I am obtaining the following equation in a regression problem: \begin{eqnarray} Z'_1Y_1\Omega^{-1}_{1}A+Z'_2Y_2\Omega^{-1}_{2}A = Z_{1}'Z_1\Pi A'\Omega^{-1}_1A + Z_{2}'Z_2\Pi A'\Omega^{-1}_2A \end{eqnarray} and would like to solve it for $\Pi$. Now, let's define the matrices: $Z_i$ is $n_i \times k$, so that $Z_i'Z_i$ is invertible. $Y_i$ is $n_i \times (1+p)$ and $\Omega_i$ is $(1+p) \times (1+p)$ and $A$ is $(1+p) \times p$, where $i \in \{1,2\}$. For invertibility, we assume $Z_i'Z_i$ and $A'\Omega^{-1}_iA$ to be invertible.
I would really appreciate your help, perhaps also a reference on how to get started.
This comes from a likelihood function that I am maximizing and I believe the solution should look similar to this: $\Pi = (Z'Z)^{-1} Z' Y \Omega^{-1} A (A' \Omega^{-1} A)^{-1}$.
I am not sure how I can 'free' the parameter $\Pi$ adequately, because multiplying with inverses of say $Z_i'Z_i$, would eliminate the matrix in front of one occurrence of $\Pi$ but not the other one.
I would really appreciate your help!
I think that this answer is not essential because copper.hat did the job ; yet this will can be better in writing it.
Your equation can be rewritten in this sylvester equation $UX+XV=C$ where $U=(Z_1^TZ_1)^{-1}Z_2^TZ_2,V=A^T\Omega_1^{-1}A(A^T\Omega_2^{-1}A)^{-1}$ ; that is $WX=C$ where $W=U\otimes I+I\otimes V^T$ if we stack matrix in vector row by row. cf. http://en.wikipedia.org/wiki/Kronecker_product
If $spectrum(U)=(\lambda_i)_i,spectrum(V)=(\mu_j)_j$, then $Spectrum(W)=(\lambda_i+\mu_j)_{i,j}$. Thus, if for every $i,j$, $\lambda_i+\lambda_j\not=0$, then $W$ is invertible and $X=W^{-1}C$. Clearly, there is no closed form for the solution (except if $U=V$) ; then you search a numerical solution. Beware, if you invert directly $W$, then the complexity is in $O(n^6)$ ; in fact there is a method in $O(n^3)$.cf. http://en.wikipedia.org/wiki/Sylvester_equation