# Rank function can change upwards but not downwards: need intuition

Let $f: \mathbb R^n \to \mathbb R^m$ be some smooth map and $J_f$ its Jacobian.

Say $x \in \mathbb R^n$ is such that

$$\operatorname{rank}{(J_f (x))} = p$$

Then there exists a neighbourhood of $x$ on which $\operatorname{rank}{(J_f (x))} \ge p$.

I am trying to understand why this is true. This does not seem to hold for say, an arbitrary smooth map $g$: If $g(x) = y$ then there does not necessarily exists a neighbourhood on which $g \ge y$ -- $x$ could be a global minimum (just consider for example the map $(x_1, x_2) \mapsto x_1^2 + x_2^2$ at the origin).

Please could someone help me gain intuitive understanding of why the map $x \mapsto \operatorname{rank}{(J_f (x))}$ can have a neighbourhood where it is greater equal a certain value but not smaller?

• This isn't really a theorem about smooth maps - it's really a theorem about the rank function on the space $\text{Mat}(\Bbb R^n,\Bbb R^m)$ (that is, that the rank function is upper semi-continuous). Try thinking about the fact that "$\text{rank } A\geq p$" is equivalent to "$A$ has a $p \times p$ minor with nonzero determinant". – user98602 Aug 16 '15 at 2:12
• The lower the rank is, the more exacting conditions you are putting on the value. Getting a lower rank is effectively the same as saying that some function = 0. Since you don't have lower rank at $x$, the value of this function misses 0 by some amount. Since everything changes continuously, you have to move some distance before it can reach 0. – Paul Sinclair Aug 16 '15 at 2:36
• I read both of the comments here but I don't understand yet. I do understand that it's not a fact about smooth maps but I thought that maybe assuming smoothness makes the explanation simpler or could give better intuition. – a student Aug 16 '15 at 6:14
• The relationship to smoothness in your problem is that given a smooth map $f$, then $J_f(p)$ is a matrix that depends continuously on $p$. Beyond that you won't get anything useful, intuitive or otherwise. – user98602 Aug 16 '15 at 7:05
• Okay, I see. But my question remains. I don't understand the comment by Paul Sinclair. – a student Aug 16 '15 at 7:08

When ${\rm rank}\bigl(J_f(a)\bigr)=p$ then all minors of $J_f(a)$ of order $>p$ are zero, but at least one minor $M(a)$ of order $p$ is nonzero. This minor belongs to a particular selection of $p$ coordinate variables in the $x$-domain and $p$ coordinate variables in the $y$-domain. Since $x\mapsto M(x)$ is a continuous function this function is nonzero in a full neighborhood of the point $a$, whence ${\rm rank}\bigl(J_f(x)\bigr)\geq p$ in this neighborhood.
On the other hand it is easily possible that in fact ${\rm rank}\bigl(J_f(x)\bigr)> p$ in points $x$ arbitrarily close to $a$: Consider the example $f: \ (x_1,x_2)\mapsto x_1^2+x_2^2$ you gave. Here $J_f(0,0)=[0 \ \ 0]$ has rank $0$, but $J_f(x_1,x_2)=[2x_1\ \ 2x_2]$ has rank $1$ for all $(x_1,x_2)\ne(0,0)$. It so happens that "by coincidence" both minors $M_1(x)=2x_1$ and $M_2(x)=2x_2$ vanish at the same time at $(0,0)$.