5

我试图找到以下矩阵的特征值/向量:

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

使用代码:

from numpy import linalg as LA
e_vals, e_vecs = LA.eig(A)

我得到这个答案:

print(e_vals)
[ 0.  1.  1.]

print(e_vecs)
[[ 0.          0.70710678  0.        ]
 [ 0.          0.          0.70710678]
 [ 1.          0.70710678  0.70710678]]

但是,我相信以下应该是答案。

[1] Real Eigenvalue = 0.00000
[1] Real Eigenvector:
0.00000
0.00000
1.00000

[2] Real Eigenvalue = 1.00000
[2] Real Eigenvector:
1.00000
0.00000
1.00000

[3] Real Eigenvalue = 1.00000
[3] Real Eigenvector:
0.00000
1.00000
1.00000

也就是说,特征值-特征向量问题表明以下应该成立:

# A * e_vecs = e_vals * e_vecs
print(A.dot(e_vecs))
[[ 0.          0.70710678  0.        ]
 [ 0.          0.          0.70710678]
 [ 0.          0.70710678  0.70710678]]

print(e_vals.dot(e_vecs))
[ 1.          0.70710678  1.41421356]
4

1 回答 1

5

返回的特征值linalg.eig是列向量,因此您需要迭代 的转置e_vecs因为在 2D 数组上的迭代默认返回行向量):

import numpy as np
import numpy.linalg as LA
A = np.array([[1, 0, 0], [0, 1, 0], [1, 1, 0]])
e_vals, e_vecs = LA.eig(A)

print(e_vals)
# [ 0.  1.  1.]
print(e_vecs)
# [[ 0.          0.          1.        ]
#  [ 0.70710678  0.          0.70710678]
#  [ 0.          0.70710678  0.70710678]]

for val, vec in zip(e_vals, e_vecs.T):
    assert np.allclose(np.dot(A, vec), val * vec)
于 2013-09-12T18:22:15.693 回答