scipy.linalg.solve_toeplitz(c_or_cr, b, check_finite=True)[source]

Solve a Toeplitz system using Levinson Recursion

The Toeplitz matrix has constant diagonals, with c as its first column and r as its first row. If r is not given, r == conjugate(c) is assumed.


c_or_cr : array_like or tuple of (array_like, array_like)

The vector c, or a tuple of arrays (c, r). Whatever the actual shape of c, it will be converted to a 1-D array. If not supplied, r = conjugate(c) is assumed; in this case, if c[0] is real, the Toeplitz matrix is Hermitian. r[0] is ignored; the first row of the Toeplitz matrix is [c[0], r[1:]]. Whatever the actual shape of r, it will be converted to a 1-D array.

b : (M,) or (M, K) array_like

Right-hand side in T x = b.

check_finite : bool, optional

Whether to check that the input matrices contain only finite numbers. Disabling may give a performance gain, but may result in problems (result entirely NaNs) if the inputs do contain infinities or NaNs.


x : (M,) or (M, K) ndarray

The solution to the system T x = b. Shape of return matches shape of b.

See also

Toeplitz matrix


The solution is computed using Levinson-Durbin recursion, which is faster than generic least-squares methods, but can be less numerically stable.


Solve the Toeplitz system T x = b, where:

    [ 1 -1 -2 -3]       [1]
T = [ 3  1 -1 -2]   b = [2]
    [ 6  3  1 -1]       [2]
    [10  6  3  1]       [5]

To specify the Toeplitz matrix, only the first column and the first row are needed.

>>> c = np.array([1, 3, 6, 10])    # First column of T
>>> r = np.array([1, -1, -2, -3])  # First row of T
>>> b = np.array([1, 2, 2, 5])
>>> from scipy.linalg import solve_toeplitz, toeplitz
>>> x = solve_toeplitz((c, r), b)
>>> x
array([ 1.66666667, -1.        , -2.66666667,  2.33333333])

Check the result by creating the full Toeplitz matrix and multiplying it by x. We should get b.

>>> T = toeplitz(c, r)
array([ 1.,  2.,  2.,  5.])