How to prove $(P^{-1} + B^T R^{-1} B)^{-1} B^T R^{-1} = PB^T(BPB^T + R)^{-1}$ 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.

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.
