Some properties of span of a vector space

In this post, we consider “span” of a vector space V. As we know, definition of span is:

Definition 1 Let {\{v_1, v_2, \dots, v_n\}} be a finite set of vectors in a vector space {V}. The subset of {V} spanned by {\{v_1, v_2, \dots, v_n\}} is the set of all linear combinations of {v_1, v_2, \dots, v_n}. This set is called the span of {\{v_1, v_2, \dots, v_n\}} and is denoted

\displaystyle \text{span}\{v_1, v_2, \dots, v_n\}.

For example, you can see

Example 1 Let {\{v_1=(1,0), v_2=(1,1)\}}, we have {span\{v_1, v_2\}=\{x(1,0) + y(1,1)| x, y \in \mathbb{R}\}}. Of course, {span\{v_1, v_2\} \subset \mathbb{R}^2}. It is easy to see that {\mathbb{R}^2 \subset span\{v_1, v_2\}} because {v = (x',y') \in \mathbb{R}^2 \Rightarrow (x',y') = x(1,0)+y(1,1) \Leftrightarrow \begin{cases} x'&= x+ y\\ y'& = y \end{cases} }. That system has always solution {x, y \in \mathbb{R}}. This implies {span\{v_1, v_2\} = \mathbb{R}^2}.

We have the following result

Theorem 2 The span of any finite set {\{v_1, v_2, \dots, v_n\}} of vectors in a vector space {V} is a subspace of {V}.

Some properties:

  1. Suppose {W} is a subspace of a vector space {V}. Prove that if {v_1, v_2, \dots, v_n \in W}, then {span\{v_1, \dots, v_n\} \subseteq W}.
  2. Suppose {v_1, \dots, v_m} and {w_1, \dots, w_n} are vectors in a vector space satisfy:
    if {w_1, \dots, w_n \in span\{v_1, \dots, v_m\}} and {v_1, \dots, v_m \in span\{w_1, \dots, w_n\}} then {span\{v_1, \dots, v_m\} = span\{w_1, \dots, w_n\}}.

References:

R. Messer, Linear algebra gateway to mathematics, Harper Collins College Publishers

D. Lay, Linear algebra and its applications, Addison-Wesley, 2012.

(cont.)

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s