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Let $\Sigma_W$ be a $D\times D$ matrix. Let $X$ also be a $N\times q$ matrix. I am trying to solve an equation for $\Sigma_W$. However, there are Kronecker products involved and I do not really know how to handle this.

Here is the equation:

($I_q \otimes \Sigma_W)^{-1} = \frac{(I_D \otimes X^T)(I_D \otimes X)}{\sigma^2} + (\tau^2I_{qD})^{-1}$

I assume there must be a way to get rid of the identity matrices but I don't know much about Kronecker products.

Can someone explain how this could be solved for $\Sigma_W^{-1}$, if that is possible?

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    $\begingroup$ Partition the matrix on the RHS into $D\times D$ blocks. Then the block in the upper left corner is your $\Sigma_W$ matrix. $\endgroup$ – greg Dec 3 '17 at 17:10
  • $\begingroup$ Thank you. Is it possible though to derive a relationship between sigma and the terms on the RHS without using any Kronecker product? $\endgroup$ – MattSt Dec 3 '17 at 17:31
  • $\begingroup$ @greg Your solution works if, in a first step, you take the inverse of the RHS ... and consider the upper left corner of this inverse. Am I right ? $\endgroup$ – Jean Marie Dec 3 '17 at 18:41
  • $\begingroup$ (Ctd) ... I understand now, you use the property $(A \otimes B)^{-1}=A ^{-1} \otimes B^{-1}$... $\endgroup$ – Jean Marie Dec 3 '17 at 18:45
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    $\begingroup$ For the current formulation of the question, the solution is $$\Sigma^{-1}=\frac{X^TX}{\sigma^2}+\frac{I_q}{\tau^2}$$ $\endgroup$ – frank Dec 3 '17 at 18:53
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Recall the rule for a mixed Kronecker-matrix product $$(A\otimes B)(C\otimes D) = (AC)\otimes(BD)$$ assuming the matrices have compatible dimensions.

Applying this to the current problem, I get $$\eqalign{ (I_D\otimes X^T)(I_D\otimes X) &= I_D\otimes X^TX \cr I_{qD} &= I_D\otimes I_q \cr\cr (I_q\otimes\Sigma_W)^{-1} &= \sigma^{-2}I_D\otimes X^TX + \tau^{-2}I_{qD} \cr I_q\otimes\Sigma_W^{-1} &= I_D\otimes\bigg(\frac{X^TX}{\sigma^2}+\frac{I_q}{\tau^2}\bigg) \cr }$$ The first $D\times D$ block on each side of the equality is $$\Sigma_W^{-1}=\frac{X^TX}{\sigma^2}+\frac{I_q}{\tau^2}$$

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  • $\begingroup$ [+1] Very clean. As you have a good know-how in Kronecker products and sums, you can be interested by the recent question (math.stackexchange.com/q/2543670) and the reference herein which uses the determinant of a Kronecker sum for obtaining the parametric expression of generalized ellipses. $\endgroup$ – Jean Marie Dec 4 '17 at 10:23

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