scipy.sparse.linalg.

qmr

scipy.sparse.linalg.qmr(A, b, x0=None, *, rtol=1e-05, atol=0.0, maxiter=None, M1=None, M2=None, callback=None)

Use Quasi-Minimal Residual iteration to solve Ax = b.

Parameters:

A{sparse array, ndarray, LinearOperator}

The real-valued 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.

bndarray

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

x0ndarray

Starting guess for the solution.

atol, rtolfloat, 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.

maxiterinteger

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

M1{sparse array, ndarray, LinearOperator}

Left preconditioner for A.

M2{sparse array, ndarray, LinearOperator}

Right preconditioner for A. Used together with the left preconditioner M1. The matrix M1@A@M2 should have better conditioned than A alone.

callbackfunction

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

Returns:

xndarray

The converged solution.

infointeger

Provides convergence information:

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

See also

LinearOperator

Examples

Try it in your browser!

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

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