4 python matrix scipy eigenvalue eigenvector
我正在尝试找到以下矩阵的特征值/向量:
A = np.array([[1, 0, 0],
[0, 1, 0],
[1, 1, 0]])
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使用代码:
from numpy import linalg as LA
e_vals, e_vecs = LA.eig(A)
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我得到这个作为答案:
print(e_vals)
[ 0. 1. 1.]
print(e_vecs)
[[ 0. 0.70710678 0. ]
[ 0. 0. 0.70710678]
[ 1. 0.70710678 0.70710678]]
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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
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也就是说,特征值-特征向量问题表明以下内容应成立:
# 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]
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通过返回的特征值linalg.eig的列向量,所以你需要将遍历转的e_vecs(在默认情况下,一个二维数组,返回行向量自迭代):
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)
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