Dpocon Dpotrf Dpotri Dpotrs Dppcon Dpptrf Dpptri Dpptrs

Dpotrf() Sub Dpotrf ( Uplo As  String, N As  Long, A() As  Double, Info As  Long  ) Cholesky factorization of a symmetric positive definite matrix Purpose This routine computes the Cholesky factorization of a real symmetric positive definite matrix A. The factorization has the form A = U^T * U, if uplo = "U", or A = L * L^T, if uplo = "L", where U is an upper triangular matrix and L is a lower triangular matrix. This is the blocked version of the algorithm, calling Level 3 BLAS. Parameters [in] Uplo = "U": Upper triangle of A is stored. = "L": Lower triangle of A is stored. [in] N Order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in,out] A() Array A(LA1 - 1, LA2 - 1) (LA1 >= N, LA2 >= N) [in] N x N symmetric positive definite matrix A. The upper or lower triangular part is to be referenced in accordance with Uplo. [out] If Info = 0, the factor U or L from the Cholesky factorization A = U^T*U or A = L*L^T. [out] Info = 0: Successful exit. = -1: The argument Uplo had an illegal value. (Uplo <> "U" nor "L") = -2: The argument N had an illegal value. (N < 0) = -3: The argument A() is invalid. = i > 0: The leading minor of order i is not positive definite, and the factorization could not be completed. Reference LAPACK Example Program Solve the system of linear equations Ax = B and estimate the reciprocal of the condition number (RCond) of A, where A is symmetric positive definite and ( 2.2 -0.11 -0.32 ) ( -1.566 ) A = ( -0.11 2.93 0.81 ), B = ( -2.8425 ) ( -0.32 0.81 2.37 ) ( -1.1765 ) Sub Ex_Dpotrf() Const N = 3 Dim A(N - 1, N - 1) As Double, B(N - 1) As Double Dim ANorm As Double, RCond As Double, Info As Long A(0, 0) = 2.2 A(1, 0) = -0.11: A(1, 1) = 2.93 A(2, 0) = -0.32: A(2, 1) = 0.81: A(2, 2) = 2.37: B(0) = -1.566: B(1) = -2.8425: B(2) = -1.1765 ANorm = Dlansy("1", "L", N, A()) Call Dpotrf("L", N, A(), Info) If Info = 0 Then Call Dpotrs("L", N, A(), B(), Info) If Info = 0 Then Call Dpocon("L", N, A(), ANorm, RCond, Info) Debug.Print "X =", B(0), B(1), B(2) Debug.Print "RCond =", RCond Debug.Print "Info =", Info End Sub Example Results X = -0.8 -0.92 -0.29 RCond = 0.446791078068956 Info = 0