I would like to ask you a little help for the following problem.

Let $\Phi$ and $\Sigma$ be two $N \times N$ matrices s.t. the inverse of $(I_{N^2}-\Phi \otimes \Phi )$ exists and $\Sigma$ is symmetric and positive definite.

Define $\Lambda=(I_{N^2}-\Phi \otimes \Phi )^{-1} vec (\Sigma)$.

Note that $\Lambda$ is identity matrix $N ^2\times N^2$.

The Jacobian of $\Lambda$ with respect to $vec (\Sigma)$ should be easy to compute and given by

$$\frac{\partial \Lambda}{\partial vec (\Sigma)'}= \left[(I_{N^2}-\Phi \otimes \Phi )^{-1} \right]^{T}.$$

What about $\dfrac{\partial \Lambda}{\partial vec (\Phi)'}$ ?


For convenience, let me define some terms (with Latin letters so I don't have to type all those Greek letters) $$\eqalign{ s &= {\rm vec}(\Sigma) \cr F &= \Phi \cr a &= \Lambda \cr M &= (I\otimes I - F\otimes F) \cr dM &= -(dF\otimes F + F\otimes dF) \cr }$$ Then the equation we are working with is $$\eqalign{ a &= M^{-1}s \cr Ma &= s \cr }$$ Taking the differential $$\eqalign{ dM\,a + M\,da &= ds \cr }$$ If we hold $M$ constant, then we can obtain your first result (let's not quibble over transposes) $$\eqalign{ M\,da &= ds \cr da &= M^{-1}\,ds \cr \frac{\partial a}{\partial s} &= M^{-1} \cr }$$ If instead we hold $s$ constant, we obtain $$\eqalign{ dM\,a + M\,da &= 0 \cr M\,da &= -dM\,a \cr M\,da &= (dF\otimes F + F\otimes dF)\,a \cr }$$ At this point we need a few tricky Kronecker-Vec relationships $$\eqalign{ {\rm vec}(A) &= a \cr (F\otimes dF)\,{\rm vec}(A) &= (FA^T\otimes I)\,{\rm vec}(dF) \cr (dF\otimes F)\,{\rm vec}(A) &= (I\otimes FA)\,{\rm vec}(dF^T) \cr &= (I\otimes FA)\,P\,\,{\rm vec}(dF) \cr }$$ where $P$ is the appropriate Kronecker Permutation for a matrix with the dimensions of $F$.

Substituting the Kronecker-Vec expressions we arrive at $$\eqalign{ M\,da &= \big((FA^T\otimes I) + (I\otimes FA)P\,\big)\,{\rm vec}(dF) \cr da &= M^{-1}\big((FA^T\otimes I) + (I\otimes FA)P\,\big)\,{\rm vec}(dF) \cr \frac{\partial a}{\partial\,{\rm vec}(F)} &= M^{-1}\big((FA^T\otimes I) + (I\otimes FA)P\,\big) \cr }$$ which, aside from my use of Latin letters, is the quantity that you asked about.

  • $\begingroup$ Many thanks for your kind reply! $\endgroup$ – Marco B Jul 26 '15 at 23:11

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