# Power Iteration and its Variants¶

In [2]:
#keep
import numpy as np
import numpy.linalg as la


Let's prepare a matrix with some random or deliberately chosen eigenvalues:

In [6]:
#keep
n = 6

if 1:
np.random.seed(70)
eigvecs = np.random.randn(n, n)
eigvals = np.sort(np.random.randn(n))
# Uncomment for near-duplicate largest-magnitude eigenvalue
# eigvals[1] = eigvals[0] + 1e-3

A = eigvecs.dot(np.diag(eigvals)).dot(la.inv(eigvecs))
print(eigvals)

else:
# Complex eigenvalues
np.random.seed(40)
A = np.random.randn(n, n)
print(la.eig(A)[0])

[-2.667651   -0.95797093 -0.33019549 -0.29151942 -0.18635343 -0.14418093]


Let's also pick an initial vector:

In [4]:
x0 = np.random.randn(n)
x0

Out[4]:
array([ 2.26930477,  0.66356156,  0.8991019 , -0.36580094,  0.46269004,
0.079874  ])

### Power iteration¶

In [5]:
#keep
x = x0


Now implement plain power iteration.

Run the below cell in-place (Ctrl-Enter) many times.

In [6]:
x = np.dot(A, x)
x

Out[6]:
array([-6.40496816,  7.1851682 ,  1.97149585,  4.77787616,  3.09099127,
4.33054803])
• What's the problem with this method?
• Does anything useful come of this?
• How do we fix it?

### Normalized power iteration¶

Back to the beginning: Reset to the initial vector.

In [7]:
#keep
x = x0/la.norm(x0)


Implement normalized power iteration.

Run this cell in-place (Ctrl-Enter) many times.

In [8]:
x = np.dot(A, x)
nrm = la.norm(x)
x = x/nrm

print(nrm)
print(x)

4.67639723405
[-0.52706768  0.59127069  0.16223527  0.39317355  0.25435905  0.35636272]

• What do you observe about the norm?
• What is the vector $x$ now?

Extensions:

• Now try the matrix variants above.
• Suggest a better way of estimating the eigenvalue. Hint

What if we want the smallest eigenvalue (by magnitude)?

Once again, reset to the beginning.

In [9]:
#keep
x = x0/la.norm(x0)


Run the cell below in-place many times.

In [10]:
x = la.solve(A, x)
nrm = la.norm(x)
x = x/nrm

print(1/nrm)
print(x)

0.0464410512401
[ 0.07260214 -0.82392938  0.17918176 -0.52684691  0.07819739  0.00898414]

• What's the computational cost per iteration?
• Can we make this method search for a specific eigenvalue?
• What is this method called?

Can we feed an estimate of the current approximate eigenvalue back into the calculation? (Hint: Rayleigh quotient)

Reset once more.

In [11]:
#keep
x = x0/la.norm(x0)


Run this cell in-place (Ctrl-Enter) many times.

In [12]:
sigma = np.dot(x, np.dot(A, x))/np.dot(x, x)
x = la.solve(A-sigma*np.eye(n), x)
x = x/la.norm(x)

print(sigma)
print(x)

-1.17969302281
[-0.06324034 -0.54447197  0.4628917  -0.37622458  0.41482856  0.41431213]

• What's this method called?
• What's a reasonable stopping criterion?
• Computational downside of this iteration?