Why doesn't small perturbations of a matrix decrease its rank?

I want to show that if $F:\mathbb{R^n}\to\mathbb{R^m}$ has a derivative of rank = r at a point $p$, then there is a neighbourhood of that point where the rank of the function doesn't decrease, i.e. $\exists N=B(p, \delta)$ such that $rk(DF_x) \geq rk(DF_p) ~ \forall x\in N$

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I don't think the result is even true unless you assume $F$ has a continuous derivative at $x$ (indeed, I think there are 1-dimensional counterexamples). – Darsh Ranjan Sep 27 '10 at 18:49

Use the continuity of the determinant. If the rank is $r$ the Jacobian matrix has an $r$-by-$r$ submatrix $A$ with nonzero determinants. Peturbing the point will perturb the Jacobian matrix and so $A$. As the determinant is a continuous function of its entries a small enough perturbation will leave the new $A$ having nonzero determinant, and so the new Jacobian matrix will have rank $\ge r$.

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Thank you. I was trying to devise a continuous map $Rank_{x,F}:\mathbb{R^n} \to \mathbb{Z}$ to solve this problem. – doxian Sep 27 '10 at 13:35

The "why doesn't ... decrease its rank" is because it is hard to decrease rank. The rank is almost everywhere at its highest possible value, with special exceptions that are destroyed by perturbation.

There will be a maximum rank $r(f) \quad$ that holds at almost all points, and possibly some lower-rank points contained in lower-dimensional pieces of the domain of $f$.

Lower rank is an unstable property. Perturbation away from a low-rank point destroys low rank and replaces it by maximum rank, unless you perturb in special directions that remain inside the lower-dimensional locus where the rank is below maximum. At the "generic point", the rank is the highest one possible, $r(f)$, and for any point, all or (in the case of low-rank points, almost all) of the nearby points to which one can perturb, will be generic.

As in Robin Chapman's answer this is because low rank requires some functions (determinants) computed from $Df$ to equal zero, while maximum rank requires them to be nonzero. The first property is unstable and the second is stable with respect to perturbation. It is also true that the rank cannot decrease through a small perturbation, because the set of determinants that are nonzero can only expand as one moves to a close enough nearby point.

http://en.wikipedia.org/wiki/Sard%27s_theorem

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Lewis 2009 (link below) shows that $rank (A(x))$ is lower semi-continuous and says this means $rank(A(x))$ does not decrease in a sufficiently small neighborhood of any point x.

http://www.mast.queensu.ca/~andrew/notes/pdf/2009a.pdf

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