# How to find the inverse of $A\otimes A + (I+D\otimes D)^{-1}(D\otimes D)$ without forming the Kronecker product?

Is there a good way to compute the inverse of Inverse of $A\otimes A + (I+D\otimes D)^{-1}(D\otimes D)$ that doesn't require forming the full Kronecker product? Here $A$ is symmetric, positive definite and $D$ is diagonal with all positive elements.

If we only had the quantity $A + (I+D)^{-1}D$, we could solve the generalized eigenvalue problem where we have a matrix $V$ and a diagonal matrix $\Lambda$ such that

• $AV = (I+D)^{-1}DV\Lambda$
• $V^{T}AV = \Lambda$
• $V^{T}((I+D)^{-1}D)V = I$

Then, $$V^{T}(A + (I+D)^{-1}D) V = \Lambda + I,$$ which implies $$A + (I+D)^{-1}D = V^{-T}(\Lambda + I)V^{-1}.$$ Hence, $$(A + (I+D)^{-1}D)^{-1} = V(\Lambda + I)^{-1}V^T.$$ I'd like to do something similar for the original quantity with the Kronecker products, but I can't figure out how to work around the identity inside of the inverse. Really, my desire is to avoid forming the full Kronecker product, which could be very large.