# Sherman Morrison Formula with a sum of outer-product

I have a question, in which I am computing the inverse of a matrix $(A + \sum_{i=1}^{k}u_iv_i^T)$. Typically $k$ is small than $5$. And I know if $k = 1$, we can apply the Sherman Morrison Formula to quickly obtain the matrix inversion. Is there any fast way to compute the inversion, when $k > 1$? Note that Woodbury formula does not work in this case since $\sum_{i=1}^{k}u_iv_i^T$ cannot be expressed as $UCV$. Thank you.

• What are $A$, $u_i$ and $v_i$? You might want to make your question clearer. – Jack Sep 30 '16 at 22:09
• $A$ is a $r\times r$ invertible matrix and $u, v$ are $r$-dim vectors. – Hongyi Xu Oct 2 '16 at 4:36

I'll write $v_i$ as $v_i^T$ just to indicate it's a row vector. I do not see any problem with expressing the sum as $UCV$: $\sum_{i=1}^k u_i v_i^T = [u_1 \; u_2 \ldots u_k] [v_1 \; v_2 \ldots v_k]^T = UCV$ with $U=[u_1 \; u_2 \ldots u_k]$, $C=I$, $V=[v_1 \; v_2 \ldots v_k]^T$.
• Thanks. However, you will have cross terms $u_i v_j^T$, where $i$ is not equal to $j$. – Hongyi Xu Oct 2 '16 at 4:33