# Is there a matrix that converts the gradient of any function to gradient of other function?

The study of hamiltonian mechanics brought me to the following question.

Let $n$ be a natural number ($n>1$).

Let $A(\mathbf{x})$ be a $n\times n$ matrix consisting of functions $a_{ij}(\mathbf{x})$ ($a_{ij}:\mathbb{R}^n\to\mathbb{R}$): $$A(\mathbf{x})= \begin{pmatrix} a_{11}(\mathbf{x})& \cdots& a_{1n}(\mathbf{x})\\ \vdots&\ddots&\vdots\\ a_{n1}(\mathbf{x})&\cdots& a_{nn}(\mathbf{x}) \end{pmatrix}.$$ Let $A(\mathbf{x})$ be so, that for any $F(\mathbf{x})$ ($F:\mathbb{R}^n\to\mathbb{R}$): $$\begin{pmatrix} a_{11}(\mathbf{x})& \cdots& a_{1n}(\mathbf{x})\\ \vdots&\ddots&\vdots\\ a_{n1}(\mathbf{x})&\cdots& a_{nn}(\mathbf{x}) \end{pmatrix} \begin{pmatrix} \frac{\partial F}{\partial x_1}\\ \vdots\\ \frac{\partial F}{\partial x_n} \end{pmatrix} = \begin{pmatrix} g_1(\mathbf{x})\\ \vdots\\ g_n(\mathbf{x}) \end{pmatrix} = \begin{pmatrix} \frac{\partial G}{\partial x_1}\\ \vdots\\ \frac{\partial G}{\partial x_n} \end{pmatrix}$$ for some $G(\mathbf{x})$ ($G:\mathbb{R}^n\to\mathbb{R})$.

In other words, if we multiply fixed $A(\mathbf{x})$ by the gradient of any $F(\mathbf{x})$ we necessarily get the gradient of some $G(\mathbf{x})$.

Can we say anything about such $A(\mathbf{x})$? I would be glad if the only opportunity is that $A(\mathbf{x})=cE$, where $E$ is the identity matrix and $c$ is some real number. Is it correct? Is it possible to prove it?

All the functions are considered to be "good enough" ("smooth enough").

• Your hope that $A(x) = cE$ is shattered, just consider $F(x) \equiv 0$. May 18, 2016 at 10:45
• @Wauzl I need a matrix $A(\mathbf{x})$ that converts the gradient of every possible $F(\mathbf{x})$ to the gradient of some $G(\mathbf{x})$. The fact that some $A(\mathbf{x})$ converts the gradient of $F(\mathbf{x})\equiv0$ to gradient does not mean that the same $A(\mathbf{x})$ will work that well for all other possible $F(\mathbf{x})$. Probably the word "any" made the whole statement ambiguous.
– rtmd
May 18, 2016 at 20:27