Module « scipy.linalg »

Fonction cholesky_banded - module scipy.linalg

Signature de la fonction cholesky_banded

def cholesky_banded(ab, overwrite_ab=False, lower=False, check_finite=True) 

Description

cholesky_banded.__doc__

    Cholesky decompose a banded Hermitian positive-definite matrix

    The matrix a is stored in ab either in lower-diagonal or upper-
    diagonal ordered form::

        ab[u + i - j, j] == a[i,j]        (if upper form; i <= j)
        ab[    i - j, j] == a[i,j]        (if lower form; i >= j)

    Example of ab (shape of a is (6,6), u=2)::

        upper form:
        *   *   a02 a13 a24 a35
        *   a01 a12 a23 a34 a45
        a00 a11 a22 a33 a44 a55

        lower form:
        a00 a11 a22 a33 a44 a55
        a10 a21 a32 a43 a54 *
        a20 a31 a42 a53 *   *

    Parameters
    ----------
    ab : (u + 1, M) array_like
        Banded matrix
    overwrite_ab : bool, optional
        Discard data in ab (may enhance performance)
    lower : bool, optional
        Is the matrix in the lower form. (Default is upper form)
    check_finite : bool, optional
        Whether to check that the input matrix contains only finite numbers.
        Disabling may give a performance gain, but may result in problems
        (crashes, non-termination) if the inputs do contain infinities or NaNs.

    Returns
    -------
    c : (u + 1, M) ndarray
        Cholesky factorization of a, in the same banded format as ab

    See also
    --------
    cho_solve_banded : Solve a linear set equations, given the Cholesky factorization
                of a banded Hermitian.

    Examples
    --------
    >>> from scipy.linalg import cholesky_banded
    >>> from numpy import allclose, zeros, diag
    >>> Ab = np.array([[0, 0, 1j, 2, 3j], [0, -1, -2, 3, 4], [9, 8, 7, 6, 9]])
    >>> A = np.diag(Ab[0,2:], k=2) + np.diag(Ab[1,1:], k=1)
    >>> A = A + A.conj().T + np.diag(Ab[2, :])
    >>> c = cholesky_banded(Ab)
    >>> C = np.diag(c[0, 2:], k=2) + np.diag(c[1, 1:], k=1) + np.diag(c[2, :])
    >>> np.allclose(C.conj().T @ C - A, np.zeros((5, 5)))
    True