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Chapter Contents
Chapter Introduction
NAG Toolbox

# NAG Toolbox: nag_lapack_dtzrzf (f08bh)

## Purpose

nag_lapack_dtzrzf (f08bh) reduces the $m$ by $n$ ($m\le n$) real upper trapezoidal matrix $A$ to upper triangular form by means of orthogonal transformations.

## Syntax

[a, tau, info] = f08bh(a, 'm', m, 'n', n)
[a, tau, info] = nag_lapack_dtzrzf(a, 'm', m, 'n', n)

## Description

The $m$ by $n$ ($m\le n$) real upper trapezoidal matrix $A$ given by
 $A = R1 R2 ,$
where ${R}_{1}$ is an $m$ by $m$ upper triangular matrix and ${R}_{2}$ is an $m$ by $\left(n-m\right)$ matrix, is factorized as
 $A = R 0 Z ,$
where $R$ is also an $m$ by $m$ upper triangular matrix and $Z$ is an $n$ by $n$ orthogonal matrix.

## 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 http://www.netlib.org/lapack/lug

## Parameters

### Compulsory Input Parameters

1:     $\mathrm{a}\left(\mathit{lda},:\right)$ – double array
The first dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
The second dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The leading $m$ by $n$ upper trapezoidal part of the array a must contain the matrix to be factorized.

### 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:     $\mathrm{n}$int64int32nag_int scalar
Default: the second dimension of the array a.
$n$, the number of columns of the matrix $A$.
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 $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
The second dimension of the array a will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The leading $m$ by $m$ upper triangular part of a contains the upper triangular matrix $R$, and elements ${\mathbf{m}}+1$ to n of the first $m$ rows of a, with the array tau, represent the orthogonal matrix $Z$ as a product of $m$ elementary reflectors (see Representation of orthogonal or unitary matrices in the F08 Chapter Introduction).
2:     $\mathrm{tau}\left(:\right)$ – double array
The dimension of the array tau will be $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The scalar factors of the elementary reflectors.
3:     $\mathrm{info}$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: n, 3: a, 4: lda, 5: tau, 6: work, 7: lwork, 8: 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

The computed factorization is the exact factorization of a nearby matrix $A+E$, where
 $E2 = O⁡ε A2$
and $\epsilon$ is the machine precision.

The total number of floating-point operations is approximately $4{m}^{2}\left(n-m\right)$.
The complex analogue of this function is nag_lapack_ztzrzf (f08bv).

## Example

This example solves the linear least squares problems
 $minx bj - Axj 2 , j=1,2$
for the minimum norm solutions ${x}_{1}$ and ${x}_{2}$, where ${b}_{j}$ is the $j$th column of the matrix $B$,
 $A = -0.09 0.14 -0.46 0.68 1.29 -1.56 0.20 0.29 1.09 0.51 -1.48 -0.43 0.89 -0.71 -0.96 -1.09 0.84 0.77 2.11 -1.27 0.08 0.55 -1.13 0.14 1.74 -1.59 -0.72 1.06 1.24 0.34 and B= 7.4 2.7 4.2 -3.0 -8.3 -9.6 1.8 1.1 8.6 4.0 2.1 -5.7 .$
The solution is obtained by first obtaining a $QR$ factorization with column pivoting of the matrix $A$, and then the $RZ$ factorization of the leading $k$ by $k$ part of $R$ is computed, where $k$ is the estimated rank of $A$. A tolerance of $0.01$ is used to estimate the rank of $A$ from the upper triangular factor, $R$.
Note that the block size (NB) of $64$ assumed in this example is not realistic for such a small problem, but should be suitable for large problems.
```function f08bh_example

fprintf('f08bh example results\n\n');

% Going to solve Least squares problem Ax = b, m>n.
m = 6;
n = 5;
a = [-0.09,  0.14, -0.46,  0.68,  1.29;
-1.56,  0.2,   0.29,  1.09,  0.51;
-1.48, -0.43,  0.89, -0.71, -0.96;
-1.09,  0.84,  0.77,  2.11, -1.27;
0.08,  0.55, -1.13,  0.14,  1.74;
-1.59, -0.72,  1.06,  1.24,  0.34];

b = [ 7.4,  2.7;
4.2, -3.0;
-8.3, -9.6;
1.8,  1.1;
8.6,  4.0;
2.1, -5.7];

% QR factorization of A with column pivoting = Q*(R1 R2 )*(P^T)
%                                                (0  R22)

[qr, jpvt, tau, info] = f08bf( ...
a, zeros(n,1,'int64'));

% QRP'X = B, => RP'X = Q'B = C
% Compute C = Q'B
[c, info] = f08ag( ...
'Left', 'Transpose', qr, tau, b);

% Determine the rank, k, of R relative to tol;
% Choose tol to reflect the relative accuracy of the input data
tol = 0.01;
k = find(abs(diag(qr)) <= tol*abs(qr(1,1)));
if numel(k) == 0
k = numel(diag(qr));
else
k = k(1)-1;
end

fprintf('Tolerance used to estimate the rank of A\n     %11.2e\n', tol);
fprintf('Estimated rank of A\n        %d\n', k);

% Compute the RZ (TZ) factorization of the first k rows of (R1 R2)
[rz, taurz, info] = f08bh( ...
qr(1:k,:));

% Now, (TZ)P'X = C on first k rows of C
% Let ZP'X = T^{-1}C = Y (on first k rows)
y = zeros(n, 2);
y(1:k, :) = inv(triu(rz(1:k,1:k)))*c(1:k,:);

% ZP'X = Y => P'X = Z^T Y = W; Form W = Z^TY.
[w, info] = f08bk( ...
'Left', 'Transpose', int64(n-k), rz, taurz, y);

% P'X = W => X = PW, '
x = zeros(n, 2);
for i=1:n
x(jpvt(i), :) = w(i, :);
end
fprintf('\nLeast-squares solution(s)\n');
disp(x);

% Compute estimates of the square roots of the residual sums of squares
rnorm = [norm(c(k+1:6,1)), norm(c(k+1:6,2))];
fprintf('Square root(s) of the residual sum(s) of squares\n');
disp(rnorm);

```
```f08bh example results

Tolerance used to estimate the rank of A
1.00e-02
Estimated rank of A
4

Least-squares solution(s)
0.6344    3.6258
0.9699    1.8284
-1.4402   -1.6416
3.3678    2.4307
3.3992    0.2818

Square root(s) of the residual sum(s) of squares
0.0254    0.0365

```