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@pete thanks a lot, yes I used that and compared to the numerical derivative and it worked fine! however, I have a bit more advanced question that I posted here, would you mind having a look at that as well? (I odn't want to extend that comment thread even more..)
@pete and when using the chain differentiation with $A = f(\theta)$, the derivative, $\frac{\partial L}{\partial \theta} = L \Phi(L^{-1}\frac{\partial A}{\partial \theta} L^{-T})$, and not following the chain-rule: $\frac{\partial L}{\partial \theta} = \frac{\partial L}{\partial A} \cdot \frac{\partial A}{\partial \theta}$?
@pete what happens when I want to take the derivative of $A^T$? when I'm using this paper, the perturbation becomes $dL^T = (dL)^T = (L \Phi(L^{-1}dA L^{-T}))^T = \Phi(L^{-1}dA L^{-T})^T L^T$, is that correct?
