If $A$ and $B$ are invertible matrices, how to show that: $(A+BB^T)^{-1}B = A^{-1}B(I+B^TA^{-1}B)^{-1}$

Question

The best I could come up with is $$ LHS = (AB^{-1}+BB^TB^{-1})^{-1} $$

$$ RHS = (B^{-1}A+B^{-1}AB^TAB^{-1})^{-1} $$

Answer 1

Your formulation of the left and right sides has a mistake. Recall that $$(UV)^{-1} = V^{-1} U^{-1}$$ so in particular, the order of the matrices has to be reversed. As such, the left-hand side is actually \begin{align*} (A + B B^T)^{-1} B &= (B^{-1} A + B^{-1} B B^T)^{-1} \\ &= (B^{-1}A + B^T)^{-1} \end{align*} and rewriting the terms on the right side will give the same expression.

Answer 2

Assuming that the inverses in the target identity exist in the first place (they won't if $A=-I$ and $B=I$, for example) you can just cross-multiply to get rid of the inverses $$(A+BB^T)^{-1}B = A^{-1}B(I+B^TA^{-1}B)^{-1} \\ \iff \\ B(I+B^TA^{-1}B) = (A+BB^T)A^{-1}B $$ and each sides of this simplifies to $B+BB^TA^{-1}B$.