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!

  • $\begingroup$ Edit I edited your post to try to fix the formatting, but I don't understand how you placed the arrows. Sometimes $B$ is written $\vec{B}$ and sometimes not. In the jacobian, $X$ has an arrow, but nowhere else. Please check that I haven't accidentally messed things up. $\endgroup$
    – saulspatz
    Nov 29 '20 at 15:16
  • $\begingroup$ that's weird, because I didn't put arrows. $\endgroup$ Nov 29 '20 at 15:20
  • $\begingroup$ The source had many places where you had written $vec(X)$ and I assumed you meant to say $\vec{X}$ which comes out as $\vec{X}$ What did you want it to look like? (Actually, it was usually $B$, not $X$.) $\endgroup$
    – saulspatz
    Nov 29 '20 at 15:22
  • $\begingroup$ I wanted just vec(X), but I think that there was a problem with the code, that changes vec(X) with \vec $\endgroup$ Nov 29 '20 at 15:26
  • $\begingroup$ Sorry. I'm not familiar with the vec operator, and I thought it was supposed to be a typesetting command. $\endgroup$
    – saulspatz
    Nov 29 '20 at 15:29

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

  • 1
    $\begingroup$ Why $dF$ its equal to that? I don't get that $BdX$ part. I thought it was $XBdX$ $\endgroup$ Nov 29 '20 at 15:29
  • $\begingroup$ Thanks for pointing that out. It was a typo. $\endgroup$
    – greg
    Nov 29 '20 at 15:57
  • $\begingroup$ Could you explain why we have that $df$ ? Wich property did you use? $\endgroup$ Nov 29 '20 at 16:02
  • $\begingroup$ $df$ denotes the differential of the vector $f={\rm vec}(F)$, just as $dF$ denotes the differential of the matrix $F$. $\endgroup$
    – greg
    Nov 29 '20 at 16:09
  • $\begingroup$ OK. thanks. In my problem, the jacobian has $\partial vec(X^{T})$ it would be equal with $\partial vec(X)$ without the transpose? $\endgroup$ Nov 29 '20 at 16:15

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