Block matrix pseudoinverse: Difference between revisions

Revision as of 21:01, 25 August 2016

Revision as of 22:44, 25 August 2016

Line 6:

:<math> [\mathbf A, \mathbf B], \qquad \mathbf A \in \reals^{m\times n}, \qquad \mathbf B \in \reals^{m\times p}, \qquad m \geq n+p.</math>

If the above matrix is full rank, the [[Moore–Penrose pseudoinverse]] matrices of it and its transpose are as follows.

:<math>

\begin{bmatrix}

Line 19:

^{+} = [\mathbf A, \mathbf B] ([\mathbf A, \mathbf B]^T [\mathbf A, \mathbf B])^{-1}.

</math>

TheThis computation of the pseudoinverse requires (''n''&nbsp;+&nbsp;''p'')-square matrix inversion and does not take advantage of the block form.

To reduce computational costs to ''n''- and ''p''-square matrix inversions and to introduce parallelism, treating the blocks separately, one derives <ref name=Baksalary>{{cite journal|author=J.K. Baksalary and O.M. Baksalary|title=Particular formulae for the Moore–Penrose inverse of a columnwise partitioned matrix|journal=Linear Algebra Appl.|volume=421|date=2007|pages=16–23|doi=10.1016/j.laa.2006.03.031}}</ref>

To reduce complexity and introduce parallelism, we derive the following decomposed formula. From a block matrix inverse<math> \mathbf ([\mathbf A, \mathbf B]^T [\mathbf A, \mathbf B])^{-1}</math>, we can have{{citation needed|date=December 2010}}{{Original research|date=December 2010}}

:<math>

\begin{bmatrix}

\mathbf A, & \mathbf B

\end{bmatrix}

^{+} =

^{+} = \left[\mathbf P_B^\perp \mathbf A( \mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}, \quad \mathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}\right]^T,

\begin{bmatrix}

^{+} = \left[\mathbf P_B^\perp \mathbf A( \mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}, \quad \mathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}\right]^T,

\\

\quadmathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^{\perp} \mathbf B)^{+-1} ]. </math>

\end{bmatrix}

=

\begin{bmatrix}

(\mathbf P_B^{\perp}\mathbf A)^{+}

\\

(\mathbf P_A^{\perp}\mathbf B)^{+}

\end{bmatrix},

</math>

:<math>

Line 32 ⟶ 43:

\mathbf A^T \\ \mathbf B^T

\end{bmatrix}

^{+} = \left[\mathbf P_B^\perp \mathbf A( \mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}, \quad \mathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}\right],

= [(\mathbf A^T \mathbf P_B^{\perp})^{+},

\quad (\mathbf AB^T \\ \mathbf BP_A^T{\perp})^{+} ],

</math>

where [[orthogonal projection]] matrices are defined by

Line 41 ⟶ 54:

</math>

Note that theThe above formulaeformulas are not necessarily valid if <math>[\mathbf A, \mathbf B]</math> does not have full rank – for example, if <math>\mathbf A \neq 0</math>, then

Interestingly, from the [[idempotence]] of projection matrix, we can verify that the pseudoinverse of block matrix consists of pseudoinverse of projected matrices:

:<math>

\begin{bmatrix}

\mathbf A, & \mathbf B

\end{bmatrix}

^{+}

=

\begin{bmatrix}

(\mathbf P_B^{\perp}\mathbf A)^{+}

\\

(\mathbf P_A^{\perp}\mathbf B)^{+}

\end{bmatrix}, </math>

:<math>

\begin{bmatrix}

\mathbf A^T \\ \mathbf B^T

\end{bmatrix}

^{+}

= [(\mathbf A^T \mathbf P_B^{\perp})^{+},

\quad (\mathbf B^T \mathbf P_A^{\perp})^{+} ]. </math>

Thus, we decomposed the block matrix pseudoinverse into two submatrix pseudoinverses, which cost ''n''- and ''p''-square matrix inversions, respectively.

Note that the above formulae are not necessarily valid if <math>[\mathbf A, \mathbf B]</math> does not have full rank – for example, if <math>\mathbf A \neq 0</math>, then

:<math>

\begin{bmatrix}