Define a "rotation of u" by $R_{A}u\doteq u\circ A $ with A an orthogonal $n\times n$ matrix and "$\circ$" means composition. Show that $\Delta (u\circ A )=(\Delta u)\circ A$. (Note: $u(x)\in C^{m\geqslant 2} (\mathbb{R}^{n}) $)
I tried to treat u as merely a number in $\mathbb{R}^{n}$ and then take the Laplacian to each entry of the matrix. Then I compared the LHS and the RHS and found they are equal. Is this correct? I felt this way is kind of trivial as you don't use the condition "orthogonal".
Hint: Write down everything explicitly. A orthogonal matrix $A$ satisfies $A A^T = I$. Thus
$$\sum_{j=1}^n a_{ij} a_{kj} = \delta_{ik},$$
where $\delta_{ik} = 1$ if $i=k$ and $0$ if $i\neq k$.
By definition,
$$(u \circ A)(x_1, \cdots , x_n ) = u(A(x_1, \cdots x_n)) = u(\sum_{i=1}^n a_{1i} x_i, \cdots , \sum_{i=1}^n a_{ni} x_i)$$
Now take $\frac{\partial }{\partial x_i}$ to the above equation and apply chain rule.