# How to obtain matrix from summation inverse equation

I have a set of square matrices $$\{A_i\}_{i \in \{1,..., n\}}$$ and another square matrix of equal size $$K$$. Under the assumption that such a matrix exists and is unique, I want to find the unique $$B$$ that satisfies the following equation:

$$\sum_{i=1}^n(A_i+B)^{-1}=K$$

All matrices are of full rank and symmetric. Does anyone know of a way, algebraically or computationally, for me to work out the value of $$B$$?

• I'm not sure I understand - what happened to the inverse terms in the product? – JDoe2 Jul 31 '20 at 0:58

This does not seem a simple matrix equation to solve. Computationally, my first attempt would be with Newton's method, even if it takes $$O(k^6)$$ per iteration, where $$k$$ is the size of the matrices. The Jacobian of the map in the LHS is $$L_B f[H] = \sum_{i=1}^n (A_i+B)^{-1}H(A_i+B)^{-1},$$ and to solve the equation $$L_Bf[H] = Y$$ for $$H$$ given $$Y$$ you need to convert it to a $$k^2 \times k^2$$ linear system (there are faster algorithms to solve this linear matrix equation for $$n=2$$, but I do not think there is anything better otherwise).
If you need to solve it for dimensions for which this is unfeasible, then I would try turning it into a fixed-point equation, for instance $$B = \left(K - \sum_{i=2}^n (A_i+B)^{-1}\right)^{-1} - A_1,$$ and then hope that the iteration $$B_{m+1} = \left(K - \sum_{i=2}^n (A_i+B_m)^{-1}\right)^{-1} - A_1,$$ converges.
• @JDoe2 The typical strategy is showing that the right-hand side $g(B) = (K-\sum_{i>1} (A_i+B)^{-1})^{-1}-A_1$ is a contraction, but that does not seem obvious to me. Another possible way is showing that $B_{m+1}\geq B_m$ in a certain ordering, for instance the positive-definite ordering or the componentwise ordering, but that requires some hypotheses on your matrices. – Federico Poloni Jul 26 '20 at 17:22