I'm reading an essay about Properties of Matrices
A useful identity involving matrix inverses is the following $$\begin{align*}(\mathbf{P}^{-1}+\mathbf{B}^{\mathrm{T}}\mathbf{R}^{-1}\mathbf{B})^{-1}\mathbf{B}^{\mathrm{T}}\mathbf{R}^{-1}=\mathbf{P}\mathbf{B}^{\mathrm{T}}(\mathbf{B}\mathbf{P}\mathbf{B}^{\mathrm{T}}+\mathbf{R})^{-1}&&(\text{C.5})\end{align*}$$ which is easily verified by right multiplying both sides by $(\mathbf{B}\mathbf{P}\mathbf{B}^{\mathrm{T}}+\mathbf{R})$. Suppose that $\mathbf{P}$ has dimensionality $N\times N$ while $\mathbf{R}$ has dimensionality $M\times M$, so that $\mathbf{B}$ is $M\times N$. Then if $M\ll N$, it will be much cheaper to evaluate the right-hand side of (C.5) than the left-hand side. A special case that sometimes arises is $$\begin{align*}(\mathbf{I}+\mathbf{A}\mathbf{B})^{-1}\mathbf{A}=\mathbf{A}(\mathbf{I}+\mathbf{B}\mathbf{A})^{-1}&&(\text{C.6})\end{align*}$$
This section mentions an easy proof by right multiplication, but how to do it?
I will start from the C.5 equality and just do reversible operations. Then you can just start at the end and work backwards. From the equality, perform the indicated right multiplication, obtaining $$(P^{-1}+B^{T}R^{-1}B)^{-1}B^{T}R^{-1}(BPB^{T}+R)=PB^{T}$$ Distribute the $R^{-1}$ through the left set of parentheses: $$(P^{-1}+B^{T}R^{-1}B)^{-1}B^{T}(R^{-1}BPB^{T}+I)=PB^{T}$$ Do the same for $B^{T}$: $$(P^{-1}+B^{T}R^{-1}B)^{-1}(B^{T}R^{-1}BPB^{T}+B^{T})=PB^{T}$$ If we now right-factor out a $B^{T}$, we get $$(P^{-1}+B^{T}R^{-1}B)^{-1}(B^{T}R^{-1}BP+I)B^{T}=PB^{T}$$ One thing we could do now is apply cancellation since we are in a vector space, obtaining $$(P^{-1}+B^{T}R^{-1}B)^{-1}(B^{T}R^{-1}BP+I)=P$$ Right multiplying by $P^{-1}$, we get $$(P^{-1}+B^{T}R^{-1}B)^{-1}(B^{T}R^{-1}B+P^{-1})=I$$ Rearrange the right set of parentheses: $$(P^{-1}+B^{T}R^{-1}B)^{-1}(P^{-1}+B^{T}R^{-1}B)=I$$ So we get $I=I$.