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I have a non-square matrix, and a method to determine the null space of the matrix (found from this thread: How to find the Null Space of a matrix in Python using numpy? ), but I have a few problems with taking this solution.

For one, I'm not sure if the values I have are correct, since I'm not too sure what I'm looking for.

Secondly, I need to find two linearly independent vectors from this null space, but I do not know the next step from here to determine this.

Finally, I need to determine whether any of the columns of the matrix are linearly independent in R3 and R4.

Any help would be greatly appreciated.

Code:

import numpy as np
import scipy as sp
from scipy import linalg

a = np.matrix(
[
    [ 3,  2, -1,  4],
    [ 1,  0,  2,  3],
    [-2, -2,  3, -1]
])


def null(A, eps=1e-15):
    u, s, vh = linalg.svd(A)
    null_mask = (s <= eps)
    null_space = sp.compress(null_mask, vh, axis=0)
    return sp.transpose(null_space)

print(null(a))

Output:

 [[ 0.8290113 ]
 [-0.2330726 ]
 [ 0.24969281]
 [-0.44279897]]

I'm assuming since the output is anything other than an empty matrix [] that there's something special about this matrix, I just don't know what it means.

Dylan Goudie
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1 Answers1

2

I would recommend using sympy in this case:

from sympy import Matrix
a  = Matrix([
    [ 3,  2, -1,  4],
    [ 1,  0,  2,  3],
    [-2, -2,  3, -1]
])
print(a.nullspace())

Output:

[Matrix([
[ -2],
[7/2],
[  1],
[  0]]),
Matrix([
[ -3],
[5/2],
[  0],
[  1]])]

You can easily check that the result indeed belongs to the nullspace by explicitly checking that it is mapped to 0 when multiplying with your matrix a:

n1, n2 = a.nullspace()
print(a*n1, a*n2)

results in:

Matrix([[0], [0], [0]]) Matrix([[0], [0], [0]])

Finally, to get the linearly independent columns of your matrix in R3 you can use the function columnspace, which returns a list of column vectors that span the columnspace of the matrix

print(a.columnspace())

results in

[Matrix([
[ 3],
[ 1],
[-2]]), Matrix([
[ 2],
[ 0],
[-2]])]

which are the first two columns of the matrix.

jdamp
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  • Thanks a lot! What would the difference be for R4? – Dylan Goudie Oct 07 '18 at 23:51
  • Oh, that would be a.rowspace(). – Dylan Goudie Oct 08 '18 at 00:12
  • I just read somewhere that matrices with more columns than rows must have linearly dependent columns, is this true? – Dylan Goudie Oct 08 '18 at 00:29
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    Yes. The dimension of the vector space spanned by the columns of a matrix (i.e. the number of linearly independent column vectors) is identical to the dimension of the space spanned by the row vectors, read for example [here](https://en.wikipedia.org/wiki/Rank_(linear_algebra)) for some further information. – jdamp Oct 08 '18 at 06:42