目前我正在尝试解决NumPy中两个对称矩阵的广义特征值问题，并且我遇到了大麻烦，因为我期望所有特征值都是正的，但eigh会返回几个非常大的特征值数字不是全部正数，而eig返回正确的预期值（但当然非常非常慢）。
在这种情况下，请注意K按照其构造预期是对称的（这是相关代码）：
# Calculate K matrix (<i|pHp|j> in the LGL-nodes basis)
for i in range(Ne):
idx_s, idx_e = i*(Np-1), i*(Np-1)+Np
K[idx_s:idx_e, idx_s:idx_e] += dmat.T.dot(diag(w*peq[idx_s:idx_e])).dot(dmat)
# Re-make matrix for efficient vector products
K = sparse.csr_matrix(K)
# Make matrix for <i|p|j> in the LGL basis as efficient diagonal sparse matrix
S = sparse.diags(peq*w_d, 0)
# Solve the generalized eigenvalue problem: Kc = lSc for hermitian matrices K and S
lQ, Q = linalg.eigh(K.todense(), S.todense())
_lQ, _Q = linalg.eig(K.todense(), S.todense())
lQ.sort()
_lQ.sort()
if not allclose(lQ, _lQ):
print('Literally why')
print(lQ)
print(_lQ)
return
为了进行测试，dmat被定义为
array([[ -896. , 1212.00631086, -484.43454844, 275.06612251,
-179.85209531, 124.26620323, -83.05199285, 32. ],
[ -205.43460499, 0. , 290.78944413, -135.17191772,
82.83085126, -55.64467829, 36.70818656, -14.07728095],
[ 50.7185076 , -179.61445086, 0. , 184.03311398,
-87.85829324, 54.08144362, -34.37053351, 13.01021241],
[ -23.81762789, 69.05246008, -152.20398294, 0. ,
152.89115899, -72.66291308, 42.31407046, -15.57316561],
[ 15.57316561, -42.31407046, 72.66291308, -152.89115899,
0. , 152.20398294, -69.05246008, 23.81762789],
[ -13.01021241, 34.37053351, -54.08144362, 87.85829324,
-184.03311398, 0. , 179.61445086, -50.7185076 ],
[ 14.07728095, -36.70818656, 55.64467829, -82.83085126,
135.17191772, -290.78944413, 0. , 205.43460499],
[ -32. , 83.05199285, -124.26620323, 179.85209531,
-275.06612251, 484.43454844, -1212.00631086, 896. ]])
所有w[i]，w_d[i]，peq[i]基本上都是任意正值数组。 w_d和w具有相同的顺序（~1e-1），peq[i]的范围为（~1e-10到1e1）
我得到的一些输出是
Literally why
[ -6.25540943e+07 -4.82660391e+07 -2.62629052e+07 ..., 1.07960873e+10
1.07967334e+10 4.26007915e+10]
[ -5.25462340e-12+0.j 4.62614812e-01+0.j 1.23357898e+00+0.j ...,
2.17613917e+06+0.j 1.07967334e+10+0.j 4.26007915e+10+0.j]
编辑：
这是一个自包含的代码版本，便于调试
import numpy as np
from math import *
from scipy import sparse, linalg
# Variable declarations and such (pre-computed)
Ne, Np = 256, 8
N = Ne*Np - Ne + 1
domain_size = 4/Ne
x = np.array([-0.015625 , -0.01362094, -0.00924532, -0.0032703 , 0.0032703 ,
0.00924532, 0.01362094, 0.015625 ])
w = np.array([ 0.00055804, 0.00329225, 0.00533004, 0.00644467, 0.00644467,
0.00533004, 0.00329225, 0.00055804])
dmat = np.array([[ -896. , 1212.00631086, -484.43454844, 275.06612251,
-179.85209531, 124.26620323, -83.05199285, 32. ],
[ -205.43460499, 0. , 290.78944413, -135.17191772,
82.83085126, -55.64467829, 36.70818656, -14.07728095],
[ 50.7185076 , -179.61445086, 0. , 184.03311398,
-87.85829324, 54.08144362, -34.37053351, 13.01021241],
[ -23.81762789, 69.05246008, -152.20398294, 0. ,
152.89115899, -72.66291308, 42.31407046, -15.57316561],
[ 15.57316561, -42.31407046, 72.66291308, -152.89115899,
0. , 152.20398294, -69.05246008, 23.81762789],
[ -13.01021241, 34.37053351, -54.08144362, 87.85829324,
-184.03311398, 0. , 179.61445086, -50.7185076 ],
[ 14.07728095, -36.70818656, 55.64467829, -82.83085126,
135.17191772, -290.78944413, 0. , 205.43460499],
[ -32. , 83.05199285, -124.26620323, 179.85209531,
-275.06612251, 484.43454844, -1212.00631086, 896. ]])
# More declarations
x_d = np.zeros(N)
w_d = np.zeros(N)
dmat_d = np.zeros((N, N))
for i in range(Ne):
x_d[i*(Np-1):i*(Np-1)+Np] = x+i*domain_size
w_d[i*(Np-1):i*(Np-1)+Np] += w
dmat_d[i*(Np-1):i*(Np-1)+Np, i*(Np-1):i*(Np-1)+Np] += dmat
peq = (np.cos((x_d-2)*pi/4))**2
# Normalization
peq = peq/np.sum(w_d*peq)
p0 = np.maximum(peq, 1e-10)
p0 /= np.sum(p0*w_d)
# Make efficient matrix that can be built
K = sparse.lil_matrix((N, N))
# Calculate K matrix (<i|pHp|j> in the LGL-nodes basis)
for i in range(Ne):
idx_s, idx_e = i*(Np-1), i*(Np-1)+Np
K[idx_s:idx_e, idx_s:idx_e] += dmat.T.dot(np.diag(w*p0[idx_s:idx_e])).dot(dmat)
# Re-make matrix for efficient vector products
K = sparse.csr_matrix(K)
# Make matrix for <i|p|j> in the LGL basis as efficient diagonal sparse matrix
S = sparse.diags(p0*w_d, 0)
# Solve the generalized eigenvalue problem: Kc = lSc for hermitian matrices K and S
lQ, Q = linalg.eigh(K.todense(), S.todense())
_lQ, _Q = linalg.eig(K.todense(), S.todense())
lQ.sort()
_lQ.sort()
if not np.allclose(lQ, _lQ):
print('Literally why')
print(lQ)
print(_lQ)
EDIT2：这真的很奇怪。在我的机器上运行所有NumPy / SciPy测试，我没有收到任何错误。但即使运行简单的测试（具有足够大的矩阵）作为
import numpy as np
from spicy import linalg
M = np.random.random((1000,1000))
M += M.T
np.allclose(sorted(linalg.eigh(M)[0]), sorted(linalg.eig(M)[0]))
在我的机器上失败。虽然使用50x50矩阵运行相同的测试确实有效 - 即使在重建SciPy / NumPy堆栈并通过所有单元测试后也是如此。
EDIT3：实际上，在群集计算机上测试后，这似乎无处不在。我不确定原因。
由于+=和.T作为视图而非操作的就地行为，上述操作失败。