= Null Space =

A '''null space''' or '''kernel''' is a subspace spanned by the vectors that map to the zero vector.

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== Definition ==

Because zero vectors have an identity property (i.e., any linear combination of them can be added to arrive at the exact same values), they have an important role in identifying the [[LinearAlgebra/GeneralSolution|general solution]] to a system of equations.

Therefore the vectors which [[LinearAlgebra/LinearMapping|map]] to a zero vector must be identified, as through '''''A'''x = 0''. Identifying all independent vectors gives a [[LinearAlgebra/Basis|basis]] for spanning a subspace called the '''null space''' or the '''kernel'''. It is notated as ''N('''A''')''. The vectors themselves are then called '''null space vectors'''.

The number of dimensions spanned by the null space vectors is called '''nullity''', and it has a direct relation to [[LinearAlgebra/Rank|rank]]. For any matrix '''''A''''' of size ''m x n'' (i.e., there are ''n'' columns), ''rank('''A''') + nullity('''A''') = n''.

Note that [[LinearAlgebra/Elimination|Elimination]] does not change the null space of a matrix. Therefore elimination is a strategy for identifying both rank and nullity.

There is always a trivial component of the null space: the zero vector itself. If a matrix is [[LinearAlgebra/Invertibility|invertible]], this is the ''only'' null space vector.

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== Solutions ==

Leaving aside the invertible case, the non-zero vectors of the null space can be solved for.

Consider the below system of equations.

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

This system is formulated as a matrix and eliminated into:

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



=== Back-Substitution Procedure ===

The number of vectors in the null space match the number of free variables. In this case, there are 2.

The null space vectors can be 'prepopulated' or 'templated' by allowing pivot variables to vary and alternating which free variable is set to 0 or 1. In this case, they are:
 * ''[w 1 y 0]''
 * ''[w 0 y 1]''

Solve '''''A'''x = 0'' using these values. For example, substituting in the first vector's values gives...

{{{
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
}}}

...a null space vector in ''[-2 1 0 0]''.



=== Linear Combinations Procedure ===

Consider combinations of the rows that sum to a zero vector. The first null space vector found was ''[-2 1 0 0]'', corresponding to -2 times the first column and 1 times the second column (and zero times the other two). It should be immediately clear that ''-2*[1 0 0] + 1*[2 0 0] = [0 0 0]''.

Following this framework, the second null space vector is ''[2 0 -2 1]'' because ''2*[1 0 0] + -2*[2 2 0] + 1*[2 4 0] = [0 0 0]''



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