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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''')''. |
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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 composed of 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. 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 == | == Solution == |
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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: | The number of vectors in the null space match the number of free variables. In this case, there are 2. |
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| {{{ [ ? 1 ? 0 ] [ ? 0 ? 1 ] }}} |
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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| The first solution can be found by rewriting the first equation from the system (with 0 as the right hand value): | 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] }}} Repeat the process for the second solution, arriving at: {{{ [2 0 -1 1] }}} |
The second null space vector is ''[2 0 -1 1]''. |
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.
Contents
Definition
In linear algebra, the null space is the subspace that satisfies Ax = 0. It is notated as N(A).
Null spaces are composed of 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.
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.
Solution
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│ └ ┘
└ ┘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].
The second null space vector is [2 0 -1 1].