Null Spaces

A null space is a particular category of subspaces. The null space of a system of equations is the set of solutions for which the dependent variables cancel out.

Definition

For a system of equations, the null space is the subspace that satisfies Ax = 0. This is useful for finding a complete solution.

Algebraically, null spaces have an identity property. Given any valid solution to Ax = b, any combination of null spaces can be added to that solution to create another valid solution, because b + 0 = b.

Solution

Identify Free Columns

Given an eliminated matrix, the solution for null space begins with identifying the free columns.

There will be a null space vector for each free column. In other words, given a matrix A with n free columns, the null space of A (sometimes notated as N(A)) has n dimensions.

As an example, consider this system:

w + 2x + 2y + 2z = a
2w + 4x + 6y + 8z = b
3w + 6x + 8y + 10z = c

The eliminated form of the augmented matrix A looks like:

┌                  ┐
│ [1] 2  2  2     a│
│  0  0 [2] 4  b-2a│
│  0  0  0  0 c-b-a│
└                  ┘

Note that the free columns are in positions 2 and 4.

Substitute

Populate each null space vector with a 1 in the corresponding free column position, and a 0 in all other free column positions. Continuing with the same example:

[? 1 ? 0]
[? 0 ? 1]

For each null space, solve the Ax = 0 for the remaining unknowns (w and y in this example). This can be done with simple algebra.

Start with the first null space vector ([? 1 ? 0]) and the first unknown (w). Any equation from the system can be used, but typically the equation holding the relevant pivot will be easiest to reduce.

w + 2x + 2y + 2z = 0
w + 2(1) + 2y + 2(0) = 0
w + 2 + 2y = 0

w = -2 - 2y

Given this solution for w in terms of y, solve for y. Again, any equation from the system can be used, but typically the equation holding the relevant pivot will be easiest to reduce.

2w + 4x + 6y + 8z = 0
2w + 4(1) + 6y + 8(0) = 0
2w + 4 + 6y = 0

2(-2 - 2y) + 4 + 6y = 0
-4 - 4y + 4 + 6y = 0
2y = 0
y = 0

Given this solution for y, solve for w.

w = -2 - 2y
w = -2 - 2(0)
w = -2

The first null space vector, and therefore the first null space solution, is:

[-2 1 0 0]

Repeat

Repeat the substitution process for each null space vector.

Continuing with the same example, the complete null space solutions are:

[-2 1  0 0]
[2  0 -1 1]

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