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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 5:51 PM
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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.

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


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