Blockwise Moore-Penrose pseudoinverse?

5.4k Views Asked by At

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.

2

There are 2 best solutions below

1
On BEST ANSWER

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.

0
On

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

$\begin{align*} Null(A) &\subset Null(B)\\\\ Null(A^{*}) &\subset Null(D^{*})\\\\ Null(H) &\subset Null(B^{*})\\\\ Null(H^{+}) &\subset Null(D)\\\\ H &= (C-BA^{+}D)^{+}.\\\\ \end{align*}$