Posted in Linear Algebra

## 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.)

Posted in Linear Algebra

## Some examples of subspaces – 1

As we knew, a subset ${W}$ of a vector space ${V}$ is called a subspace if it is nonempty and closed under addition and scalar multiplication.

• The vector spaces ${\mathbb{R}, \mathbb{R}^2, \mathbb{R}^3}$ are the usual Euclidean spaces of analytic geometry. There are three types of subspaces of ${\mathbb{R}^2}$: ${\{\theta\},}$ a line through the origin, and ${\mathbb{R}^2}$ itself. There are four types of subspaces of ${\mathbb{R}^3}$: ${\{\theta\}}$, a line through the origin, and ${\mathbb{R}^3}$ itself.
• Let ${P[x]}$ be the set of all polynomials in the single variable ${x}$, with coefficients from ${\mathbb{R}}$ and ${P_n[x]}$ be the subset of ${P[x]}$ consisting of all polynomials of degree ${n}$ or less. We have a result: ${P_n[x]}$ is a subspace of ${P[x]}$.
• When ${n > 0}$, the set of all polynomials of degree exactly ${n}$ is not a subspace of ${P[x]}$.
• Let ${\mathcal{F}(X, F)}$ be the set of all functions ${f: X \rightarrow F}$ (${F}$ is field ${\mathbb{R}}$ or ${\mathbb{C}}$). ${\mathcal{F}(X, F)}$ is a vector space. Why?
• ${\mathcal{F}(X, F)}$ have many subspaces, for example, we consider the case of ${X \subset \mathbb{R}^n}$ and ${F = \mathbb{R}}$. The set ${C(X)}$ of all continuous functions ${f: X \rightarrow \mathbb{R}}$ is a subspace of ${\mathcal{F}(X, \mathbb{R})}$. The set ${\mathcal{D}(X)}$ of all differentiable functions ${f: X \rightarrow \mathbb{R}}$ is a subspace ${C(X)}$ and also of ${\mathcal{F}(X, \mathbb{R})}$. Prove that?

References

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

Nguyen Huu Viet Hung, Linear algebra (Vietnamese), VNU Publisher .