Hi Changting Xu,

I found this: (I + AB)^-1 = A*(I + BA)^1*A^-1

Best,

Thank you, Anton Vrdoljak! I have tried this identity but ended up with the inverse of (I+BA), which is equivalently challenging problem.

https://en.wikipedia.org/wiki/Neumann_series might be useful.

Thank you, it is possible to use Neumann series, but the results are cumbersome. So an explicit inverse is preferable. Nils Wagner