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# NAG Toolbox: nag_lapack_zcgesv (f07aq)

## Purpose

nag_lapack_zcgesv (f07aq) computes the solution to a complex system of linear equations
 AX = B , $AX=B ,$
where A$A$ is an n$n$ by n$n$ matrix and X$X$ and B$B$ are n$n$ by r$r$ matrices.

## Syntax

[a, ipiv, x, iter, info] = f07aq(a, b, 'n', n, 'nrhs_p', nrhs_p)
[a, ipiv, x, iter, info] = nag_lapack_zcgesv(a, b, 'n', n, 'nrhs_p', nrhs_p)

## Description

nag_lapack_zcgesv (f07aq) first attempts to factorize the matrix in single precision and use this factorization within an iterative refinement procedure to produce a solution with double precision accuracy. If the approach fails the method switches to a double precision factorization and solve.
The iterative refinement process is stopped if
 iter > itermax , $iter>itermax ,$
where iter is the number of iterations carried out thus far and itermax$\mathit{itermax}$ is the maximum number of iterations allowed, which is fixed at 30$30$ iterations. The process is also stopped if for all right-hand sides we have
 ‖resid‖ < sqrt(n) ‖x‖ ‖A‖ ε , $‖resid‖ < n ‖x‖ ‖A‖ ε ,$
where resid$‖\mathit{resid}‖$ is the $\infty$-norm of the residual, x$‖x‖$ is the $\infty$-norm of the solution, A$‖A‖$ is the $\infty$-operator-norm of the matrix A$A$ and ε$\epsilon$ is the machine precision returned by nag_machine_precision (x02aj).
The iterative refinement strategy used by nag_lapack_zcgesv (f07aq) can be more efficient than the corresponding direct full precision algorithm. Since this strategy must perform iterative refinement on each right-hand side, any efficiency gains will reduce as the number of right-hand sides increases. Conversely, as the matrix size increases the cost of these iterative refinements become less significant relative to the cost of factorization. Thus, any efficiency gains will be greatest for a very small number of right-hand sides and for large matrix sizes. The cut-off values for the number of right-hand sides and matrix size, for which the iterative refinement strategy performs better, depends on the relative performance of the reduced and full precision factorization and back-substitution. For now, nag_lapack_zcgesv (f07aq) always attempts the iterative refinement strategy first; you are advised to compare the performance of nag_lapack_zcgesv (f07aq) with that of its full precision counterpart nag_lapack_zgesv (f07an) to determine whether this strategy is worthwhile for your particular problem dimensions.

## 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
Buttari A, Dongarra J, Langou J, Langou J, Luszczek P and Kurzak J (2007) Mixed precision iterative refinement techniques for the solution of dense linear systems International Journal of High Performance Computing Applications
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore

## Parameters

### Compulsory Input Parameters

1:     a(lda, : $:$) – complex array
The first dimension of the array a must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
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 n$n$ by n$n$ coefficient matrix A$A$.
2:     b(ldb, : $:$) – complex array
The first dimension of the array b must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
The n$n$ by r$r$ right-hand side matrix B$B$.

### Optional Input Parameters

1:     n – int64int32nag_int scalar
Default: The first dimension of the array b and the second dimension of the array a. (An error is raised if these dimensions are not equal.)
n$n$, the number of linear equations, i.e., the order of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.
2:     nrhs_p – int64int32nag_int scalar
Default: The second dimension of the array b.
r$r$, the number of right-hand sides, i.e., the number of columns of the matrix B$B$.
Constraint: nrhs0${\mathbf{nrhs}}\ge 0$.

### Input Parameters Omitted from the MATLAB Interface

lda ldb ldx work swork rwork

### Output Parameters

1:     a(lda, : $:$) – complex array
The first dimension of the array a will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
ldamax (1,n)$\mathit{lda}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
If iterative refinement has been successfully used (i.e., if ${\mathbf{INFO}}={\mathbf{0}}$ and iter0${\mathbf{iter}}\ge 0$), then A$A$ is unchanged. If double precision factorization has been used (when ${\mathbf{INFO}}={\mathbf{0}}$ and iter < 0${\mathbf{iter}}<0$), A$A$ contains the factors L$L$ and U$U$ from the factorization A = PLU$A=PLU$; the unit diagonal elements of L$L$ are not stored.
2:     ipiv(n) – int64int32nag_int array
If no constraints are violated, the pivot indices that define the permutation matrix P$P$; at the i$i$th step row i$i$ of the matrix was interchanged with row ipiv(i)${\mathbf{ipiv}}\left(i\right)$. ipiv(i) = i${\mathbf{ipiv}}\left(i\right)=i$ indicates a row interchange was not required. ipiv${\mathbf{ipiv}}$ corresponds either to the single precision factorization (if ${\mathbf{INFO}}={\mathbf{0}}$ and iter0${\mathbf{iter}}\ge 0$) or to the double precision factorization (if ${\mathbf{INFO}}={\mathbf{0}}$ and iter < 0${\mathbf{iter}}<0$).
3:     x(ldx, : $:$) – complex array
The first dimension of the array x will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be max (1,nrhs)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{nrhs}}\right)$
ldxmax (1,n)$\mathit{ldx}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
If ${\mathbf{INFO}}={\mathbf{0}}$, the n$n$ by r$r$ solution matrix X$X$.
4:     iter – int64int32nag_int scalar
If iter > 0${\mathbf{iter}}>0$, iterative refinement has been successfully used and iter is the number of iterations carried out.
If iter < 0${\mathbf{iter}}<0$, iterative refinement has failed for one of the reasons given below and double precision factorization has been carried out instead.
iter = 1${\mathbf{iter}}=-1$
Taking into account machine parameters, and the values of n and nrhs_p, it is not worth working in single precision.
iter = 2${\mathbf{iter}}=-2$
Overflow of an entry occurred when moving from double to single precision.
iter = 3${\mathbf{iter}}=-3$
An intermediate single precision factorization failed.
iter = 31${\mathbf{iter}}=-31$
The maximum permitted number of iterations was exceeded.
5:     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

Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_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: n, 2: nrhs_p, 3: a, 4: lda, 5: ipiv, 6: b, 7: ldb, 8: x, 9: ldx, 10: work, 11: swork, 12: rwork, 13: iter, 14: 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.
W INFO > 0${\mathbf{INFO}}>0$
If info = i${\mathbf{info}}=i$, uii${u}_{ii}$ is exactly zero. The factorization has been completed, but the factor U$U$ is exactly singular, so the solution could not be computed.

## Accuracy

The computed solution for a single right-hand side, $\stackrel{^}{x}$, satisfies the equation of the form
 (A + E) x̂ = b , $(A+E) x^=b ,$
where
 ‖E‖1 = O(ε) ‖A‖1 $‖E‖1 = O(ε) ‖A‖1$
and ε $\epsilon$ is the machine precision. An approximate error bound for the computed solution is given by
 (‖x̂ − x‖1)/(‖x‖1) ≤ κ(A) (‖E‖1)/(‖A‖1) $‖ x^ - x ‖1 ‖ x ‖1 ≤ κ(A) ‖ E ‖1 ‖ A ‖1$
where κ(A) = A11 A1 $\kappa \left(A\right)={‖{A}^{-1}‖}_{1}{‖A‖}_{1}$, the condition number of A $A$ with respect to the solution of the linear equations. See Section 4.4 of Anderson et al. (1999) for further details.

## Further Comments

The real analogue of this function is nag_lapack_dsgesv (f07ac).

## Example

```function nag_lapack_zcgesv_example
a = [ -1.34 + 2.55i,  0.28 + 3.17i,  -6.39 - 2.2i,  0.72 - 0.92i;
-0.17 - 1.41i,  3.31 - 0.15i,  -0.15 + 1.34i,  1.29 + 1.38i;
-3.29 - 2.39i,  -1.91 + 4.42i,  -0.14 - 1.35i,  1.72 + 1.35i;
2.41 + 0.39i,  -0.56 + 1.47i,  -0.83 - 0.69i,  -1.96 + 0.67i];
b = [ 26.26 + 51.78i;
6.43 - 8.68i;
-5.75 + 25.31i;
1.16 + 2.57i];
[a, ipiv, x, iter, info] = nag_lapack_zcgesv(a, b)
```
```

a =

-1.3400 + 2.5500i   0.2800 + 3.1700i  -6.3900 - 2.2000i   0.7200 - 0.9200i
-0.1700 - 1.4100i   3.3100 - 0.1500i  -0.1500 + 1.3400i   1.2900 + 1.3800i
-3.2900 - 2.3900i  -1.9100 + 4.4200i  -0.1400 - 1.3500i   1.7200 + 1.3500i
2.4100 + 0.3900i  -0.5600 + 1.4700i  -0.8300 - 0.6900i  -1.9600 + 0.6700i

ipiv =

3
2
3
4

x =

1.0000 + 1.0000i
2.0000 - 3.0000i
-4.0000 - 5.0000i
0.0000 + 6.0000i

iter =

2

info =

0

```
```function f07aq_example
a = [ -1.34 + 2.55i,  0.28 + 3.17i,  -6.39 - 2.2i,  0.72 - 0.92i;
-0.17 - 1.41i,  3.31 - 0.15i,  -0.15 + 1.34i,  1.29 + 1.38i;
-3.29 - 2.39i,  -1.91 + 4.42i,  -0.14 - 1.35i,  1.72 + 1.35i;
2.41 + 0.39i,  -0.56 + 1.47i,  -0.83 - 0.69i,  -1.96 + 0.67i];
b = [ 26.26 + 51.78i;
6.43 - 8.68i;
-5.75 + 25.31i;
1.16 + 2.57i];
[a, ipiv, x, iter, info] = f07aq(a, b)
```
```

a =

-1.3400 + 2.5500i   0.2800 + 3.1700i  -6.3900 - 2.2000i   0.7200 - 0.9200i
-0.1700 - 1.4100i   3.3100 - 0.1500i  -0.1500 + 1.3400i   1.2900 + 1.3800i
-3.2900 - 2.3900i  -1.9100 + 4.4200i  -0.1400 - 1.3500i   1.7200 + 1.3500i
2.4100 + 0.3900i  -0.5600 + 1.4700i  -0.8300 - 0.6900i  -1.9600 + 0.6700i

ipiv =

3
2
3
4

x =

1.0000 + 1.0000i
2.0000 - 3.0000i
-4.0000 - 5.0000i
0.0000 + 6.0000i

iter =

2

info =

0

```

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