TITLE: Jacobian matrix and Hessian matrix identity
QUESTION [7 upvotes]: I am trying understand the following identity in two dimensions:
$$ H_{XY} = J^TH_{xy}J$$
Here $x,y$ and $X,Y$ are different coordinates for $\mathbb R^2$, the $J$ is the Jacobian matrix and the $H$ are the Hessian matrices in the coordinates. 
My thoughts are the following: the $J$ is locally a coordinate transformation from $XY$ to $xy$ coordinates. At least is how I thought of Jacobians until today. The problem is that then $J^T$ should be $J^{-1}$. I checked and Jacobians are not necessarily unitary. Now I wonder: how do I understand the equation above and in general the Jacobian matrix? In particular why is it $J^T$ and not $J^{-1}$?

REPLY [3 votes]: Concerning the hessian, there is no simple formula for the change of variable, EXCEPT when we are on a critical point. Let denote by $\phi$ the function "change of variable" $X\rightarrow x$. First, we have a look on the case $n=1$. $(f\circ \phi)'(X)=f'(x)\phi'(X)$ and $(f\circ \phi)''(X)=f''(x)\phi'^2(X)+f'(x)\phi''(X)$. The previous expression has an interesting form only if $f'(x)=0$ ($x$ is a critical point and $X$ also since $(f\circ \phi)'(X)=0$). Now , in the general case, it is more complicated because we must consider the first derivative (linear) and the second derivative (bilinear and then, associated to a symmetric matrix). $D(f\circ \phi)_X:H\rightarrow Df_x(D\phi_X(H))$ and $D^2(f\circ \phi)_X:(H,K)\rightarrow D^2f_x(D\phi_X(H),D\phi_X(K))+Df_x(D^2\phi_X(H,K))$. Here $x$ is a critical point, that is $Df_x=0$ and $D^2(f\circ\phi)_X(H,K)=D^2f_x(D\phi_X(H),D\phi_X(K))$. With the blue's notation, $D\phi_X=J$, we obtain the equality between symmetric matrices $D^2(f\circ\phi)_X=J^TD^2f_xJ$. This formula permits to study the critical points of $f$; are they really an extremum of the function $f$ ?