Let $\otimes$ denote the Kronecker product. I know that for two matrices $A$ and $B$, $(A \otimes B)^\dagger = A^\dagger \otimes B^\dagger =: \Omega^\dagger$, where the $\dagger$-superscript denotes the Moore-Penrose inverse. Now let $E$ be a diagonal 0-1 matrix (arbitrary positions) conformable with $\Omega$. Hence, $E$ is an orthogonal projection. Given $A$ and $B$ are invertible, what can we say about $(E \Omega E)^\dagger?$

It seems tempting to somehow exploit the property $E^\dagger = E$. Yet clearly, we have $(E \Omega E)^\dagger \neq(E\Omega^\dagger E)$. So far, I could not find something useful on this case.

In case nothing can be said under this generality, is any combination of the following additional qualifications useful:

- $A$ is symmetric and circulant
- $A$ equals $I - \gamma G_c$, where $I$ is the identity matrix, $\gamma > 0$ is some parameter, and $G_c$ is the adjacency matrix of the complete graph
- $B$ is symmetric and positive definite
- $B$ is eventually positive

"Eventual positivity" is defined as $\exists k \in \mathcal{N}: B^l \geq 0 \, \forall\, l \geq k$, where the matrix inequality is entrywise.