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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 $\alpha^{-1}$.

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    $\begingroup$ 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. $\endgroup$ – Ethan Bolker Feb 25 '18 at 13:46
  • $\begingroup$ Thanks @EthanBolker, I edited the question. $\endgroup$ – Bruno Feb 25 '18 at 13:50
  • $\begingroup$ @DonAntonio, by $\alpha$ positive I mean strictly positive and $\mathbf{A}$ i think is always invertible when it is positive definite. Right? $\endgroup$ – Bruno Feb 25 '18 at 13:54
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    $\begingroup$ 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 ${\bf A} = {\bf P \Lambda P}^{-1}$. $\endgroup$ – Travis Feb 25 '18 at 13:55
  • $\begingroup$ @Gio Yes, you're right. I missed that part. $\endgroup$ – DonAntonio Feb 25 '18 at 13:56

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