|
Size: 2391
Comment:
|
Size: 2919
Comment: Reorganized
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 3: | Line 3: |
| 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. |
| Line 11: | Line 11: |
| == Utility == | == Definition == |
| Line 13: | Line 13: |
| Defining the null space of a system is useful for defining the '''complete solution'''. | 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]]. |
| Line 15: | Line 15: |
| 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''. |
| Line 21: | Line 21: |
| == Solving == | == Solution == === Identify Free Columns === |
| Line 25: | Line 29: |
| Null spaces will follow a pattern: * 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. * 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. |
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. |
| Line 40: | Line 39: |
| The eliminated form of the augmented matrix ''A'' looks like: | The eliminated form of the augmented matrix '''''A''''' looks like: |
| Line 50: | Line 49: |
| From this point, the null space of ''A'' can be computed. | Note that the free columns are in positions 2 and 4. |
| Line 52: | Line 51: |
| Note that the free columns are in positions 2 and 4. Therefore, the null space solutions begin like: | === 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: |
| Line 59: | Line 61: |
| The first solution can be found by rewriting the first equation from the system (with 0 as the right hand value): | 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. |
| Line 65: | Line 69: |
| Line 68: | Line 73: |
| Substitute this into the second equation: | 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 74: | Line 79: |
| Line 80: | Line 86: |
| Substitute this into the first equation again: | Given this solution for ''y'', solve for ''w''. |
| Line 88: | Line 94: |
| The first null space solution is: | The first null space vector, and therefore the first null space solution, is: |
| Line 94: | Line 100: |
| Repeat the process for the second solution, arriving at: | === Repeat === Repeat the substitution process for each null space vector. Continuing with the same example, the complete null space solutions are: |
| Line 97: | Line 109: |
| [2 0 -1 1] | [-2 1 0 0] [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.
Definition
For a system of equations, the null space is the subspace that satisfies Ax = 0. This is useful for finding a complete solution.
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
Identify Free Columns
Given an 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 Ax = 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.
w + 2x + 2y + 2z = 0 w + 2(1) + 2y + 2(0) = 0 w + 2 + 2y = 0 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.
2w + 4x + 6y + 8z = 0 2w + 4(1) + 6y + 8(0) = 0 2w + 4 + 6y = 0 2(-2 - 2y) + 4 + 6y = 0 -4 - 4y + 4 + 6y = 0 2y = 0 y = 0
Given this solution for y, solve for w.
w = -2 - 2y w = -2 - 2(0) w = -2
The first null space vector, and therefore the first null space solution, is:
[-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]