I have $\Sigma$ a symmetric $2 \times 2$ matrix, and $\Sigma^{-1}$ is its inverse.
Now, $\tilde{\Sigma}^{-1}=\Sigma^{-1} \otimes I_{n \times n}$ (Kronecker product).
I have a function $Y=f(\tilde{\Sigma}^{-1})$ that gives a value in $\mathbb R$.
Let's define $\Phi_{\Sigma}=vech(\Sigma)$
Now, I am trying to get $\frac{\partial Y}{\partial \Phi_{\Sigma}}$.
So far I have
$\frac{\partial Y}{\partial \Phi_{\Sigma}^T} = \Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T \Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$
I've been working on this an got that $\Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$ is a vector with $n \times n$ elements. Now, working with the first part of the derivative
$\Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T = \Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial vec(\Sigma)^T} D_2 \Bigg)^T = \Bigg( vec \Big( \frac{\partial \tilde{\Sigma}^{-1}}{\partial \Sigma}\Big) D_2 \Bigg)^T = \Bigg( vec \Big( \frac{\partial \tilde{\Sigma}^{-1}}{\partial \Sigma^{-1}} \frac{\partial \Sigma^{-1}}{\partial \Sigma} \Big) D_2 \Bigg)^T$
$= \Bigg( vec \Big( (I_2 \otimes I_n) (-\Sigma^{-1} \Sigma^{-1}) \Big) D_2 \Bigg)^T$
where $D_2$ is the duplication matrix
However, the matrices $(I_2 \otimes I_n)$ and $-\Sigma^{-1} \Sigma^{-1}$ are not conformable. So it is wrong. Also, since $\Bigg( \frac{\partial vec(Y)}{\partial vec(\tilde{\Sigma}^{-1})^T} \Bigg)$ is a vector with $n \times n$ elements, and $\frac{\partial Y}{\partial \Phi_{\Sigma}}$ is $3 \times 1$, so $\Bigg( \frac{\partial vec(\tilde{\Sigma}^{-1})}{\partial \Phi_{\Sigma}^T} \Bigg)^T$ should be $3 \times (n \times n)$. May I ask for advice on solving this task?
\mathbb R
for $\mathbb R$. $\endgroup$