If $A$ is an arbitrary matrix, does there exist a function $f:\mathbb{R}^n\to\mathbb{R}$ such that $\nabla f(x) = Ax$?

Question

I know when $A$ is a symmetric matrix then $\nabla \frac{1}{2}\langle x,Ax\rangle = Ax$ but this only works when $A$ is symmetric. I am curious about the other case, if I have some arbitrary matrix (i.e., not necessarily symmetric) $A$ then is there always a function $f$ so that $\nabla f(x) = Ax$?

Answer 1

Simple Anwer. No. For example, if $\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, then the line integral of $x \mapsto \mathbf{A}x$ along the counter-clockwise oriented unit circle $\gamma(t) = (\cos t, \sin t)$, $0 \leq t \leq 2\pi$, is

$$ \oint_{\gamma} \mathbf{A}x \cdot \mathrm{d}l = \int_{0}^{2\pi} (\mathbf{A}\gamma(t)) \cdot \gamma'(t) \, \mathrm{d}t = - \int_{0}^{2\pi} \sin^2 t \, \mathrm{d}t = - \pi \neq 0. $$

General Answer. In general, for an $n \times n$ matrix $\mathbf{A}$, the vector field $F = \mathbf{A}x$ in $\mathbb{R}^n$ has a scalar potential if and only if the corresponding differential $1$-form

$$ \omega = \mathbf{A}_1 x \, \mathrm{d}x^1 + \cdots + \mathbf{A}_n x \, \mathrm{d}x^n $$

is closed (because every closed $1$-form on $\mathbb{R}^n$ is exact). Since

$$ \mathrm{d}\omega = \sum_{i,j} \mathbf{A}_{ij} \mathrm{d}x^j \wedge \mathrm{d}x^i = \sum_{i < j} (\mathbf{A}_{ji} - \mathbf{A}_{ij}) \mathrm{d}x^i \wedge \mathrm{d}x^j, $$

it follows that $\mathrm{d}\omega = 0$ if and only if $\mathbf{A}_{ji} - \mathbf{A}_{ij} = 0$ for all $i < j$, or equivalently, $\mathbf{A}$ is symmetric.

Answer 2

In this context $f$ would usually be called the Scalar potential of $Ax$. The most prominent condition for a scalar field to exist is $\nabla\times(Ax)=0.$

For a counter example you can simple try the two dimensional case: $$ \nabla f=\begin{pmatrix} a&&b\\c&&d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}\\ \Rightarrow\partial_x f=ax+by\\ \Rightarrow f=ax^2/2+bxy+C_1(y)\\ \Rightarrow\partial_y f=bx+C_1'(y)\stackrel{!}{=}cx+dy $$ Where the last equation can only hold if $b=c$.