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← Revision 23 as of 2026-02-08 19:00:35 ⇥
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| = Null Spaces = | = Null Space = |
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| The '''null space''' of a system of equations is the set of solutions for which the dependent variables 'cancel out'. In other words, all values of ''x'' such that ''Ax = 0''. | A '''null space''' or '''kernel''' is a subspace spanned by the vectors that map to the zero vector. |
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| == Utility == | == Definition == |
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| Defining the null space of a system is useful for defining the '''complete solution'''. | 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. |
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| 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''. | 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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| == Solving == | == Solutions == |
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| Given an [[LinearAlgebra/Elimination|eliminated]] matrix, the solution for null space begins with identifying the '''free columns'''. | Leaving aside the invertible case, the non-zero vectors of the null space can be solved for. |
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| Null spaces will follow a pattern: * There will be a null space for each free column. * Populate each vector with the corresponding free column position holding a one, and all other free column positions holding a zero. * Solve the system of equation given these values and given a right hand side value of 0. As an example, consider this system: |
Consider the below system of equations. |
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| w + 2x + 2y + 2z = a 2w + 4x + 6y + 8z = b 3w + 6x + 8y + 10z = c |
w + 2x + 2y + 2z = 1 2w + 4x + 6y + 8z = 5 3w + 6x + 8y + 10z = 6 |
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| This is eliminated into the following augmented matrix: | This system is formulated as a matrix and eliminated into: |
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| ┌ ┐ │ [1] 2 2 2 a│ │ 0 0 [2] 4 b-2a│ │ 0 0 0 0 c-b-a│ └ ┘ |
┌ ┐ ┌ ┐ ┌ ┐ │[1] 2 2 2│ │ w│ │ 1│ │ 0 0 [2] 4│ │ x│ = │ 3│ │ 0 0 0 0│ │ y│ │ 0│ └ ┘ │ z│ └ ┘ └ ┘ |
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| The free columns are 2 and 4. Therefore, the null space solutions begin like: | |
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| {{{ [? 1 ? 0] [? 0 ? 1] }}} |
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| The first solution can be found by rewriting the first equation from the system (with 0 as the right hand value): | === 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... |
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| }}} | |
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| Substitute this into the second equation: {{{ |
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| 2w + 4 + 6y = 0 |
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| }}} | |
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| Substitute this into the first equation again: {{{ |
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| The first null space solution is: {{{ [-2 1 0 0] }}} Repeat the process for the second solution, arriving at: {{{ [2 0 -1 1] }}} ---- |
...a null space vector in ''[-2 1 0 0]''. |
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| == Full Rank Matrices == | === Linear Combinations Procedure === |
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| If a matrix has no free columns, it is said to be '''full rank''' and the ''only'' null space is the zero vector (i.e. ''[0 ... 0]''). | 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]'' |
Null Space
A null space or kernel is a subspace spanned by the vectors that map to the zero vector.
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 general solution to a system of equations.
Therefore the vectors which map to a zero vector must be identified, as through Ax = 0. Identifying all independent vectors gives a 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 rank. For any matrix A of size m x n (i.e., there are n columns), rank(A) + nullity(A) = n.
Note that 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 invertible, this is the only null space vector.
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 Ax = 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]