Inverse of the sum of matrices

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 $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and $$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

From this lemma, we can take a general $A+B$ that is invertible and write it as $A+B = A + B_1+B_2+\cdots+B_r$, where $B_i$ each have rank $1$ and such that each $A+B_1+\cdots+B_k$ is invertible (such a decomposition always exists if $A+B$ is invertible and $\mathrm{rank}(B)=r$). Then you get:

Theorem. Let $A$ and $A+B$ be nonsingular matrices, and let $B$ have rank $r\gt 0$. Let $B=B_1+\cdots+B_r$, where each $B_i$ has rank $1$, and each $C_{k+1} = A+B_1+\cdots+B_k$ is nonsingular. Setting $C_1 = A$, then $$C_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$ where $g_k = \frac{1}{1 + \mathrm{trace}(C_k^{-1}B_k)}$. In particular, $$(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$

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

It is shown in On Deriving the Inverse of a Sum of Matrices that

$(A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}$.

This equation cannot be used to calculate $(A+B)^{-1}$, but it is useful for perturbation analysis where $B$ is a perturbation of $A$. There are several other variations of the above form (see equations (22)-(26) in this paper).

This result is good because it only requires $A$ and $A+B$ to be nonsingular. As a comparison, the SMW identity or Ken Miller's paper (as mentioned in the other answers) requires some nonsingualrity or rank conditions of $B$.

This I found accidentally.

Suppose given $A$, and $B$, where $A$ and $A+B$ are invertible. Now we want to know the expression of $(A+B)^{-1}$ without imposing the all inverse. Now we follow the intuition like this. Suppose that we can express $(A+B)^{-1} = A^{-1} + X$, next we will present simple straight forward method to compute $X$ \begin{equation} (A+B)^{-1} = A^{-1} + X \end{equation} \begin{equation} (A^{-1} + X) (A + B) = I \end{equation} \begin{equation} A^{-1} A + X A + A^{-1} B + X B = I \end{equation} \begin{equation} X(A + B) = - A^{-1} B \end{equation} \begin{equation} X = - A^{-1} B ( A + B)^{-1} \end{equation} \begin{equation} X = - A^{-1} B (A^{-1} + X) \end{equation} \begin{equation} (I + A^{-1}B) X = - A^{-1} B A^{-1} \end{equation} \begin{equation} X = - (I + A^{-1}B)^{-1} A^{-1} B A^{-1} \end{equation}

This lemma is simplification of lemma presented by Ken Miller, 1981