Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ twice differentiable and $\Delta f:\mathbb{R}^n \rightarrow \mathbb{R}$, defined by
$$ \Delta(f(x))=\text{trace}(\text{hess}f(x)) = \sum_{i=1}^n\frac{\partial^2 f}{\partial {x_i}^2}(x).$$ where hess denote de hessian matrix of $f$.
The question is:
Prove that if $T \in \text{L}(\mathbb{R}^n)$ is a orthogonal operator, then $\Delta(f \circ T)=\Delta f \circ T$.
My attempt
I was abble to show ( by the chain rule) using only the fact that $f \circ T$ is twice differentiable that
$$\Delta(f \circ T)(x) = \sum_{i=1}^{n}f''(T(x))(T(e_i),T(e_i))$$ where we are interpreting $f''$ as a bilinear form and $\{e_i\}$ is the canonical basis of $\mathbb{R}^n$. Knowing that
$$\Delta f(T(x)) = \sum_{i=1}^{n} f''(T(x))(e_i,e_i)$$
I want to know how to use the orthogonality of $T$ to show that
$$\sum_{i=1}^{n}f''(T(x))(T(e_i),T(e_i)) = \sum_{i=1}^{n} f''(T(x))(e_i,e_i)$$
Let $H$ be the matrix of $f''_{T(x)}$ relative to the standard basis of $\Bbb{R^n}$, and let $A$ be the matrix of $T$ relative to the standard basis. The idea is to write the trace of the bilinear forms as traces of the associated matrices, and then to use the cyclic property of the trace: \begin{align} \text{Tr}(f''_{T(x)}[T(\cdot), T(\cdot)]) &= \text{Tr}(A^t H A) \\ &= \text{Tr}(H A \cdot A^t) \hspace{.25in}\text{(cyclic property of Tr)} \\ &= \text{Tr}(H \cdot I_n) \hspace{.4in}\text{($T$ and hence $A$ orthogonal)} \\ &= \text{Tr}(H) \\ &= \text{Tr}(f''_{T(x)}[\cdot, \cdot]) \end{align}
I'll leave it to you to see that the trace of the bilinear forms and their associated matrices are related as I have stated.