Potential for $A\,\nabla V$

Question

Let $V:\mathbb R^n\to \mathbb R$ smooth, let $A$ invertible real matrix of dimension $n\times n$. Consider the following vector field $F:\mathbb R^n\to\mathbb R^ n$, $$ F(x) \equiv A\,\nabla V(x) \,.$$ Is it possible to determine some condition on $A$ such that $F$ has a potential, namely there exists $U:\mathbb R^n\to \mathbb R$ such that $$ \nabla U(x) = F(x) \;\;?$$ In particular I am interested in the case $A$ diagonal matrix.

Answer 1

This is helplessly false, even in the diagonal case. Here are two examples (following up on my comments).

(1) Take $V(x,y)=\frac12(x^2+y^2)$ and $A=\begin{bmatrix} 0 & -1\\1& 0\end{bmatrix}$. Then $$A\nabla V = \begin{bmatrix} 0 & -1\\1& 0\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix} = \begin{bmatrix} -y\\x\end{bmatrix}$$ is not conservative.

Note, however, that if we stick with the rotation matrix and take $V$ to be harmonic, then $A\nabla V$ will again have curl $0$. For example, with $V(x,y)=\frac12(x^2-y^2)$, we get $A\nabla V = \begin{bmatrix} y\\x \end{bmatrix} = \nabla(xy)$. No coincidence that we get the harmonic conjugate here.

(2) Take $V(x,y) = xy$ and $A=\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}$. Then $$A\nabla V = \begin{bmatrix} a & 0\\ 0 & b \end{bmatrix}\begin{bmatrix} y\\x\end{bmatrix} = \begin{bmatrix} ay\\bx\end{bmatrix}$$ is not conservative unless $a=b$.