subroutine htribk(nm,n,ar,ai,tau,m,zr,zi)
c
integer i,j,k,l,m,n,nm
double precision ar(nm,n),ai(nm,n),tau(2,n),zr(nm,m),zi(nm,m)
double precision h,s,si
c
c this subroutine is a translation of a complex analogue of
c the algol procedure trbak1, num. math. 11, 181-195(1968)
c by martin, reinsch, and wilkinson.
c handbook for auto. comp., vol.ii-linear algebra, 212-226(1971).
c
c this subroutine forms the eigenvectors of a complex hermitian
c matrix by back transforming those of the corresponding
c real symmetric tridiagonal matrix determined by htridi.
c
c on input
c
c nm must be set to the row dimension of two-dimensional
c array parameters as declared in the calling program
c dimension statement.
c
c n is the order of the matrix.
c
c ar and ai contain information about the unitary trans-
c formations used in the reduction by htridi in their
c full lower triangles except for the diagonal of ar.
c
c tau contains further information about the transformations.
c
c m is the number of eigenvectors to be back transformed.
c
c zr contains the eigenvectors to be back transformed
c in its first m columns.
c
c on output
c
c zr and zi contain the real and imaginary parts,
c respectively, of the transformed eigenvectors
c in their first m columns.
c
c note that the last component of each returned vector
c is real and that vector euclidean norms are preserved.
c
c questions and comments should be directed to burton s. garbow,
c mathematics and computer science div, argonne national laboratory
c
c this version dated august 1983.
c
c ------------------------------------------------------------------
c
if (m .eq. 0) go to 200
c .......... transform the eigenvectors of the real symmetric
c tridiagonal matrix to those of the hermitian
c tridiagonal matrix. ..........
do 50 k = 1, n
c
do 50 j = 1, m
zi(k,j) = -zr(k,j) * tau(2,k)
zr(k,j) = zr(k,j) * tau(1,k)
50 continue
c
if (n .eq. 1) go to 200
c .......... recover and apply the householder matrices ..........
do 140 i = 2, n
l = i - 1
h = ai(i,i)
if (h .eq. 0.0d0) go to 140
c
do 130 j = 1, m
s = 0.0d0
si = 0.0d0
c
do 110 k = 1, l
s = s + ar(i,k) * zr(k,j) - ai(i,k) * zi(k,j)
si = si + ar(i,k) * zi(k,j) + ai(i,k) * zr(k,j)
110 continue
c .......... double divisions avoid possible underflow ..........
s = (s / h) / h
si = (si / h) / h
c
do 120 k = 1, l
zr(k,j) = zr(k,j) - s * ar(i,k) - si * ai(i,k)
zi(k,j) = zi(k,j) - si * ar(i,k) + s * ai(i,k)
120 continue
c
130 continue
c
140 continue
c
200 return
end