scipy.linalg.eig是否给出正确的左特征向量? [英] Is scipy.linalg.eig giving the correct left eigenvectors?

问题描述

我对scipy.linalg.eig如何计算左右特征向量有疑问.也许我误解了一切，但事情似乎不适合我...

I have a question regarding the way how scipy.linalg.eig computes left and right eigenvectors. Maybe I misunderstood everything, but things seem not to be right to me...

从一开始.为了获得特征值和两个特征向量，我使用以下代码:

From the beginning. To get eigenvalues and both eigenvectors I used the following:

ev, left_v, right_v = scipy.linalg.eig(A, left=True)

根据手册，在调用函数时设置left=True，我应该期望将左特征向量作为left_v的列，其中第i列引用第i个特征值.但是，结果不是我预期的，所以我做了一个简单的检查.

According to the manual, after setting left=True while calling the function I should expect to get left eigenvectors as columns of left_v where the ith column refers to the ith eigenvalue. However, the results were not what i anticipated so I did a simple check.

我计算了两次调用该函数的左右特征向量(有关详细信息，请参见此处):

I computed right and left eigenvectors invoking the function twice (look here for details):

right_ev, right_v_2 = scipy.linalg.eig(A)
left_ev, left_v_2 = scipy.linalg.eig(A.T)

其中，left_v_2的列是与left_ev中的对应值关联的特征向量. 值得强调的是，right_ev_2和left_ev_2都给出相同的特征值，但是它们的顺序不同，需要加以考虑.

where columns ofleft_v_2 are eigenvectors associated with corresponding values in left_ev. Worth stressing that both right_ev_2 and left_ev_2 give the same eigenvalues, however they are in different order, which needs to be accounted for.

比较left_ev和left_ev_2(在对特征值进行重新排序之后)，可以迅速发现前者是后者的共轭，因此从scipy.linalg.eig与left=True得出的left_ev无效.左特征向量.

Comparing left_ev and left_ev_2 (after reordering with respect to eigenvalues) one can quickly spot that the former is the conjugate of the latter and therefore left_ev obtained from scipy.linalg.eig with left=True is not a valid left eigenvector.

可以基于以下事实对特征向量的有效性进行另一次检查:对于任意实方矩阵，左右特征向量都是正交的，即:

Another check on the validity of the eigenvectors can be done based on the fact that for an arbitrary real square matrix left and right eigenvectors are biorthogonal, i.e.:

left_v.T.dot(right_v)应该给出一个对角矩阵，但是没有， 直到我将其更改为:left_v.T.conj().dot(right_v)，

left_v.T.dot(right_v) should give a diagonal matrix, but it doesn't, until i change it to: left_v.T.conj().dot(right_v),

同时:

left_v_2.T.dot(right_v_2)给出了预期的对角矩阵.

left_v_2.T.dot(right_v_2) gives an anticipated diagonal matrix.

有人以前遇到过类似的问题吗?我说的对吗?在描述eig时，科学手册是否有点不准确?你能给点建议吗?

Did anyone encounter similar problem before? Am I right with what I say? Is the sciPy manual a bit imprecise while describing eig? Can you give any advice?

非常感谢！

推荐答案

关于vl，eig文档字符串说:

a.H vl[:,i] = w[i].conj() b.H vl[:,i]

或者，采用两侧的共轭转置(即埃尔米特转置)(即.H的意思)，并假设b是同一性，

Or, taking the conjugate transpose (i.e. Hermitian transpose) of both sides (which is what .H means), and assuming b is the identity,

vl[:,i].H a = w[i] vl[:,i].H

所以vl的共轭转置的行是a的实际左特征向量.

So the rows of the conjugate transpose of vl are the actual left eigenvectors of a.

Numpy数组实际上没有.H属性，因此必须使用.conj().T.

Numpy arrays don't actually have the .H attribute, so you must use .conj().T.

这是一个验证计算的脚本:

Here's a script to verify the calculation:

import numpy as np
from scipy.linalg import eig

# This only affects the printed output.
np.set_printoptions(precision=4)

a = np.array([[6, 2],
              [-1, 4]])

w, vl, vr = eig(a, left=True)

print "eigenvalues:", w
print

# check the left eigenvectors one-by-one:
for k in range(a.shape[0]):
    val = w[k]
    # Use a slice to maintain shape; vec is a 2x1 array.
    # That allows a meaningful transpose using .T.
    vec = vl[:, k:k+1]
    # rowvec is 1x2; it is the conjugate transpose of vec.
    # This should be the left eigenvector.
    rowvec = vec.conj().T
    # Verify that rowvec is a left eigenvector
    lhs = rowvec.dot(a)
    rhs = val * rowvec
    print "Compare", lhs, "to", rhs
    print rowvec, "is",
    if not np.allclose(lhs, rhs):
        print "*NOT*",
    print "a left eigenvector for eigenvalue", val

print
print "Matrix version:"
print "This"
print vl.conj().T.dot(a)
print "should equal this"
print np.diag(w).dot(vl.conj().T)

输出:

eigenvalues: [ 5.+1.j  5.-1.j]

Compare [[ 1.6330+2.4495j  4.0825+0.8165j]] to [[ 1.6330+2.4495j  4.0825+0.8165j]]
[[ 0.4082+0.4082j  0.8165-0.j    ]] is a left eigenvector for eigenvalue (5+1j)
Compare [[ 1.6330-2.4495j  4.0825-0.8165j]] to [[ 1.6330-2.4495j  4.0825-0.8165j]]
[[ 0.4082-0.4082j  0.8165+0.j    ]] is a left eigenvector for eigenvalue (5-1j)

Matrix version:
This
[[ 1.6330+2.4495j  4.0825+0.8165j]
 [ 1.6330-2.4495j  4.0825-0.8165j]]
should equal this
[[ 1.6330+2.4495j  4.0825+0.8165j]
 [ 1.6330-2.4495j  4.0825-0.8165j]]

现在，eig文档字符串在返回值的描述中也说:

Now, the eig docstring also says in the description of the return values:

vl : double or complex ndarray
    The normalized left eigenvector corresponding to the eigenvalue
    ``w[i]`` is the column v[:,i]. Only returned if ``left=True``.
    Of shape ``(M, M)``.

，这可能会引起误解，因为左特征向量的常规定义(例如 http://mathworld. wolfram.com/LeftEigenvector.html 或 http://en.wikipedia.org/wiki /Eigenvalues_and_eigenvectors#Left_and_right_eigenvectors )是一个行向量，因此vl列的共轭转置实际上是左特征向量.

and that is potentially misleading, since the conventional definition of a left eigenvector (e.g. http://mathworld.wolfram.com/LeftEigenvector.html or http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Left_and_right_eigenvectors) is a row vector, so it is the conjugate transpose of the column of vl that is actually the left eigenvector.

这篇关于scipy.linalg.eig是否给出正确的左特征向量?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！