Two forms of $\big(A^{-1} + B^{-1}\big)^{-1}$ I am trying to calculate $\big(A^{-1} + B^{-1}\big)^{-1}$ and I find that there are two ways to get the answers:

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(A^{-1}(I + AB^{-1})\big)^{-1}
 = \big(A^{-1}(B+A)B^{-1}\big)^{-1} = B(A+B)^{-1}A$$


$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(B^{-1}(BA^{-1} + I)\big)^{-1}
 = \big(B^{-1}(B + A)A^{-1}\big)^{-1}
 = A(A+B)^{-1}B$$
So my question is that: is there a way to prove that
$$B(A+B)^{-1}A = A(A+B)^{-1}B$$
without proving that they are both equal to the inverse of $(A^{-1}+B^{-1})$ ?
$$\begin{split}
B(A+B)^{-1}A&amp;=B(A+B)^{-1}(A+B-B)\\&amp;= B-B(A+B)^{-1}B\\&amp;=B-(A+B-A)(A+B)^{-1}B
\\&amp;= B - B + A(A+B)^{-1}B\\&amp; = A(A+B)^{-1}B
\end{split}$$
You only need that $A+B$ is invertible.

If $A$ and $B$ are both invertible, then as you demonstrated, it follows that $A^{-1}+B^{-1}$ is also invertible and $$B(A+B)^{-1}A = (A^{-1}+B^{-1})^{-1} =A(A+B)^{-1}B$$

If $A$ or $B$ is not invertible, then since $\sigma(A)$ and $\sigma(B)$ are finite, there exists $r&gt; 0$ such that for all $\varepsilon \in \langle0,r\rangle$ we have that $A_\varepsilon := A-\varepsilon I$ and $B_\varepsilon := B+\varepsilon I$ are both invertible. Now from the first case it follows that
$$B_\varepsilon(A+B)^{-1}A_\varepsilon = B_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}A_\varepsilon = A_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}B_\varepsilon = A_\varepsilon(A+B)^{-1}B_\varepsilon.$$
Letting $\varepsilon \to 0$ from continuity of matrix operations we get $B(A+B)^{-1}A =A(A+B)^{-1}B$.