The following is a well known theorem on the inverse of $(A+B)$. (Link to the paper: http://www.jstor.org/stable/2690437)
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 $(A+B)^{-1}=A^{-1}$).
Now my question is: can we instead of decomposing $B$ into $r$ matrices $B_1,B_2,\dots,B_r$, just consider columns of $B$ each by each i.e. if $B$ is a $n\times n$ matrix, then $B=B_1+B_2+\dots+B_n$ where each of $B_i$ is the $i'th$ column of $B$.
Any comment is highly appreciated.