
    Stopping tolerances. `lsqr` continues iterations until a
    certain backward error estimate is smaller than some quantity
    depending on atol and btol.  Let ``r = b - Ax`` be the
    residual vector for the current approximate solution ``x``.
    If ``Ax = b`` seems to be consistent, `lsqr` terminates
    when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
    Otherwise, `lsqr` terminates when ``norm(A^H r) <=
    atol * norm(A) * norm(r)``.  If both tolerances are 1.0e-6 (default),
    the final ``norm(r)`` should be accurate to about 6
    digits. (The final ``x`` will usually have fewer correct digits,
    depending on ``cond(A)`` and the size of LAMBDA.)  If `atol`
    or `btol` is None, a default value of 1.0e-6 will be used.
    Ideally, they should be estimates of the relative error in the
    entries of ``A`` and ``b`` respectively.  For example, if the entries
    of ``A`` have 7 correct digits, set ``atol = 1e-7``. This prevents
    the algorithm from doing unnecessary work beyond the
    uncertainty of the input data.
conlim : float, optional
    Another stopping tolerance.  lsqr terminates if an estimate of
    ``cond(A)`` exceeds `conlim`.  For compatible systems ``Ax =
    b``, `conlim` could be as large as 1.0e+12 (say).  For
    least-squares problems, conlim should be less than 1.0e+8.
    Maximum precision can be obtained by setting ``atol = btol =
    conlim = zero``, but the number of iterations may then be
    excessive. Default is 1e8.
iter_lim : int, optional
    Explicit limitation on number of iterations (for safety).
show : bool, optional
    Display an iteration log. Default is False.
calc_var : bool, optional
    Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
x0 : array_like, shape (n,), optional
    Initial guess of x, if None zeros are used. Default is None.

    .. versionadded:: 1.0.0

Returns
-------
x : ndarray of float
    The final solution.
istop : int
    Gives the reason for termination.
    1 means x is an approximate solution to Ax = b.
    2 means x approximately solves the least-squares problem.
itn : int
    Iteration number upon termination.
r1norm : float
    ``norm(r)``, where ``r = b - Ax``.
r2norm : float
    ``sqrt( norm(r)^2  +  damp^2 * norm(x - x0)^2 )``.  Equal to `r1norm`
    if ``damp == 0``.
anorm : float
    Estimate of Frobenius norm of ``Abar = [[A]; [damp*I]]``.
acond : float
    Estimate of ``cond(Abar)``.
arnorm : float
    Estimate of ``norm(A'@r - damp^2*(x - x0))``.
xnorm : float
    ``norm(x)``
var : ndarray of float
    If ``calc_var`` is True, estimates all diagonals of
    ``(A'A)^{-1}`` (if ``damp == 0``) or more generally ``(A'A +
    damp^2*I)^{-1}``.  This is well defined if A has full column
    rank or ``damp > 0``.  (Not sure what var means if ``rank(A)
    < n`` and ``damp = 0.``)

Notes
-----
LSQR uses an iterative method to approximate the solution.  The
number of iterations required to reach a certain accuracy depends
strongly on the scaling of the problem.  Poor scaling of the rows
or columns of A should therefore be avoided where possible.

For example, in problem 1 the solution is unaltered by
row-scaling.  If a row of A is very small or large compared to
the other rows of A, the corresponding row of ( A  b ) should be
scaled up or down.

In problems 1 and 2, the solution x is easily recovered
following column-scaling.  Unless better information is known,
the nonzero columns of A should be scaled so that they all have
the same Euclidean norm (e.g., 1.0).

In problem 3, there is no freedom to re-scale if damp is
nonzero.  However, the value of damp should be assigned only
after attention has been paid to the scaling of A.

The parameter damp is intended to help regularize
ill-conditioned systems, by preventing the true solution from
being very large.  Another aid to regularization is provided by
the parameter acond, which may be used to terminate iterations
before the computed solution becomes very large.

If some initial estimate ``x0`` is known and if ``damp == 0``,
one could proceed as follows:

  1. Compute a residual vector ``r0 = b - A@x0``.
  2. Use LSQR to solve the system  ``A@dx = r0``.
  3. Add the correction dx to obtain a final solution ``x = x0 + dx``.

This requires that ``x0`` be available before and after the call
to LSQR.  To judge the benefits, suppose LSQR takes k1 iterations
to solve A@x = b and k2 iterations to solve A@dx = r0.
If x0 is "good", norm(r0) will be smaller than norm(b).
If the same stopping tolerances atol and btol are used for each
system, k1 and k2 will be similar, but the final solution x0 + dx
should be more accurate.  The only way to reduce the total work
is to use a larger stopping tolerance for the second system.
If some value btol is suitable for A@x = b, the larger value
btol*norm(b)/norm(r0)  should be suitable for A@dx = r0.

Preconditioning is another way to reduce the number of iterations.
If it is possible to solve a related system ``M@x = b``
efficiently, where M approximates A in some helpful way (e.g. M -
A has low rank or its elements are small relative to those of A),
LSQR may converge more rapidly on the system ``A@M(inverse)@z =
b``, after which x can be recovered by solving M@x = z.

If A is symmetric, LSQR should not be used!

Alternatives are the symmetric conjugate-gradient method (cg)
and/or SYMMLQ.  SYMMLQ is an implementation of symmetric cg that
applies to any symmetric A and will converge more rapidly than
LSQR.  If A is positive definite, there are other implementations
of symmetric cg that require slightly less work per iteration than
SYMMLQ (but will take the same number of iterations).

References
----------
.. [1] C. C. Paige and M. A. Saunders (1982a).
       "LSQR: An algorithm for sparse linear equations and
       sparse least squares", ACM TOMS 8(1), 43-71.
.. [2] C. C. Paige and M. A. Saunders (1982b).
       "Algorithm 583.  LSQR: Sparse linear equations and least
       squares problems", ACM TOMS 8(2), 195-209.
.. [3] M. A. Saunders (1995).  "Solution of sparse rectangular
       systems using LSQR and CRAIG", BIT 35, 588-604.

Examples
--------
>>> import numpy as np
>>> from scipy.sparse import csc_array
>>> from scipy.sparse.linalg import lsqr
>>> A = csc_array([[1., 0.], [1., 1.], [0., 1.]], dtype=float)

The first example has the trivial solution ``[0, 0]``

>>> b = np.array([0., 0., 0.], dtype=float)
>>> x, istop, itn, normr = lsqr(A, b)[:4]
>>> istop
0
>>> x
array([ 0.,  0.])

The stopping code ``istop=0`` returned indicates that a vector of zeros was
found as a solution. The returned solution `x` indeed contains
``[0., 0.]``. The next example has a non-trivial solution:

>>> b = np.array([1., 0., -1.], dtype=float)
>>> x, istop, itn, r1norm = lsqr(A, b)[:4]
>>> istop
1
>>> x
array([ 1., -1.])
>>> itn
1
>>> r1norm
4.440892098500627e-16

As indicated by ``istop=1``, `lsqr` found a solution obeying the tolerance
limits. The given solution ``[1., -1.]`` obviously solves the equation. The
remaining return values include information about the number of iterations
(`itn=1`) and the remaining difference of left and right side of the solved
equation.
The final example demonstrates the behavior in the case where there is no
solution for the equation:

>>> b = np.array([1., 0.01, -1.], dtype=float)
>>> x, istop, itn, r1norm = lsqr(A, b)[:4]
>>> istop
2
>>> x
array([ 1.00333333, -0.99666667])
>>> A.dot(x)-b
array([ 0.00333333, -0.00333333,  0.00333333])
>>> r1norm
0.005773502691896255

`istop` indicates that the system is inconsistent and thus `x` is rather an
approximate solution to the corresponding least-squares problem. `r1norm`
contains the norm of the minimal residual that was found.
r