Laplacian of $ \nabla^2 f( A x)$ in terms of Laplacian of $ \nabla^2_y f( y)$ How to find the Laplacian
\begin{align}
 \nabla^2_x f( A x)
\end{align}
where $n \times m$ matrix and we know know the laplacian of
\begin{align}
 \nabla^2_y f( y)
\end{align}
This question is about chain rule for the Laplacian.
I search but could not find a formula for it. 
I think that the answer is 
\begin{align}
 \nabla^2_x f( A x)= {\rm Tr}( AA^T )  \nabla^2_y f( y) |_{y=Ax} 
\end{align} 
 A: The gradient of the function $f(y)$ is the vector $g$ which satisfies
$$df = g^Tdy$$
and the hessian is the matrix $H$ which satisfies
$$dg = H\,dy$$
and the laplacian is the trace of the hessian $\,\,L={\rm Tr}(H)$.
Suppose a new independent vector $x$ is introduced with $\,y=Ax$.
It really helps to consider the case where $A$ is rectangular, so that $x$ and $y$ have different dimensions. Then when you start to wonder why $A$ was transposed or why it was pre-multiplied or post-multiplied, you'll see that it's the only way to make the dimensions come out correctly.
To find the gradient and hessian wrt $x$, one wants modifie${\rm d^*}$ quantities which satisfy 
$$df=g_*^Tdx,\,\,\,\,\,\,dg_*=H_*\,dx,\,\,\,\,\,\,L_*={\rm Tr}(H_*)$$ 
Working these out is straightforward
$$\eqalign{
 df &= g^Tdy = g^TA\,dx = (A^Tg)^Tdx = g_*^Tdx \cr
 g_* &= A^Tg \cr\cr
 dg_* &= A^Tdg = A^T(H\,(dy)) = A^TH(A\,dx) = H_*\,dx \cr
 H_* &= A^THA \cr
 L_* &= {\rm Tr}(A^THA) = {\rm Tr}(AA^TH) \cr
}$$
The last expression is the laplacian in terms of the new variable.
Unfortunately, knowing the value of ${\rm Tr}(H)$ is of no help in evaluating ${\rm Tr}(AA^TH)$.
