Two forms of $\big(A^{-1} + B^{-1}\big)^{-1}$

Question

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

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

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

Answer 1

$$\begin{split} B(A+B)^{-1}A&=B(A+B)^{-1}(A+B-B)\\&= B-B(A+B)^{-1}B\\&=B-(A+B-A)(A+B)^{-1}B \\&= B - B + A(A+B)^{-1}B\\& = A(A+B)^{-1}B \end{split}$$

Answer 2

You only need that $A+B$ is invertible.

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

2. If $A$ or $B$ is not invertible, then since $\sigma(A)$ and $\sigma(B)$ are finite, there exists $r> 0$ 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 $B(A+B)^{-1}A =A(A+B)^{-1}B$.