# Generic deformation of matrix

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.

• 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! – Ilya Bogdanov Nov 19 '18 at 22:54
• @IlyaBogdanov That is correct. One has to choose the coeffiecients iteratively to preserve rank. My pardon, going to change the question. – user131561 Nov 19 '18 at 23:08