scipy.sparse.linalg.bicg(A, b, x0=None, *, tol=<object object>, maxiter=None, M=None, callback=None, atol=0.0, rtol=1e-05)[source]#

Use BIConjugate Gradient iteration to solve Ax = b.

A{sparse matrix, ndarray, LinearOperator}

The real or complex N-by-N matrix of the linear system. Alternatively, A can be a linear operator which can produce Ax and A^T x using, e.g., scipy.sparse.linalg.LinearOperator.


Right hand side of the linear system. Has shape (N,) or (N,1).


Starting guess for the solution.

rtol, atolfloat, optional

Parameters for the convergence test. For convergence, norm(b - A @ x) <= max(rtol*norm(b), atol) should be satisfied. The default is atol=0. and rtol=1e-5.


Maximum number of iterations. Iteration will stop after maxiter steps even if the specified tolerance has not been achieved.

M{sparse matrix, ndarray, LinearOperator}

Preconditioner for A. The preconditioner should approximate the inverse of A. Effective preconditioning dramatically improves the rate of convergence, which implies that fewer iterations are needed to reach a given error tolerance.


User-supplied function to call after each iteration. It is called as callback(xk), where xk is the current solution vector.

tolfloat, optional, deprecated

Deprecated since version 1.12.0: bicg keyword argument tol is deprecated in favor of rtol and will be removed in SciPy 1.14.0.


The converged solution.

Provides convergence information:

0 : successful exit >0 : convergence to tolerance not achieved, number of iterations <0 : parameter breakdown


>>> import numpy as np
>>> from scipy.sparse import csc_matrix
>>> from scipy.sparse.linalg import bicg
>>> A = csc_matrix([[3, 2, 0], [1, -1, 0], [0, 5, 1.]])
>>> b = np.array([2., 4., -1.])
>>> x, exitCode = bicg(A, b, atol=1e-5)
>>> print(exitCode)  # 0 indicates successful convergence
>>> np.allclose(, b)