Registered:
7/15/05


Re: Matrices of rank at least k
Posted:
Nov 28, 2012 6:05 PM


quasi wrote: >quasi 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 it, but there surely is a >polynomial P_k  just not the one I described above.
Ok, here's the fix ...
An m x n matrix A has rank <= k
iff every (k+1) x (k+1) submatrix of A has determinant 0,
iff the sum of the squares of the determinants of all (k+1) x (k+1) submatrices of A has determinant 0,
quasi

