TITLE: What is inverse of $I+A$?
QUESTION [14 upvotes]: Assume $A$ is a square invertible matrix and we have $A^{-1}$. If we know that $I+A$ is also invertible, do we have a close form for $(I+A)^{-1}$ in terms of $A^{-1}$ and $A$?
Does it make it any easier if we know that sum of all rows are equal?

REPLY [13 votes]: I know I'am a bit late, but I wanted to add the following for future readers.
If you know the Eigendecomposition of your matrix $A = Q \Lambda Q^{-1}$ the result is the following:
\begin{equation}
\left( A + I\right)^{-1}
=\left( Q \Lambda Q^{-1} + I\right)^{-1}
=\left( Q \left(\Lambda + I \right) Q^{-1}\right)^{-1}
= Q \left(\Lambda + I \right)^{-1} Q^{-1}
\end{equation}
I know that this is a very trivial calculation and the result is maybe not helpful to many of you, but I wasted countless hours using the Shermann-Morrison-Theorem because I forgot about the Eigendecomposition.
Cheers,
Lukas