Inverse of the sum of matrices

I have two square matrices: A and B. A1 is known and I want to calculate (A+B)1. Are there theorems that help with calculating the inverse of the sum of matrices? In general case B1 is not known, but if it is necessary then it can be assumed that B1 is also known.

Answer

In general, A+B need not be invertible, even when A and B are. But one might ask whether you can have a formula under the additional assumption that A+B is invertible.

As noted by Adrián Barquero, there is a paper by Ken Miller published in the Mathematics Magazine in 1981 that addresses this.

He proves the following:

Lemma. If A and A+B are invertible, and B has rank 1, then let g=trace(BA1). Then g1 and
(A+B)1=A111+gA1BA1.

From this lemma, we can take a general A+B that is invertible and write it as A+B=A+B1+B2++Br, where Bi each have rank 1 and such that each A+B1++Bk is invertible (such a decomposition always exists if A+B is invertible and rank(B)=r). Then you get:

Theorem. Let A and A+B be nonsingular matrices, and let B have rank r>0. Let B=B1++Br, where each Bi has rank 1, and each Ck+1=A+B1++Bk is nonsingular. Setting C1=A, then
C1k+1=C1kgkC1kBkC1k
where gk=11+trace(C1kBk). In particular,
(A+B)1=C1rgrC1rBrC1r.

(If the rank of B is 0, then B=0, so (A+B)1=A1).

Attribution
Source : Link , Question Author : Tomek Tarczynski , Answer Author : Michael Hardy

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