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On the wikipedia page "Overdetermined System," I came across the claim that an overdetermined system is almost always inconsistent when constructed with random coefficients. Is anyone familiar with a proof for this or somewhere that I could read more about it? I am interested in the case of linear systems with real coefficients.

Matt R
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  • For an overdetermined system to have a solution, everything needs to happen just right. It's like throwing three sticks up in the air and having all of them come down on the same place like a six-spoked wheel, or flipping a coin and having it come down on its edge. – Neal Aug 19 '20 at 20:58
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    We need to be careful with what we mean by "random." This is true whenever coefficients are selected following an absolutely continuous distribution as a consequence of the fact that the set of matrices with deficient rank has Lebesgue measure $0$. – Ben Grossmann Aug 19 '20 at 21:00
  • To show that the set in question has measure zero, it suffices to note that the set of rank-deficient matrices is the common zero set to a set of polynomials (namely matrix minors of maximal size). From there, this argument suffices. – Ben Grossmann Aug 19 '20 at 21:10

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By Rouche-Capelli’s theorem the system Ax=b is inconsistent iff rkA <rk(A|b). Since if the joint distribution of the entries of a matrix is absolutely continuous, it is known that it has full rank almost surely (a.s.), it follows that for a matrix A of dimensions mxn, m>n, rkA=n=min(m,n)<m a.s. and rk(A|b)=n+1=min(n+1,m)<=m a.s. Hence the two ranks are different almost surely. That an absolutely continuous matrix has full rank a.s. follows from the set of zeroes of a non-constant polynomial having Lebesgue measure zero. If M is the set of minors of A of size n, for example, the polynomial in the entries of A constructed by adding up all the squares of these minors’ determinants is a non constant polynomial, as A varies. A is rankdeficient iff this polynomial is zero, hence it is rank deficient with its entries belonging to a Lebesgue set of measure zero. Hence with probability zero by absolute continuity. So the matrix is almost surely full rank. Same goes for (A|b) with minor adjustments.

xyz
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