Subspace And All Solutions to Ax=b-蒲公英云

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

v + w v+w v+w is in the subspace
c v cv cv in the subspace

Every subspace must contain the zero vector. for example：

The plane in R 3 R^3 R3 has to go through ( 0, 0, 0)
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

r	A	solutions
r=m and r=n	square and invertible	Ax=b has 1 solution
r=m and r<n	short and wide	Ax=b has ∞ \infty ∞ solutions
r<m and r=n	tall and thin	Ax=b has 0 or 1 solution
r<m and r<n	not full rank	Ax=b has 0 or ∞ \infty ∞ solutions