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This question is about differentiation in $\mathbb{R}^3$.

Let $V : \mathbb{R}^3 \to \mathbb{R}$ be a smooth enough function, $f:\mathbb{R}^3\to\mathbb{R}^3$ its gradient, and $M$ a $3\times 3$ real matrix.

Does there exist a function $W : \mathbb{R}^3 \to \mathbb{R}$ , whose gradient is $Mf$? What would be an expression of $W$?

Thanks!

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I don't think so. Take the 2-dimensional case for simplicity. Let $f(x,y)=x^2+y$ and $M=[0, 1; 1 ,0]$, that is, $M$ is the reflection across $y=x$. Then you want $\nabla g = (1,2x)$. From $g_x=1$ we conclude $g(x,y)=x+h(y)$. Then g_y=h'(y) and we need h'(y)=2x, which cannot be.

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    Hm, that's true.2011-07-13