Dpbcon Dpbtrf Dptcon Dpttrf Dpttrs

Dpbtrf() Sub Dpbtrf ( Uplo As  String, N As  Long, Kd As  Long, Ab() As  Double, Info As  Long  ) Cholesky factorization of a symmetric positive definite band matrix Purpose This routine computes the Cholesky factorization of a real symmetric positive definite band 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. Parameters [in] Uplo = "U": Upper triangle of A is stored. = "L": Lower triangle of A is stored. [in] N Number of linear equations, i.e., order of the matrix A. (N >= 0) (If N = 0, returns without computation) [in] Kd Number of super-diagonals of the matrix A if Uplo = "U", or number of sub-diagonals if Uplo = "L". (Kd >= 0) [in,out] Ab() Array Ab(LAb1 - 1, LAb2 - 1) (LAb1 >= Kd + 1, LAb2 >= N) [in] N x N symmetric band matrix A in Kd+1 x N symmetric band matrix form. Upper or lower part is to be stored in accordance with uplo. See below for further details. [out] If Info = 0, the triangular factor U or L from the Cholesky factorization A = U^T*U or A = L*L^T of the band matrix A, in the same storage format as A. [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 Kd had an illegal value. (Kd < 0) = -4: The argument Ab() is invalid. = i > 0: The leading minor of order i of A is not positive definite, so the factorization could not be completed. Further Details The symmetric band matrix form is illustrated by the following example, when N = 6, Kd = 2, and Uplo = "U": On entry: * * a13 a24 a35 a46 * a12 a23 a34 a45 a56 a11 a22 a33 a44 a55 a66 On exit: * * u13 u24 u35 u46 * u12 u23 u34 u45 u56 u11 u22 u33 u44 u55 u66 Similarly, if uplo = "L" the format of A is as follows: On entry: a11 a22 a33 a44 a55 a66 a21 a32 a43 a54 a65 * a31 a42 a53 a64 * * On exit: l11 l22 l33 l44 l55 l66 l21 l32 l43 l54 l65 * l31 l42 l53 l64 * * Array elements marked * are not used by the routine. 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 banded symmetric positive definite and ( 0.61 0.79 0 ) ( 0.3034 ) A = ( 0.79 2.23 0.25 ), B = ( 0.8537 ) ( 0 0.25 2.87 ) ( 0.8000 ) Sub Ex_Dpbtrf() Const N As Long = 3, Kd = 1 Dim Ab(Kd, N - 1) As Double, B(N - 1) As Double Dim ANorm As Double, RCond As Double, Info As Long Ab(0, 0) = 0.61: Ab(0, 1) = 2.23: Ab(0, 2) = 2.87 Ab(1, 0) = 0.79: Ab(1, 1) = 0.25 B(0) = 0.3034: B(1) = 0.8537: B(2) = 0.8 ANorm = Dlansb("1", "L", N, Kd, Ab()) Call Dpbtrf("L", N, Kd, Ab(), Info) If Info = 0 Then Call Dpbtrs("L", N, Kd, Ab(), B(), Info) If Info = 0 Then Call Dpbcon("L", N, Kd, Ab(), 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 = 6.99999999999999E-02 0.33 0.25 RCond = 7.20810157140908E-02 Info = 0