MIT线性代数笔记-第七讲

computing the null space

A=⎡⎣⎢1232462682810⎤⎦⎥−>⎡⎣⎢100200222244⎤⎦⎥−>⎡⎣⎢100200220240⎤⎦⎥=U$A=\left[\begin{array}{cccc}1& 2& 2& 2\\ 2& 4& 6& 8\\ 3& 6& 8& 10\end{array}\right]->\left[\begin{array}{cccc}1& 2& 2& 2\\ 0& 0& 2& 4\\ 0& 0& 2& 4\end{array}\right]->\left[\begin{array}{cccc}1& 2& 2& 2\\ 0& 0& 2& 4\\ 0& 0& 0& 0\end{array}\right]=U$

注意到，A有两个pivot

rank

rank of A = counts of pivot

第一列和第三列含有pivot的列被称为pivot column,而没有pivot的列被称为free column.Free column means I can assign any number freely to the variable(在A为X2和X4)

由于X2，X4可以取任意值，那么我们先将X2设为1，X4设为0，则X为:
[−2100]$\left[\begin{array}{cccc}-2& 1& 0& 0\end{array}\right]$

注意到，我们可以将X乘以任意系数，结果仍是AX = 0的解，即:

c[−2100]$c\left[\begin{array}{cccc}-2& 1& 0& 0\end{array}\right]$

我们还可以将x4 = 1, x2 = 0,得到
[20−21]$\left[\begin{array}{cccc}2& 0& -2& 1\end{array}\right]$

这两个X被称作AX = 0的spectial solutions,AX = 0的解为spection solutions的线性组合，即
c1⎡⎣⎢⎢⎢−2100⎤⎦⎥⎥⎥+c1⎡⎣⎢⎢⎢20−21⎤⎦⎥⎥⎥$c1\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+c1\left[\begin{array}{c}2\\ 0\\ -2\\ 1\end{array}\right]$

AX = 0有多少个spectial solution?
为free variable的数量，即n - rank.

remember,when I do the elimination,the null space of A don’t change

R = reduced row echelon form

U=⎡⎣⎢100200220240⎤⎦⎥$U=\left[\begin{array}{cccc}1& 2& 2& 2\\ 0& 0& 2& 4\\ 0& 0& 0& 0\end{array}\right]$

对U再做些变换，之前我们做elimination一直都是downwards，也就是向下，现在我们试试upwards,向上消元，也就是说，R矩阵，在pivot上下的值都为0
U=⎡⎣⎢100200220240⎤⎦⎥−>⎡⎣⎢100200020−240⎤⎦⎥$U=\left[\begin{array}{cccc}1& 2& 2& 2\\ 0& 0& 2& 4\\ 0& 0& 0& 0\end{array}\right]->\left[\begin{array}{cccc}1& 2& 0& -2\\ 0& 0& 2& 4\\ 0& 0& 0& 0\end{array}\right]$

我们还可以再多做一步
⎡⎣⎢100200020−240⎤⎦⎥−>⎡⎣⎢100200010−220⎤⎦⎥=R=rref(A)(用matlab中的这个命令可以得到R)$\left[\begin{array}{cccc}1& 2& 0& -2\\ 0& 0& 2& 4\\ 0& 0& 0& 0\end{array}\right]->\left[\begin{array}{cccc}1& 2& 0& -2\\ 0& 0& 1& 2\\ 0& 0& 0& 0\end{array}\right]=R=rref(A)(用matlab中的这个命令可以得到R)$

R能带给我们哪些信息？
pivot row and pivot columns are one and three,notice I is in the pivot rows and columns
[1001]$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

之前我们用将AX = 0转换为UX = 0来解X，现在我们可以用RX = 0来解X

对比一下之前解得的AX = 0的两个特殊解与I 和 free columns:
c1⎡⎣⎢⎢⎢−2100⎤⎦⎥⎥⎥+c1⎡⎣⎢⎢⎢20−21⎤⎦⎥⎥⎥$c1\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+c1\left[\begin{array}{c}2\\ 0\\ -2\\ 1\end{array}\right]$

I=[1001]andF=[20−22]$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]andF=\left[\begin{array}{cc}2& -2\\ 0& 2\end{array}\right]$
注意到spectial solution中的值即为I和F(多了个负号)的组合

rref form

[I0F0]$\left[\begin{array}{cc}I& F\\ 0& 0\end{array}\right]$

RX = 0,RN = 0
Null Space matrix(columns are the spectial solutions)

N=[−FI]$N=\left[\begin{array}{c}-F\\ I\end{array}\right]$