Backward Stability by Example¶

Copyright (C) 2024 Andreas Kloeckner

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

A numerical method for solving linear systems¶

Here's code for (mostly) straightforward Gauss-Jordan elimination to solve a linear system:

In [3]:
def mylinsolve(A, b):
    n = len(A)
    Ab = np.concatenate([A, b.reshape(-1, 1)], axis=1)
    
    # reverse the rows
    Ab = Ab[::-1]
    
    for col in range(n):
        Ab[col] /= Ab[col, col]
    
        for row in range(n):
            if col == row:
                continue
            Ab[row] -= Ab[col]*Ab[row, col]

    return Ab[:, -1]

Test that it solves linear systems:

In [4]:
Atest = np.random.randn(5, 5)
btest = np.random.randn(5)

xtest = la.solve(Atest, btest)

la.norm(mylinsolve(Atest, btest) - xtest, 2) / la.norm(xtest, 2)
Out[4]:
np.float64(2.3637620827446447e-15)

Setting up a "bad" problem¶

Make an example of a very poorly conditioned matrix, called the Vandermonde matrix.

In [5]:
n = 20
nodes = np.linspace(0, 1, n)
A = nodes.reshape(-1, 1) ** np.arange(n)

Check the condition number:

In [6]:
la.cond(A, 2)
Out[6]:
np.float64(1.156327775325343e+16)

Set up a "true" solution, and a right-hand side for a linear system:

In [7]:
xtrue = np.random.randn(n)
b = A @ xtrue

Scenario 1: "Backward Stability"¶

Solve it computationally, call the solution xcomp:

In [39]:
xcomp = mylinsolve(A, b)

Compute the relative forward error (i.e. the error in the solution):

In [40]:
la.norm(xcomp - xtrue, 2)/la.norm(xtrue, 2)
Out[40]:
np.float64(11.09453699621568)

Compute the

  • relative backward error (assign to bw_err)
  • i.e. the relative error in the right-hand side
  • also known as the relative residual.
In [41]:
bw_err = la.norm(A@xcomp - b, 2)/la.norm(b, 2)
bw_err
Out[41]:
np.float64(1.6518461918240131e-10)

Compare the forward error with the guarantee from the conditioning bound.

In [42]:
bw_err * la.cond(A, 2)
Out[42]:
np.float64(1910075.632171501)

  • Repeat this scenario with la.solve.

  • Why is this not exactly the backward error scenario from class (hence the quotes above)?

    Solution Because we are unable to apply the "true" inverse (i.e. matrix-vector product), i.e. a solve undisturbed by numerical error.

Scenario 2: Input Perturbation¶

Now intentionally perturb the right-hand side and repeat the experiment

In [28]:
delta_b = np.random.randn(n) * 1e-14 
b_perturbed = b + delta_b
xcomp_perturbed = mylinsolve(A, b_perturbed)

Find the relative forward perturbation vs xtrue:

In [29]:
la.norm(xcomp_perturbed - xtrue, 2)/la.norm(xtrue, 2)
Out[29]:
np.float64(32.86580758319449)

Compare with the perturbation bound obtained from conditioning:

In [30]:
la.norm(delta_b, 2)/la.norm(b, 2) * la.cond(A, 2)
Out[30]:
np.float64(100.89391277573719)
In [ ]: