How to find linearly independent rows from a matrix

I edited the code for Cauchy-Schwartz inequality which scales better with dimension: the inputs are the matrix and its dimension, while the output is a new rectangular matrix which contains along its rows the linearly independent columns of the starting matrix. This works in the assumption that the first column in never null, but can be readily generalized in order to implement this case too. Another thing that I observed is that 1e-5 seems to be a “sloppy” threshold, since some particular pathologic vectors were found to be linearly dependent in that case: 1e-4 doesn’t give me the same problems. I hope this could be of some help: it was pretty difficult for me to find a really working routine to extract li vectors, and so I’m willing to share mine. If you find some bug, please report them!!

from numpy import dot, zeros
from numpy.linalg import matrix_rank, norm

def find_li_vectors(dim, R):

    r = matrix_rank(R) 
    index = zeros( r ) #this will save the positions of the li columns in the matrix
    counter = 0
    index[0] = 0 #without loss of generality we pick the first column as linearly independent
    j = 0 #therefore the second index is simply 0

    for i in range(R.shape[0]): #loop over the columns
        if i != j: #if the two columns are not the same
            inner_product = dot( R[:,i], R[:,j] ) #compute the scalar product
            norm_i = norm(R[:,i]) #compute norms
            norm_j = norm(R[:,j])

            #inner product and the product of the norms are equal only if the two vectors are parallel
            #therefore we are looking for the ones which exhibit a difference which is bigger than a threshold
            if absolute(inner_product - norm_j * norm_i) > 1e-4:
                counter += 1 #counter is incremented
                index[counter] = i #index is saved
                j = i #j is refreshed
            #do not forget to refresh j: otherwise you would compute only the vectors li with the first column!!

    R_independent = zeros((r, dim))

    i = 0
    #now save everything in a new matrix
    while( i < r ):
        R_independent[i,:] = R[index[i],:] 
        i += 1

    return R_independent

I know this was asked a while ago, but here is a very simple (although probably inefficient) solution. Given an array, the following finds a set of linearly independent vectors by progressively adding a vector and testing if the rank has increased:

from numpy.linalg import matrix_rank

def LI_vecs(dim,M):
    LI=[M[0]]
    for i in range(dim):
        tmp=[]
        for r in LI:
            tmp.append(r)
        tmp.append(M[i])                #set tmp=LI+[M[i]]
        if matrix_rank(tmp)>len(LI):    #test if M[i] is linearly independent from all (row) vectors in LI
            LI.append(M[i])             #note that matrix_rank does not need to take in a square matrix
    return LI                           #return set of linearly independent (row) vectors

#Example
mat=[[1,2,3,4],[4,5,6,7],[5,7,9,11],[2,4,6,8]]
LI_vecs(4,mat)

With sympy you can find the linear independant rows using: sympy.Matrix.rref:

>>> import sympy 
>>> import numpy as np
>>> mat = np.array([[0,1,0,0],[0,0,1,0],[0,1,1,0],[1,0,0,1]])  # your matrix
>>> _, inds = sympy.Matrix(mat).T.rref()   # to check the rows you need to transpose!
>>> inds
[0, 1, 3]

Which basically tells you the rows 0, 1 and 3 are linear independant while row 2 isn’t (it’s a linear combination of row 0 and 1).

Then you could remove these rows with slicing:

>>> mat[inds]
array([[0, 1, 0, 0],
       [0, 0, 1, 0],
       [1, 0, 0, 1]])

This also works well for rectangular (not only for quadratic) matrices.

I interpret the problem as finding rows that are linearly independent from other rows.
That is equivalent to finding rows that are linearly dependent on other rows.

Gaussian elimination and treat numbers smaller than a threshold as zeros can do that. It is faster than finding eigenvalues of a matrix, testing all combinations of rows with Cauchy-Schwarz inequality, or singular value decomposition.

See:
https://math.stackexchange.com/questions/1297437/using-gauss-elimination-to-check-for-linear-dependence

Problem with floating point numbers:

http://numpy-discussion.10968.n7.nabble.com/Reduced-row-echelon-form-td16486.html

With regards to the following discussion:

Find dependent rows/columns of a matrix using Matlab?

from sympy import *
A = Matrix([[1,1,1],[2,2,2],[1,7,5]])
print(A.nullspace())

It is obvious that the first and second row are multiplication of each other.
If we execute the above code we get [-1/3, -2/3, 1]. The indices of the zero elements in the null space show independence. But why is the third element here not zero? If we multiply the A matrix with the null space, we get a zero column vector. So what’s wrong?

The answer which we are looking for is the null space of the transpose of A.

B = A.T
print(B.nullspace())

Now we get the [-2, 1, 0], which shows that the third row is independent.

Two important notes here:

1. Consider whether we want to check the row dependencies or the column
   dependencies.

2. Notice that the null space of a matrix is not equal to the null
   space of the transpose of that matrix unless it is symmetric.

You can basically find the vectors spanning the columnspace of the matrix by using SymPy library‘s columnspace() method of Matrix object. Automatically, they are the linearly independent columns of the matrix.

import sympy as sp 
import numpy as np 

M  = sp.Matrix([[0, 1, 0, 0],
                [0, 0, 1, 0],
                [1, 0, 0, 1]])


for i in M.columnspace():
    print(np.array(i))
    print()

# The output is following.

# [[0]
#  [0]
#  [1]]

# [[1]
#  [0]
#  [0]]

# [[0]
#  [1]
#  [0]]

A (possibly) shorter solution:

def cartesian_product(vec1, vec2):
    return np.transpose(np.array([np.repeat(vec1, len(vec2)), np.tile(vec2, len(vec1))]))

def check_dependencies(matrix_):
    def rows_are_dependent(indices):
        if indices[0] == indices[1]:
            return False
        return np.unique(matrix_[indices[0],] / matrix_[indices[1],]).shape[0] == 1

    return np.any(np.apply_along_axis(rows_are_dependent, 1, cartesian_product(np.arange(matrix_.shape[0]), np.arange(matrix_.shape[0]))))