TITLE: Find function whose gradient is Ax, with A antisymmetric
QUESTION [5 upvotes]: I have two variables $x_1,x_2\in\mathbb{R}$, and the vector function $g(x)=[x_2,-x_1]^T=Ax$, where A=[0,1;-1,0], and $x=[x_1,x_2]^T$. Is there any way to find a function $f:\mathbb{R^2}\rightarrow \mathbb{R}$, such that $\nabla f(x)=g(x)$?

REPLY [3 votes]: Alternatively, suppose the contrary that $\nabla f=g$ on $\mathbb R^2$. Then $(\nabla f(\mathbf x))\cdot\mathbf x\equiv0$. That is, the directional derivative of $f$ along every line pointing out from the origin is zero. Therefore $f$ is constant on every such line and in turn, $f$ is constant on $\mathbb R^2$. Hence $\nabla f=0$, contradicting that $\nabla f=g\ne0$.