subspace

A subspace of a vector space $V$ is a subset $H$ of $V$ that has two properties.

$H$ is closed under vector addition. That is, for each $u$ and $v$ in $H$ , the sum $u+v$ is in $H$ .
$H$ is closed under multiplication by scalars. That is, for each $u$ in $H$ and each scalar $c$ , the vector $cu$ is in $H$ .

Note

In some materials, $H$ has the third property that The zero vector of $V$ is in $H$ . This property could be inclued in Property 2 while $c=0$ .

For example,

A line through the origin in $R^2$ is a subspace.
The $xy$ -plane in $R^3$ is a subspace.
The set of all polynomials $P$ is a subspace of $C[0,1]$ .

Proof

如何证明一个空间是subspace？

只需要根据定义，将该空间的向量 $x, y$ 代入两个条件验证即可。

Span

Span（线性生成空间）is the set of all linear combination of specialized vectors.

For example,

$span(v_1, v_2, ... v_n) = \{ \sum_{i=1}^{n}\alpha_iv_i|\alpha_i \in \Re \}$

Means span( $v_1, v_2, ... v_n$ ) is generated / spanned by $v_1, v_2, ... v_n$ .

Theorem

span( $v_1, v_2, ... v_n$ ) is a subspace of $V$ .

Null space

Given a matrix $A \in \Re^{m \times n}$ . The null space or kernel of $A$ is defined as

$Nul A = Ker A = \{ x \in \Re^n | Ax = 0\}$

Column space

Given a matrix $A \in \Re^{m \times n}$ . The column space or image of $A$ is defined as the set of all linear combination of the colums of $A$ .

$col A = ImA=\{b \in \Re^m | b = Ax \quad \mathrm{for \quad some} \quad x \in \Re^n\}$

Computation

But how to compute the null space and column space of a speified matrix?

For example,

$A = \begin{bmatrix} 1 \quad 2 \quad 1 \quad 3 \quad 2 \\ 3 \quad 4 \quad 9 \quad 0 \quad 7 \\ 2 \quad 3 \quad 5 \quad 1 \quad 8 \\ 2 \quad 2 \quad 8 \quad -3 \quad 5 \end{bmatrix}$

how to get $Nul(A)$ and $Col(A)$ ?

Answer

Convert matrix $A$ to the row minimal matrix. The specific calculation process is omitted here.

$A=> \begin{bmatrix} 1 \quad 0 \quad 7 \quad 0 \quad 85 \\ 0 \quad 1 \quad -3 \quad 0 \quad 31 \\ 0 \quad 0 \quad 0 \quad 1 \quad -7 \\ \end{bmatrix}$

We know that the pivot number represents the location of dependent variables and the remaining variables are called free variables. In this row minimal matrix, the pivot numbers are 1, 2, 4.（非零行的第一个非零元所在的位置）Hence we could get two basis of the vector space:

$\xi_1 = \begin{bmatrix} x_1 \\ y_1 \\ 0 \\ z_1 \\ 1 \end{bmatrix} \quad \xi_2 = \begin{bmatrix} x_2 \\ y_2 \\ 1 \\ z_2 \\ 0 \end{bmatrix}$

let $A\xi_1 = 0$ and $A\xi_2 = 0$ , we get the values of $x, y, z$ respectively.

$\xi_1 = \begin{bmatrix} -85 \\ -31 \\ 0 \\ 7 \\ 1 \end{bmatrix} \quad \xi_2 = \begin{bmatrix} -7 \\ 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

Null space

Use the basis to represent null space

$\mathrm{Nul}A = \mathrm{span}\{\xi_1, \xi_2\}$

Column space

After we get pivot number, we get the column space as well since column space consists of the column vectors of pivot number in $A$ .

$\mathrm{Col}A=\mathrm{span}\{ \begin{bmatrix} 1 \\ 3 \\ 2 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 4 \\ 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ 1 \\ -3 \end{bmatrix} \}$

Subspace, Null Space and Col Space in Linear Algebra

subspace

Proof

Span

Null space

Column space

Computation

References