Given an $n\times n$ invertible matrix $\mathbf A$ and two column vectors $\mathbf u$, $\mathbf v\in\mathbb R^n$, suppose that $1 + {\mathbf v}^T {\mathbf A}^{-1}\mathbf u \neq 0$. Then the Sherman-Morrison formula states that \begin{equation*} (\mathbf A + \mathbf u \mathbf v^T)^{-1} = \mathbf A^{-1} - {\mathbf A^{-1}\mathbf u\mathbf v^T \mathbf A^{-1} \over 1 + \mathbf v^T \mathbf A^{-1}\mathbf u}. \end{equation*}

Question: I'm wondering whether we have a similar formula when the inverse in the Sherman-Morrison formula is replaced by the Moore-Penrose Pseudoinverse in case that $\mathbf A$ is singular matrix.

Given an $n\times n$ invertible matrix $\mathbf A$ and two column vectors $\mathbf u$, $\mathbf v\in\mathbb R^n$, suppose that $1 + {\mathbf v}^T {\mathbf A}^{-1}\mathbf u \neq 0$. Then the Sherman-Morrison formula states that \begin{equation*} (\mathbf A + \mathbf u \mathbf v^T)^{-1} = \mathbf A^{-1} - {\mathbf A^{-1}\mathbf u\mathbf v^T \mathbf A^{-1} \over 1 + \mathbf v^T \mathbf A^{-1}\mathbf u}. \end{equation*}

Question: I'm wondering whether we have a similar formula when the inverse in the Sherman-Morrison formula is replaced by the Moore-Penrose pseudoinverse in case that $\mathbf A$ is singular matrix.