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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").

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  • $\begingroup$ Your hope that $A(x) = cE$ is shattered, just consider $F(x) \equiv 0$. $\endgroup$ May 18, 2016 at 10:45
  • $\begingroup$ @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. $\endgroup$
    – rtmd
    May 18, 2016 at 20:27

1 Answer 1

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I have found the solution (with the help of Mathoverflow users).

See here: https://mathoverflow.net/questions/239275/is-there-a-matrix-that-converts-the-gradient-of-every-possible-function-to-gradi

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  • $\begingroup$ This is not an answer. $\endgroup$
    – user147263
    May 19, 2016 at 22:44
  • $\begingroup$ @sandwich Do you mean that the solution on Mathoverflow is wrong, or the message here is not informative? If latter, how can I give the link to the discussion on Mathoverflow? I wouldn't like to delete the question, because it may be useful for further users. $\endgroup$
    – rtmd
    May 20, 2016 at 7:52

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