Let $A(x)$ be a $m \times n$ matrix, whose entries are real polynomials $f_{i,j}:\mathbb{R}^S \to \mathbb{R}$. Denote the ith row by $f_i$ And let $rk:\mathbb{R}^S \to \mathbb{N}$ be the rank function that assigns to each point $x_0\in \mathbb{R}^S$, the rank of $A(x_0)$. This means, that at the point $x_0$ there are $rk(x_0)$ rows $f_i(x_0)$ which are linearly independent vectors.

Now I consider deformations of $A(x)$ of the following type: I take the first row $f_1$ and I choose $m-1$ small real numbers $\epsilon_{1,2}, \ldots \epsilon_{1,m}$ and substitute the row $f_1$ by $ \tilde{f_1}:= f_1 + \sum_{j} \epsilon_{i,j} f_j$. Then I have a new matrix where the rank function has not changed since I only performed elementary row operations. Now I do the same for the second row (on this new matrix) ... and so on.

Observe, that the function rank is trivially the same for the new matrix. Can I assume that there is a suitable choice of $\epsilon$ numbers such that at each point $x_0$ any subset formed by $rk(x_0)$ rows is a system of linearly independent vectors?

Intuitively, I think that the answer is yes and should follow from some transversality argument. Also I think that this might be done somewhere so I'd rather put a reference than spend half a page of my paper proving a linear algebra result. In any case, I am asking for a reference where this or a similar problem is solved.

  • $\begingroup$ The rank function does not seem to stay the same. If you take some $x$ near the discontinuity point of the initial rank, then the rank at that point may decrease! $\endgroup$ – Ilya Bogdanov Nov 19 '18 at 22:54
  • $\begingroup$ @IlyaBogdanov That is correct. One has to choose the coeffiecients iteratively to preserve rank. My pardon, going to change the question. $\endgroup$ – user131561 Nov 19 '18 at 23:08

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