Arnoldi Iteration¶
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 [1]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt
Let us make a matrix with a defined set of eigenvalues and eigenvectors, given by eigvals and eigvecs.
In [2]:
np.random.seed(40)
# Generate matrix with eigenvalues 1...25
n = 25
eigvals = np.linspace(1., n, n)
eigvecs = np.random.randn(n, n)
print(eigvals)
A = la.solve(eigvecs, np.dot(np.diag(eigvals), eigvecs))
print(la.eig(A)[0])
Initialization¶
Set up $Q$ and $H$:
In [3]:
Q = np.zeros((n, n))
H = np.zeros((n, n))
k = 0
Pick a starting vector, normalize it
In [4]:
x0 = np.random.randn(n)
x0 = x0/la.norm(x0)
# Poke it into the first column of Q
Q[:, k] = x0
del x0
Make a list to save arrays of Ritz values:
In [5]:
ritz_values = []
Algorithm¶
Carry out one iteration of Arnoldi iteration.
Run this cell in-place (Ctrl-Enter) until H is filled.
In [30]:
print(k)
u = A @ Q[:, k]
# Carry out Gram-Schmidt on u against Q
for j in range(k+1):
qj = Q[:, j]
H[j,k] = qj @ u
u = u - H[j,k]*qj
if k+1 < n:
H[k+1, k] = la.norm(u)
Q[:, k+1] = u/H[k+1, k]
k += 1
pt.spy(H)
ritz_values.append(la.eig(H)[0])
Check that $Q^T A Q =H$:
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la.norm(Q.T @ A @ Q - H)/ la.norm(A)
Out[31]:
Check that Q is orthogonal:
In [32]:
la.norm(Q.T @ Q - np.eye(n))
Out[32]:
Plot convergence of Ritz values¶
Enable the Ritz value collection above to make this work.
In [33]:
for i, rv in enumerate(ritz_values):
pt.plot([i] * len(rv), rv, "x")
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