# Does the Sherman-Morrison formula apply with "rank-one" block matrices?

Consider the matrix

$$B = A + \begin{bmatrix} R_{1} & R_{2} \\ R_{1} & R_{2} \end{bmatrix} = A + \begin{bmatrix} I \\ I\end{bmatrix} \begin{bmatrix} R_{1} & R_{2} \end{bmatrix}$$ where $$A$$ is an invertible square matrix, and $$R_1$$ and $$R_2$$ are symmetric matrices. Notice that the block matrix $$R$$ is akin to a rank-one matrix, in the sense that the two rows are the same.

If $$R_1$$ and $$R_2$$ were scalar, we could use the Sherman–Morrison formula to compute $$B^{-1}$$ in a simple way. Is there a version of that formula that would apply when $$R_1$$ and $$R_2$$ are matrices?

I think you are looking for something like the Woodbury matrix identity. In the link, choose: $$U,V^{-1}$$ to be eigenvectors corresponding to non-zero eigenvalues of the given matrix
$$UDU^{-1} = \begin{pmatrix} R_1 & R_2 \\ R_1 & R_2 \end{pmatrix}$$
$$V = U^{-1}$$, $$C = D$$ (in the link). Use only those columns in $$U$$ with non-zero eigenvalues in $$D$$.
• Thanks for your help! I'll give this a try, but I suspect that doing the spectral decomposition of $R$ won't be simple in my case. May 26 at 13:42
• You can also use/try $U = [I,I]^T$, $C = I$, $V = [R_1,R_2]$. May 26 at 14:11