# 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? Jul 31, 2020 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.

The scalar version of this equation is a secular equation, but searching for this term I found nothing interesting for matrix arguments.

• Hmm... I see - just to clarify - I am a little unfamiliar with fixed-point equations. How would I be able to prove convergence for a fixed point equation? Jul 26, 2020 at 6:26
• @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. Jul 26, 2020 at 17:22
• In practice, I suggest to implement it first, and then if it converges you can worry about proving theoretical properties. Jul 26, 2020 at 17:23

If $$n\geq 2$$, generically, you are unlikely to encounter such an equation with only one solution.

Let $$d$$ be the dimension of our symmetric matrices.

$$\bullet$$ Consider the case $$n=2$$. Then the algebraic resolution of the system leads to obtaining the roots of a real polynomial $$P$$ of degree $$2^d$$ (generically). Our system admits a unique solution iff $$P$$ admits only one real root. Since $$degree(P)$$ is even, this is not possible.

Then we must assume that $$n\geq 3$$.

$$\bullet$$ Note that a random polynomial of degree $$p$$ has

$$O(\log(p))$$ real zeros on average -when the coefficients are independent standard normals-

or $$O(\sqrt{p})$$ -when the variance varies with the index of the coefficient-.

If $$A_1,A_2,K$$ are randomly chosen, then $$P$$ can be roughly considered as random and the probability that $$P$$ admits a single real root is very small when $$p$$ is large.

$$\bullet$$ For example, if $$d=2,n=3$$, we obtain -generically- a polynomial of degree $$11$$. I did some tests and got $$5,7,9$$ or $$11$$ real roots.

Finally, the existence of a unique solution is a very special case.