# 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.

Thanks in advance!