scipy.linalg.solve_circulant(c, b, singular='raise', tol=None, caxis=-1, baxis=0, outaxis=0)[source]#

Solve C x = b for x, where C is a circulant matrix.

C is the circulant matrix associated with the vector c.

The system is solved by doing division in Fourier space. The calculation is:

x = ifft(fft(b) / fft(c))

where fft and ifft are the fast Fourier transform and its inverse, respectively. For a large vector c, this is much faster than solving the system with the full circulant matrix.


The coefficients of the circulant matrix.


Right-hand side matrix in a x = b.

singularstr, optional

This argument controls how a near singular circulant matrix is handled. If singular is “raise” and the circulant matrix is near singular, a LinAlgError is raised. If singular is “lstsq”, the least squares solution is returned. Default is “raise”.

tolfloat, optional

If any eigenvalue of the circulant matrix has an absolute value that is less than or equal to tol, the matrix is considered to be near singular. If not given, tol is set to:

tol = abs_eigs.max() * abs_eigs.size * np.finfo(np.float64).eps

where abs_eigs is the array of absolute values of the eigenvalues of the circulant matrix.


When c has dimension greater than 1, it is viewed as a collection of circulant vectors. In this case, caxis is the axis of c that holds the vectors of circulant coefficients.


When b has dimension greater than 1, it is viewed as a collection of vectors. In this case, baxis is the axis of b that holds the right-hand side vectors.


When c or b are multidimensional, the value returned by solve_circulant is multidimensional. In this case, outaxis is the axis of the result that holds the solution vectors.


Solution to the system C x = b.


If the circulant matrix associated with c is near singular.

See also


circulant matrix


For a 1-D vector c with length m, and an array b with shape (m, ...),

solve_circulant(c, b)

returns the same result as

solve(circulant(c), b)

where solve and circulant are from scipy.linalg.

Added in version 0.16.0.


>>> import numpy as np
>>> from scipy.linalg import solve_circulant, solve, circulant, lstsq
>>> c = np.array([2, 2, 4])
>>> b = np.array([1, 2, 3])
>>> solve_circulant(c, b)
array([ 0.75, -0.25,  0.25])

Compare that result to solving the system with scipy.linalg.solve:

>>> solve(circulant(c), b)
array([ 0.75, -0.25,  0.25])

A singular example:

>>> c = np.array([1, 1, 0, 0])
>>> b = np.array([1, 2, 3, 4])

Calling solve_circulant(c, b) will raise a LinAlgError. For the least square solution, use the option singular='lstsq':

>>> solve_circulant(c, b, singular='lstsq')
array([ 0.25,  1.25,  2.25,  1.25])

Compare to scipy.linalg.lstsq:

>>> x, resid, rnk, s = lstsq(circulant(c), b)
>>> x
array([ 0.25,  1.25,  2.25,  1.25])

A broadcasting example:

Suppose we have the vectors of two circulant matrices stored in an array with shape (2, 5), and three b vectors stored in an array with shape (3, 5). For example,

>>> c = np.array([[1.5, 2, 3, 0, 0], [1, 1, 4, 3, 2]])
>>> b = np.arange(15).reshape(-1, 5)

We want to solve all combinations of circulant matrices and b vectors, with the result stored in an array with shape (2, 3, 5). When we disregard the axes of c and b that hold the vectors of coefficients, the shapes of the collections are (2,) and (3,), respectively, which are not compatible for broadcasting. To have a broadcast result with shape (2, 3), we add a trivial dimension to c: c[:, np.newaxis, :] has shape (2, 1, 5). The last dimension holds the coefficients of the circulant matrices, so when we call solve_circulant, we can use the default caxis=-1. The coefficients of the b vectors are in the last dimension of the array b, so we use baxis=-1. If we use the default outaxis, the result will have shape (5, 2, 3), so we’ll use outaxis=-1 to put the solution vectors in the last dimension.

>>> x = solve_circulant(c[:, np.newaxis, :], b, baxis=-1, outaxis=-1)
>>> x.shape
(2, 3, 5)
>>> np.set_printoptions(precision=3)  # For compact output of numbers.
>>> x
array([[[-0.118,  0.22 ,  1.277, -0.142,  0.302],
        [ 0.651,  0.989,  2.046,  0.627,  1.072],
        [ 1.42 ,  1.758,  2.816,  1.396,  1.841]],
       [[ 0.401,  0.304,  0.694, -0.867,  0.377],
        [ 0.856,  0.758,  1.149, -0.412,  0.831],
        [ 1.31 ,  1.213,  1.603,  0.042,  1.286]]])

Check by solving one pair of c and b vectors (cf. x[1, 1, :]):

>>> solve_circulant(c[1], b[1, :])
array([ 0.856,  0.758,  1.149, -0.412,  0.831])