Linear Algebra Lecture 6: Vectors spaces and subspaces
Column space of A: solving Ax = b Null space of A
Vector space:
v+w and cv are in the space or all vector linear combations cv + dw are in the space
Subspace
子空间的定义：1. 必须过零点， 2 线性组合仍然在此子空间内。
Plane through (0 0 0) is a subspace - P
Line through (0 0 0) is a subspace L
Is $P \cup L$ a subspace? no
because if I add vector from P and a vector from L is not on the subspace
Is $P \cap L $ s subspace? yes
$P \cap L $ is zero
Generally Subspace S and T, $S \cap T$ is subspace.
Example
Columne space of A, A is subspace of $R^4$
举一个非独立矩阵的例子， 下面的例子，
这个矩阵含有3个4维向量， 第3列向量是前两列的线性组合， 是没有用的。。所以它的列空间（列向量的线性组合）不能长成整个4维空间。
所以对于b不为零，是不是都有解，但解不是子空间，因为他的解构成的子空间不通过0
$$
A =
\begin{bmatrix}
1 & 1 & 2 \\
2 & 1 & 3 \\
3 & 1 & 4 \\
4 & 1 & 5
\end{bmatrix}
$$
then the columns of A is in subsapce. All the combinations of the columns are in the subspace.
Is the subspace full file the 4-D space.
Does Ax=b have a solution for every b? No 4 equations with 3 unknowns
2 & 1 & 3 \\
3 & 1 & 4 \\
4 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\
x_2 \\
x_3 \end{bmatrix} ＝ \begin{bmatrix} b_1 \\
b_2 \\
b_3 \\
b_4 \end{bmatrix} $$
I can solve Ax=b exactly when b is in the column space.
Can I solve b = [ 1 2 3 4 ]?
X = [1 0 0] solves it.
Can I sovel b = [ 1 1 1 1]?
X = [0 1 0] solves it.
b = [0 0 0 0 ] is always OK.
*I can solve Ax = b exact when b is a comninations of the columns of A – C(A)
Does each column contribute something new or not?
For the above example, column 3 is col1 + col2, and does not contribute to the subspace.
so we called column 1 and 2 independent.
And A is 2-D subspace of $R^4$
Null Space
所有x使得Ax=0所构成的空间叫做零空间。
找出符合条件的x， 上面的例子，x很好找，比如（1， 1， -1）， 那么$\lambda x $ 都是Ax=0的解，$\lambda x $ 为矩阵A的零空间。
因为$\lambda x $ 经过A的变换都变成零向量了。
all solutions
$
X =
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
$
to Ax = 0 in $R^3$
then we have X = c $\begin{bmatrix}
1 \\
1 \\
-1 \\
\end{bmatrix}
$ would be the solution.
think of geometry , null space is line through zero.
Check the solutions to Ax = 0 always givce a subspace.
然后证明$\lambda x $ 符合子空间定义。
If Av = 0 , and Aw = 0. then A(v+w) = 0** because A(v+w) = Av + Aw
and A(12V) = 0 becuase A(12v) = 12Av
然后说明哪些不构成子空间， 重点在强调子空间的定义的第一点：必须过原点。
Do the X solutions form a subspace when b is not zero? No
because zero does not solve the solution, [1 0 0] is solution [0 -1 1]. it’s like a plane does not go thru. origion or line does not go thru the origin.
So X does not form a subspace.
Subspace has to go thru. the origin.
Essential of Subspace :
- columns space: few vectors makes a subspace
- the equations has to be satisfied also defines a subspace
