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$A$ is a $d\times n$ matrix and $\mu>0$. I'm trying to show that $$(AA^T + \mu I)^{-1} A = A(A^T A+\mu I)^{-1}.$$

The only way I've thought about doing this was by the brute force method of checking whether the components of the matrices on both sides were equal. Is there a better way?

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Hint: Left-multiply both sides with $(AA^T+\mu I)$ and right-multiply both sides with $(A^TA+\mu I)$. (These are reversible operations.) By associativity and the definition of the inverse, reduce the problem to showing that:

$$A(A^TA+\mu I)=(AA^T + \mu I) A$$

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  • $\begingroup$ Wow I had a similar idea but I tried doing it by finding the inverse of $(AA^T + \mu I)^{-1} A$. *Facepalm. $\endgroup$
    – Kashif
    Sep 26, 2014 at 23:06

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