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!

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    $\begingroup$ Apparently here $A$ acts on a vector space $V$ and then $B$ acts on another vector space $U$ such that $V\otimes V=U\oplus U$.Surely the answer depends on the choice of $U$ inside the tensor product? I don't see a way to tackle the question without having this specified first. That may be just my limitation though :-) $\endgroup$ – Jyrki Lahtonen Jul 14 '18 at 5:36
  • $\begingroup$ Above we do need two distinct isomorphic copies of $U$. The sum of a subspace with itself is never direct (ok, there is an uninteresting counterexample:-) $\endgroup$ – Jyrki Lahtonen Jul 14 '18 at 18:59

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