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I have the following matrix form,

$$\tilde{R}=[(I_T\otimes V)+(I_T\otimes\Sigma(\theta))K^{-1}(I_T\otimes\Sigma(\theta))']$$

where $\theta=[\theta_1,...\theta_N]$ is a vector, so each element of the $M\times P$ matrix $\Sigma(\theta)$ is a (possibily different) function of the vector, $\Sigma_{ij}=f_{ij}(\theta)$. I need to compute each $\frac{\partial\tilde{R}}{\partial\theta_n }$

Then,

$$\frac{\partial\tilde{R}}{\partial\theta_n }=\frac{\partial(I_T\otimes\Sigma(\theta))K^{-1}(I_T\otimes\Sigma(\theta))'}{\partial\theta_n}=2\frac{\partial(I_T\otimes\Sigma(\theta))}{\partial\theta_n}K^{-1}(I_T\otimes\Sigma(\theta))'$$

So I just need to compute the first derivative. Just a couple of things,

1) Could someone tell me if I am on the right track so far? 2) How would I compute the derivative that involves the Kronecker product? I would say that in this case I can just compute,

$$\frac{\partial\Sigma(\theta_n))}{\theta_n}$$

which would be also a $M\times P$ matrix. Then, the derivative would be, $$I_T\otimes \frac{\partial\Sigma(\theta_n))}{\theta_n}$$ but I am not sure if this is right... Many thanks in advance!!!

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  • $\begingroup$ Note that $$d(XYX^T)=dX\,YX^T+XY\,dX^T\neq 2dX\,YX^T$$ Other than that your approach seems correct, since the scalar derivative of a Kronecker product satisfies $$d(X\otimes Y)=dX\otimes Y+X\otimes dY$$ $\endgroup$ – lynn Mar 6 '17 at 17:55
  • $\begingroup$ Oh ok! You are right! Thank you so much!!! $\endgroup$ – Chucky Mar 7 '17 at 6:54

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