python中的左侧特征向量？

Question

如何计算python中的左侧特征向量？

    >>> import from numpy as np
    >>> from scipy.linalg import eig
    >>> np.set_printoptions(precision=4)
    >>> T = np.mat("0.2 0.4 0.4;0.8 0.2 0.0;0.8 0.0 0.2")
    >>> print "T\n", T
    T
    [[ 0.2  0.4  0.4]
     [ 0.8  0.2  0. ]
     [ 0.8  0.   0.2]]
    >>> w, vl, vr = eig(T, left=True)
    >>> vl
    array([[ 0.8165,  0.8165,  0.    ],
           [ 0.4082, -0.4082, -0.7071],
           [ 0.4082, -0.4082,  0.7071]])

这似乎不正确，谷歌对此并不友好！

Answer 1

Your result is correct to my understanding.

However, you might be misinterpreting it. The numpy docs are a bit clearer on what the left eigenvectors should be.

Finally, it is emphasized that v consists of the right (as in right-hand side) eigenvectors of a. A vector y satisfying dot(y.T, a) = z * y.T for some number z is called a left eigenvector of a, and, in general, the left and right eigenvectors of a matrix are not necessarily the (perhaps conjugate) transposes of each other.

即您需要转置vl. vl[:,i].T是第 i 个左特征向量。如果我对此进行测试，我会得到结果是正确的。

>>> import numpy as np
>>> from scipy.linalg import eig
>>> np.set_printoptions(precision=4)
>>> T = np.mat("0.2 0.4 0.4;0.8 0.2 0.0;0.8 0.0 0.2")
>>> print "T\n", T
T
[[ 0.2  0.4  0.4]
 [ 0.8  0.2  0. ]
 [ 0.8  0.   0.2]]
>>> w, vl, vr = eig(T, left=True)
>>> vl
array([[ 0.8165,  0.8165,  0.    ],
       [ 0.4082, -0.4082, -0.7071],
       [ 0.4082, -0.4082,  0.7071]])
>>> [ np.allclose(np.dot(vl[:,i].T, T), w[i]*vl[:,i].T) for i in range(3) ]
[True, True, True]