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!
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 $$f={\rm vec}(F),\quad x={\rm vec}(X)$$