Is there a standard solution for Gauss elimination in Python?

Question

Is there somewhere in the cosmos of scipy/numpy/... a standard method for Gauss-elimination of a matrix?

One finds many snippets via google, but I would prefer to use "trusted" modules if possible.

Answer 1

I finally found, that it can be done using LU decomposition. Here the U matrix represents the reduced form of the linear system.

from numpy import array from scipy.linalg import lu  a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])  pl, u = lu(a, permute_l=True) 

Then u reads

array([[ 2.,  4.,  4.,  4.],        [ 0.,  2.,  1.,  2.],        [ 0.,  0.,  1.,  1.],        [ 0.,  0.,  0.,  0.]]) 

Depending on the solvability of the system this matrix has an upper triangular or trapezoidal structure. In the above case a line of zeros arises, as the matrix has only rank 3.

Answer 2

One function that can be worth checking is _remove_redundancy, if you wish to remove repeated or redundant equations:

import numpy as np import scipy.optimize  a = np.array([[1.,1.,1.,1.],               [0.,0.,0.,1.],               [0.,0.,0.,2.],               [0.,0.,0.,3.]]) print(scipy.optimize._remove_redundancy._remove_redundancy(a, np.zeros_like(a[:, 0]))[0]) 

which gives:

[[1. 1. 1. 1.]  [0. 0. 0. 3.]] 

As a note to @flonk answer, using a LU decomposition might not always give the desired reduced row matrix. Example:

import numpy as np import scipy.linalg  a = np.array([[1.,1.,1.,1.],               [0.,0.,0.,1.],               [0.,0.,0.,2.],               [0.,0.,0.,3.]])  _,_, u = scipy.linalg.lu(a) print(u) 

gives the same matrix:

[[1. 1. 1. 1.]  [0. 0. 0. 1.]  [0. 0. 0. 2.]  [0. 0. 0. 3.]] 

even though the last 3 rows are linearly dependent.