# Pseudo-inverse of “sandwiched” Kronecker product

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.