First, the definition steering algorithm
In Section VI spoke space, column space, defined null space, which is the main section explains how to find these spaces, that is, to solve the $ Ax = 0 $ is the process of how to process, in the following matrix $ A $ A Case Study :( here are the main long phalanx)
$A=\left[\begin{array}{llll}{1} & {2} & {2} & {2} \\ {2} & {4} & {6} & {8} \\ {3} & {6} & {8} & {10}\end{array}\right]$
Careful observation matrix above will find the relationship between the ranks: 2 times the first column is the second column, the third row is the first two rows and
1) Elimination: null space will not change, because the solution of the equation unchanged
The first stage (the first to find a principal component): $ \ left [\ begin {array} {llll} {1} & {2} & {2} & {2} \\ {0} & {0} & { 2} & {4} \\ {0} & {0} & {2} & {4} \ end {array} \ right] $
The second stage (second pivot find found to be 0, so that the position can not be through the back of the exchange line is not zero, we will later 2 0 as principal components, continue the line of the following elimination)
$\left[\begin{array}{llll}{0} & {2} & {2} & {2} \\ {0} & {0} & {2} & {4} \\ {0} & {0} & {0} & {0}\end{array}\right]=U$
To sum up: we have found two main yuan 2, we know that the number of PCA is 2, so the rank of the matrix is 1 and 2 here, the number of primary element matrix of rank (rank) is the matrix
We had requested $ Ax = 0 $, now after elimination, solving equations unchanged, now only requires $ Ux = 0 $
2) a main column and the column consisting of
The column for the main pivot column (e.g., first column and third column), an additional column is free (as in the second and fourth columns), so-called free column that we can assign the corresponding coefficient of any unknown variables as $ x_2 = 1, x_4 = 0 $
$x_{1}+2 x_{2}+2 x_{3}+2 x_{4}=0$
$2 x_{3}+4 x_{4}=0$
The equations as: $ x = \ left [\ begin {array} {c} {- 2} \\ {1} \\ {0} \\ {0} \ end {array} \ right] $
Of course, there are also the following solution: $ x = C \ left [\ begin {array} {c} {- 2} \\ {1} \\ {0} \\ {0} \ end {array} \ right] $
We have said that there is freedom of the column so that we can arbitrarily assign values for the free variables, so of course $ x_2 = 0, x_4 = 1 $ are possible:
The equations as: $ x = \ left [\ begin {array} {c} {2} \\ {0} \\ {-2} \\ {1} \ end {array} \ right] $
Of course, there are also the following solution: $ x = d \ left [\ begin {array} {c} {2} \\ {0} \\ {-2} \\ {1} \ end {array} \ right] $
With the above two particular solutions, we can obtain all x, i.e., the null space matrix: $ X = C \ left [\ Array the begin {C} {} {- 2} {\\ \\ {0}. 1 } \\ {0} \ end { array} \ right] + d \ left [\ begin {array} {c} {2} \\ {0} \\ {-2} \\ {1} \ end {array } \ right] $
Therefore, the zero vector is a linear combination of the particular solution matrix
3) simplify U: Let U be simplified to obtain R
Simplified procedure: Starting from the U, the vertical pivot all become 0, i.e., $ \ left [\ begin {array} {llll} {1} & {2} & {0} & {-2} \\ {0 } & {0} & {2} & {4} \\ {0} & {0} & {0} & {0} \ end {array} \ right] $, then the main element 1 becomes simplified i.e. row echelon form R
$\left[\begin{array}{cccc}{1} & {2} & {0} & {-2} \\ {0} & {0} & {1} & {2} \\ {0} & {0} & {0} & {0}\end{array}\right]=R$
Seen from the above process, the all-zero row is a linear combination of the other row
The whole process from solving $ Ax = 0 $, elimination becomes $ Ux = 0 $, becomes simplified $ Rx = 0 $, familiar with the whole process, Matlab is very easy to implement the process $ R = rref (A) $
We observed from the $ R $ matrix (pivot row and pivot column intersection income) and matrix-free (free variables of rows and columns) :( Note the zero line did not write)
Matrix $ I $: $ \ left [\ begin {array} {cccc} {1} & {0} \\ {0} & {1} \ end {array} \ right] $ matrix and consisting of $ F $: $ \ left [\ begin {array} {cccc} {2} & {- 2} \\ {0} & {2} \ end {array} \ right] $
Compare two Special Solution 2) above, we find that the null space matrix can be seen directly from $ $ R & lt matrix, i.e. null space $ x = \ left [\ begin {array} {cccc} {- F} \ \ {I} \ end {array } \ right] $