compute the infinity norm of the difference between the two solutions

Question

In the following code I have been able to:

1. Implement Gaussian elimination with no pivoting for a general square linear system.
2. I have tested it by solving Ax=b, where A is a random 100x100 matrix and b is a random 100x1 vector.
3. I have compared my solution against the solution obtained using numpy.linalg.solve

However in the final task I need to compute the infinity norm of the difference between the two solutions. I know the infinity norm is the greatest absolute row sum of a matrix. But how can I do this to compute the infinity norm of the difference between the two solutions, my solution and the numpy.linalg.solve. Looking for some help with this!

import numpy as np
def GENP(A, b):
    '''
    Gaussian elimination with no pivoting.
    % input: A is an n x n nonsingular matrix
    %        b is an n x 1 vector
    % output: x is the solution of Ax=b.
    % post-condition: A and b have been modified. 
    '''
    n =  len(A)
    if b.size != n:
        raise ValueError("Invalid argument: incompatible sizes between A & b.", b.size, n)
    for pivot_row in range(n-1):
        for row in range(pivot_row+1, n):
            multiplier = A[row][pivot_row]/A[pivot_row][pivot_row]
            #the only one in this column since the rest are zero
            A[row][pivot_row] = multiplier
            for col in range(pivot_row + 1, n):
                A[row][col] = A[row][col] - multiplier*A[pivot_row][col]
            #Equation solution column
            b[row] = b[row] - multiplier*b[pivot_row]
    x = np.zeros(n)
    k = n-1
    x[k] = b[k]/A[k,k]
    while k >= 0:
        x[k] = (b[k] - np.dot(A[k,k+1:],x[k+1:]))/A[k,k]
        k = k-1
    return x
if __name__ == "__main__":
    A = np.round(np.random.rand(100, 100)*10)
    b =  np.round(np.random.rand(100)*10)
    print (GENP(np.copy(A), np.copy(b)))

for example this code gives the following output for task 1 listed above:

[-6.61537666  0.95704368  1.30101768 -3.69577873 -2.51427519 -4.56927017
 -1.61201589  2.88242622  1.67836096  2.18145556  2.60831672  0.08055869
 -2.39347903  2.19672137 -0.91609732 -1.17994959 -3.87309152 -2.53330865
  5.97476318  3.74687301  5.38585146 -2.71597978  2.0034079  -0.35045844
  0.43988439 -2.2623829  -1.82137544  3.20545721 -4.98871738 -6.94378666
 -6.5076601   3.28448129  3.42318453 -1.63900434  4.70352047 -4.12289961
 -0.79514656  3.09744616  2.96397264  2.60408589  2.38707091  8.72909353
 -1.33584905  1.30879264 -0.28008339  0.93560728 -1.40591226  1.31004142
 -1.43422946  0.41875924  3.28412668  3.82169545  1.96675247  2.76094378
 -0.90069455  1.3641636  -0.60520103  3.4814196  -1.43076816  5.01222382
  0.19160657  2.23163261  2.42183726 -0.52941262 -7.35597457 -3.41685057
 -0.24359225 -5.33856181 -1.41741354 -0.35654736 -1.71158503 -2.24469314
 -3.26453092  1.0932765   1.58333208  0.15567584  0.02793548  1.59561909
  0.31732915 -1.00695954  3.41663177 -4.06869021  3.74388762 -0.82868155
  1.49789582 -1.63559124  0.2741194  -1.11709237  1.97177449  0.66410154
  0.48397714 -1.96241854  0.34975886  1.3317751   2.25763568 -6.80055066
 -0.65903682 -1.07105965 -0.40211347 -0.30507635]

then for task two my code gives the following:

my_solution = GENP(np.copy(A), np.copy(b))
numpy_solution = np.linalg.solve(A, b)
print(numpy_solution)

resulting in:

[-6.61537666  0.95704368  1.30101768 -3.69577873 -2.51427519 -4.56927017
-1.61201589  2.88242622  1.67836096  2.18145556  2.60831672  0.08055869
-2.39347903  2.19672137 -0.91609732 -1.17994959 -3.87309152 -2.53330865
 5.97476318  3.74687301  5.38585146 -2.71597978  2.0034079  -0.35045844
 0.43988439 -2.2623829  -1.82137544  3.20545721 -4.98871738 -6.94378666
-6.5076601   3.28448129  3.42318453 -1.63900434  4.70352047 -4.12289961
-0.79514656  3.09744616  2.96397264  2.60408589  2.38707091  8.72909353
-1.33584905  1.30879264 -0.28008339  0.93560728 -1.40591226  1.31004142
-1.43422946  0.41875924  3.28412668  3.82169545  1.96675247  2.76094378
 -0.90069455  1.3641636  -0.60520103  3.4814196  -1.43076816  5.01222382
 0.19160657  2.23163261  2.42183726 -0.52941262 -7.35597457 -3.41685057
-0.24359225 -5.33856181 -1.41741354 -0.35654736 -1.71158503 -2.24469314
-3.26453092  1.0932765   1.58333208  0.15567584  0.02793548  1.59561909
 0.31732915 -1.00695954  3.41663177 -4.06869021  3.74388762 -0.82868155
 1.49789582 -1.63559124  0.2741194  -1.11709237  1.97177449  0.66410154
 0.48397714 -1.96241854  0.34975886  1.3317751   2.25763568 -6.80055066
-0.65903682 -1.07105965 -0.40211347 -0.30507635]

finally for task 3:

if np.allclose(my_solution, numpy_solution):
    print("These solutions agree")
else:
   print("These solutions do not agree")

resulting in:

These solutions agree

Answer 1

In general if you have some norm $||\cdot||$ then you also have a metric $d(x,y) = ||x-y||$ so if you're comparing to row reduced matrices $A$ and $A'$ it will be $||A-A'||$.