Block matrix pseudoinverse

In mathematics, block matrix pseudoinverse is a formula of pseudoinverse of a partitioned matrix. This is useful for decomposing or approximating many algorithms updating parameters in signal processing, which are based on least squares method.

Derivation

Consider a column-wise partitioned matrix:

$[\mathbf {A} ,\mathbf {B} ],\qquad \mathbf {A} \in \mathbb {R} ^{m\times n},\qquad \mathbf {B} \in \mathbb {R} ^{m\times p},\qquad m\geq n+p.$

If the above matrix is full rank, the pseudoinverse matrices of it and its transpose are as follows.

${\begin{bmatrix}\mathbf {A} ,&\mathbf {B} \end{bmatrix}}^{+}=([\mathbf {A} ,\mathbf {B} ]^{T}[\mathbf {A} ,\mathbf {B} ])^{-1}[\mathbf {A} ,\mathbf {B} ]^{T},$

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

The pseudoinverse requires (n + p)-square matrix inversion.

To reduce complexity and introduce parallelism, we derive the following decomposed formula. From a block matrix inverse$\mathbf {(} [\mathbf {A} ,\mathbf {B} ]^{T}[\mathbf {A} ,\mathbf {B} ])^{-1}$, we can have

${\begin{bmatrix}\mathbf {A} ,&\mathbf {B} \end{bmatrix}}^{+}=\left[\mathbf {P} _{B}^{\perp }\mathbf {A} (\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1},\quad \mathbf {P} _{A}^{\perp }\mathbf {B} (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\right]^{T},$

${\begin{bmatrix}\mathbf {A} ^{T}\\\mathbf {B} ^{T}\end{bmatrix}}^{+}=\left[\mathbf {P} _{B}^{\perp }\mathbf {A} (\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1},\quad \mathbf {P} _{A}^{\perp }\mathbf {B} (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\right],$

where orthogonal projection matrices are defined by

${\begin{aligned}\mathbf {P} _{A}^{\perp }&=\mathbf {I} -\mathbf {A} (\mathbf {A} ^{T}\mathbf {A} )^{-1}\mathbf {A} ^{T},\\\mathbf {P} _{B}^{\perp }&=\mathbf {I} -\mathbf {B} (\mathbf {B} ^{T}\mathbf {B} )^{-1}\mathbf {B} ^{T}.\end{aligned}}$

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

${\begin{bmatrix}\mathbf {A} ,&\mathbf {B} \end{bmatrix}}^{+}={\begin{bmatrix}(\mathbf {P} _{B}^{\perp }\mathbf {A} )^{+}\\(\mathbf {P} _{A}^{\perp }\mathbf {B} )^{+}\end{bmatrix}},$

${\begin{bmatrix}\mathbf {A} ^{T}\\\mathbf {B} ^{T}\end{bmatrix}}^{+}=[(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp })^{+},\quad (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp })^{+}].$

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 $[\mathbf {A} ,\mathbf {B} ]$ does not have full rank – for example, if $\mathbf {A} \neq 0$, then

${\begin{bmatrix}\mathbf {A} ,&\mathbf {A} \end{bmatrix}}^{+}={\frac {1}{2}}{\begin{bmatrix}\mathbf {A} ^{+}\\\mathbf {A} ^{+}\end{bmatrix}}\neq {\begin{bmatrix}(\mathbf {P} _{A}^{\perp }\mathbf {A} )^{+}\\(\mathbf {P} _{A}^{\perp }\mathbf {A} )^{+}\end{bmatrix}}=0$

Application to least squares problems

Given the same matrices as above, we consider the following least squares problems, which appear as multiple objective optimizations or constrained problems in signal processing. Eventually, we can implement a parallel algorithm for least squares based on the following results.

Column-wise partitioning in over-determined least squares

Suppose a solution $\mathbf {x} ={\begin{bmatrix}\mathbf {x} _{1}\\\mathbf {x} _{2}\\\end{bmatrix}}$ solves an over-determined system:

${\begin{bmatrix}\mathbf {A} ,&\mathbf {B} \end{bmatrix}}{\begin{bmatrix}\mathbf {x} _{1}\\\mathbf {x} _{2}\\\end{bmatrix}}=\mathbf {d} ,\qquad \mathbf {d} \in \mathbb {R} ^{m\times 1}.$

Using the block matrix pseudoinverse, we have

$\mathbf {x} ={\begin{bmatrix}\mathbf {A} ,&\mathbf {B} \end{bmatrix}}^{+}\,\mathbf {d} ={\begin{bmatrix}(\mathbf {P} _{B}^{\perp }\mathbf {A} )^{+}\\(\mathbf {P} _{A}^{\perp }\mathbf {B} )^{+}\end{bmatrix}}\mathbf {d} .$

Therefore, we have a decomposed solution:

$\mathbf {x} _{1}=(\mathbf {P} _{B}^{\perp }\mathbf {A} )^{+}\,\mathbf {d} ,\qquad \mathbf {x} _{2}=(\mathbf {P} _{A}^{\perp }\mathbf {B} )^{+}\,\mathbf {d} .$

Row-wise partitioning in under-determined least squares

[Comment re. below: According to section "Derivation" above this method of calculating [A, B]+ for m>=n+p. Can it then be used for an underdetermined system where by definition m (size of x and equal to number of variables) > n+p (number of equations)?]

Suppose a solution $\mathbf {x}$ solves an under-determined system:

${\begin{bmatrix}\mathbf {A} ^{T}\\\mathbf {B} ^{T}\end{bmatrix}}\mathbf {x} ={\begin{bmatrix}\mathbf {e} \\\mathbf {f} \end{bmatrix}},\qquad \mathbf {e} \in \mathbb {R} ^{n\times 1},\qquad \mathbf {f} \in \mathbb {R} ^{p\times 1}.$

The minimum-norm solution is given by

$\mathbf {x} ={\begin{bmatrix}\mathbf {A} ^{T}\\\mathbf {B} ^{T}\end{bmatrix}}^{+}\,{\begin{bmatrix}\mathbf {e} \\\mathbf {f} \end{bmatrix}}.$

Using the block matrix pseudoinverse, we have

$\mathbf {x} =[(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp })^{+},\quad (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp })^{+}]{\begin{bmatrix}\mathbf {e} \\\mathbf {f} \end{bmatrix}}=(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp })^{+}\,\mathbf {e} +(\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp })^{+}\,\mathbf {f} .$

Comments on matrix inversion

Instead of $\mathbf {(} [\mathbf {A} ,\mathbf {B} ]^{T}[\mathbf {A} ,\mathbf {B} ])^{-1}$, we need to calculate directly or indirectly

$\quad (\mathbf {A} ^{T}\mathbf {A} )^{-1},\quad (\mathbf {B} ^{T}\mathbf {B} )^{-1},\quad (\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1},\quad (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}.$

In a dense and small system, we can use singular value decomposition, QR decomposition, or Cholesky decomposition to replace the matrix inversions with numerical routines. In a large system, we may employ iterative methods such as Krylov subspace methods.

Considering parallel algorithms, we can compute $(\mathbf {A} ^{T}\mathbf {A} )^{-1}$ and $(\mathbf {B} ^{T}\mathbf {B} )^{-1}$ in parallel. Then, we finish to compute $(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1}$ and $(\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}$ also in parallel.

Block matrix inversion

Let a block matrix be

${\begin{bmatrix}A&B\\C&D\end{bmatrix}}.$

We can get an inverse formula by combining the previous results in.

${\begin{bmatrix}A&B\\C&D\end{bmatrix}}^{-1}={\begin{bmatrix}(A-BD^{-1}C)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\\-D^{-1}C(A-BD^{-1}C)^{-1}&(D-CA^{-1}B)^{-1}\end{bmatrix}}={\begin{bmatrix}S_{D}^{-1}&-A^{-1}BS_{A}^{-1}\\-D^{-1}CS_{D}^{-1}&S_{A}^{-1}\end{bmatrix}},$

where $S_{A}$ and $S_{D}$, respectively, Schur complements of $A$ and $D$, are defined by $S_{A}=D-CA^{-1}B$, and $S_{D}=A-BD^{-1}C$. This relation is derived by using Block Triangular Decomposition. It is called simple block matrix inversion.

Now we can obtain the inverse of the symmetric block matrix:

${\begin{bmatrix}\mathbf {A} ^{T}\mathbf {A} &\mathbf {A} ^{T}\mathbf {B} \\\mathbf {B} ^{T}\mathbf {A} &\mathbf {B} ^{T}\mathbf {B} \end{bmatrix}}^{-1}={\begin{bmatrix}(\mathbf {A} ^{T}\mathbf {A} -\mathbf {A} ^{T}\mathbf {B} (\mathbf {B} ^{T}\mathbf {B} )^{-1}\mathbf {B} ^{T}\mathbf {A} )^{-1}&-(\mathbf {A} ^{T}\mathbf {A} )^{-1}\mathbf {A} ^{T}\mathbf {B} (\mathbf {B} ^{T}\mathbf {B} -\mathbf {B} ^{T}\mathbf {A} (\mathbf {A} ^{T}\mathbf {A} )^{-1}\mathbf {A} ^{T}\mathbf {B} )^{-1}\\-(\mathbf {B} ^{T}\mathbf {B} )^{-1}\mathbf {B} ^{T}\mathbf {A} (\mathbf {A} ^{T}\mathbf {A} -\mathbf {A} ^{T}\mathbf {B} (\mathbf {B} ^{T}\mathbf {B} )^{-1}\mathbf {B} ^{T}\mathbf {A} )^{-1}&(\mathbf {B} ^{T}\mathbf {B} -\mathbf {B} ^{T}\mathbf {A} (\mathbf {A} ^{T}\mathbf {A} )^{-1}\mathbf {A} ^{T}\mathbf {B} )^{-1}\end{bmatrix}}$

$={\begin{bmatrix}(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1}&-(\mathbf {A} ^{T}\mathbf {A} )^{-1}\mathbf {A} ^{T}\mathbf {B} (\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\\-(\mathbf {B} ^{T}\mathbf {B} )^{-1}\mathbf {B} ^{T}\mathbf {A} (\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1}&(\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\end{bmatrix}}$

Since the block matrix is symmetric, we also have

${\begin{bmatrix}\mathbf {A} ^{T}\mathbf {A} &\mathbf {A} ^{T}\mathbf {B} \\\mathbf {B} ^{T}\mathbf {A} &\mathbf {B} ^{T}\mathbf {B} \end{bmatrix}}^{-1}={\begin{bmatrix}(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1}&-(\mathbf {A} ^{T}\mathbf {P} _{B}^{\perp }\mathbf {A} )^{-1}\mathbf {A} ^{T}\mathbf {B} (\mathbf {B} ^{T}\mathbf {B} )^{-1}\\-(\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\mathbf {B} ^{T}\mathbf {A} (\mathbf {A} ^{T}\mathbf {A} )^{-1}&(\mathbf {B} ^{T}\mathbf {P} _{A}^{\perp }\mathbf {B} )^{-1}\end{bmatrix}}.$

Then, we can see how the Schur complements are connected to the projection matrices of the symmetric, partitioned matrix.

See also

• Invertible matrix#Blockwise inversion

External links

• The Matrix Reference Manual by Mike Brookes
• Linear Algebra Glossary by John Burkardt
• The Matrix Cookbook by Kaare Brandt Petersen
• Lecture 8: Least-norm solutions of undetermined equations by Stephen P. Boyd