Div$f$ is invariant under an orthogonal change of coordinates Let $f: \mathbb{R^n} \to \mathbb{R^n}$ and $Df$ exists. I need to show that div$f$ is invariant under an orthogonal change of coordinates. Let $T:\mathbb{R^n} \to \mathbb{R^n}$ be an orthogonal transformation. Then Suppose that $$f(x_1,x_2,..,x_n)=(f_1(x_1,x_2,..,x_n),f_2(x_1,x_2,..,x_n),..,f_n(x_1,x_2,..,x_n))$$
Also suppose that $T(x_1,x_2,..,x_n)=(x'_1,x'_2,..,x'_n),T^{t}T=TT^{t}=I$
Let $X=(x_1,x_2,..x_n)$ and $X'=(x'_1,x'_2,..,x'_n)$
Now I want to use the fact that $divf=trace(Df)$. So I let $$g(X')=f(T(X))$$(considering them as a function of $X$ and $X'$ respectively)
This is where I am stuck. Somehow on differentiating two sides I should have $DT$ and $(DT)^{-1}$ and that should do it for me.
Thanks for the help!!
 A: Lef $f:\mathbb{R}^n\longrightarrow \mathbb{R}^n$ a function such that $df$ exists. Let $J_f$ the jacobin matrix associate to $f$. 
Let now $\phi:(\mathbb{R}^n, y_1,...,y_n)\longrightarrow (\mathbb{R}^n,x_1...,x_n)$ an orthogonal change of coordinates. So $x_i=\phi_i(x_1,...,x_n)$, and define $J_\phi$ matrix associate to $\phi$.
Set $g:\phi\circ f\circ \phi^{-1}:(\mathbb{R}^n, y_1,...,y_n)\longrightarrow (\mathbb{R}^n,y_1...,y_n)$; the jacobian matrix associate to $g$ is $J_g$. Clearly $J_g=J_\phi J_f J_\phi^{-1}$.
Now $div g=tr(J_g)=tr(J_\phi J_f J_\phi^{-1})=tr(J_f)=div f$ So, the divergence is invariant under orthogonal transformation.
A: So you have $X^\prime=T.X$. Therefore $$g(X^\prime)=T.f[T^t.X^\prime]$$
You shoudn't forget to multiply $f[T^t.X^\prime]$ on the left by $T$ to move the result of $f$ in the "new" coordinate system.
Then you have using chain rule and the fact that the derivative of a linear map is itself $$Dg = T.Df.T^t$$ You can then apply the trace. Knowing that $tr(ABC)=tr(ACB)$ you get the requested result.
