Is there a simple way to swap rows in a matrix to form identity matrix on left side

user_13952307 0
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Given a matrix, A, that can be square, but, is most likely rectangular; and given that an identity matrix can be formed on the Rank(A) left most columns, is there a nice way to row swap to achieve this identity matrix?

For example, starting with

import numpy as np
row1 = np.array([1, 0, 0, 2, 3])
row2 = np.array([0, 0, 1, 4, 5])
row3 = np.array([0, 1, 0, 6, 7])
A = np.array([row1, row2, row3])

I would like to have

row1 = np.array([1, 0, 0, 2, 3])
row2 = np.array([0, 1, 0, 6, 7])
row3 = np.array([0, 0, 1, 4, 5])
A = np.array([row1, row2, row3])

I would like for this to be general, more than 1 row may need to be swapped, and there is no limit to the size of the matrix. Nor is there any guarantee the swap will be neighboring rows. Time does not matter i.e., efficiency is not important. Columns will not need to be swapped. The columns making up the Rank(A) identity matrix will on the left side.

A harder "test case" is

row1 = np.array([0, 0, 0, 0, 1, 5, 6, 7])
row2 = np.array([0, 0, 1, 0, 0, 1, 4, 2])
row3 = np.array([0, 0, 0, 1, 0, 6, 8, 4])
row4 = np.array([0, 1, 0, 0, 0, 1, 1, 1])
row5 = np.array([1, 0, 0, 0, 0, 6, 8, 4])
A = np.array([row1, row2, row3, row4, row5])
python-3.x numpy sorting python
3 Answers

Note that the shuffled-identity part of A is a unitary matrix.

And because of properties of permutations of the identity matrix, that means you can simply multiply A by A[:rows, :rows].T (the identity perm part):

rows = 3 # the number of rows

A = A[:rows, :rows].T.dot(A)

Result; A = :

array([[1, 0, 0, 2, 3],
       [0, 1, 0, 6, 7],
       [0, 0, 1, 4, 5]])

Who's rows are switched to form the identity.

user_101663930

I have a solution, but it might be (very) unoptimised. Keeping in mind that this solution is dependent on the number of rows corresponding to the length of the identity matrix:

def find_RankA(arr):
    identity_len = arr.shape[0]
    new_arr = np.empty_like(arr)
    for row in arr:
        replacement_index = row[:identity_len].tolist().index(1)
        new_arr[replacement_index] = row
    return new_arr

if __name__ == '__main__':
    row1 = np.array([0, 0, 0, 0, 1, 5, 6, 7])
    row2 = np.array([0, 0, 1, 0, 0, 1, 4, 2])
    row3 = np.array([0, 0, 0, 1, 0, 6, 8, 4])
    row4 = np.array([0, 1, 0, 0, 0, 1, 1, 1])
    row5 = np.array([1, 0, 0, 0, 0, 6, 8, 4])
    A = np.array([row1, row2, row3, row4, row5])
    print(find_RankA(A))

(Feedback welcome)

We can shorten the function to:

def find_RankA(arr):
    new_arr = np.empty_like(arr)
    for row in arr:
        new_arr[row[:arr.shape[0]].tolist().index(1)] = row
    return new_arr
user_87630970

Obligatory sorting based method:

A[np.argsort(np.argmax(A[:,:len(A)], 1))]
user_71388140
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