ZCGESV(3F)

NAME
ZCGESV - the solution to a real system of linear equations A * X = B,

SYNOPSIS

SUBROUTINE ZCGESV(
    N, NRHS, A, LDA, IPIV, B, LDB, X, LDX, WORK,

+   SWORK, ITER, INFO)

INTEGER
    INFO,ITER,LDA,LDB,LDX,N,NRHS

INTEGER
    IPIV(*)

COMPLEX
    SWORK(*)

COMPLEX*16
    A(LDA,*),B(LDB,*),WORK(N,*),X(LDX,*)

PURPOSE
ZCGESV computes the solution to a real system of linear equations
   A * X = B, where A is an N-by-N matrix and X and B are N-by-NRHS matrices. ZCGESV first attempts to factorize the matrix in SINGLE COMPLEX PRECISION and use this factorization within an iterative refinement procedure to produce a solution with DOUBLE COMPLEX PRECISION normwise backward error quality (see below). If the approach fails the method switches to a DOUBLE COMPLEX PRECISION factorization and solve.
The iterative refinement is not going to be a winning strategy if the ratio SINGLE PRECISION performance over DOUBLE PRECISION performance is too small. A reasonable strategy should take the number of right-hand sides and the size of the matrix into account. This might be done with a call to ILAENV in the future. Up to now, we always try iterative refinement. The iterative refinement process is stopped if
    ITER > ITERMAX
or for all the RHS we have:
    RNRM < SQRT(N)*XNRM*ANRM*EPS*BWDMAX
where
    o ITER is the number of the current iteration in the iterative
      refinement process
    o RNRM is the infinity-norm of the residual
    o XNRM is the infinity-norm of the solution
    o ANRM is the infinity-operator-norm of the matrix A
    o EPS is the machine epsilon returned by DLAMCH('Epsilon') The value ITERMAX and BWDMAX are fixed to 30 and 1.0D+00 respectively.

ARGUMENTS

N       (input) INTEGER
    The number of linear equations, i.e., the order of the matrix A. N >= 0.

NRHS    (input) INTEGER
    The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0.

A       (input or input/ouptut) COMPLEX*16 array,
    dimension (LDA,N) On entry, the N-by-N coefficient matrix A. On exit, if iterative refinement has been successfully used (INFO.EQ.0 and ITER.GE.0, see description below), then A is unchanged, if double precision factorization has been used (INFO.EQ.0 and ITER.LT.0, see description below), then the array A contains the factors L and U from the factorization A = P*L*U; the unit diagonal elements of L are not stored.

LDA     (input) INTEGER
    The leading dimension of the array A. LDA >= max(1,N).

IPIV    (output) INTEGER array, dimension (N)
    The pivot indices that define the permutation matrix P; row i of the matrix was interchanged with row IPIV(i). Corresponds either to the single precision factorization (if INFO.EQ.0 and ITER.GE.0) or the double precision factorization (if INFO.EQ.0 and ITER.LT.0).

B       (input) COMPLEX*16 array, dimension (LDB,NRHS)
    The N-by-NRHS matrix of right hand side matrix B.

LDB     (input) INTEGER
    The leading dimension of the array B. LDB >= max(1,N).

X       (output) COMPLEX*16 array, dimension (LDX,NRHS)
    If INFO = 0, the N-by-NRHS solution matrix X.

LDX     (input) INTEGER
    The leading dimension of the array X. LDX >= max(1,N).

WORK    (workspace) COMPLEX*16 array, dimension (N*NRHS)
    This array is used to hold the residual vectors.

SWORK   (workspace) COMPLEX array, dimension (N*(N+NRHS))
    This array is used to use the single precision matrix and the right-hand sides or solutions in single precision.

ITER    (output) INTEGER
    < 0: iterative refinement has failed, double precision factorization has been performed -1 : taking into account machine parameters, N, NRHS, it is a priori not worth working in SINGLE PRECISION -2 : overflow of an entry when moving from double to SINGLE PRECISION -3 : failure of SGETRF
    -31: stop the iterative refinement after the 30th iterations > 0: iterative refinement has been sucessfully used. Returns the number of iterations

INFO    (output) INTEGER
    = 0: successful exit
    < 0: if INFO = -i, the i-th argument had an illegal value
    > 0: if INFO = i, U(i,i) computed in DOUBLE PRECISION is exactly zero. The factorization has been completed, but the factor U is exactly singular, so the solution could not be computed. =========