dpoequ.f

          SUBROUTINE DPOEQU( N, A, LDA, S, SCOND, AMAX, INFO )
    *
    *  -- LAPACK routine (version 3.1) --
    *     Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
    *     November 2006
    *
    *     .. Scalar Arguments ..
          INTEGER            INFO, LDA, N
          DOUBLE PRECISION   AMAX, SCOND
    *     ..
    *     .. Array Arguments ..
          DOUBLE PRECISION   A( LDA, * ), S( * )
    *     ..
    *
    *  Purpose
    *  =======
    *
    *  DPOEQU computes row and column scalings intended to equilibrate a
    *  symmetric positive definite matrix A and reduce its condition number
    *  (with respect to the two-norm).  S contains the scale factors,
    *  S(i) = 1/sqrt(A(i,i)), chosen so that the scaled matrix B with
    *  elements B(i,j) = S(i)*A(i,j)*S(j) has ones on the diagonal.  This
    *  choice of S puts the condition number of B within a factor N of the
    *  smallest possible condition number over all possible diagonal
    *  scalings.
    *
    *  Arguments
    *  =========
    *
    *  N       (input) INTEGER
    *          The order of the matrix A.  N >= 0.
    *
    *  A       (input) DOUBLE PRECISION array, dimension (LDA,N)
    *          The N-by-N symmetric positive definite matrix whose scaling
    *          factors are to be computed.  Only the diagonal elements of A
    *          are referenced.
    *
    *  LDA     (input) INTEGER
    *          The leading dimension of the array A.  LDA >= max(1,N).
    *
    *  S       (output) DOUBLE PRECISION array, dimension (N)
    *          If INFO = 0, S contains the scale factors for A.
    *
    *  SCOND   (output) DOUBLE PRECISION
    *          If INFO = 0, S contains the ratio of the smallest S(i) to
    *          the largest S(i).  If SCOND >= 0.1 and AMAX is neither too
    *          large nor too small, it is not worth scaling by S.
    *
    *  AMAX    (output) DOUBLE PRECISION
    *          Absolute value of largest matrix element.  If AMAX is very
    *          close to overflow or very close to underflow, the matrix
    *          should be scaled.
    *
    *  INFO    (output) INTEGER
    *          = 0:  successful exit
    *          < 0:  if INFO = -i, the i-th argument had an illegal value
    *          > 0:  if INFO = i, the i-th diagonal element is nonpositive.
    *
    *  =====================================================================
    *
    *     .. Parameters ..
          DOUBLE PRECISION   ZERO, ONE
          PARAMETER          ( ZERO = 0.0D+0, ONE = 1.0D+0 )
    *     ..
    *     .. Local Scalars ..
          INTEGER            I
          DOUBLE PRECISION   SMIN
    *     ..
    *     .. External Subroutines ..
          EXTERNAL           XERBLA
    *     ..
    *     .. Intrinsic Functions ..
          INTRINSIC          MAX, MIN, SQRT
    *     ..
    *     .. Executable Statements ..
    *
    *     Test the input parameters.
    *
          INFO = 0
          IF( N.LT.0 ) THEN
             INFO = -1
          ELSE IF( LDA.LT.MAX( 1, N ) ) THEN
             INFO = -3
          END IF
          IF( INFO.NE.0 ) THEN
             CALL XERBLA( 'DPOEQU', -INFO )
             RETURN
          END IF
    *
    *     Quick return if possible
    *
          IF( N.EQ.0 ) THEN
             SCOND = ONE
             AMAX = ZERO
             RETURN
          END IF
    *
    *     Find the minimum and maximum diagonal elements.
    *
          S( 1 ) = A( 1, 1 )
          SMIN = S( 1 )
          AMAX = S( 1 )
          DO 10 I = 2, N
             S( I ) = A( I, I )
             SMIN = MIN( SMIN, S( I ) )
             AMAX = MAX( AMAX, S( I ) )
       10 CONTINUE
    *
          IF( SMIN.LE.ZERO ) THEN
    *
    *        Find the first non-positive diagonal element and return.
    *
             DO 20 I = 1, N
                IF( S( I ).LE.ZERO ) THEN
                   INFO = I
                   RETURN
                END IF
       20    CONTINUE
          ELSE
    *
    *        Set the scale factors to the reciprocals
    *        of the diagonal elements.
    *
             DO 30 I = 1, N
                S( I ) = ONE / SQRT( S( I ) )
       30    CONTINUE
    *
    *        Compute SCOND = min(S(I)) / max(S(I))
    *
             SCOND = SQRT( SMIN ) / SQRT( AMAX )
          END IF
          RETURN
    *
    *     End of DPOEQU
    *
          END