def sqrtm(A, disp=True, blocksize=64):
    """
    Matrix square root.

    Parameters
    ----------
    A : (N, N) array_like
        Matrix whose square root to evaluate
    disp : bool, optional
        Print warning if error in the result is estimated large
        instead of returning estimated error. (Default: True)
    blocksize : integer, optional
        If the blocksize is not degenerate with respect to the
        size of the input array, then use a blocked algorithm. (Default: 64)

    Returns
    -------
    sqrtm : (N, N) ndarray
        Value of the sqrt function at `A`

    errest : float
        (if disp == False)

        Frobenius norm of the estimated error, ||err||_F / ||A||_F

    References
    ----------
    .. [1] Edvin Deadman, Nicholas J. Higham, Rui Ralha (2013)
           "Blocked Schur Algorithms for Computing the Matrix Square Root,
           Lecture Notes in Computer Science, 7782. pp. 171-182.

    Examples
    --------
    >>> from scipy.linalg import sqrtm
    >>> a = np.array([[1.0, 3.0], [1.0, 4.0]])
    >>> r = sqrtm(a)
    >>> r
    array([[ 0.75592895,  1.13389342],
           [ 0.37796447,  1.88982237]])
    >>> r.dot(r)
    array([[ 1.,  3.],
           [ 1.,  4.]])

    """
    A = _asarray_validated(A, check_finite=True, as_inexact=True)
    if len(A.shape) != 2:
        raise ValueError("Non-matrix input to matrix function.")
    if blocksize < 1:
        raise ValueError("The blocksize should be at least 1.")
    keep_it_real = np.isrealobj(A)
    if keep_it_real:
        T, Z = schur(A)
        if not np.array_equal(T, np.triu(T)):
            T, Z = rsf2csf(T, Z)
    else:
        T, Z = schur(A, output='complex')
    failflag = False
    try:
        R = _sqrtm_triu(T, blocksize=blocksize)
        ZH = np.conjugate(Z).T
        X = Z.dot(R).dot(ZH)
    except SqrtmError:
        failflag = True
        X = np.empty_like(A)
        X.fill(np.nan)

    if disp:
        nzeig = np.any(np.diag(T) == 0)
        if nzeig:
            print("Matrix is singular and may not have a square root.")
        elif failflag:
            print("Failed to find a square root.")
        return X
    else:
        try:
            arg2 = norm(X.dot(X) - A, 'fro')**2 / norm(A, 'fro')
        except __HOLE__:
            # NaNs in matrix
            arg2 = np.inf

        return X, arg2
dataset/ETHPy150Open scipy/scipy/scipy/linalg/_matfuncs_sqrtm.py/sqrtm