TITLE: Blockwise Moore-Penrose pseudoinverse?
QUESTION [14 upvotes]: 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.

REPLY [3 votes]: This is quite a late answer, but I had been searching for this for a while as well and finally found something I thought was more satisfying. There is a paper "The Moore-Penrose inverse of a partitioned matrix $M  = \begin{bmatrix} A & D \\\  B & C \end{bmatrix} $" in Linear Algebra and its Applications by Hung & Markham from 1975. They give a more general formula for the pseudoinverse of a matrix, and then give a corollary with necessary and sufficient conditions that cause the formula to reduce to one that looks familiar. I'll record it here for progeny, but take a look at the paper.
Theorem Let $M=\begin{bmatrix}A&D\\\ B&C\end{bmatrix}$. Then
$M^{+} = \begin{bmatrix}K^{+}(A^{*}-EF) & K^{+}(B^{*}-EH)\\\ F & H\end{bmatrix}$,
where
$
\begin{align*}
K &= A^{*}A + B^{*}B\\\\
E &= A^{*}D + B^{*}C\\\\
R &= D - AK^{+}E\\\\
S &= C - BK^{+}E\\\\
L &= R^{*}R+S^{*}S\\\\
T &= K^{+}E(I-L^{+}L)\\\\
F &= L^{+}R^{*}+(I - L^{+}L)(I+T^{*}T)^{-1}(K^{+}E)^{*}K^{+}(A^{*}-EL^{+}R^{*})\\\\
H &= L^{+}S^{*}+(I-L^{+}L)(I+T^{*}T)^{-1}(K^{+}E)^{*}K^{+}(B^{*}-EL^{+}S^{*})
\end{align*}
$
Corollary $M^{+} = 
\begin{bmatrix}
A^{+}+A^{+}DHBA^{+} & -A^{+}DH\\\\
-HBA^{+} & H
\end{bmatrix}
$
if and only if
$\begin{align*}
Null(A) &\subset Null(B)\\\\
Null(A^{*}) &\subset Null(D^{*})\\\\
Null(H) &\subset Null(B^{*})\\\\
Null(H^{+}) &\subset Null(D)\\\\
H &= (C-BA^{+}D)^{+}.\\\\
\end{align*}$