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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''')''. |
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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''. | 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''. |
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| == Solving == | == Zero Vector as a Solution == |
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| Given an [[LinearAlgebra/Elimination|eliminated]] matrix, the solution for null space begins with identifying the '''free columns'''. | All systems of linear equations have a null space containing the '''zero vector'''. |
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| Null spaces will follow a pattern: | For invertible matrices, the zero vector is the ''only'' null space solution. |
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| * 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. |
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| As an example, consider this system: | == Solution == Leaving aside the invertible case, the remaining vectors of the null space can be solved for. === Introduction === 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 rewritten as a linear system 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: | === 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: |
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| [ ? 1 ? 0 ] [ ? 0 ? 1 ] |
[w 1 y 0] [w 0 y 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): | 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. Solve '''''A'''x = 0'' using these values. For example, the solution using the 1st vector: |
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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: | Leads to the solution ''[-2 1 0 0]''. |
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| {{{ [-2 1 0 0] }}} Repeat the process for the second solution, arriving at: {{{ [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].