Applying span twice to a set of vectors

Question

Consider a line $\ell$ in $\mathbb R^2$ that passes through the origin. What happens when one applies span twice? Particularly, what is $\operatorname{span} \{\operatorname{span} \{\ell \} \}?$

Answer 1

Claim: for every nonzero vector $\mathbf v$ of a $k$-vector space $V,$ we have that $\operatorname{span} \{\operatorname{span} \{\mathbf v \} \} = \operatorname{span} \{\mathbf v \}.$

Proof. By definition, we have that $\operatorname{span} \{\mathbf v \} = \{c \mathbf v \,|\, c \in k \}.$ Consequently, we have that $\operatorname{span} \{\operatorname{span} \{\mathbf v \} \} = \{d \mathbf w \,|\, d \in k \text{ and } \mathbf w \in \operatorname{span} \{\mathbf v \} \} = \{cd \mathbf v \,|\, c, d \in k \} = \{e \mathbf v \,|\, e \in k \} = \operatorname{span} \{\mathbf v \},$ where the third equality holds by assumption that $k$ is a field.

Given a line $\ell$ in $\mathbb R^2$ that passes through the origin, there are two cases.

(1.) If $\ell$ is the line $x = 0,$ i.e., the $y$-axis, then every point on $\ell$ is of the form $(0, y) = y(0, 1)$ for some real number $y,$ hence we can identify $\ell$ with $\operatorname{span}_{\mathbb R} \{ \langle 0, 1 \rangle \}.$ Consequently, by the above fact, we have that $\operatorname{span} \{ \ell \} = \operatorname{span}\{ \operatorname{span}_{\mathbb R} \{ \langle 0, 1 \rangle \} \} = \operatorname{span}_{\mathbb R} \{ \langle 0, 1 \rangle \}$ and $\operatorname{span} \{ \operatorname{span} \{ \ell \} \} = \operatorname{span} \{\operatorname{span}_{\mathbb R} \{ \langle 0, 1 \rangle \} \} = \operatorname{span}_{\mathbb R} \{\langle 0, 1 \rangle \} = \operatorname{span} \{ \ell \}.$

(2.) If $\ell$ is the line $y = mx$ for some real number $m,$ then every point on $\ell$ is of the form $(x, mx) = x(1, m)$ for some real number $x,$ hence we can identify $\ell$ with $\operatorname{span}_{\mathbb R} \{ \langle 1, m \rangle \}.$ Once again, we find that $\operatorname{span} \{ \operatorname{span} \{ \ell \} \} = \operatorname{span} \{ \ell \}$ by the above fact using a similar argument as (1.).