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| = Null Spaces = | = Null Space = |
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| For a system of equations, the '''null space''' is the subspace that satisfies '''''A'''x = 0''. This is useful for finding a [[LinearAlgebra/Solution|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 '''''A'''x = 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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| == Zero Vector as a Solution == | == Solutions == |
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| For all systems of equations, there is a null space. The zero vector is always such a solution. | Leaving aside the invertible case, the non-zero vectors of the null space can be solved for. |
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| If a matrix '''''A''''' is invertible, the zero vector is the only null space solution. | Consider the below system of equations. |
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| ---- | {{{ 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│ └ ┘ └ ┘ }}} |
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| == Solution == | === Back-Substitution Procedure === |
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| Leaving aside the invertible '''''A''''' case, the remaining vectors of the null space can be solved for. | The number of vectors in the null space match the number of free variables. In this case, there are 2. |
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| 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]'' |
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=== Identify Free Columns === Given an [[LinearAlgebra/Elimination|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 '''''A'''x = 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. |
Solve '''''A'''x = 0'' using these values. For example, substituting in the first vector's values gives... |
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| w = -2 - 2y | |
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| 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. {{{ |
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| 2w + 4 + 6y = 0 | |
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| }}} | |
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| Given this solution for ''y'', solve for ''w''. {{{ |
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| The first null space vector, and therefore the first null space solution, is: {{{ [-2 1 0 0] }}} |
...a null space vector in ''[-2 1 0 0]''. |
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| === Repeat === | === Linear Combinations Procedure === |
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| Repeat the substitution process for each null space vector. | 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]''. |
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| Continuing with the same example, the complete null space solutions are: {{{ [-2 1 0 0] [2 0 -1 1] }}} |
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]