scipy.linalg.solveh_banded#

scipy.linalg.solveh_banded(ab, b, overwrite_ab=False, overwrite_b=False, lower=False, check_finite=True)[source]#

Solve equation a x = b. a is Hermitian positive-definite banded 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 *   *

Cells marked with * are not used.

Parameters

ab(u + 1, M) array_like: Banded matrix
b(M,) or (M, K) array_like: Right-hand side
overwrite_abbool, optional: Discard data in ab (may enhance performance)
overwrite_bbool, optional: Discard data in b (may enhance performance)
lowerbool, optional: Is the matrix in the lower form. (Default is upper form)
check_finitebool, optional: Whether to check that the input matrices contain 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

x(M,) or (M, K) ndarray: The solution to the system a x = b. Shape of return matches shape of b.

Examples

Solve the banded system A x = b, where:

    [ 4  2 -1  0  0  0]       [1]
    [ 2  5  2 -1  0  0]       [2]
A = [-1  2  6  2 -1  0]   b = [2]
    [ 0 -1  2  7  2 -1]       [3]
    [ 0  0 -1  2  8  2]       [3]
    [ 0  0  0 -1  2  9]       [3]

>>> from scipy.linalg import solveh_banded

ab contains the main diagonal and the nonzero diagonals below the main diagonal. That is, we use the lower form:

>>> ab = np.array([[ 4,  5,  6,  7, 8, 9],
...                [ 2,  2,  2,  2, 2, 0],
...                [-1, -1, -1, -1, 0, 0]])
>>> b = np.array([1, 2, 2, 3, 3, 3])
>>> x = solveh_banded(ab, b, lower=True)
>>> x
array([ 0.03431373,  0.45938375,  0.05602241,  0.47759104,  0.17577031,
        0.34733894])

Solve the Hermitian banded system H x = b, where:

    [ 8   2-1j   0     0  ]        [ 1  ]
H = [2+1j  5     1j    0  ]    b = [1+1j]
    [ 0   -1j    9   -2-1j]        [1-2j]
    [ 0    0   -2+1j   6  ]        [ 0  ]

In this example, we put the upper diagonals in the array hb:

>>> hb = np.array([[0, 2-1j, 1j, -2-1j],
...                [8,  5,    9,   6  ]])
>>> b = np.array([1, 1+1j, 1-2j, 0])
>>> x = solveh_banded(hb, b)
>>> x
array([ 0.07318536-0.02939412j,  0.11877624+0.17696461j,
        0.10077984-0.23035393j, -0.00479904-0.09358128j])