How To Arrive At The Unit Matrix From Numpy.dot(a, A_inv)

I prepare a matrix of random numbers, calculate its inverse and matrix multiply it with the original matrix. This, in theory, gives the unit matrix. How can I let numpy do that for

Solution 1:

While getting True would be didactically appealing, it would also be divorced from the realities of floating-point computations.

When dealing with the floating point, one necessarily has to be prepared not only for inexact results, but for all manner of other numerical issues that arise.

I highly recommend reading What Every Computer Scientist Should Know About Floating-Point Arithmetic.

In your particular case, to ensure that A * inv(A) is close enough to the identity matrix, you could compute a matrix norm of numpy.dot(A, A_inv) - E and ensure that it is small enough.

As a side note, you don't have to use a loop to populate A and E. Instead, you could just use

A = numpy.random.randint(0, 10, (size,size))
E = numpy.eye(size)

Solution 2:

Agreeing with most of the points already made. However I would suggest that rather than looking at the individual off-diagonal elements, you take their rms sum; this reflects in some sense the "energy" that leaked into the off-diagonal terms as a result of imperfect calculations. If you then divide this RMS number by the sum of the diagonal terms, you get a metric of just how well the inverse worked. For example, the following code:

import numpy
import matplotlib.pyplot as plt
from numpy import mean, sqrt
N = 1000
R = numpy.zeros(N)

for size inrange(50,N,50):

  A = numpy.zeros((size, size))
  E = numpy.zeros((size, size))

  for i inrange(size):
      for j inrange(size):
          A[i][j]+=numpy.random.randint(10)
          if i == j:
              E[i][j]=1

  A_inv = numpy.linalg.linalg.inv(A)
  D = numpy.dot(A, A_inv) - E
  S = sqrt(mean(D**2))
  R[size] = S/size
  print"size: ", size, "; rms is ",  S/size

plt.plot(range(50,N,50), R[range(50, N, 50)])
plt.ylabel('RMS fraction')
plt.show()

Shows that the rms error is pretty stable with size of the array all the way up to a size of 950x950 (it does slow down a bit...). However, it's never "exact", and there are some outliers (presumably when the matrix is more nearly singular - this can happen with random matrices.)

Example plot (every time you run it, it will look a bit different):

Solution 3:

In the end, you can round your answer with

m = np.round(m, decimals=10)

or check to see if they're very different:

np.abs(A*A.I - i).mean() < 1e-10

if you want to kill off the tiny numbers.

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I would implement this with the numpy.matrix class.

import numpy

size = 100
A = numpy.matrix(numpy.random.randint(0,10,(size,)*2))
E = numpy.eye(size)

print A * A.I
print np.abs(A * A.I - E).mean() < 1e-10

Solution 4:

Your problem can be reduced to a common float-comparison problem. The correct way to compare such arrays would be:

EPS = 1e-8# for example
(np.abs(numpy.dot(A, A_inv) - E) < EPS).all()