|
Size: 3265
Comment: Zero vector
|
Size: 2465
Comment: Rank-nullity theorem
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 1: | Line 1: |
| = Null Spaces = | = Null Space = |
| Line 13: | Line 13: |
| 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''')''. |
| Line 15: | Line 15: |
| 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 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. 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''. |
| Line 21: | Line 25: |
| == Zero Vector as a Solution == | == Solutions == |
| Line 23: | Line 27: |
| 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. |
| Line 25: | Line 29: |
| If a matrix '''''A''''' is invertible, the zero vector is the only null space solution. ---- == Solution == Leaving aside the invertible '''''A''''' case, the remaining vectors of the null space can be solved for. === 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: |
Consider the below system of equations. |
| Line 46: | Line 32: |
| 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 |
| Line 51: | Line 37: |
| The eliminated form of the augmented matrix '''''A''''' looks like: | This system is formulated as a matrix and eliminated into: |
| Line 54: | Line 40: |
| ┌ ┐ │ [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│ └ ┘ └ ┘ |
| Line 61: | Line 48: |
| Note that the free columns are in positions 2 and 4. | The number of vectors in the null space match the number of free variables. In this case, there are 2. |
| Line 63: | Line 50: |
| 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]'' |
|
| Line 64: | Line 54: |
| === 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... |
| Line 81: | Line 60: |
| w = -2 - 2y | |
| Line 82: | Line 62: |
| 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. {{{ |
|
| Line 92: | Line 66: |
| 2w + 4 + 6y = 0 | |
| Line 96: | Line 71: |
| }}} | |
| Line 98: | Line 72: |
| Given this solution for ''y'', solve for ''w''. {{{ |
|
| Line 106: | Line 77: |
| The first null space vector, and therefore the first null space solution, is: | ...a null space vector in ''[-2 1 0 0]''. |
| Line 108: | Line 79: |
| {{{ [-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] }}} |
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.
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│ └ ┘
└ ┘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].