NAG Toolbox: nag_lapack_dggrqf (f08zf)

Purpose

Syntax

Description

References

Parameters

Compulsory Input Parameters

1:     $\mathrm{a}\left(\mathit{lda},:\right)$ – double array

The first dimension of the array a must be at least $\max \left(1,\mathbf{m}\right)$.

The second dimension of the array a must be at least $\max \left(1,\mathbf{n}\right)$.

The $m$ by $n$ matrix $A$.

2:     $\mathrm{b}\left(\mathit{ldb},:\right)$ – double array

The first dimension of the array b must be at least $\max \left(1,\mathbf{p}\right)$.

The second dimension of the array b must be at least $\max \left(1,\mathbf{n}\right)$.

The $p$ by $n$ matrix $B$.

Optional Input Parameters

1:     $\mathrm{m}$ – int64int32nag_int scalar

Default: the first dimension of the array a.

$m$, the number of rows of the matrix $A$.

Constraint: $\mathbf{m}\ge 0$.

2:     $\mathrm{p}$ – int64int32nag_int scalar

Default: the first dimension of the array b.

$p$, the number of rows of the matrix $B$.

Constraint: $\mathbf{p}\ge 0$.

3:     $\mathrm{n}$ – int64int32nag_int scalar

Default: the second dimension of the arrays a, b.

$n$, the number of columns of the matrices $A$ and $B$.

Constraint: $\mathbf{n}\ge 0$.

Output Parameters

1:     $\mathrm{a}\left(\mathit{lda},:\right)$ – double array

The first dimension of the array a will be $\max \left(1,\mathbf{m}\right)$.

The second dimension of the array a will be $\max \left(1,\mathbf{n}\right)$.

If $m\le n$, the upper triangle of the subarray $\mathbf{a}\left(1:m,n-m+1:n\right)$ contains the $m$ by $m$ upper triangular matrix $R_{12}$.

If $m\ge n$, the elements on and above the $\left(m-n\right)$th subdiagonal contain the $m$ by $n$ upper trapezoidal matrix $R$; the remaining elements, with the array taua, represent the orthogonal matrix $Q$ as a product of $\min \left(m,n\right)$ elementary reflectors (see Representation of orthogonal or unitary matrices in the F08 Chapter Introduction).

2:     $\mathrm{taua}\left(\min \left(\mathbf{m},\mathbf{n}\right)\right)$ – double array

The scalar factors of the elementary reflectors which represent the orthogonal matrix $Q$.

3:     $\mathrm{b}\left(\mathit{ldb},:\right)$ – double array

The first dimension of the array b will be $\max \left(1,\mathbf{p}\right)$.

The second dimension of the array b will be $\max \left(1,\mathbf{n}\right)$.

The elements on and above the diagonal of the array contain the $\min \left(p,n\right)$ by $n$ upper trapezoidal matrix $T$ ($T$ is upper triangular if $p\ge n$); the elements below the diagonal, with the array taub, represent the orthogonal matrix $Z$ as a product of elementary reflectors (see Representation of orthogonal or unitary matrices in the F08 Chapter Introduction).

4:     $\mathrm{taub}\left(\min \left(\mathbf{p},\mathbf{n}\right)\right)$ – double array

The scalar factors of the elementary reflectors which represent the orthogonal matrix $Z$.

5:     $\mathrm{info}$ – int64int32nag_int scalar

$\mathbf{info}=0$ unless the function detects an error (see Error Indicators and Warnings).

Error Indicators and Warnings

   $\mathbf{info}=-i$

If $\mathbf{info}=-i$, parameter $i$ had an illegal value on entry. The parameters are numbered as follows:

1: m, 2: p, 3: n, 4: a, 5: lda, 6: taua, 7: b, 8: ldb, 9: taub, 10: work, 11: lwork, 12: info.

It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.

Accuracy

Further Comments

Example

Open in the MATLAB editor: f08zf_example

function f08zf_example


fprintf('f08zf example results\n\n');

% Find x that minimizes norm(c-Ax) subject to Bx = d .

m = int64(6);
n = int64(4);
p = int64(2);
a = [-0.57, -1.28, -0.39,  0.25;
     -1.93,  1.08, -0.31, -2.14;
      2.30,  0.24,  0.40, -0.35;
     -1.93,  0.64, -0.66,  0.08;
      0.15,  0.30,  0.15, -2.13;
     -0.02,  1.03, -1.43,  0.50];
b = [1, 0, -1, 0;
     0, 1, 0, -1];
c = [-1.50; -2.14;  1.23; -0.54; -1.68; 0.82];
d = [0; 0];

% Compute the generalized RQ factorization of (B,A) as
% A = ZRQ, B = TQ
[TQ, taub, ZR, taua, info] = f08zf( ...
                                    b, a);

% Set Qx = y. The problem reduces to:
% minimize (Ry - Z^Tc) subject to Ty = d

% Update c = Z^T*c -> minimize (Ry-c)
[cup, info] = f08ag( ...
                     'Left','Transpose',ZR,taua,c);

% Solve Ty = d for last p elements
T12 = triu(TQ(1:p,n-p+1:n));

[y2, info] = f07te( ...
                    'Upper', 'No transpose', 'Non-unit', T12, d);

% (from Ry-c) R11*y1 + R12*y2 = c1 --> R11*y1 = c1 - R12*y2
% Update c1
c1 = cup(1:n-p) - ZR(1:n-p,n-p+1:n)*y2;

% Solve R11*y1 = c1 for y1
R11 = triu(ZR(1:n-p,1:n-p));
[y1, info] = f07te( ...
                    'Upper', 'No transpose', 'Non-unit', R11, c1);

% Contruct y and backtransform for x = Q^Ty
y = [y1;y2];
[~, x, info] = f08ck( ...
                      'Left', 'Transpose', TQ, taub, y);

fprintf('Constrained least squares solution\n');
disp(x);
res = a*x - c;
fprintf('Residual norm\n%12.2e\n', norm(res));

f08zf example results

Constrained least squares solution
    0.4890
    0.9975
    0.4890
    0.9975

Residual norm
    2.51e-02