问题
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))
The last assertion fails. Why?
回答1:
Just play with small numbers to debug your problem.
Start with A=np.random.randn(3,2)
instead of your much larger matrix with size (50,20)
In my random case, I find that
v1 = array([[-0.33872745, 0.94088454],
[-0.94088454, -0.33872745]])
and for v2
:
v2 = array([[ 0.33872745, -0.94088454],
[ 0.94088454, 0.33872745]])
they only differ for a sign, and obviously, even if normalized to have unit module, the vector can differ for a sign.
Now if you try the trick
assert np.all(np.isclose(V1,-1*V2))
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.
A correct way to check for equality between the vectors is:
assert allclose(abs((V1*V2).sum(0)),1.)
and indeed, to get a feeling of how this works you can print this quantity:
(V1*V2).sum(0)
that indeed is either +1
or -1
depending on the vector:
array([ 1., -1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
1., -1., 1., 1., 1., -1., -1.])
EDIT: 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:
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
来源:https://stackoverflow.com/questions/27781872/eigenvectors-computed-with-numpys-eigh-and-svd-do-not-match