quasi
Posts:
9,076
Registered:
7/15/05
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Re: Matrices of rank at least k
Posted:
Nov 28, 2012 6:05 PM
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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_(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 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
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