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$A$ 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!

user1551
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Sam A
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3 Answers3

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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*}

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As already commented, why not check?:

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

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

DonAntonio
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  • I am not sure if this is valid statement : $(A^{-1}+B^{-1}).B(A+B)^{-1}A=A^{-1}B(A+B)^{-1}A+B^{-1}B(A+B)^{-1}A = (A^{-1}B+I)(A+B)^{-1}A$ – Sam A Dec 23 '16 at 01:22
  • @SamA What exactly you think isn't valid? And I recognize the first step in my answer but not the second one... – DonAntonio Dec 23 '16 at 05:14
  • #DonAntonia, I meant the validity of my solution. Thank you for participating. – Sam A Dec 24 '16 at 01:31
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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.

user1551
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