How to find linearly independent rows from a matrix

Question

How to identify the linearly independent rows from a matrix? For instance,

Answer 1

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