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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''')''. |
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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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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│ └ ┘ └ ┘ }}} |
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| Given an [[LinearAlgebra/Elimination|eliminated]] matrix, the solution for null space begins with identifying the '''free columns'''. | For a matrix with ''n'' free columns, the null space has ''n'' dimensions and has ''n'' solutions. |
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| 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. | Identify the columns with a pivot in the eliminated form. The remaining columns represent '''free variables'''. |
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| 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. |
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| 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: | Because 2 solutions are expected in this example's null space, the solution vectors are pre-populated as: |
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| [? 1 ? 0] [? 0 ? 1] |
[w 1 y 0] [w 0 y 1] |
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| 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. | 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. |
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| 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, the solution using the 1st vector: |
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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: | Leads to the solution ''[-2 1 0 0]''. |
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| {{{ [-2 1 0 0] }}} === Repeat === Repeat the substitution process for each null space vector. Continuing with the same example, the complete null space solutions are: {{{ [-2 1 0 0] [2 0 -1 1] }}} |
Repeat for each vector. The second solution is ''[2 0 -1 1]''. |
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.
Contents
Definition
In linear algebra, the null space is the subspace that satisfies Ax = 0. It is notated as N(A).
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.
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.
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 nth vector has a 1 in the nth free variable, and a 0 in all other free variables. The pivot variables are left to vary.
Solve Ax = 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].