Subspace And All Solutions to Ax=b

红太狼 2022-05-16 15:54 290阅读 0赞

## Subspace ##

### The Definition Of Subspace ###

*  A subspace of a vector space is a set of vectors (including **0**) that satisfied two requirments: if  w w w and  v v v are verctor in the subspace and  c c c is any sclar,then

1.   v + w v+w v\+w is in the subspace
2.   c v cv cv in the subspace

*   Every subspace must contain the zero vector. for example：

1.  The plane in  R 3 R^3 R3 has to go through ( 0, 0, 0)
2.  Lines through the origin are also subspace.

### The Column Space of  A A A ###

*   The Column space consists of all linear combinations of the columns. The combinations are all possible vectors  A x Ax Ax. They fill the column space  C ( A ) C(A) C(A)

*  The system  A x = b Ax=b Ax=b is solvable if and only if b is in the column space of  A A A

*   Suppose  A A A is a m by n matrix , columns belongs to  R m R^m Rm （每个column 是m个元素）, the column space of  A A A is a subspace of  R m R^m Rm

### The Row Space of  A A A ###

*  Like column space , row space is a linear conbinations of rows，rows belongs to  R n R^n Rn（每一行有n个元素）,所以row space 是  R n R^n Rn的subspace

### The Null Space of  A A A ###

*   There are nonzero solutions to  A x = 0 Ax=0 Ax=0 , each solution x belongs to the nullspace of A, which is denoted by N( A A A) .  
      
    
 *   The solution vector x have n components. they are vectors in  R n R^n Rn , so nullspace is a subspace of  R n R^n Rn

*   Solving  A x = 0 Ax=0 Ax=0 by elimination

A x = 0          − >       \[ 1 1 2 3 2 2 8 10 3 3 10 13 \] \[ x 1 x 2 x 3 x 4 \] = \[ 0 0 0 0 \] Ax=0 \\space\\space\\space\\space\\space\\space\\space\\space->\\space\\space\\space\\space\\space\\left\[ \\begin\{matrix\} 1 & 1&2 & 3 \\\\ 2 & 2 & 8 &10\\\\ 3 & 3 & 10& 13 \\end\{matrix\} \\right\] \\left\[ \\begin\{matrix\} x1 \\\\ x2 \\\\ x3 \\\\ x4 \\end\{matrix\} \\right\]= \\left\[ \\begin\{matrix\} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\end\{matrix\} \\right\] Ax=0        −>     ⎣⎡123123281031013⎦⎤⎣⎢⎢⎡x1x2x3x4⎦⎥⎥⎤=⎣⎢⎢⎡0000⎦⎥⎥⎤  
  
  
Get Triangular U by eliminating A：  U = \[ 1 1 2 3 0 0 4 4 0 0 0 0 \] U= \\left\[ \\begin\{matrix\} 1 & 1&2 & 3 \\\\ 0 & 0 & 4 &4\\\\ 0 & 0 & 0& 0 \\end\{matrix\} \\right\] U=⎣⎡100100240340⎦⎤

The pivot variables are  x 1 x1 x1 and  x 3 x3 x3 : column1 and 3 contains pivots  
 The free variables are  x 2 x2 x2 and  x 4 x4 x4: column2 and 4 have no pivots

Special solutions:

*  set  x 2 = 1 , x 4 = 0 x2=1 ,x4=0 x2=1,x4=0 , by back substitution  x 3 = 0 x3=0 x3=0 ,then  x 1 = − 1 x1=-1 x1=−1
 *  set  x 2 = 0 , x 4 = 1 x2=0 ,x4=1 x2=0,x4=1 , by back substitution  x 3 = − 1 x3=-1 x3=−1 ,then  x 1 = − 1 x1=-1 x1=−1

Complete solutions to  A x = 0 Ax=0 Ax=0：   x 2 x2 x2 and  x 4 x4 x4 can be any multiplies  
 x = x 2 \[ − 1 1 0 0 \] + x 4 \[ − 1 0 − 1 1 \] x=x2\\left\[ \\begin\{matrix\} -1 \\\\ 1 \\\\ 0 \\\\ 0 \\end\{matrix\} \\right\]+x4\\left\[ \\begin\{matrix\} -1 \\\\ 0 \\\\ -1 \\\\ 1 \\end\{matrix\} \\right\] x=x2⎣⎢⎢⎡−1100⎦⎥⎥⎤\+x4⎣⎢⎢⎡−10−11⎦⎥⎥⎤

*   With n>m ,there is at least one free variable . The system  A x = 0 Ax=0 Ax=0 has at least one nonzero solution. the nulllspace dimension is the number of free variables(at least n-m).

### The Rank And Reduced Echelon Form R ###

*  The rank of  A A A is the number of pivots ,this number is r

*  The rank r is the “dimension” of the column space. It is also the dimension of the row space . The great thing is that r also reveals the dimetion of nullspace(n-r).

*  R：the conponents above the pivots are 0 in the pivot columns (column 1 and column 3 )  
     A = \[ 1 3 0 2 − 1 0 0 1 4 − 3 1 3 1 6 − 4 \] y i e l d s   R = \[ 1 3 0 2 − 1 0 0 1 4 − 3 0 0 0 0 0 \] A= \\left\[ \\begin\{matrix\} 1 & 3&0 & 2&-1\\\\ 0 & 0 & 1 &4&-3\\\\ 1 & 3 & 1& 6&-4 \\end\{matrix\} \\right\] yields \\space R=\\left\[ \\begin\{matrix\} 1 & 3&0 & 2&-1\\\\ 0 & 0 & 1 &4&-3\\\\ 0 & 0 & 0& 0&0 \\end\{matrix\} \\right\] A=⎣⎡101303011246−1−3−4⎦⎤yields R=⎣⎡100300010240−1−30⎦⎤

The form of R is like : \[ I F 0 0 \] \\left\[ \\begin\{matrix\} I &F \\\\ 0 & 0\\\\ \\end\{matrix\} \\right\] \[I0F0\]  
Nullspace matrix form is like : \[ − F I \] \\left\[ \\begin\{matrix\} -F \\\\ I\\\\ \\end\{matrix\} \\right\] \[−FI\]

Nullspace for the A（5-2=3） is： \[ − 3 − 2 1 1 0 0 0 − 4 3 0 1 0 0 0 1 \] \[ x 1   n o t   f r e e x 2   f r e e x 3   n o t   f r e e x 4   f r e e x 5   f r e e \] \\left\[ \\begin\{matrix\} -3 & -2&1 \\\\ 1 & 0 & 0 \\\\ 0 & -4 & 3\\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end\{matrix\} \\right\]\\left\[ \\begin\{matrix\} x1 \\space not\\space free \\\\ x2 \\space free \\\\ x3 \\space not\\space free \\\\ x4 \\space free \\\\ x5 \\space free \\\\ \\end\{matrix\} \\right\] ⎣⎢⎢⎢⎢⎡−31000−20−41010301⎦⎥⎥⎥⎥⎤⎣⎢⎢⎢⎢⎡x1 not freex2 freex3 not freex4 freex5 free⎦⎥⎥⎥⎥⎤

--------------------

## All solutions to  A x = b Ax=b Ax=b ##

### Ax=b ###

\[ 1 3 0 2 0 0 1 4 1 3 1 6 \] \[ x 1 x 2 x 3 x 4 \] = \[ 1 6 7 \] \\left\[ \\begin\{matrix\} 1 & 3&0&2\\\\ 0 & 0 & 1 &4\\\\ 1 & 3 & 1& 6 \\end\{matrix\} \\right\]\\left\[ \\begin\{matrix\} x1 \\\\ x2\\\\ x3 \\\\ x4 \\end\{matrix\} \\right\]=\\left\[ \\begin\{matrix\} 1 \\\\ 6\\\\ 7 \\end\{matrix\} \\right\] ⎣⎡101303011246⎦⎤⎣⎢⎢⎡x1x2x3x4⎦⎥⎥⎤=⎣⎡167⎦⎤

augmented matrix：  
 \[ A   b \] = \[ 1 3 0 2 1 0 0 1 4 6 1 3 1 6 7 \] \[A \\space b\]=\\left\[ \\begin\{matrix\} 1 & 3&0&2&1\\\\ 0 & 0 & 1 &4&6\\\\ 1 & 3 & 1& 6&7 \\end\{matrix\} \\right\] \[A b\]=⎣⎡101303011246167⎦⎤

after elimination：  
 \[ 1 3 0 2 1 0 0 1 4 6 0 0 0 0 0 \] = \[ R   d \] \\left\[ \\begin\{matrix\} 1 & 3&0&2&1\\\\ 0 & 0 & 1 &4&6\\\\ 0 & 0 &0& 0&0 \\end\{matrix\} \\right\]=\[R\\space d\] ⎣⎡100300010240160⎦⎤=\[R d\]

particular solution：set x2=x4=0 then Xp=(1, 0, 6, 0)  
the solutions to Ax=0： X n = \[ − 3 − 2 1 0 0 − 4 0 1 \] Xn=\\left\[ \\begin\{matrix\} -3& -2\\\\ 1 & 0 \\\\ 0 & -4 \\\\ 0&1 \\end\{matrix\} \\right\] Xn=⎣⎢⎢⎡−3100−20−41⎦⎥⎥⎤

Complete solutions：X=Xp+Xn  
 X = \[ 1 0 6 0 \] + x 2 \[ − 3 1 0 0 \] + x 4 \[ − 2 0 − 4 1 \] X=\\left\[ \\begin\{matrix\} 1\\\\ 0 \\\\ 6 \\\\ 0 \\end\{matrix\} \\right\]+x2\\left\[ \\begin\{matrix\} -3\\\\ 1 \\\\ 0 \\\\ 0 \\end\{matrix\} \\right\]+x4\\left\[ \\begin\{matrix\} -2\\\\ 0 \\\\ -4 \\\\ 1 \\end\{matrix\} \\right\] X=⎣⎢⎢⎡1060⎦⎥⎥⎤\+x2⎣⎢⎢⎡−3100⎦⎥⎥⎤\+x4⎣⎢⎢⎡−20−41⎦⎥⎥⎤

### four possibilities of solutions depend on rank r ###

<table> 
 <thead> 
  <tr> 
   <th>r</th> 
   <th>A</th> 
   <th>solutions</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>r=m and r=n</td> 
   <td>square and invertible</td> 
   <td>Ax=b has 1 solution</td> 
  </tr> 
  <tr> 
   <td>r=m and r&lt;n</td> 
   <td>short and wide</td> 
   <td>Ax=b has <span><span><span> ∞ \infty </span><span><span><span style="height: 0.43056em; vertical-align: 0em;"></span><span>∞</span></span></span></span></span> solutions</td> 
  </tr> 
  <tr> 
   <td>r&lt;m and r=n</td> 
   <td>tall and thin</td> 
   <td>Ax=b has 0 or 1 solution</td> 
  </tr> 
  <tr> 
   <td>r&lt;m and r&lt;n</td> 
   <td>not full rank</td> 
   <td>Ax=b has 0 or <span><span><span> ∞ \infty </span><span><span><span style="height: 0.43056em; vertical-align: 0em;"></span><span>∞</span></span></span></span></span> solutions</td> 
  </tr> 
 </tbody> 
</table>