ZTRSYL(3) solves the complex Sylvester matrix equation

SYNOPSIS

SUBROUTINE ZTRSYL(
TRANA, TRANB, ISGN, M, N, A, LDA, B, LDB, C, LDC, SCALE, INFO )

    
CHARACTER TRANA, TRANB

    
INTEGER INFO, ISGN, LDA, LDB, LDC, M, N

    
DOUBLE PRECISION SCALE

    
COMPLEX*16 A( LDA, * ), B( LDB, * ), C( LDC, * )

PURPOSE

ZTRSYL solves the complex Sylvester matrix equation:
   op(A)*X + X*op(B) = scale*C or

   op(A)*X - X*op(B) = scale*C,
where op(A) = A or A**H, and A and B are both upper triangular. A is M-by-M and B is N-by-N; the right hand side C and the solution X are M-by-N; and scale is an output scale factor, set <= 1 to avoid overflow in X.

ARGUMENTS

TRANA (input) CHARACTER*1
Specifies the option op(A):
= 'N': op(A) = A (No transpose)
= 'C': op(A) = A**H (Conjugate transpose)
TRANB (input) CHARACTER*1

Specifies the option op(B):
= 'N': op(B) = B (No transpose)
= 'C': op(B) = B**H (Conjugate transpose)
ISGN (input) INTEGER

Specifies the sign in the equation:
= +1: solve op(A)*X + X*op(B) = scale*C
= -1: solve op(A)*X - X*op(B) = scale*C
M (input) INTEGER
The order of the matrix A, and the number of rows in the matrices X and C. M >= 0.
N (input) INTEGER
The order of the matrix B, and the number of columns in the matrices X and C. N >= 0.
A (input) COMPLEX*16 array, dimension (LDA,M)
The upper triangular matrix A.
LDA (input) INTEGER
The leading dimension of the array A. LDA >= max(1,M).
B (input) COMPLEX*16 array, dimension (LDB,N)
The upper triangular matrix B.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= max(1,N).
C (input/output) COMPLEX*16 array, dimension (LDC,N)
On entry, the M-by-N right hand side matrix C. On exit, C is overwritten by the solution matrix X.
LDC (input) INTEGER
The leading dimension of the array C. LDC >= max(1,M)
SCALE (output) DOUBLE PRECISION
The scale factor, scale, set <= 1 to avoid overflow in X.
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
= 1: A and B have common or very close eigenvalues; perturbed values were used to solve the equation (but the matrices A and B are unchanged).