# Inverse of the sum of matrices

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

In general, $$A+BA+B$$ need not be invertible, even when $$AA$$ and $$BB$$ are. But one might ask whether you can have a formula under the additional assumption that $$A+BA+B$$ 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 $$AA$$ and $$A+BA+B$$ are invertible, and $$BB$$ has rank $$11$$, then let $$g=trace(BA−1)g=\operatorname{trace}(BA^{-1})$$. Then $$g≠−1g\neq -1$$ and
$$(A+B)−1=A−1−11+gA−1BA−1.(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

From this lemma, we can take a general $$A+BA+B$$ that is invertible and write it as $$A+B=A+B1+B2+⋯+BrA+B = A + B_1+B_2+\cdots+B_r$$, where $$BiB_i$$ each have rank $$11$$ and such that each $$A+B1+⋯+BkA+B_1+\cdots+B_k$$ is invertible (such a decomposition always exists if $$A+BA+B$$ is invertible and $$rank(B)=r\mathrm{rank}(B)=r$$). Then you get:

Theorem. Let $$AA$$ and $$A+BA+B$$ be nonsingular matrices, and let $$BB$$ have rank $$r>0r\gt 0$$. Let $$B=B1+⋯+BrB=B_1+\cdots+B_r$$, where each $$BiB_i$$ has rank $$11$$, and each $$Ck+1=A+B1+⋯+BkC_{k+1} = A+B_1+\cdots+B_k$$ is nonsingular. Setting $$C1=AC_1 = A$$, then
$$C−1k+1=C−1k−gkC−1kBkC−1kC_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$
where $$gk=11+trace(C−1kBk)g_k = \frac{1}{1 + \operatorname{trace}(C_k^{-1}B_k)}$$. In particular,
$$(A+B)−1=C−1r−grC−1rBrC−1r.(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$

(If the rank of $$BB$$ is $$00$$, then $$B=0B=0$$, so $$(A+B)−1=A−1(A+B)^{-1}=A^{-1}$$).