Python Inverse of a Matrix
For those like me, who were looking for a pure Python solution without pandas
or numpy
involved, check out the following GitHub project: https://github.com/ThomIves/MatrixInverse.
It generously provides a very good explanation of how the process looks like "behind the scenes". The author has nicely described the step-by-step approach and presented some practical examples, all easy to follow.
This is just a little code snippet from there to illustrate the approach very briefly (AM
is the source matrix, IM
is the identity matrix of the same size):
def invert_matrix(AM, IM):
for fd in range(len(AM)):
fdScaler = 1.0 / AM[fd][fd]
for j in range(len(AM)):
AM[fd][j] *= fdScaler
IM[fd][j] *= fdScaler
for i in list(range(len(AM)))[0:fd] + list(range(len(AM)))[fd+1:]:
crScaler = AM[i][fd]
for j in range(len(AM)):
AM[i][j] = AM[i][j] - crScaler * AM[fd][j]
IM[i][j] = IM[i][j] - crScaler * IM[fd][j]
return IM
But please do follow the entire thing, you'll learn a lot more than just copy-pasting this code! There's a Jupyter notebook as well, btw.
Hope that helps someone, I personally found it extremely useful for my very particular task (Absorbing Markov Chain) where I wasn't able to use any non-standard packages.
You should have a look at numpy if you do matrix manipulation. This is a module mainly written in C, which will be much faster than programming in pure python. Here is an example of how to invert a matrix, and do other matrix manipulation.
from numpy import matrix
from numpy import linalg
A = matrix( [[1,2,3],[11,12,13],[21,22,23]]) # Creates a matrix.
x = matrix( [[1],[2],[3]] ) # Creates a matrix (like a column vector).
y = matrix( [[1,2,3]] ) # Creates a matrix (like a row vector).
print A.T # Transpose of A.
print A*x # Matrix multiplication of A and x.
print A.I # Inverse of A.
print linalg.solve(A, x) # Solve the linear equation system.
You can also have a look at the array module, which is a much more efficient implementation of lists when you have to deal with only one data type.
Make sure you really need to invert the matrix. This is often unnecessary and can be numerically unstable. When most people ask how to invert a matrix, they really want to know how to solve Ax = b where A is a matrix and x and b are vectors. It's more efficient and more accurate to use code that solves the equation Ax = b for x directly than to calculate A inverse then multiply the inverse by B. Even if you need to solve Ax = b for many b values, it's not a good idea to invert A. If you have to solve the system for multiple b values, save the Cholesky factorization of A, but don't invert it.
See Don't invert that matrix.
It is a pity that the chosen matrix, repeated here again, is either singular or badly conditioned:
A = matrix( [[1,2,3],[11,12,13],[21,22,23]])
By definition, the inverse of A when multiplied by the matrix A itself must give a unit matrix. The A chosen in the much praised explanation does not do that. In fact just looking at the inverse gives a clue that the inversion did not work correctly. Look at the magnitude of the individual terms - they are very, very big compared with the terms of the original A matrix...
It is remarkable that the humans when picking an example of a matrix so often manage to pick a singular matrix!
I did have a problem with the solution, so looked into it further. On the ubuntu-kubuntu platform, the debian package numpy does not have the matrix and the linalg sub-packages, so in addition to import of numpy, scipy needs to be imported also.
If the diagonal terms of A are multiplied by a large enough factor, say 2, the matrix will most likely cease to be singular or near singular. So
A = matrix( [[2,2,3],[11,24,13],[21,22,46]])
becomes neither singular nor nearly singular and the example gives meaningful results... When dealing with floating numbers one must be watchful for the effects of inavoidable round off errors.