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There exists a convenient formula for computing the inverse of a block matrix consisting of 4 matrices $\mathbf{A, B, C, D}$

$ \begin{bmatrix}\mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix} ^{-1}$

the inverse can be written as a function of $A^{-1}$ and $(A-B D^{-1}C)^{-1}$ (wikipedia)

$\begin{bmatrix} \mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \end{bmatrix}$

I wonder if a similar formula exists for the pseudo-inverse of non-invertible block matrices.

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Someone might check whether or not the formula given above, with all ${}^{-1}$s replaced with $^+$s, satisfies the conditions for the MP pseudoinverse. That someone will not be me... – anon Sep 6 '11 at 9:15
Ha, I tried that on an example. Certainly doesn't work in general. (but I'll try again just in case I made a mistake) – Wouter Sep 6 '11 at 9:20
The biggest obstacle is due to the fact that, as you found out, pseudoinverses are not substitutes for the regular inverses. Please check the definition on elsewhere, there is no place that says $AA^\dagger = I$. Therefore, the woodbury-sherman-morrison does not roll out directly from such a construction. – user13838 Sep 6 '11 at 11:11
@percusse: the formula given above is quite a different beast from SMW... – J. M. Sep 6 '11 at 11:23
@J.M. : If you take the Schur complement of a matrix and invert the three block you will get this formula. How is it different from this? – user13838 Sep 6 '11 at 11:34
up vote 4 down vote accepted

Using the fabulous search engine netted me quite a lot of references on the pseudoinverses of partitioned matrices. Miao's paper gives a pretty general formula, Cline's paper concentrates on column-partitioned matrices $\mathbf A=[\mathbf U\quad\mathbf V]$, Rohde's paper handles the pseudoinverse of a partitioned Hermitian matrix, Hall and Hartwig's paper give formulae for column-partitioned matrices and row-partitioned matrices $\mathbf A=\bigl[\begin{smallmatrix}\mathbf U\\\mathbf V\end{smallmatrix}\bigr]$, and this paper by Hartwig gives an expression for the pseudoinverse of a bordered matrix, $\mathbf A=\bigl[\begin{smallmatrix}\mathbf B&\mathbf u\\\mathbf v^T&c\end{smallmatrix}\bigr]$.

As you can probably tell from the number of papers that needed to treat special cases (depending usually on the ranks of the blocks and such), the expression for the pseudoinverse of a block matrix can be quite complicated. I will not try to reproduce those formulae here, and instead ask you to consult those papers and their references.

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Thank you, I will read those papers. – Wouter Sep 6 '11 at 11:04

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