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.

1 Derivation
2 Application to least squares problems
- 2.1 Column-wise partitioning in over-determined least squares
- 2.2 Row-wise partitioning in under-determined least squares
3 Comments on matrix inversion
4 Proof of block matrix inversion
5 Notes and references
6 See also
7 External links

Derivation

Consider a column-wise partitioned matrix:

$[\mathbf A, \mathbf B], \qquad \mathbf A \in \reals^{m\times n}, \qquad \mathbf B \in \reals^{m\times p}, \qquad m \geq n%2Bp.$

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

$\begin{bmatrix} \mathbf A, & \mathbf B \end{bmatrix} ^{%2B} = ([\mathbf A, \mathbf B]^T [\mathbf A, \mathbf B])^{-1} [\mathbf A, \mathbf B]^T,$

$\begin{bmatrix} \mathbf A^T \\ \mathbf B^T \end{bmatrix} ^{%2B} = [\mathbf A, \mathbf B] ([\mathbf A, \mathbf B]^T [\mathbf A, \mathbf B])^{-1}.$

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

$\begin{bmatrix} \mathbf A, & \mathbf B \end{bmatrix} ^{%2B} = \left[\mathbf P_B^\perp \mathbf A( \mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}, \quad \mathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}\right]^T,$

$\begin{bmatrix} \mathbf A^T \\ \mathbf B^T \end{bmatrix} ^{%2B} = \left[\mathbf P_B^\perp \mathbf A( \mathbf A^T \mathbf P_B^\perp \mathbf A)^{-1}, \quad \mathbf P_A^\perp \mathbf B(\mathbf B^T \mathbf P_A^\perp \mathbf B)^{-1}\right],$

where orthogonal projection matrices are defined by

$\begin{align} \mathbf P_A^\perp & = \mathbf I - \mathbf A (\mathbf A^T \mathbf A)^{-1} \mathbf A^T, \\ \mathbf P_B^\perp & = \mathbf I - \mathbf B (\mathbf B^T \mathbf B)^{-1} \mathbf B^T. \end{align}$

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

$\begin{bmatrix} \mathbf A, & \mathbf B \end{bmatrix} ^{%2B} = \begin{bmatrix} (\mathbf P_B^{\perp}\mathbf A)^{%2B} \\ (\mathbf P_A^{\perp}\mathbf B)^{%2B} \end{bmatrix},$

$\begin{bmatrix} \mathbf A^T \\ \mathbf B^T \end{bmatrix} ^{%2B} = [(\mathbf A^T \mathbf P_B^{\perp})^{%2B}, \quad (\mathbf B^T \mathbf P_A^{\perp})^{%2B} ].$

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

$\begin{bmatrix} \mathbf A, & \mathbf A \end{bmatrix} ^{%2B} = \frac{1}{2} \begin{bmatrix} \mathbf A^{%2B} \\ \mathbf A^{%2B} \end{bmatrix} \neq \begin{bmatrix} (\mathbf P_A^{\perp}\mathbf A)^{%2B} \\ (\mathbf P_A^{\perp}\mathbf A)^{%2B} \end{bmatrix} = 0$

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:

$\begin{bmatrix} \mathbf A, & \mathbf B \end{bmatrix} \begin{bmatrix} \mathbf x_1 \\ \mathbf x_2 \\ \end{bmatrix} = \mathbf d , \qquad \mathbf d \in \reals^{m\times 1}.$

Using the block matrix pseudoinverse, we have

$\mathbf x = \begin{bmatrix} \mathbf A, & \mathbf B \end{bmatrix} ^{%2B}\, \mathbf d = \begin{bmatrix} (\mathbf P_B^{\perp} \mathbf A)^{%2B}\\ (\mathbf P_A^{\perp} \mathbf B)^{%2B} \end{bmatrix} \mathbf d .$

Therefore, we have a decomposed solution:

$\mathbf x_1 = (\mathbf P_B^{\perp} \mathbf A)^{%2B}\, \mathbf d , \qquad \mathbf x_2 = (\mathbf P_A^{\perp} \mathbf B)^{%2B} \, \mathbf d .$

Row-wise partitioning in under-determined least squares

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

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

The minimum-norm solution is given by

$\mathbf x = \begin{bmatrix} \mathbf A^T \\ \mathbf B^T \end{bmatrix} ^{%2B}\, \begin{bmatrix} \mathbf e \\ \mathbf f \end{bmatrix}.$

Using the block matrix pseudoinverse, we have

$\mathbf x = [(\mathbf A^T\mathbf P_B^{\perp})^{%2B}, \quad (\mathbf B^T\mathbf P_A^{\perp})^{%2B} ] \begin{bmatrix} \mathbf e \\ \mathbf f \end{bmatrix} = (\mathbf A^T\mathbf P_B^{\perp})^{%2B}\,\mathbf e %2B (\mathbf B^T\mathbf P_A^{\perp} )^{%2B}\,\mathbf f .$

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.

Proof of 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 ^[1].

$\begin{bmatrix} A & B \\ C & D \end{bmatrix}^{-1} = \begin{bmatrix} (A - BD^{-1}C)^{-1} & -A^{-1}B(D - CA^{-1}B)^{-1} \\ -D^{-1}C(A - BD^{-1}C)^{-1} & (D - CA^{-1}B)^{-1} \end{bmatrix} = \begin{bmatrix} S^{-1}_D & -A^{-1}BS^{-1}_A \\ -D^{-1}CS^{-1}_D & S^{-1}_A \end{bmatrix} ,$

where $S_A$ and $S_D$ , respectively, Schur complements of $A$ and $D$ , are defined by $S_A = D - C A^{-1}B$ , and $S_D = A - BD^{-1}C$ . This relation is derived by using Block Triangular Decomposition. It is called simple block matrix inversion.^[2]

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.

Notes and references

External links

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