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 $h(x) = \sqrt[3]{x}$ in $x = 0$ is an easier example for the last paragraph.

  • 1
    $\begingroup$ Thanks, this is very helpful $\endgroup$ – alfalfa Sep 12 at 16:51

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