Question

我正在使用numpy来计算圆形矩阵的特征值和特征向量。这是我的代码（j = 1,2 ... 6的Hji是预定义的）：

>>> import numpy as np
>>> H = np.array([H1i, H2i, H3i, H4i, H5i, H6i])
>>> H
array([[ 0.,  1.,  0.,  0.,  0.,  1.],
       [ 1.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  1.,  0.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  0.,  1.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  1.],
       [ 1.,  0.,  0.,  0.,  1.,  0.]])
>>> from numpy import linalg as LA
>>> w, v = LA.eig(H)

>>> w
array([-2.,  2.,  1., -1., -1.,  1.])
>>> v
array([[ 0.40824829, -0.40824829, -0.57735027,  0.57732307,  0.06604706,
         0.09791921],
       [-0.40824829, -0.40824829, -0.28867513, -0.29351503, -0.5297411 ,
        -0.4437968 ],
       [ 0.40824829, -0.40824829,  0.28867513, -0.28380804,  0.46369403,
        -0.54171601],
       [-0.40824829, -0.40824829,  0.57735027,  0.57732307,  0.06604706,
        -0.09791921],
       [ 0.40824829, -0.40824829,  0.28867513, -0.29351503, -0.5297411 ,
         0.4437968 ],
       [-0.40824829, -0.40824829, -0.28867513, -0.28380804,  0.46369403,
         0.54171601]])

特征值是正确的。然而，对于特征向量，我发现它们不是线性独立的

>>> V = np.zeros((6,6))
>>> for i in range(6):
...     for j in range(6):
...         V[i,j] = np.dot(v[:,i], v[:,j])
... 

>>> V
array([[  1.00000000e+00,  -2.77555756e-17,  -2.49800181e-16,
         -3.19189120e-16,  -1.11022302e-16,   2.77555756e-17],
       [ -2.77555756e-17,   1.00000000e+00,  -1.24900090e-16,
         -1.11022302e-16,  -8.32667268e-17,   0.00000000e+00],
       [ -2.49800181e-16,  -1.24900090e-16,   1.00000000e+00,
         -1.52655666e-16,   8.32667268e-17,  -1.69601044e-01],
       [ -3.19189120e-16,  -1.11022302e-16,  -1.52655666e-16,
          1.00000000e+00,   1.24034735e-01,  -8.32667268e-17],
       [ -1.11022302e-16,  -8.32667268e-17,   8.32667268e-17,
          1.24034735e-01,   1.00000000e+00,  -1.66533454e-16],
       [  2.77555756e-17,   0.00000000e+00,  -1.69601044e-01,
         -8.32667268e-17,  -1.66533454e-16,   1.00000000e+00]])
>>>

你可以看到有非对角线项（检查V [2,5] = -1.69601044e-01），这意味着它们不是线性独立向量。由于这是一个Hermitian矩阵，它的特征向量如何变得依赖？

顺便说一句，我也使用matlab来计算它并返回正确的值

V =

    0.4082   -0.2887   -0.5000    0.5000    0.2887   -0.4082
   -0.4082   -0.2887    0.5000    0.5000   -0.2887   -0.4082
    0.4082    0.5774         0         0   -0.5774   -0.4082
   -0.4082   -0.2887   -0.5000   -0.5000   -0.2887   -0.4082
    0.4082   -0.2887    0.5000   -0.5000    0.2887   -0.4082
   -0.4082    0.5774         0         0    0.5774   -0.4082


D =

   -2.0000         0         0         0         0         0
         0   -1.0000         0         0         0         0
         0         0   -1.0000         0         0         0
         0         0         0    1.0000         0         0
         0         0         0         0    1.0000         0
         0         0         0         0         0    2.0000

Answer 1

eig返回的结果非常好。

可以看出这一点

np.allclose(v.dot(np.diag(w)).dot(LA.inv(v)),H)

True

请注意，eig的输出对应于v * diag(w) * inv(v)形式的输入矩阵的分解，其适用于通用的可对角化矩阵。由于eig将H视为没有特殊结构，因此预期返回的特征向量不具有特殊结构，例如正交。（不要将正交性与线性独立性混淆 - v的列确实是线性独立的，因为可以通过非零LA.det(v)进行简单验证。）

函数eigh知道输入矩阵是hermitian并返回一组更方便的，即正交的特征向量。

Answer 2

对于Hermitian和对称矩阵，您应该使用另一个函数：eigh。

import numpy as np
from numpy import linalg as LA

H = np.array([[ 0.,  1.,  0.,  0.,  0.,  1.],
       [ 1.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  1.,  0.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  0.,  1.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  1.],
       [ 1.,  0.,  0.,  0.,  1.,  0.]])

w, v = LA.eigh(H)

V = np.zeros((6,6))
for i in range(6):
    for j in range(6):
        V[i,j] = np.dot(v[:,i], v[:,j])

w
Out[19]: array([-2., -1., -1.,  1.,  1.,  2.])

v
Out[20]: 
array([[-0.40824829, -0.57735027,  0.        ,  0.        ,  0.57735027,
         0.40824829],
       [ 0.40824829,  0.28867513, -0.5       , -0.5       ,  0.28867513,
         0.40824829],
       [-0.40824829,  0.28867513,  0.5       , -0.5       , -0.28867513,
         0.40824829],
       [ 0.40824829, -0.57735027,  0.        ,  0.        , -0.57735027,
         0.40824829],
       [-0.40824829,  0.28867513, -0.5       ,  0.5       , -0.28867513,
         0.40824829],
       [ 0.40824829,  0.28867513,  0.5       ,  0.5       ,  0.28867513,
         0.40824829]])

V
Out[21]: 
array([[  1.00000000e+00,   8.32667268e-17,   2.77555756e-17,
          8.32667268e-17,  -2.08166817e-16,   0.00000000e+00],
       [  8.32667268e-17,   1.00000000e+00,   5.55111512e-17,
          5.55111512e-17,  -2.22044605e-16,  -1.11022302e-16],
       [  2.77555756e-17,   5.55111512e-17,   1.00000000e+00,
          0.00000000e+00,   2.77555756e-17,   1.11022302e-16],
       [  8.32667268e-17,   5.55111512e-17,   0.00000000e+00,
          1.00000000e+00,   8.32667268e-17,   5.55111512e-17],
       [ -2.08166817e-16,  -2.22044605e-16,   2.77555756e-17,
          8.32667268e-17,   1.00000000e+00,   0.00000000e+00],
       [  0.00000000e+00,  -1.11022302e-16,   1.11022302e-16,
          5.55111512e-17,   0.00000000e+00,   1.00000000e+00]])

numpy似乎为圆形矩阵

2 个答案: