Is there a way to invert $(A\otimes A)(B\oplus B)^{-1}(A\otimes A)+I$ without forming the Kronecker product? Here, both $A\succ 0$ and $B\succ0$.
Generally speaking, I would say that finding the spectral decomposition of $(A\otimes A)(B\oplus B)^{-1}(A\otimes A)$ would be the way to go since we could rewrite the above quantity as $V(D+I)V^T$ where $VDV^T=(A\otimes A)(B\oplus B)^{-1}(A\otimes A)$. Then, the inverse is simply $V(D+I)^{-1}V^T$. Part of the reason I gravitate to a spectral decomposition is that is that sometimes we can find the decomposition of Kronecker products and sums without forming the Kronecker product or sum itself. That said, I don't see how that works in this case.
Thanks for the help!