Power Iteration and its Variants

Copyright (C) 2020 Andreas Kloeckner

MIT License Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
In [104]:
import numpy as np

import numpy.linalg as la

np.set_printoptions(precision=3)

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

In [105]:
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 [106]:
#clear

x0 = np.random.randn(n)

x0
Out[106]:
array([ 2.269,  0.664,  0.899, -0.366,  0.463,  0.08 ])

Power iteration

In [107]:
x = x0

Now implement plain power iteration.

In [127]:
#clear

for i in range(20):

    x = A @ x

    print(x)

[ 1.329e+12 -4.740e+12 -7.728e+11  1.018e+12 -8.466e+11 -2.213e+12]
[-3.545e+12  1.264e+13  2.062e+12 -2.716e+12  2.258e+12  5.903e+12]
[ 9.457e+12 -3.373e+13 -5.500e+12  7.245e+12 -6.025e+12 -1.575e+13]
[-2.523e+13  8.998e+13  1.467e+13 -1.933e+13  1.607e+13  4.201e+13]
[ 6.730e+13 -2.400e+14 -3.914e+13  5.156e+13 -4.287e+13 -1.121e+14]
[-1.795e+14  6.403e+14  1.044e+14 -1.375e+14  1.144e+14  2.989e+14]
[ 4.789e+14 -1.708e+15 -2.785e+14  3.669e+14 -3.051e+14 -7.975e+14]
[-1.278e+15  4.557e+15  7.430e+14 -9.788e+14  8.139e+14  2.127e+15]
[ 3.408e+15 -1.216e+16 -1.982e+15  2.611e+15 -2.171e+15 -5.675e+15]
[-9.092e+15  3.243e+16  5.288e+15 -6.965e+15  5.792e+15  1.514e+16]
[ 2.425e+16 -8.650e+16 -1.411e+16  1.858e+16 -1.545e+16 -4.039e+16]
[-6.470e+16  2.308e+17  3.763e+16 -4.957e+16  4.122e+16  1.077e+17]
[ 1.726e+17 -6.156e+17 -1.004e+17  1.322e+17 -1.100e+17 -2.874e+17]
[-4.604e+17  1.642e+18  2.678e+17 -3.527e+17  2.933e+17  7.667e+17]
[ 1.228e+18 -4.381e+18 -7.143e+17  9.410e+17 -7.825e+17 -2.045e+18]
[-3.277e+18  1.169e+19  1.906e+18 -2.510e+18  2.087e+18  5.456e+18]
[ 8.741e+18 -3.117e+19 -5.083e+18  6.696e+18 -5.569e+18 -1.455e+19]
[-2.332e+19  8.316e+19  1.356e+19 -1.786e+19  1.486e+19  3.883e+19]
[ 6.220e+19 -2.219e+20 -3.618e+19  4.765e+19 -3.963e+19 -1.036e+20]
[-1.659e+20  5.918e+20  9.650e+19 -1.271e+20  1.057e+20  2.763e+20]

  • 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 [128]:
x = x0/la.norm(x0)

Implement normalized power iteration.

In [129]:
#clear

for i in range(10):

    x = A @ x

    nrm = la.norm(x)

    x = x/nrm

    print(x)



print(nrm)

[-0.285  0.819  0.106 -0.172  0.149  0.431]
[ 0.243 -0.842 -0.13   0.173 -0.156 -0.402]
[-0.238  0.845  0.135 -0.177  0.153  0.396]
[ 0.237 -0.845 -0.137  0.18  -0.152 -0.395]
[-0.237  0.845  0.137 -0.181  0.151  0.395]
[ 0.237 -0.845 -0.138  0.181 -0.151 -0.394]
[-0.237  0.845  0.138 -0.181  0.151  0.394]
[ 0.237 -0.845 -0.138  0.181 -0.151 -0.394]
[-0.237  0.845  0.138 -0.181  0.151  0.394]
[ 0.237 -0.845 -0.138  0.181 -0.151 -0.394]
2.667662268743778

  • What do you observe about the norm?

  • What about the sign?

  • What is the vector $x$ now?

Extensions:

  • Now try the matrix variants above.

  • Suggest a better way of estimating the eigenvalue. Hint

Inverse iteration

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

Once again, reset to the beginning.

In [143]:
x = x0/la.norm(x0)
In [144]:
#clear

for i in range(30):

    x = la.solve(A, x)

    nrm = la.norm(x)

    x = x/nrm

    print(x)

[ 0.344 -0.55  -0.094  0.23  -0.05  -0.718]
[-0.527  0.236  0.046 -0.123 -0.066  0.803]
[ 0.599 -0.01  -0.037 -0.031  0.127 -0.789]
[-0.617 -0.128  0.043  0.173 -0.153  0.74 ]
[ 0.611  0.213 -0.052 -0.284  0.161 -0.687]
[-0.597 -0.265  0.061  0.365 -0.161  0.64 ]
[ 0.583  0.299 -0.067 -0.424  0.16  -0.601]
[-0.57  -0.322  0.072  0.465 -0.157  0.571]
[ 0.558  0.337 -0.076 -0.496  0.155 -0.547]
[-0.55  -0.348  0.078  0.517 -0.153  0.529]
[ 0.543  0.356 -0.08  -0.533  0.151 -0.515]
[-0.537 -0.362  0.082  0.545 -0.15   0.504]
[ 0.533  0.366 -0.083 -0.554  0.149 -0.496]
[-0.529 -0.369  0.084  0.561 -0.148  0.49 ]
[ 0.527  0.371 -0.084 -0.566  0.147 -0.485]
[-0.525 -0.373  0.085  0.57  -0.147  0.481]
[ 0.523  0.375 -0.085 -0.573  0.146 -0.478]
[-0.522 -0.376  0.086  0.575 -0.146  0.476]
[ 0.521  0.376 -0.086 -0.577  0.146 -0.474]
[-0.521 -0.377  0.086  0.578 -0.146  0.473]
[ 0.52   0.377 -0.086 -0.579  0.146 -0.472]
[-0.52  -0.378  0.086  0.58  -0.145  0.471]
[ 0.519  0.378 -0.086 -0.581  0.145 -0.471]
[-0.519 -0.378  0.086  0.581 -0.145  0.47 ]
[ 0.519  0.379 -0.086 -0.582  0.145 -0.47 ]
[-0.519 -0.379  0.086  0.582 -0.145  0.47 ]
[ 0.519  0.379 -0.086 -0.582  0.145 -0.469]
[-0.518 -0.379  0.086  0.582 -0.145  0.469]
[ 0.518  0.379 -0.086 -0.582  0.145 -0.469]
[-0.518 -0.379  0.086  0.583 -0.145  0.469]

  • What's the computational cost per iteration?

  • Can we make this method search for a specific eigenvalue?

Inverse Shift 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]:
#clear

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]:
#clear

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?