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Assume that we have matrices $A, B, C, X$ such that both $X$ and $B$ are symmetric and the dimension of all those matrices agree for multiplications. I need to find the derivative of

$\frac{log\,|B^{-1}|}{\partial x_{ij}}$, such that

$x_{ij}$ indicates the elements of $X$, and $A$ and $B$ are constant of $X$

$B=\big(X \otimes A +C\big)^{-1}$

Based on my reading I think the answer is

$\frac{log\,|B^{-1}|}{\partial x_{ij}}= \text{tr}\big((B^{-1})^{-1} \frac{\partial\,B^{-1}}{\partial x_{ij}}\big),$

$\qquad\quad\,\,=\text{tr}\Big[B\,\, \frac{\partial\big(X\otimes A\,+C\big)}{\partial x_{ij}}\Big]$

$\qquad\quad\,\,=\text{tr}\Big[B\,\, \big(\frac{\partial\,X}{\partial x_{ij}}\otimes A\big)\Big]$

It is known that $\frac{\partial\,X}{\partial x_{ij}}=(2-\delta_{ij})\,E_{ij}$, where for a symmetric matrix $\delta_{ij}$ is one for $i=j$ and zero otherwise, while $E_{ij}$ is the matrix with 1 on the $ij^{th}$ position and zero elsewhere.

My question is how can I expand $\Big[B\big(\frac{\partial\,X}{\partial x_{ij}}\otimes A\big)\Big]$, so which matrix should $E_{ij}$ takes its $ij^{th}$ element.

N. I. ElZayat
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  • Your answer looks good, except for the last part where you enforce the symmetry constraint. The constrained derivative is not what you've written, instead it's $$\frac{\partial X}{\partial X_{ij}} = E_{ij} + E_{ji} - I\circ E_{ij} $$ where $\circ$ is the elementwise/Hadamard product. – greg Jun 11 '18 at 11:23
  • for the symmetry constrained in may reference it is written as $E_{ij}+E_{ji}-\delta_{ij}E_{ii}$ and regards my question, can you answer me "should I substitute for this derivative rule in the Kronecker product statement" – N. I. ElZayat Jun 11 '18 at 14:56

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