I need to find the derivative with respect to $\mathbf{\Omega}$ of

$$ Tr\left(\left(\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}\right)\mathbf{S}\right) $$

In the above, $\mathbf{\Omega}^{-1}$ is symmetric, $\mathbf{C}$ is symmetric, $\mathbf{I}$ is the identity matrix and $\mathbf{S}$ is symmetric.

I understand that writing this using $:$ notation might help, which if I'm correct allows me to write

$$ \begin{align} \phi&=\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}:\mathbf{S}\\ &=\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}:\mathbf{I}\circ\mathbf{S} \end{align} $$

However from there I'm unsure - do I need a version of the product rule to deal with the right hand side - in which case how is this written when using $:$ notation? Or using the rule that $d\mathbf{\Omega^{-1}}=-\mathbf{\Omega^{-1}}d\mathbf{\Omega}\,\,\mathbf{\Omega^{-1}}$?


You're on the right track, that is the correct product rule.

Here is a useful relationship when $A,B$ are symmetric matrices $$\eqalign{ {\rm Sym}(X) &\doteq \tfrac 12\left(X+X^T\right) \\ d(BAB) &= dB\,AB + BA\,dB \\ &= dB\,AB + (dB\,AB)^T \\ &= 2\;{\rm Sym}(dB\,AB) \\ }$$

Putting the pieces together yields $$\eqalign{ \phi &= (\Omega^{-1}C\Omega^{-1}\circ I):S \\ &= (I\circ S):\Omega^{-1}C\Omega^{-1} \\\\ d\phi &= (I\circ S):d\left(\Omega^{-1}C\Omega^{-1}\right) &\big({\rm sym\,rule}\big) \\ &= (I\circ S):2\;{\rm Sym}\left(d\Omega^{-1}C\Omega^{-1}\right) \\ &= 2\;{\rm Sym}(I\circ S):d\Omega^{-1}C\Omega^{-1} \\ &= 2\;(I\circ S):d\Omega^{-1}C\Omega^{-1} \\ &= 2\;(I\circ S)\Omega^{-1}C:d\Omega^{-1} &\big({\rm product\,rule}\big) \\ &= -2\;(I\circ S)\Omega^{-1}C:\Omega^{-1}d\Omega\,\Omega^{-1} \\ &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1}:d\Omega \\\\ \frac{\partial\phi}{\partial\Omega} &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1} &\big({\rm gradient\,matrix}\big) \\\\ }$$ Here is another identity which was used above $$\eqalign{ X:{\rm Sym}(Y) &= X:\tfrac 12(Y+Y^T) \\ &= \tfrac 12X:Y + \tfrac 12X:Y^T \\ &= \tfrac 12X:Y + \tfrac 12X^T:Y \\ &= \tfrac 12(X+X^T):Y \\ &= {\rm Sym}(X):Y \\ }$$ A similar identity exists for the function $${\rm Skew}(X) = \tfrac 12\left(X-X^T\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.