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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_lapack_zgglse (f08zn)

## Purpose

nag_lapack_zgglse (f08zn) solves a complex linear equality-constrained least squares problem.

## Syntax

[a, b, c, d, x, info] = f08zn(a, b, c, d, 'm', m, 'n', n, 'p', p)
[a, b, c, d, x, info] = nag_lapack_zgglse(a, b, c, d, 'm', m, 'n', n, 'p', p)

## Description

nag_lapack_zgglse (f08zn) solves the complex linear equality-constrained least squares (LSE) problem
 minimize ‖c − Ax‖2  subject to  Bx = d x
$minimize x ‖c-Ax‖2 subject to Bx=d$
where A$A$ is an m$m$ by n$n$ matrix, B$B$ is a p$p$ by n$n$ matrix, c$c$ is an m$m$ element vector and d$d$ is a p$p$ element vector. It is assumed that pnm + p$p\le n\le m+p$, rank(B) = p$\mathrm{rank}\left(B\right)=p$ and rank(E) = n$\mathrm{rank}\left(\mathrm{E}\right)=n$, where E =
 ( A ) B
$E=\left(\begin{array}{c}A\\ B\end{array}\right)$. These conditions ensure that the LSE problem has a unique solution, which is obtained using a generalized RQ$RQ$ factorization of the matrices B$B$ and A$A$.

## References

Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia
Anderson E, Bai Z and Dongarra J (1992) Generalized QR factorization and its applications Linear Algebra Appl. (Volume 162–164) 243–271
Eldèn L (1980) Perturbation theory for the least squares problem with linear equality constraints SIAM J. Numer. Anal. 17 338–350

## Parameters

### Compulsory Input Parameters

1:     a(lda, : $:$) – complex array
The first dimension of the array a must be at least max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The m$m$ by n$n$ matrix A$A$.
2:     b(ldb, : $:$) – complex array
The first dimension of the array b must be at least max (1,p)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{p}}\right)$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The p$p$ by n$n$ matrix B$B$.
3:     c(m) – complex array
m, the dimension of the array, must satisfy the constraint m0${\mathbf{m}}\ge 0$.
The right-hand side vector c$c$ for the least squares part of the LSE problem.
4:     d(p) – complex array
p, the dimension of the array, must satisfy the constraint 0pnm + p$0\le {\mathbf{p}}\le {\mathbf{n}}\le {\mathbf{m}}+{\mathbf{p}}$.
The right-hand side vector d$d$ for the equality constraints.

### Optional Input Parameters

1:     m – int64int32nag_int scalar
Default: The dimension of the array c and the first dimension of the array a. (An error is raised if these dimensions are not equal.)
m$m$, the number of rows of the matrix A$A$.
Constraint: m0${\mathbf{m}}\ge 0$.
2:     n – int64int32nag_int scalar
Default: The second dimension of the arrays a, b.
n$n$, the number of columns of the matrices A$A$ and B$B$.
Constraint: n0${\mathbf{n}}\ge 0$.
3:     p – int64int32nag_int scalar
Default: The dimension of the array d and the first dimension of the array b. (An error is raised if these dimensions are not equal.)
p$p$, the number of rows of the matrix B$B$.
Constraint: 0pnm + p$0\le {\mathbf{p}}\le {\mathbf{n}}\le {\mathbf{m}}+{\mathbf{p}}$.

### Input Parameters Omitted from the MATLAB Interface

lda ldb work lwork

### Output Parameters

1:     a(lda, : $:$) – complex array
The first dimension of the array a will be max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
2:     b(ldb, : $:$) – complex array
The first dimension of the array b will be max (1,p)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{p}}\right)$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
3:     c(m) – complex array
The residual sum of squares for the solution vector x$x$ is given by the sum of squares of elements c(np + 1),c(np + 2),,c(m)${\mathbf{c}}\left({\mathbf{n}}-{\mathbf{p}}+1\right),{\mathbf{c}}\left({\mathbf{n}}-{\mathbf{p}}+2\right),\dots ,{\mathbf{c}}\left({\mathbf{m}}\right)$; the remaining elements are overwritten.
4:     d(p) – complex array
5:     x(n) – complex array
The solution vector x$x$ of the LSE problem.
6:     info – int64int32nag_int scalar
info = 0${\mathbf{info}}=0$ unless the function detects an error (see Section [Error Indicators and Warnings]).

## Error Indicators and Warnings

info = i${\mathbf{info}}=-i$
If info = i${\mathbf{info}}=-i$, parameter i$i$ had an illegal value on entry. The parameters are numbered as follows:
1: m, 2: n, 3: p, 4: a, 5: lda, 6: b, 7: ldb, 8: c, 9: d, 10: x, 11: work, 12: lwork, 13: 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.
INFO = 1${\mathbf{INFO}}=1$
The upper triangular factor R$R$ associated with B$B$ in the generalized RQ$RQ$ factorization of the pair (B,A)$\left(B,A\right)$ is singular, so that rank(B) < p$\mathrm{rank}\left(B\right); the least squares solution could not be computed.
INFO = 2${\mathbf{INFO}}=2$
The (NP)$\left(N-P\right)$ by (NP)$\left(N-P\right)$ part of the upper trapezoidal factor T$T$ associated with A$A$ in the generalized RQ$RQ$ factorization of the pair (B,A)$\left(B,A\right)$ is singular, so that the rank of the matrix (E$E$) comprising the rows of A$A$ and B$B$ is less than n$n$; the least squares solutions could not be computed.

## Accuracy

For an error analysis, see Anderson et al. (1992) and Eldèn (1980). See also Section 4.6 of Anderson et al. (1999).

When mn = p$m\ge n=p$, the total number of real floating point operations is approximately (8/3)n2(6m + n)$\frac{8}{3}{n}^{2}\left(6m+n\right)$; if pn$p\ll n$, the number reduces to approximately (8/3)n2(3mn)$\frac{8}{3}{n}^{2}\left(3m-n\right)$.

## Example

```function nag_lapack_zgglse_example
a = [ 0.96 - 0.81i,  -0.03 + 0.96i,  -0.91 + 2.06i,  -0.05 + 0.41i;
-0.98 + 1.98i,  -1.2 + 0.19i,  -0.66 + 0.42i,  -0.81 + 0.56i;
0.62 - 0.46i,  1.01 + 0.02i,  0.63 - 0.17i,  -1.11 + 0.6i;
0.37 + 0.38i,  0.19 - 0.54i,  -0.98 - 0.36i,  0.22 - 0.2i;
0.83 + 0.51i,  0.2 + 0.01i,  -0.17 - 0.46i,  1.47 + 1.59i;
1.08 - 0.28i,  0.2 - 0.12i,  -0.07 + 1.23i,  0.26 + 0.26i];
b = [complex(1),  0 + 0i,  -1 + 0i,  0 + 0i;
0 + 0i,  1 + 0i,  0 + 0i,  -1 + 0i];
c = [ -2.54 + 0.09i;
1.65 - 2.26i;
-2.11 - 3.96i;
1.82 + 3.3i;
-6.41 + 3.77i;
2.07 + 0.66i];
d = [complex(0);
0 + 0i];
% Solve the equality-constrained least-squares problem
% minimize ||c - A*x|| (in the 2-norm) subject to B*x = D
[a, b, c, d, x, info] = nag_lapack_zgglse(a, b, c, d);

fprintf('\nConstrained least-squares solution\n');
disp(transpose(x));

rnorm = norm(c(3:6));
fprintf('Square root of the residual sum of squares\n');
disp(rnorm);
```
```

Constrained least-squares solution
1.0874 - 1.9621i  -0.7409 + 3.7297i   1.0874 - 1.9621i  -0.7409 + 3.7297i

Square root of the residual sum of squares
0.1587

```
```function f08zn_example
a = [ 0.96 - 0.81i,  -0.03 + 0.96i,  -0.91 + 2.06i,  -0.05 + 0.41i;
-0.98 + 1.98i,  -1.2 + 0.19i,  -0.66 + 0.42i,  -0.81 + 0.56i;
0.62 - 0.46i,  1.01 + 0.02i,  0.63 - 0.17i,  -1.11 + 0.6i;
0.37 + 0.38i,  0.19 - 0.54i,  -0.98 - 0.36i,  0.22 - 0.2i;
0.83 + 0.51i,  0.2 + 0.01i,  -0.17 - 0.46i,  1.47 + 1.59i;
1.08 - 0.28i,  0.2 - 0.12i,  -0.07 + 1.23i,  0.26 + 0.26i];
b = [complex(1),  0 + 0i,  -1 + 0i,  0 + 0i;
0 + 0i,  1 + 0i,  0 + 0i,  -1 + 0i];
c = [ -2.54 + 0.09i;
1.65 - 2.26i;
-2.11 - 3.96i;
1.82 + 3.3i;
-6.41 + 3.77i;
2.07 + 0.66i];
d = [complex(0);
0 + 0i];
% Solve the equality-constrained least-squares problem
% minimize ||c - A*x|| (in the 2-norm) subject to B*x = D
[a, b, c, d, x, info] = f08zn(a, b, c, d);

fprintf('\nConstrained least-squares solution\n');
disp(transpose(x));

rnorm = norm(c(3:6));
fprintf('Square root of the residual sum of squares\n');
disp(rnorm);
```
```

Constrained least-squares solution
1.0874 - 1.9621i  -0.7409 + 3.7297i   1.0874 - 1.9621i  -0.7409 + 3.7297i

Square root of the residual sum of squares
0.1587

```