Subspace
The Definition Of Subspace
- A subspace of a vector space is a set of vectors (including 0) that satisfied two requirments: if www and vvv are verctor in the subspace and ccc is any sclar,then
- v+wv+wv+w is in the subspace
- cvcvcv in the subspace
- Every subspace must contain the zero vector. for example:
- The plane in R3R^3R3 has to go through ( 0, 0, 0)
- Lines through the origin are also subspace.
The Column Space of AAA
- The Column space consists of all linear combinations of the columns. The combinations are all possible vectors AxAxAx. They fill the column space C(A)C(A)C(A)
- The system Ax=bAx=bAx=b is solvable if and only if b is in the column space of AAA
- Suppose AAA is a m by n matrix , columns belongs to RmR^mRm (每个column 是m个元素), the column space of AAA is a subspace of RmR^mRm
The Row Space of AAA
- Like column space , row space is a linear conbinations of rows,rows belongs to RnR^nRn(每一行有n个元素),所以row space 是 RnR^nRn的subspace
The Null Space of AAA
- There are nonzero solutions to Ax=0Ax=0Ax=0 , each solution x belongs to the nullspace of A, which is denoted by N(AAA) .
- The solution vector x have n components. they are vectors in RnR^nRn , so nullspace is a subspace of RnR^nRn
- Solving Ax=0Ax=0Ax=0 by elimination
Ax=0 −> [112322810331013][x1x2x3x4]=[0000]
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=[112300440000]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 x1x1x1 and x3x3x3 : column1 and 3 contains pivots
The free variables are x2x2x2 and x4x4x4: column2 and 4 have no pivots
Special solutions:
- set x2=1,x4=0x2=1 ,x4=0x2=1,x4=0 , by back substitution x3=0x3=0x3=0 ,then x1=−1x1=-1x1=−1
- set x2=0,x4=1x2=0 ,x4=1x2=0,x4=1 , by back substitution x3=−1x3=-1x3=−1 ,then x1=−1x1=-1x1=−1
Complete solutions to Ax=0Ax=0Ax=0: x2x2x2 and x4x4x4 can be any multiplies
x=x2[−1100]+x4[−10−11]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 Ax=0Ax=0Ax=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 AAA 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=[1302−10014−31316−4]yields R=[1302−10014−300000]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 :[IF00]\left[
\begin{matrix}
I &F \\
0 & 0\\
\end{matrix}
\right][I0F0]
Nullspace matrix form is like :[−FI]\left[
\begin{matrix}
-F \\
I\\
\end{matrix}
\right][−FI]
Nullspace for the A(5-2=3) is:[−3−211000−43010001][x1 not freex2 freex3 not freex4 freex5 free]\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 Ax=bAx=bAx=b
Ax=b
[130200141316][x1x2x3x4]=[167]\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]=[130210014613167][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:
[130210014600000]=[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:Xn=[−3−2100−401]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=[1060]+x2[−3100]+x4[−20−41]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 |