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I have what appears to be a simple equation:

$$ C_u = H(C_y \otimes R)H^T $$

where the matrix dimensions are compatible, i.e. $\dim(C_u)=\dim(C_y)=n \times n$, $\dim(H)=n \times np$, $\dim(R)=p \times p$, and the unknown matrix is $C_y$. ($\otimes$ denotes the Kronecker product.)

What is the closed-form solution for $C_y$? I am expecting that operations such as matrix multiplication, $\otimes$, and $vec$ should be enough.

Many thanks in advance.

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For convenience, I rename $C_u$ and $C_y$ as $U$ and $Y$. So, we have the equation $H(Y\otimes R)H^T=U$, where $Y$ is the unknown. Partition $H$ as $(H_1,H_2,\ldots,H_n)$, where each $H_i$ is an $n\times p$ matrix. Then our equation can be rewritten as \begin{align*} \sum_{i,j} y_{ij} H_i R H_j^T &= U,\\ \sum_{i,j} y_{ij} \operatorname{vec}(H_i R H_j^T) &= \operatorname{vec}(U). \end{align*} This is a usual linear system of equations in the form of $A\operatorname{vec}(Y)=\operatorname{vec}(U)$, where \begin{align*} A &= (v_{11},v_{21},\ldots,v_{n1},v_{12},v_{22},\ldots,v_{n2},\ldots,v_{1n},v_{2n},\ldots,v_{nn}),\\ v_{ij} &= \operatorname{vec}(H_i R H_j^T). \end{align*}

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  • $\begingroup$ I will check this in detail, many thanks for the quick reply! $\endgroup$ – user80268 May 30 '13 at 23:34
  • $\begingroup$ user1551, you have been very helpful. So here is a related question: Given $H(Y\otimes R)H^T=U$, is there a way to connect $\det(Y)$ with $\det(U)$? (The matrix $R$ may be assumed to be symmetric positive definite.) $\endgroup$ – user80268 May 30 '13 at 23:53
  • $\begingroup$ ...and $U,Y$ are positive definite as well. $\endgroup$ – user80268 May 31 '13 at 0:18
  • $\begingroup$ @user80268 I don't think so. If I am not mistaken, the $A$ in my answer is singular. Therefore the solutions $Y$ are not unique and it is unlikely that you can infer $\det Y$ from $\det U$ (or the other way round). Consider, e.g. $n=p=2$, $R=U=I_2$ and $H=\pmatrix{1&0&0&1\\ 0&1&1&0}$. If my hand calculations are correct, every matrix of the form $Y=\pmatrix{a\\ &1-a}$ (with $0<a<1$) is a +ve def. solution. Now $\det U=1$ but $\det Y=a(1-a)$ depends on the exact value of $a$. $\endgroup$ – user1551 May 31 '13 at 0:33
  • $\begingroup$ I think you are right about the determinants. I have also tried all kinds of formulas related to Kronecker products, vec, etc, but didn't go anywhere. At any rate, you have given me a lot of insight, so I will try to look at the original problem that led to the above formulation from a different angle, hopefully more fruitful. Best. $\endgroup$ – user80268 May 31 '13 at 1:14

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