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numpyeigenvalueeigenvector
Ale
Ale

Reputation: 21

Eigenvector normalization in numpy

I'm using the linalg in numpy to compute eigenvalues and eigenvectors of matrices of signed reals. I've read this previous question but still don't grasp the normalization of eigenvectors. Here is an example straight off Wikipedia:

    import numpy as np
    from numpy import linalg as la

    a = np.matrix([[2, 1], [1, 2]], dtype=np.float)

    eigh_vals, eigh_vects = np.linalg.eig(a)

    print 'eigen_values='
    print eigh_vals

    print 'eigen_vectors='
    print eigh_vects

The eigenvalues are 1 and 3. For eigenvectors we expect scalar multiples of [1, -1] and [1, 1], which I get:

    eig_vals=
    [ 3.  1.]
    eig_vets=
    [[ 0.70710678 -0.70710678]
    [ 0.70710678  0.70710678]]

I understand the 1/sqrt(2) factor is to have the norm=1 but why? Can normalization be 'switched off'? Thanks!

Upvotes: 2

Views: 10327

Answers (3)

OChicken
OChicken

Reputation: 91

import numpy as np
import sympy as sp
v = sp.Matrix([[2, 1], [1, 2]])
v_vec  = v.eigenvects()

v_vec is a list contains 2 tuples:

[(1, 1, [Matrix([
   [-1],
   [ 1]])]), (3, 1, [Matrix([
   [1],
   [1]])])]

1 and 3 is the two eigenvalues. The '1' behind 1 & 3 is the number of the eigenvalues. In each tuple, the third element is the eigenvector of each eigenvalue. It is a Matrix object in sp. You can convert a Matrix object to the np array.

v_vec1 = np.array(v_vec[0][2], dtype=float)
v_vec2 = np.array(v_vec[1][2], dtype=float)
print('v_vec1 =', v_vec1)
print('v_vec2 =', v_vec2)

Here is the normalized eigenvectors you would get:

v_vec1 = [[-1.  1.]]
v_vec2 = [[1. 1.]]

Upvotes: 1

Paul Panzer
Paul Panzer

Reputation: 53119

If sympy is an option for you, it appears to normalize less aggressively:

import sympy
a = sympy.Matrix([[2, 1], [1, 2]])
a.eigenvects()
# [(1, 1, [Matrix([
# [-1],
# [ 1]])]), (3, 1, [Matrix([
# [1],
# [1]])])]

Upvotes: 0

Nils Werner
Nils Werner

Reputation: 36859

The key message for the first eigenvector in the Wikipedia article is

Any non-zero vector with v1 = −v2 solves this equation.

So the actual solution is V1 = [x, -x]. Picking the vector V1 = [1, -1] may be pleasing to the human eye, but it is just as aritrary as picking a vector V1 = [104051, -104051] or any other real value.

Actually, picking V1 = [1, -1] / sqrt(2) is the least arbitrary. Of all the possible vectors for V1, it's the only one that is of unit length.

However if instead of unit length you prefer the first value to be 1, you can do

eigh_vects /= eigh_vects[:, 0]

Upvotes: 1

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