Prove $\operatorname{div}(A \circ f \circ A^{-1}) (p) = \operatorname{div}(f)(A^{-1}p)$ with $A \in \operatorname{GL}_3(\mathbb{R})$

Question

Let $A \in \operatorname{GL}_3(\mathbb{R})$ and $f: \mathbb{R}^3 \to \mathbb{R}^3$ a continuously differentiable vector field. We define the vector field $f_A := A \circ f \circ A^{-1}$

Prove that:

1. $\operatorname{div}(f_A) (p) = \operatorname{div}(f)(A^{-1}p)$ for all $p \in \mathbb{R^3}$
2. $\operatorname{rot}(f_A) (p) = \operatorname{rot}(f)(A^{-1}p)$ for all $p \in \mathbb{R^3}$, if $A\in \operatorname{SO}_3(\mathbb{R})$

How I tried solving it

by brute force:

$\operatorname{div}(f_A) (p)$ = ($\partial_1 A_{1,}(f_1(A_{1,}^{-1})) + \partial_2 A_{2,}(f_2(A_{2,}^{-1})) + \partial_3 A_{3,}(f_3(A_{3,}^{-1})))(p)$ However I do not really know how one continues from here.

By using the definition

Our textbook defines $\operatorname{div}(f)(p) = \lim_{h\to 0} \frac{1}{h^3} \int_{\partial(p+[0,h]^3)} f \cdot d\vec{n}$

Similary $\operatorname{rot}(f)(p) = \lim_{r\to 0} \frac{1}{\pi r^2} \int_{\partial B_r(p)} f \cdot d\vec{s}$

These definitions seem very technical so I am not sure how they should be used.

Your help is greatly appreciated since I don't even know where to start.

Answer 1

For the divergence.

Applying the chain rule, the Fréchet dérivative of $f_A$ at $p$ is $$f_A^\prime(p)=A^\prime[(f \circ A^{-1})(p)] \circ f^\prime(A^{-1}(p)) \circ \left(A^{-1}\right)^\prime(p).$$

As $A$ and $A^\prime$ are linear applications, their Fréchet derivatives at any point are themselves. Hence $A^\prime[(f \circ A^{-1})(p)]=A$ and $\left(A^{-1}\right)^\prime(p)= A^{-1}$ which implies $$f_A^\prime(p)=A \circ f^\prime(A^{-1}(p)) \circ A^{-1}.$$

As for any $h: \mathbb{R}^3 \to \mathbb{R}^3$ continuously differentiable, $\operatorname{div}(h)(p) = \operatorname{Tr}(h^\prime(p))$ you get

$$\begin{align} \operatorname{div(f_A)}(p) &= \operatorname{Tr}(A \circ f^\prime(A^{-1}(p)) \circ A^{-1}) = \operatorname{Tr}(A \circ A^{-1} \circ f^\prime(A^{-1}(p)))\\ &=\operatorname{Tr}(f^\prime(A^{-1}(p))) = \operatorname{div}(f)(A^{-1}(p)) \end{align}$$

For the curl equality

Use the equality $$\operatorname{div}(X_0 \wedge F) = - \operatorname{rot}F \cdot X_0$$ where $X_0$ is any constant vector field and the identity above for $\operatorname{div}$.