Inverse of sum of two matrix inverses

Question

$A$ is a positive definite matrix and $B$ is a positive diagonal matrix. We want to figure out the inverse of the sum of inverses of $A+B$, namely $(A^{-1}+B^{-1})^{-1}$. Is it correct to apply the following general matrices identity? $$(A^{-1}+B^{-1})^{-1}=B(A+B)^{-1}A.$$ Thank you!

Answer 1

One can easily prove the identity $A^{-1} + B^{-1} = A^{-1}(A+B)B^{-1}$ (pre- multiply with $A$ and post-multiply with $B$ and recall the matrix addition is commutative). Now since $A$ is positive definite it is invertible (definition) and since $B$ is positive diagonal its determinant $\text{det}(B) = \prod_i b_{ii}$ is bigger than zero hence it also invertible. One thing to note is that the sum $A+B$ does not have to be invertible (take $A = -I_n$ and $B = I_n$), but for these $A$ and $B$ it is. Since $A+B$ is invertible, $(A^{-1} + B^{-1})^{-1}$ is also because it is the product of invertible matrices. The inverse is then given by \begin{align*} (A^{-1} + B^{-1})^{-1} = \big( A^{-1}(A+B)B^{-1} \big)^{-1} = B(A+B)^{-1}A. \end{align*}

Answer 2

Yes. A positive diagonal matrix is positive definite too. Therefore $A+B$ is positive definite and in turn invertible. So, $(A+B)^{-1}$ exists and the identity is applicable.

Answer 3

As already commented, why not check?:

$$\left(A^{-1}+B^{-1}\right)\cdot B(A+B)^{-1}A=A^{-1}B(A+B)^{-1}A+\overbrace{(A+B)^{-1}A}^{=A^{-1}A(A+B)^{-1}A}=$$

$$A^{-1}\left(B+A\right)(A+B)^{-1}A=A^{-1}IA=I$$