Rank of derivative matrix of vector-valued function

Suppose I have a function $$f : \mathbb R^n \to \mathbb R^m$$ and I consider the derivative matrix $$Df = \begin{bmatrix}\frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n} \end{bmatrix}.$$

What can I say about the rank of $$Df$$? In particular, I'm very interested in determining the invertibility of $$Df$$ when $$n=m$$.

I'm new to these things that connect calculus and linear algebra and I'm really not sure how to even go about approaching this other than by looking at examples. Like I was wondering if $$f$$ is a bijection, maybe with a continuous inverse, then does that imply $$Df$$ is invertible?

Some examples I've looked at: if $$f(x) = Ax$$ for some matrix $$A \in \mathbb R^{m\times n}$$ with rows which I'll denote by $$a_1,\dots,a_m$$, then $$(Df)_{ij} = \frac{\partial a_i^Tx}{\partial x_j} = a_{ij}$$ so $$Df = A$$ therefore $$\text{rank }Df = \text{rank } A$$.

Another example: $$n=m=2$$ and $$f(x,y) = (x^2,xy)$$. Then $$Df = \begin{bmatrix}2x & 0 \\ y & x\end{bmatrix}$$ so $$\det Df = 2x^2$$ so $$Df$$ is invertible iff $$x\neq 0$$.

In the most general case, you can a priori not say anything about the rank of $$Df$$ but the usual properties as $$1 \leq \operatorname{rank}(Df) \leq \min \{m, n \}$$.

Suppose $$m = n$$. What you always have is the inverse function theorem: If $$Df$$ is continuous and invertible at some point $$p \in \mathbb{R}^n$$, then $$f$$ is locally invertible at $$p$$. With your $$f(x, y) = (x^2, xy)$$ one concludes that $$f$$ is locally invertible at all points $$(x, y)$$ with $$x \neq 0$$. Note that the inverse function theorem does not ensure global invertibility, a counterexample is the function $$g(x, y) = (\mathrm{e}^x \cos(y), \mathrm{e}^x \sin(y))$$.

Concerning your other question, consider

$$h: [0, 2] \longrightarrow [0, 3],\ x \longmapsto \begin{cases} x & ,0 \leq x < 1 \\ 2x - 1 & ,1 \leq x \leq 2. \end{cases}$$

Then you can show that $$f$$ is a bijection and has a continuous inverse, but $$Df$$ is not invertible (at $$1$$) since it does not exist at $$1$$, i. e., $$f$$ is not differentiable at $$1$$.

Edit: Probably $$h(x) = \sqrt{x}$$ in $$x = 0$$ is an easier example for the last paragraph.

• Thanks, this is very helpful – alfalfa Sep 12 at 16:51