I'm trying to prove the following:
Let $X_{m\times m}$ be a symmetric matrix of variables. Find the Jacobian, defined as:
$$\frac{\partial}{\partial vec{X'}}vec{F}$$
where $F(X)=XBX$ , where $B_{m\times m}$ is a symmetric matrix of constants. Answer: ($XB\otimes I_m$)+($I_m\otimes XB$). Where $\otimes$ refers to the Kronecker product.
My attempt:
\begin{align*}
vec{F} &= vec{XBX} \\
&= (X'\otimes X)vec{B} \\
&= (X\otimes X)vec{B}
\end{align*}
Where "vec" refers to the vec operator.
So, the jacobian is given by:
$$
\frac{\partial}{\partial(X)'}(X\otimes X)vec{B}
$$
But, I don't know what to do next. Any suggestion please? Thanks!
Best Answer
Calculate the differential before applying the vec operator. $$\eqalign{ dF &= dX\,BX + XB\,dX \\ df &= \big((BX)^T\otimes I\big)\,dx + \big(I\otimes XB\big)\,dx \\ &= \big((XB\otimes I) + (I\otimes XB)\big)\,dx \\ \frac{\partial f}{\partial x} &= \big((XB\otimes I) + (I\otimes XB)\big) \\ }$$ where $$f={\rm vec}(F),\quad x={\rm vec}(X)$$