用numpy的eigh和svd计算的特征向量不匹配 [英] Eigenvectors computed with numpy's eigh and svd do not match

问题描述

考虑奇异值分解M = USV *.然后，M * M的特征值分解给出M * M = V(S * S)V * = VS * U * USV *.我希望通过numpy验证这种相等性，方法是显示eigh函数返回的特征向量与svd函数返回的特征向量相同:

Consider singular value decomposition M=USV*. Then the eigenvalue decomposition of M* M gives M* M= V (S* S) V*=VS* U* USV*. I wish to verify this equality with numpy by showing that the eigenvectors returned by eigh function are the same as those returned by svd function:

import numpy as np
np.random.seed(42)
# create mean centered data
A=np.random.randn(50,20)
M= A-np.array(A.mean(0),ndmin=2)

# svd
U1,S1,V1=np.linalg.svd(M) 
S1=np.square(S1)
V1=V1.T  

# eig
S2,V2=np.linalg.eigh(np.dot(M.T,M))
indx=np.argsort(S2)[::-1]
S2=S2[indx]
V2=V2[:,indx]

# both Vs are in orthonormal form
assert np.all(np.isclose(np.linalg.norm(V1,axis=1), np.ones(V1.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V1,axis=0), np.ones(V1.shape[1])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=1), np.ones(V2.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=0), np.ones(V2.shape[1])))

assert np.all(np.isclose(S1,S2))
assert np.all(np.isclose(V1,V2))

最后一个断言失败.为什么?

The last assertion fails. Why?

推荐答案

只需玩少量数字即可调试问题.

Just play with small numbers to debug your problem.

以A=np.random.randn(3,2)开头，而不是尺寸为(50,20)

Start with A=np.random.randn(3,2) instead of your much larger matrix with size (50,20)

在我的随机情况下，我发现

In my random case, I find that

v1 = array([[-0.33872745,  0.94088454],
   [-0.94088454, -0.33872745]])

以及v2:

v2 = array([[ 0.33872745, -0.94088454],
   [ 0.94088454,  0.33872745]])

它们仅在符号上有所不同，显然，即使标准化为具有单位模块，矢量在符号上也可能不同.

they only differ for a sign, and obviously, even if normalized to have unit module, the vector can differ for a sign.

现在，如果您尝试把戏

assert np.all(np.isclose(V1,-1*V2))

对于您原来的大矩阵，它失败了……再次，这没关系.发生的情况是，某些向量已与-1相乘，而另一些则没有.

for your original big matrix, it fails... again, this is OK. What happens is that some vectors have been multiplied by -1, some others haven't.

检查向量之间是否相等的正确方法是:

A correct way to check for equality between the vectors is:

assert allclose(abs((V1*V2).sum(0)),1.)

实际上，要了解其工作原理，可以打印以下数量:

and indeed, to get a feeling of how this works you can print this quantity:

(V1*V2).sum(0)

根据向量，确实是+1或-1:

array([ 1., -1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,
    1., -1.,  1.,  1.,  1., -1., -1.])

在大多数情况下，尤其是从随机矩阵开始时，会发生这种情况.但是请注意，如@Sven Marnach在下面的评论中指出的那样，如果一个或多个特征值的特征空间的维数大于1，则该测试可能会失败.

This will happen in most cases, especially if starting from a random matrix. Notice however that this test will likely fail if one or more eigenvalues has an eigenspace of dimension larger than 1, as pointed out by @Sven Marnach in his comment below:

除了向量乘以-1之外，可能还有其他差异. 如果任何特征值具有多维特征空间，则您 可能会获得该特征空间的任意正交基础，并且 这样的基地可以通过套期交易彼此相对旋转 一元矩阵

There might be other differences than just vectors multiplied by -1. If any of the eigenvalues has a multi-dimensional eigenspace, you might get an arbitrary orthonormal basis of that eigenspace, and to such bases might be rotated against each other by an arbitraty unitarian matrix

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