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"""Frechet derivative of the matrix exponential."""
import numpy as np
import scipy.linalg
__all__ = ['expm_frechet', 'expm_cond']
def expm_frechet(A, E, method=None, compute_expm=True, check_finite=True):
"""
Frechet derivative of the matrix exponential of A in the direction E.
Parameters
----------
A : (N, N) array_like
Matrix of which to take the matrix exponential.
E : (N, N) array_like
Matrix direction in which to take the Frechet derivative.
method : str, optional
Choice of algorithm. Should be one of
- `SPS` (default)
- `blockEnlarge`
compute_expm : bool, optional
Whether to compute also `expm_A` in addition to `expm_frechet_AE`.
Default is True.
check_finite : bool, optional
Whether to check that the input matrix contains only finite numbers.
Disabling may give a performance gain, but may result in problems
(crashes, non-termination) if the inputs do contain infinities or NaNs.
Returns
-------
expm_A : ndarray
Matrix exponential of A.
expm_frechet_AE : ndarray
Frechet derivative of the matrix exponential of A in the direction E.
For ``compute_expm = False``, only `expm_frechet_AE` is returned.
See Also
--------
expm : Compute the exponential of a matrix.
Notes
-----
This section describes the available implementations that can be selected
by the `method` parameter. The default method is *SPS*.
Method *blockEnlarge* is a naive algorithm.
Method *SPS* is Scaling-Pade-Squaring [1]_.
It is a sophisticated implementation which should take
only about 3/8 as much time as the naive implementation.
The asymptotics are the same.
.. versionadded:: 0.13.0
References
----------
.. [1] Awad H. Al-Mohy and Nicholas J. Higham (2009)
Computing the Frechet Derivative of the Matrix Exponential,
with an application to Condition Number Estimation.
SIAM Journal On Matrix Analysis and Applications.,
30 (4). pp. 1639-1657. ISSN 1095-7162
Examples
--------
>>> import numpy as np
>>> from scipy import linalg
>>> rng = np.random.default_rng()
>>> A = rng.standard_normal((3, 3))
>>> E = rng.standard_normal((3, 3))
>>> expm_A, expm_frechet_AE = linalg.expm_frechet(A, E)
>>> expm_A.shape, expm_frechet_AE.shape
((3, 3), (3, 3))
Create a 6x6 matrix containing [[A, E], [0, A]]:
>>> M = np.zeros((6, 6))
>>> M[:3, :3] = A
>>> M[:3, 3:] = E
>>> M[3:, 3:] = A
>>> expm_M = linalg.expm(M)
>>> np.allclose(expm_A, expm_M[:3, :3])
True
>>> np.allclose(expm_frechet_AE, expm_M[:3, 3:])
True
"""
if check_finite:
A = np.asarray_chkfinite(A)
E = np.asarray_chkfinite(E)
else:
A = np.asarray(A)
E = np.asarray(E)
if A.ndim != 2 or A.shape[0] != A.shape[1]:
raise ValueError('expected A to be a square matrix')
if E.ndim != 2 or E.shape[0] != E.shape[1]:
raise ValueError('expected E to be a square matrix')
if A.shape != E.shape:
raise ValueError('expected A and E to be the same shape')
if method is None:
method = 'SPS'
if method == 'SPS':
expm_A, expm_frechet_AE = expm_frechet_algo_64(A, E)
elif method == 'blockEnlarge':
expm_A, expm_frechet_AE = expm_frechet_block_enlarge(A, E)
else:
raise ValueError(f'Unknown implementation {method}')
if compute_expm:
return expm_A, expm_frechet_AE
else:
return expm_frechet_AE
def expm_frechet_block_enlarge(A, E):
"""
This is a helper function, mostly for testing and profiling.
Return expm(A), frechet(A, E)
"""
n = A.shape[0]
M = np.vstack([
np.hstack([A, E]),
np.hstack([np.zeros_like(A), A])])
expm_M = scipy.linalg.expm(M)
return expm_M[:n, :n], expm_M[:n, n:]
"""
Maximal values ell_m of ||2**-s A|| such that the backward error bound
does not exceed 2**-53.
"""
ell_table_61 = (
None,
# 1
2.11e-8,
3.56e-4,
1.08e-2,
6.49e-2,
2.00e-1,
4.37e-1,
7.83e-1,
1.23e0,
1.78e0,
2.42e0,
# 11
3.13e0,
3.90e0,
4.74e0,
5.63e0,
6.56e0,
7.52e0,
8.53e0,
9.56e0,
1.06e1,
1.17e1,
)
# The b vectors and U and V are copypasted
# from scipy.sparse.linalg.matfuncs.py.
# M, Lu, Lv follow (6.11), (6.12), (6.13), (3.3)
def _diff_pade3(A, E, ident):
b = (120., 60., 12., 1.)
A2 = A.dot(A)
M2 = np.dot(A, E) + np.dot(E, A)
U = A.dot(b[3]*A2 + b[1]*ident)
V = b[2]*A2 + b[0]*ident
Lu = A.dot(b[3]*M2) + E.dot(b[3]*A2 + b[1]*ident)
Lv = b[2]*M2
return U, V, Lu, Lv
def _diff_pade5(A, E, ident):
b = (30240., 15120., 3360., 420., 30., 1.)
A2 = A.dot(A)
M2 = np.dot(A, E) + np.dot(E, A)
A4 = np.dot(A2, A2)
M4 = np.dot(A2, M2) + np.dot(M2, A2)
U = A.dot(b[5]*A4 + b[3]*A2 + b[1]*ident)
V = b[4]*A4 + b[2]*A2 + b[0]*ident
Lu = (A.dot(b[5]*M4 + b[3]*M2) +
E.dot(b[5]*A4 + b[3]*A2 + b[1]*ident))
Lv = b[4]*M4 + b[2]*M2
return U, V, Lu, Lv
def _diff_pade7(A, E, ident):
b = (17297280., 8648640., 1995840., 277200., 25200., 1512., 56., 1.)
A2 = A.dot(A)
M2 = np.dot(A, E) + np.dot(E, A)
A4 = np.dot(A2, A2)
M4 = np.dot(A2, M2) + np.dot(M2, A2)
A6 = np.dot(A2, A4)
M6 = np.dot(A4, M2) + np.dot(M4, A2)
U = A.dot(b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident)
V = b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident
Lu = (A.dot(b[7]*M6 + b[5]*M4 + b[3]*M2) +
E.dot(b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident))
Lv = b[6]*M6 + b[4]*M4 + b[2]*M2
return U, V, Lu, Lv
def _diff_pade9(A, E, ident):
b = (17643225600., 8821612800., 2075673600., 302702400., 30270240.,
2162160., 110880., 3960., 90., 1.)
A2 = A.dot(A)
M2 = np.dot(A, E) + np.dot(E, A)
A4 = np.dot(A2, A2)
M4 = np.dot(A2, M2) + np.dot(M2, A2)
A6 = np.dot(A2, A4)
M6 = np.dot(A4, M2) + np.dot(M4, A2)
A8 = np.dot(A4, A4)
M8 = np.dot(A4, M4) + np.dot(M4, A4)
U = A.dot(b[9]*A8 + b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident)
V = b[8]*A8 + b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident
Lu = (A.dot(b[9]*M8 + b[7]*M6 + b[5]*M4 + b[3]*M2) +
E.dot(b[9]*A8 + b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident))
Lv = b[8]*M8 + b[6]*M6 + b[4]*M4 + b[2]*M2
return U, V, Lu, Lv
def expm_frechet_algo_64(A, E):
n = A.shape[0]
s = None
ident = np.identity(n)
A_norm_1 = scipy.linalg.norm(A, 1)
m_pade_pairs = (
(3, _diff_pade3),
(5, _diff_pade5),
(7, _diff_pade7),
(9, _diff_pade9))
for m, pade in m_pade_pairs:
if A_norm_1 <= ell_table_61[m]:
U, V, Lu, Lv = pade(A, E, ident)
s = 0
break
if s is None:
# scaling
s = max(0, int(np.ceil(np.log2(A_norm_1 / ell_table_61[13]))))
A = A * 2.0**-s
E = E * 2.0**-s
# pade order 13
A2 = np.dot(A, A)
M2 = np.dot(A, E) + np.dot(E, A)
A4 = np.dot(A2, A2)
M4 = np.dot(A2, M2) + np.dot(M2, A2)
A6 = np.dot(A2, A4)
M6 = np.dot(A4, M2) + np.dot(M4, A2)
b = (64764752532480000., 32382376266240000., 7771770303897600.,
1187353796428800., 129060195264000., 10559470521600.,
670442572800., 33522128640., 1323241920., 40840800., 960960.,
16380., 182., 1.)
W1 = b[13]*A6 + b[11]*A4 + b[9]*A2
W2 = b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident
Z1 = b[12]*A6 + b[10]*A4 + b[8]*A2
Z2 = b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident
W = np.dot(A6, W1) + W2
U = np.dot(A, W)
V = np.dot(A6, Z1) + Z2
Lw1 = b[13]*M6 + b[11]*M4 + b[9]*M2
Lw2 = b[7]*M6 + b[5]*M4 + b[3]*M2
Lz1 = b[12]*M6 + b[10]*M4 + b[8]*M2
Lz2 = b[6]*M6 + b[4]*M4 + b[2]*M2
Lw = np.dot(A6, Lw1) + np.dot(M6, W1) + Lw2
Lu = np.dot(A, Lw) + np.dot(E, W)
Lv = np.dot(A6, Lz1) + np.dot(M6, Z1) + Lz2
# factor once and solve twice
lu_piv = scipy.linalg.lu_factor(-U + V)
R = scipy.linalg.lu_solve(lu_piv, U + V)
L = scipy.linalg.lu_solve(lu_piv, Lu + Lv + np.dot((Lu - Lv), R))
# squaring
for k in range(s):
L = np.dot(R, L) + np.dot(L, R)
R = np.dot(R, R)
return R, L
def vec(M):
"""
Stack columns of M to construct a single vector.
This is somewhat standard notation in linear algebra.
Parameters
----------
M : 2-D array_like
Input matrix
Returns
-------
v : 1-D ndarray
Output vector
"""
return M.T.ravel()
def expm_frechet_kronform(A, method=None, check_finite=True):
"""
Construct the Kronecker form of the Frechet derivative of expm.
Parameters
----------
A : array_like with shape (N, N)
Matrix to be expm'd.
method : str, optional
Extra keyword to be passed to expm_frechet.
check_finite : bool, optional
Whether to check that the input matrix contains only finite numbers.
Disabling may give a performance gain, but may result in problems
(crashes, non-termination) if the inputs do contain infinities or NaNs.
Returns
-------
K : 2-D ndarray with shape (N*N, N*N)
Kronecker form of the Frechet derivative of the matrix exponential.
Notes
-----
This function is used to help compute the condition number
of the matrix exponential.
See Also
--------
expm : Compute a matrix exponential.
expm_frechet : Compute the Frechet derivative of the matrix exponential.
expm_cond : Compute the relative condition number of the matrix exponential
in the Frobenius norm.
"""
if check_finite:
A = np.asarray_chkfinite(A)
else:
A = np.asarray(A)
if len(A.shape) != 2 or A.shape[0] != A.shape[1]:
raise ValueError('expected a square matrix')
n = A.shape[0]
ident = np.identity(n)
cols = []
for i in range(n):
for j in range(n):
E = np.outer(ident[i], ident[j])
F = expm_frechet(A, E,
method=method, compute_expm=False, check_finite=False)
cols.append(vec(F))
return np.vstack(cols).T
def expm_cond(A, check_finite=True):
"""
Relative condition number of the matrix exponential in the Frobenius norm.
Parameters
----------
A : 2-D array_like
Square input matrix with shape (N, N).
check_finite : bool, optional
Whether to check that the input matrix contains only finite numbers.
Disabling may give a performance gain, but may result in problems
(crashes, non-termination) if the inputs do contain infinities or NaNs.
Returns
-------
kappa : float
The relative condition number of the matrix exponential
in the Frobenius norm
See Also
--------
expm : Compute the exponential of a matrix.
expm_frechet : Compute the Frechet derivative of the matrix exponential.
Notes
-----
A faster estimate for the condition number in the 1-norm
has been published but is not yet implemented in SciPy.
.. versionadded:: 0.14.0
Examples
--------
>>> import numpy as np
>>> from scipy.linalg import expm_cond
>>> A = np.array([[-0.3, 0.2, 0.6], [0.6, 0.3, -0.1], [-0.7, 1.2, 0.9]])
>>> k = expm_cond(A)
>>> k
1.7787805864469866
"""
if check_finite:
A = np.asarray_chkfinite(A)
else:
A = np.asarray(A)
if len(A.shape) != 2 or A.shape[0] != A.shape[1]:
raise ValueError('expected a square matrix')
X = scipy.linalg.expm(A)
K = expm_frechet_kronform(A, check_finite=False)
# The following norm choices are deliberate.
# The norms of A and X are Frobenius norms,
# and the norm of K is the induced 2-norm.
A_norm = scipy.linalg.norm(A, 'fro')
X_norm = scipy.linalg.norm(X, 'fro')
K_norm = scipy.linalg.norm(K, 2)
kappa = (K_norm * A_norm) / X_norm
return kappa
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