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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''' 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. |
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| == Utility == | == Definition == |
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| Defining the null space of a system is useful for defining the '''complete solution'''. | In linear algebra, the '''null space''' is the subspace that satisfies '''''A'''x = 0''. It is notated as ''N('''A''')''. [[LinearAlgebra/Elimination|Elimination]] does not change the null space of a matrix. |
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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''. | Null spaces are spanned by null space vectors. There is always a '''zero vector''' in the null space. If a matrix is [[LinearAlgebra/Invertibility|invertible]] however, the zero vector is the ''only'' null space vector. Identifying the set of null space vectors that are not the zero vector gives a [[LinearAlgebra/Basis|basis]] for the null space. 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''. Null space vectors 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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| == 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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| The eliminated form of the augmented matrix ''A'' looks like: | 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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| From this point, the null space of ''A'' (sometimes notated as ''N(A)'') can be computed. | |
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| Note that the free columns are in positions 2 and 4. Therefore, the null space solutions begin like: | |
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| {{{ [? 1 ? 0] [? 0 ? 1] }}} |
=== Back-Substitution Procedure === |
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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): | 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: | ...a null space vector in ''[-2 1 0 0]''. |
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| {{{ [-2 1 0 0] }}} |
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| Repeat the process for the second solution, arriving at: | |
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| {{{ [2 0 -1 1] }}} |
=== 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]'' |
Null Space
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
In linear algebra, the null space is the subspace that satisfies Ax = 0. It is notated as N(A). Elimination does not change the null space of a matrix.
Null spaces are spanned by null space vectors. There is always a zero vector in the null space. If a matrix is invertible however, the zero vector is the only null space vector. Identifying the set of null space vectors that are not the zero vector gives a basis for the null space.
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.
Null space vectors 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.
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]