# Inverse Matrix of a sum

Let $\mathbf{A}$ a $n\times n$ symmetric positive definite matrix and $\alpha$ a positive constant. I want to simplify the following expression: $$\left(\mathbf{A} + \alpha \, \mathbf{I}\right)^{-1},$$ where $\mathbf{I}$ is the identity matrix of order $n$.

I looked in the Matrix Cookbook in order to find an identity, but I found only more general formulas like the The Woodbury identity''. Do you know how to find a simpler identity for this easy case?

Thanks a lot!

Edit: My goal is to obtain an easy expression in terms of $\mathbf{A}^{-1}$ and $\alpha^{-1}$.

• You could try looking at it as a geometric series. If you edit the question to tell us what you want to do with this matrix perhaps we can help you find another path to your goal. – Ethan Bolker Feb 25 '18 at 13:46
• Thanks @EthanBolker, I edited the question. – Bruno Feb 25 '18 at 13:50
• @DonAntonio, by $\alpha$ positive I mean strictly positive and $\mathbf{A}$ i think is always invertible when it is positive definite. Right? – Bruno Feb 25 '18 at 13:54
• Assuming that for you positive definite means symmetric, $A$ is diagonalizable and so one could be reasonably explicit about the form of $({\bf A} + \alpha {\bf I})^{-1}$ in terms of a diagonalization ${\bf A} = {\bf P \Lambda P}^{-1}$. – Travis Feb 25 '18 at 13:55
• @Gio Yes, you're right. I missed that part. – DonAntonio Feb 25 '18 at 13:56