cpftrs (l)  Linux Man Pages
cpftrs: solves a system of linear equations A*X = B with a Hermitian positive definite matrix A using the Cholesky factorization A = U**H*U or A = L*L**H computed by CPFTRF
NAME
CPFTRS  solves a system of linear equations A*X = B with a Hermitian positive definite matrix A using the Cholesky factorization A = U**H*U or A = L*L**H computed by CPFTRFSYNOPSIS
 SUBROUTINE CPFTRS(
 TRANSR, UPLO, N, NRHS, A, B, LDB, INFO )
 CHARACTER TRANSR, UPLO
 INTEGER INFO, LDB, N, NRHS
 COMPLEX A( 0: * ), B( LDB, * )
PURPOSE
CPFTRS solves a system of linear equations A*X = B with a Hermitian positive definite matrix A using the Cholesky factorization A = U**H*U or A = L*L**H computed by CPFTRF.ARGUMENTS
 TRANSR (input) CHARACTER

= aqNaq: The Normal TRANSR of RFP A is stored;
= aqCaq: The Conjugatetranspose TRANSR of RFP A is stored.  UPLO (input) CHARACTER

= aqUaq: Upper triangle of RFP A is stored;
= aqLaq: Lower triangle of RFP A is stored.  N (input) INTEGER
 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) COMPLEX array, dimension ( N*(N+1)/2 );
 The triangular factor U or L from the Cholesky factorization of RFP A = U**H*U or RFP A = L*L**H, as computed by CPFTRF. See note below for more details about RFP A.
 B (input/output) COMPLEX array, dimension (LDB,NRHS)
 On entry, the right hand side matrix B. On exit, the solution matrix X.
 LDB (input) INTEGER
 The leading dimension of the array B. LDB >= max(1,N).
 INFO (output) INTEGER

= 0: successful exit
< 0: if INFO = i, the ith argument had an illegal value
FURTHER DETAILS
We first consider Standard Packed Format when N is even.We give an example where N = 6.
Let TRANSR = aqNaq. RFP holds AP as follows:
For UPLO = aqUaq the upper trapezoid A(0:5,0:2) consists of the last three columns of AP upper. The lower triangle A(4:6,0:2) consists of conjugatetranspose of the first three columns of AP upper. For UPLO = aqLaq the lower trapezoid A(1:6,0:2) consists of the first three columns of AP lower. The upper triangle A(0:2,0:2) consists of conjugatetranspose of the last three columns of AP lower. To denote conjugate we place  above the element. This covers the case N even and TRANSR = aqNaq.
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