Generalized RQ Factorization

The generalized RQ (GRQ) factorization of an m-by-n matrix A and a p-by-n matrix B is given by the pair of factorizations

$$A = R Q \quad \mbox{and} \quad B = Z T Q$$

where Q and Z are respectively n-by-n and p-by-p orthogonal matrices (or unitary matrices if A and B are complex). R has the form

$$R = \bordermatrix{ & n-m & m \cr m & 0 & R_{12} }, \quad \mbox{if} \quad m \leq n$$

or

$$R = \bordermatrix{ & n \cr m-n & R_{11} \cr \hfill n & R_{21} }, \quad \mbox{if} \quad m > n,$$

where R₁₂ or R₂₁ is upper triangular. T has the form

$$T = \bordermatrix{ & n \cr \hfill n & T_{11} \cr p-n & 0 }, \quad \mbox{if} \quad p \geq n$$

or

$$T = \bordermatrix{ & p & n-p \cr p & T_{11} & T_{12} }, \quad \mbox{if} \quad p < n$$

where T₁₁ is upper triangular.

Note that if B is square and nonsingular, the GRQ factorization of A and B implicitly gives the RQ factorization of the matrix AB^-1:

A B^-1 = ( R T^-1 ) Z^T

without explicitly computing the matrix inverse B^-1 or the product AB^-1.

The routine xGGRQF computes the GRQ factorization by first computing the RQ factorization of A and then the QR factorization of BQ^T. The orthogonal (or unitary) matrices Q and Z can either be formed explicitly or just used to multiply another given matrix in the same way as the orthogonal (or unitary) matrix in the RQ factorization (see section 2.4.2).

The GRQ factorization can be used to solve the linear equality-constrained least squares problem (LSE) (see (2.2) and [55, page 567]). We use the GRQ factorization of B and A (note that B and A have swapped roles), written as

$$B = T Q \quad \mbox{and} \quad A = Z R Q.$$

We write the linear equality constraints Bx = d as:

T Q x = d

which we partition as:

$$\bordermatrix{ & n-p & p \cr p & 0 & T_{12} } \left( \begin... ...t( \begin{array}{c} x_1 \\ x_2 \end{array} \right) \equiv Q x.$$

Therefore x₂ is the solution of the upper triangular system

T₁₂ x₂ = d

Furthermore,

$$\begin{eqnarray*} \Vert A x - c \Vert _2 & = & \Vert Z^T A x - Z^T c \Vert _2 \\ & = & \Vert R Q x - Z^T c \Vert _2. \end{eqnarray*}$$

We partition this expression as:

$$\bordermatrix{ & n-p & p \cr \hfill n-p & R_{11} & R_{12} \... ...ght) - \left( \begin{array}{c} c_1 \\ c_2 \end{array} \right)$$

where $\left( \begin{array}{c} c_1 \\ c_2 \end{array} \right) \equiv Z^T c$ , which can be computed by xORMQR (or xUNMQR).

To solve the LSE problem, we set

R₁₁ x₁ + R₁₂ x₂ - c₁ = 0

which gives x₁ as the solution of the upper triangular system

R₁₁ x₁ = c₁ - R₁₂ x₂.

Finally, the desired solution is given by

$$x = Q^T \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) ,$$

which can be computed by xORMRQ (or xUNMRQ).