Block inversion when Schur-complement is zero

Question

Let $\pmb{M} \in \mathbb{C}^{N \times N}$ be a square symmetric matrix be partitioned as follows: \begin{equation} \pmb{M} = \left[ \begin{array}{c|c} \pmb{A} & \pmb{b} \\ \hline \pmb{b}^{\text{H}} & \beta \end{array} \right] \end{equation} where $\pmb{A} \in \mathbb{C}^{(N-1) \times (N-1)}$, $\pmb{b} \in \mathbb{C}^{(N-1) \times 1}$ and $\beta$ is a scalar.

Now, $\pmb{A}$ is invertible but the Schur complement of $\pmb{A}$ in $\pmb{M}$ is zero, namely

\begin{equation} \pmb{M}/\pmb{A} = \beta - \pmb{b}^{\text{H}}\pmb{A}^{-1}\pmb{b} = 0 \end{equation}

How can I compute a "pseudo-block-inverse" of $\pmb{M}$ in terms of $\pmb{A}$, $\pmb{b}$ and $\beta$?

Thanks a lot.

Answer 1

if you need $\mathbf M^{-1}:=\left[\begin{matrix}\mathbf X&\mathbf y\\\mathbf y^H&\alpha\end{matrix}\right]$, then it can by found from the equation $ \mathbf M^{-1}\mathbf M=\mathbf I. $ You will get (if I am not mistaken): $\mathbf X = (\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1}$, $y=-\frac 1 \beta (\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1})\mathbf b$, $\alpha = \frac 1\beta (1-\frac{1}{\beta}\mathbf b^H(\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1}\mathbf b)$

Answer 2

Here is a partial answer. If $\beta$ also happens to be zero, we get $$ M = \pmatrix{A&b\\ b^H&0} = \pmatrix{I\\ x^H}A\pmatrix{I&x} $$ where $x=A^{-1}b$. In this case it is straightforward to verify that the matrix $M^+$ below satisfies the four defining properties of Moore-Penrose pseudo-inverse of $M$: $$ M^+=\pmatrix{I\\ x^H}(I+xx^H)^{-1}A^{-1}(I+xx^H)^{-1}\pmatrix{I&x}. $$