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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.

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2 Answers 2

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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 $M$: $$ M^+=\pmatrix{I\\ x^H}(I+xx^H)^{-1}A^{-1}(I+xx^H)^{-1}\pmatrix{I&x}. $$

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  • $\begingroup$ Thank you for your satisfying comment.. I'm wondering what happens if $\beta = 0$ $\endgroup$ Commented Aug 23, 2016 at 13:31
  • $\begingroup$ in the above comment i meant $\beta \neq 0$ (typo).. By the way, your expression is very close to the one where $\beta \neq 0$ in epubs.siam.org/doi/abs/10.1137/0131003 .. A paper by R. E. Hartwig $\endgroup$ Commented Aug 23, 2016 at 14:46
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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): $\mathbf X = (\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1}$, $y=-\frac 1 \beta (\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1})\mathbf b$, $\alpha = \frac 1\beta (1-\frac{1}{\beta}\mathbf b^H(\mathbf A-\frac 1\beta\mathbf b\mathbf b^H)^{-1}\mathbf b)$

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  • $\begingroup$ Well, what you said is not true. Because as mentioned above \begin{equation} \beta = \pmb{b}^{\text{H}} \pmb{A}^{-1} \pmb{b} \end{equation} Therefore the matrix $\pmb{A} - \frac{1}{\beta}\pmb{b}\pmb{b}^{\text{H}}$ is singular because there exists a vector $\pmb{\alpha} = \pmb{A}^{-1}\pmb{b}$ such that $(\pmb{A} - \frac{1}{\beta}\pmb{b}\pmb{b}^{\text{H}})\pmb{\alpha} = \pmb{0}$. In other words, the matrix $\pmb{A} - \frac{1}{\beta}\pmb{b}\pmb{b}^{\text{H}}$ admits a null space $\endgroup$ Commented Aug 23, 2016 at 13:17
  • $\begingroup$ Another way of saying why this inverse doesn't exist is because $\pmb{A} - \frac{1}{\beta}\pmb{b}\pmb{b}^{\text{H}}$ is the Schur complement of $\alpha$ in $\pmb{M}$.. Therefore, it is also singular $\endgroup$ Commented Aug 23, 2016 at 13:24
  • $\begingroup$ I see, I missed the condition M/A=.... What do you means by ``pseudo-block inverse of square'' matrix $\mathbf M$? $\endgroup$ Commented Aug 23, 2016 at 13:29
  • $\begingroup$ It means a partitioned matrix with block entries in terms of $\pmb{A}$, $\pmb{b}$ and $\beta$..This explains the term "block" and "pseudo-inverse" since the inverse of $\pmb{M}$ does not exist. $\endgroup$ Commented Aug 23, 2016 at 13:33
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    $\begingroup$ The answer could possibly be found here math.stackexchange.com/questions/62249/… $\endgroup$ Commented Aug 23, 2016 at 13:47

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