Is this a valid thing to do, if P and R are both square matrices whose inverse exists?: P * inverse( I + R*P ) = inverse( inverse(P) + R )
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Sure: $\displaystyle P(I+RP)^{-1} = \left(P^{-1}\right)^{-1}(I+RP)^{-1} = \underbrace{\left[(I+RP)P^{-1}\right]^{-1}}_{B^{-1}A^{-1} = (AB)^{-1}} $ $\displaystyle = \left[P^{-1} + RPP^{-1}\right]^{-1} = \left[P^{-1} + R\right]^{-1} $
Thank you. That $\displaystyle {B^{-1}A^{-1} = (AB)^{-1}}$ part was what I didn't know about.
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