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Given a block matrix of the form

\begin{pmatrix} A & B^* \\ B & 0 \end{pmatrix}

where $A$ is singular (otherwise one could simply use the well-known block matrix inverse), is there a simpler formula for the Moore-Penrose inverse than the general one?

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  • $\begingroup$ @Eric What's wrong with pseudoinverse? This is hardly the only question about such, and since the matrix is not necessarily invertible, inverse alone is not exactly correct $\endgroup$
    – Tobias Kienzler
    Nov 26, 2014 at 18:29
  • $\begingroup$ I went back and forth on this one (I first saw it Nov 23); I'm not opposed to this tag per se, but I am opposed to the proliferation of unused tags. The fact that there are 69 questions in the search results for "pseudoinverse", all of which get along just fine without the tag pseudoinverse, was ultimately what tipped me. But I admit it's an equally compelling argument to go the other way — just that for me that's not an option because I'm not going to edit 69 questions just for retagging. $\endgroup$
    – Eric Stucky
    Nov 26, 2014 at 23:11
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    $\begingroup$ @EricStucky You wouldn't have to tag them all by yourself. Anyway, the two of us are too small a sample to make a community-decision, so let's discuss this on meta ;) $\endgroup$
    – Tobias Kienzler
    Nov 27, 2014 at 9:28

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This question is merely a special case of another one with $C=B^*$ (actually, $B$ and $B^*$ are swapped, but I'll implicitly do that). So user1551's answer can be adapted:

If you are looking for a closed-form formula in terms of $A,B$ and $C$, I am very skeptical about its usefulness. Yet that doesn't mean there isn't one: since $X$ is invertible, $$ X^{-1} = (X^TX)^{-1}X^T > =\begin{bmatrix}A^TA+C^TC & A^TB\\ B^TA & B^TB\end{bmatrix}^{-1} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}. $$ As $B$ has full column rank, $B^TB$ is invertible. Therefore you can use the formula for Schur complement to calculate the inverse of the block matrix on the RHS above, and the result is $$ X^{-1}= \begin{bmatrix} S^{-1} & -S^{-1} A^TB (B^TB)^{-1} \\ -(B^TB)^{-1} B^TA S^{-1} & (B^TB)^{-1} + (B^TB)^{-1} B^TA S^{-1} A^TB (B^TB)^{-1} \end{bmatrix} \begin{bmatrix}A^T & C^T\\ B^T & 0\end{bmatrix}, $$ where $S=A^TA+C^TC-A^TB(B^TB)^{-1}B^TA$.

In this case that means:

\begin{align} S &= A^TA + B^TB - A^TB^*(\bar B B^*)^{-1}\bar BA\quad \text{where}\ \bar B=(B^*)^T\text{, i.e. the complex conjugate}, \\ X^{-1} &= \begin{bmatrix} S^{-1} & -S^{-1}A^TB^*(\bar BB^*)^{-1} \\ -(\bar BB^T)^{-1}\bar BAS^{-1} & (\bar BB^*)^{-1}+(\bar BB^*)^{-1}\bar BAS^{-1}A^TB^*(\bar BB^*)^{-1} \end{bmatrix}\begin{bmatrix} A^T & B^T \\ \bar B & 0 \end{bmatrix}. \end{align}

That is unfortunately not really simpler...

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