Power Iteration and its Variants¶

Copyright (C) 2020 Andreas Kloeckner

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In [9]:
import numpy as np
import numpy.linalg as la
np.set_printoptions(precision=3, linewidth=120)

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

In [95]:
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])

[-2.668 -0.958 -0.33  -0.292 -0.186 -0.144]

Let's also pick an initial vector:

In [96]:
x0 = np.random.randn(n)
x0
Out[96]:
array([ 2.269,  0.664,  0.899, -0.366,  0.463,  0.08 ])

Power iteration¶

In [97]:
x = x0

Now implement plain power iteration.

In [98]:
for i in range(20):
    x = A @ x
    print(x)

[-7.705 22.151  2.865 -4.648  4.043 11.651]
[ 19.254 -66.76  -10.275  13.745 -12.354 -31.851]
[-50.872 180.931  28.982 -38.003  32.878  84.804]
[ 135.725 -484.052  -78.458  103.027  -87.053 -226.183]
[-362.275 1292.214  210.269 -276.615  231.404  603.437]
[  966.69  -3447.945  -561.808   739.696  -616.452 -1609.861]
[-2579.07   9198.61   1499.54  -1974.989  1643.638  4294.662]
[  6880.334 -24539.336  -4001.041   5270.259  -4383.84  -11456.774]
[-18354.595  65463.009  10674.137 -14060.82   11693.775  30562.786]
[  48963.909 -174633.058  -28475.596   37510.904  -31194.161  -81530.956]
[-130618.864  465860.62    75963.646 -100067.48    83214.419  217496.239]
[  348445.777 -1242754.094  -202645.16    266946.529  -221986.341  -580204.159]
[-929531.949 3315234.721  540587.199 -712121.531  592181.424 1547782.299]
[ 2479667.045 -8843889.704 -1442098.59   1899693.011 -1579732.736 -4128943.082]
[-6614886.466 23592411.656  3847016.323 -5067719.196  4214175.002 11014579.211]
[ 17646208.664 -62936320.902 -10262497.484  13518907.353 -11241947.561 -29383053.282]
[-4.707e+07  1.679e+08  2.738e+07 -3.606e+07  2.999e+07  7.838e+07]
[ 1.256e+08 -4.479e+08 -7.303e+07  9.621e+07 -8.000e+07 -2.091e+08]
[-3.350e+08  1.195e+09  1.948e+08 -2.566e+08  2.134e+08  5.578e+08]
[ 8.936e+08 -3.187e+09 -5.197e+08  6.846e+08 -5.693e+08 -1.488e+09]
  • 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 [99]:
x0 = np.random.randn(n)
x = x0

Implement normalized power iteration.

In [100]:
for i in range(10):
    x = A @ x
    nrm = la.norm(x)
    x = x/nrm
    print(la.solve(eigvecs, x))
    #print(x)

print(nrm)

[ 0.242  0.297 -0.081  0.058 -0.045 -0.038]
[-0.259 -0.114  0.011 -0.007  0.003  0.002]
[ 2.709e-01  4.282e-02 -1.380e-03  7.734e-04 -2.481e-04 -1.241e-04]
[-2.757e-01 -1.565e-02  1.738e-04 -8.599e-05  1.763e-05  6.822e-06]
[ 2.774e-01  5.653e-03 -2.165e-05  9.455e-06 -1.239e-06 -3.710e-07]
[-2.780e-01 -2.035e-03  2.686e-06 -1.036e-06  8.678e-08  2.010e-08]
[ 2.782e-01  7.312e-04 -3.327e-07  1.133e-07 -6.067e-09 -1.087e-09]
[-2.783e-01 -2.627e-04  4.119e-08 -1.238e-08  4.239e-10  5.877e-11]
[ 2.783e-01  9.433e-05 -5.099e-09  1.353e-09 -2.962e-11 -3.178e-12]
[-2.783e-01 -3.388e-05  6.312e-10 -1.479e-10  2.070e-12  1.724e-13]
2.6675529342865976

Checking convergence¶

In [101]:
x = x0
errors = []
coeffs = la.solve(eigvecs, x0)

for i in range(10):
    x = A @ x
    errors.append(
        la.norm(x/eigvals[0]**(i+1) - coeffs[0]*eigvecs[:,0], 2))
    print("coefficients:", la.solve(eigvecs, x/la.norm(x,2)))

conv_factor = eigvals[1]/eigvals[0]

errors = np.array(errors)
for i in range(len(errors)-1):
    print(f"{i=}: {errors[i]=:.6e}, {errors[i+1]/errors[i]=:.6g}, {conv_factor=:.6g}")

coefficients: [ 0.242  0.297 -0.081  0.058 -0.045 -0.038]
coefficients: [-0.259 -0.114  0.011 -0.007  0.003  0.002]
coefficients: [ 2.709e-01  4.282e-02 -1.380e-03  7.734e-04 -2.481e-04 -1.241e-04]
coefficients: [-2.757e-01 -1.565e-02  1.738e-04 -8.599e-05  1.763e-05  6.822e-06]
coefficients: [ 2.774e-01  5.653e-03 -2.165e-05  9.455e-06 -1.239e-06 -3.710e-07]
coefficients: [-2.780e-01 -2.035e-03  2.686e-06 -1.036e-06  8.678e-08  2.010e-08]
coefficients: [ 2.782e-01  7.312e-04 -3.327e-07  1.133e-07 -6.067e-09 -1.087e-09]
coefficients: [-2.783e-01 -2.627e-04  4.119e-08 -1.238e-08  4.239e-10  5.877e-11]
coefficients: [ 2.783e-01  9.433e-05 -5.099e-09  1.353e-09 -2.962e-11 -3.178e-12]
coefficients: [-2.783e-01 -3.388e-05  6.312e-10 -1.479e-10  2.071e-12  1.736e-13]
i=0: errors[i]=3.370824e+00, errors[i+1]/errors[i]=0.446913, conv_factor=0.359107
i=1: errors[i]=1.506465e+00, errors[i+1]/errors[i]=0.374136, conv_factor=0.359107
i=2: errors[i]=5.636223e-01, errors[i+1]/errors[i]=0.361846, conv_factor=0.359107
i=3: errors[i]=2.039444e-01, errors[i+1]/errors[i]=0.359503, conv_factor=0.359107
i=4: errors[i]=7.331854e-02, errors[i+1]/errors[i]=0.35911, conv_factor=0.359107
i=5: errors[i]=2.632940e-02, errors[i+1]/errors[i]=0.359074, conv_factor=0.359107
i=6: errors[i]=9.454210e-03, errors[i+1]/errors[i]=0.359087, conv_factor=0.359107
i=7: errors[i]=3.394882e-03, errors[i+1]/errors[i]=0.359097, conv_factor=0.359107
i=8: errors[i]=1.219094e-03, errors[i+1]/errors[i]=0.359103, conv_factor=0.359107
  • Now try the matrix variants above.

Inverse iteration¶

What if we want the eigenvalue closest to a give value $\sigma$?

Once again, reset to the beginning.

In [148]:
x = x0/la.norm(x0)
In [149]:
sigma = 1
A_sigma = A-sigma*np.eye(A.shape[0])
for i in range(30):
    x = la.solve(A_sigma, x)
    nrm = la.norm(x)
    x = x/nrm
    print(x)

[ 0.011 -0.809 -0.191  0.223 -0.186 -0.474]
[-0.127  0.775  0.149 -0.232  0.16   0.531]
[ 0.192 -0.728 -0.123  0.246 -0.135 -0.582]
[-0.246  0.674  0.103 -0.261  0.108  0.628]
[ 0.295 -0.616 -0.085  0.271 -0.08  -0.669]
[-0.338  0.556  0.07  -0.277  0.052  0.702]
[ 0.376 -0.497 -0.057  0.277 -0.026 -0.728]
[-0.41   0.441  0.047 -0.273  0.002  0.749]
[ 0.44  -0.387 -0.038  0.264  0.019 -0.765]
[-0.465  0.338  0.031 -0.252 -0.039  0.777]
[ 0.488 -0.291 -0.026  0.237  0.056 -0.786]
[-0.508  0.249  0.022 -0.219 -0.071  0.792]
[ 0.525 -0.209 -0.019  0.199  0.084 -0.796]
[-0.54   0.172  0.017 -0.178 -0.095  0.798]
[ 0.553 -0.138 -0.016  0.155  0.105 -0.799]
[-0.565  0.107  0.016 -0.132 -0.114  0.799]
[ 0.575 -0.078 -0.016  0.108  0.122 -0.798]
[-0.584  0.05   0.016 -0.084 -0.129  0.796]
[ 0.591 -0.025 -0.017  0.06   0.135 -0.792]
[-0.597  0.001  0.018 -0.035 -0.14   0.789]
[ 0.603  0.021 -0.02   0.011  0.144 -0.784]
[-0.607 -0.041  0.022  0.013 -0.148  0.779]
[ 0.611  0.061 -0.024 -0.036  0.151 -0.774]
[-0.614 -0.079  0.025  0.059 -0.154  0.768]
[ 0.616  0.096 -0.027 -0.082  0.156 -0.761]
[-0.618 -0.112  0.03   0.103 -0.158  0.755]
[ 0.619  0.126 -0.032 -0.124  0.16  -0.748]
[-0.619 -0.14   0.034  0.145 -0.161  0.741]
[ 0.619  0.154 -0.036 -0.164  0.162 -0.734]
[-0.619 -0.166  0.038  0.183 -0.163  0.726]

Rayleigh quotient iteration¶

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

Reset once more.

In [153]:
x = x0/la.norm(x0)

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

In [154]:
for i in range(10):
    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(x)

[ 0.063 -0.792 -0.173  0.23  -0.173 -0.505]
[-0.191  0.726  0.126 -0.248  0.133  0.585]
[ 0.521 -0.254 -0.051  0.121  0.058 -0.802]
[-0.544 -0.35   0.081  0.533 -0.15   0.519]
[ 0.52   0.378 -0.086 -0.58   0.146 -0.472]
[-0.518 -0.379  0.086  0.583 -0.145  0.469]
[ 0.518  0.379 -0.086 -0.583  0.145 -0.469]
[-0.518 -0.379  0.086  0.583 -0.145  0.469]
[-0.518 -0.379  0.086  0.583 -0.145  0.469]
[ 0.518  0.379 -0.086 -0.583  0.145 -0.469]
  • What's a reasonable stopping criterion?
  • Computational downside of this iteration?