## Determining linear independence of a set of vectors

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Matlab post Occasionally we have a set of vectors and we need to determine whether the vectors are linearly independent of each other. This may be necessary to determine if the vectors form a basis, or to determine how many independent equations there are, or to determine how many independent reactions there are.

Reference: Kreysig, Advanced Engineering Mathematics, sec. 7.4

Matlab provides a rank command which gives you the number of singular values greater than some tolerance. The numpy.rank function, unfortunately, does not do that. It returns the number of dimensions in the array. We will just compute the rank from singular value decomposition.

The default tolerance used in Matlab is max(size(A))*eps(norm(A)). Let us break that down. eps(norm(A)) is the positive distance from abs(X) to the next larger in magnitude floating point number of the same precision as X. Basically, the smallest significant number. We multiply that by the size of A, and take the largest number. We have to use some judgment in what the tolerance is, and what “zero” means.

import numpy as np
v1 = [6, 0, 3, 1, 4, 2];
v2 = [0, -1, 2, 7, 0, 5];
v3 = [12, 3, 0, -19, 8, -11];

A = np.row_stack([v1, v2, v3])

# matlab definition
eps = np.finfo(np.linalg.norm(A).dtype).eps
TOLERANCE = max(eps * np.array(A.shape))

U, s, V = np.linalg.svd(A)
print s
print np.sum(s > TOLERANCE)

TOLERANCE = 1e-14
print np.sum(s > TOLERANCE)

>>> >>> >>> >>> >>> >>> ... >>> >>> >>> >>> [  2.75209239e+01   9.30584482e+00   1.42425400e-15]
3
>>> >>> 2


You can see if you choose too small a TOLERANCE, nothing looks like zero. the result with TOLERANCE=1e-14 suggests the rows are not linearly independent. Let us show that one row can be expressed as a linear combination of the other rows.

The number of rows is greater than the rank, so these vectors are not independent. Let's demonstrate that one vector can be defined as a linear combination of the other two vectors. Mathematically we represent this as:

$$x_1 \mathit{v1} + x_2 \mathit{v2} = v3$$

or

$$[x_1 x_2][v1; v2] = v3$$

This is not the usual linear algebra form of Ax = b. To get there, we transpose each side of the equation to get:

[v1.T v2.T][x_1; x_2] = v3.T

which is the form Ax = b. We solve it in a least-squares sense.

A = np.column_stack([v1, v2])
x = np.linalg.lstsq(A, v3)
print x[0]

>>> [ 2. -3.]


This shows that v3 = 2*v1 - 3*v2

## 1 another example

#Problem set 7.4 #17
import numpy as np

v1 = [0.2, 1.2, 5.3, 2.8, 1.6]
v2 = [4.3, 3.4, 0.9, 2.0, -4.3]

A = np.row_stack([v1, v2])
U, s, V = np.linalg.svd(A)
print s

[ 7.57773162  5.99149259]


You can tell by inspection the rank is 2 because there are no near-zero singular values.

## 2 Near deficient rank

the rank command roughly works in the following way: the matrix is converted to a reduced row echelon form, and then the number of rows that are not all equal to zero are counted. Matlab uses a tolerance to determine what is equal to zero. If there is uncertainty in the numbers, you may have to define what zero is, e.g. if the absolute value of a number is less than 1e-5, you may consider that close enough to be zero. The default tolerance is usually very small, of order 1e-15. If we believe that any number less than 1e-5 is practically equivalent to zero, we can use that information to compute the rank like this.

import numpy as np

A = [[1, 2, 3],
[0, 2, 3],
[0, 0, 1e-6]]

U, s, V = np.linalg.svd(A)
print s
print np.sum(np.abs(s) > 1e-15)
print np.sum(np.abs(s) > 1e-5)

[  5.14874857e+00   7.00277208e-01   5.54700196e-07]
3
2


## 3 Application to independent chemical reactions.

reference: Exercise 2.4 in Chemical Reactor Analysis and Design Fundamentals by Rawlings and Ekerdt.

The following reactions are proposed in the hydrogenation of bromine:

Let this be our species vector: v = [H2 H Br2 Br HBr].T

the reactions are then defined by M*v where M is a stoichometric matrix in which each row represents a reaction with negative stoichiometric coefficients for reactants, and positive stoichiometric coefficients for products. A stoichiometric coefficient of 0 is used for species not participating in the reaction.

import numpy as np

#    [H2  H Br2 Br HBr]
M = [[-1,  0, -1,  0,  2],  # H2 + Br2 == 2HBR
[ 0,  0, -1,  2,  0],  # Br2 == 2Br
[-1,  1,  0, -1,  1],  # Br + H2 == HBr + H
[ 0, -1, -1,  1,  1],  # H + Br2 == HBr + Br
[ 1, -1,  0,  1,  -1], # H + HBr == H2 + Br
[ 0,  0,  1, -2,  0]]  # 2Br == Br2

U, s, V = np.linalg.svd(M)
print s
print np.sum(np.abs(s) > 1e-15)

import sympy
M = sympy.Matrix(M)
reduced_form, inds = M.rref()

print reduced_form

labels = ['H2',  'H', 'Br2', 'Br', 'HBr']
for row in reduced_form.tolist():
s = '0 = '
for nu,species in zip(row,labels):
if nu != 0:

s += ' {0:+d}{1}'.format(int(nu), species)
if s != '0 = ': print s

[  3.84742803e+00   3.32555975e+00   1.46217301e+00   1.73313660e-16
8.57422679e-17]
3
[1, 0, 0,  2, -2]
[0, 1, 0,  1, -1]
[0, 0, 1, -2,  0]
[0, 0, 0,  0,  0]
[0, 0, 0,  0,  0]
[0, 0, 0,  0,  0]
0 =  +1H2 +2Br -2HBr
0 =  +1H +1Br -1HBr
0 =  +1Br2 -2Br


6 reactions are given, but the rank of the matrix is only 3. so there are only three independent reactions. You can see that reaction 6 is just the opposite of reaction 2, so it is clearly not independent. Also, reactions 3 and 5 are just the reverse of each other, so one of them can also be eliminated. finally, reaction 4 is equal to reaction 1 minus reaction 3.

There are many possible independent reactions. In the code above, we use sympy to put the matrix into reduced row echelon form, which enables us to identify three independent reactions, and shows that three rows are all zero, i.e. they are not independent of the other three reactions. The choice of independent reactions is not unique.