= 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.

<<TableOfContents>>

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== Definition ==

In linear algebra, the '''null space''' is the subspace that satisfies '''''A'''x = 0''. It is notated as ''N('''A''')''.

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

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== Zero Vector as a Solution ==

All systems of linear equations have a null space containing the '''zero vector'''.

For invertible matrices, the zero vector is the ''only'' null space solution.

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== Solution ==

Leaving aside the invertible case, the remaining vectors of the null space can be solved for.



=== Introduction ===

Consider the below system of equations.

{{{
w + 2x + 2y + 2z = 1
2w + 4x + 6y + 8z = 5
3w + 6x + 8y + 10z = 6
}}}

This system is rewritten as a linear system and eliminated into:

{{{
┌           ┐ ┌  ┐   ┌  ┐
│[1] 2  2  2│ │ w│   │ 1│
│ 0  0 [2] 4│ │ x│ = │ 3│
│ 0  0  0  0│ │ y│   │ 0│
└           ┘ │ z│   └  ┘
              └  ┘
}}}



=== Identify Free Columns ===

For a matrix with ''n'' free columns, the null space has ''n'' dimensions and has ''n'' solutions.

Identify the columns with a pivot in the eliminated form. The remaining columns represent '''free variables'''.



=== Substitute ===

Because 2 solutions are expected in this example's null space, the solution vectors are pre-populated as:

{{{
[w 1 y 0]
[w 0 y 1]
}}}

The ''n''th vector has a 1 in the ''n''th free variable, and a 0 in all other free variables. The pivot variables are left to vary.

Solve '''''A'''x = 0'' using these values.

For example, the solution using the 1st vector:

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

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

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

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

Leads to the solution ''[-2 1 0 0]''.

Repeat for each vector. The second solution is ''[2 0 -1 1]''.



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