Prove $(P^{-1} + B^T R^{-1} B)^{-1} B^T R^{-1} = PB^T(BPB^T + R)^{-1}$

Question

It is Equation C.5 from https://www.seas.upenn.edu/~cis520/papers/bishop_appendix_C.pdf

I tried right multiply both sides with $(BPB^T + R)$, but not sure how to continue from there.

Answer 1

We have: \begin{align*} (P^{-1}+B^TR^{-1}B)^{-1}B^TR^{-1}&=PB^T(BPB^T+R)^{-1}\\ (P^{-1}+B^TR^{-1}B)^{-1}B^TR^{-1}(BPB^T+R)&=PB^T(BPB^T+R)^{-1}(BPB^T+R)\\ (P^{-1}+B^TR^{-1}B)^{-1}(B^TR^{-1}BPB^T+B^TR^{-1}R)&=PB^TI\\ (P^{-1}+B^TR^{-1}B)^{-1}(B^TR^{-1}BPB^T+B^T)&=PB^T\\ (P^{-1}+B^TR^{-1}B)^{-1}(B^TR^{-1}BPB^T+P^{-1}PB^T)&=PB^T\\ (P^{-1}+B^TR^{-1}B)^{-1}(B^TR^{-1}B+P^{-1})PB^T&=PB^T\\ IPB^T&=PB^T, \end{align*} which checks out.