Let v1→=[23] and v1→=[46] what is the **span** of the vector space defined by v1→...

span ${\displaystyle \left(\{\overrightarrow{v_1},\overrightarrow{v_2}\}\right)=\{\lambda \overrightarrow{v_1}\mid \lambda \in F\}}$
Explanation: Typically we talk about the span of a set of vectors, rather than of an entire vector space. We will proceed, then, in examining the span of ${\displaystyle \{\overrightarrow{v_1},\overrightarrow{v_2}\}}$ within a given vector space.
The span of a set of vectors in a vector space is the set of all finite linear combinations of those vectors. That is, given a subset S of a vector space over a field F, we have
${\displaystyle span\left(S\right)=\{\sum _{i=1}^k{\lambda }_k{s}_k\mid n\in \mathbb{N},s_i\in S,{\lambda }_i\in F\}}$
(the set of any finite sum with each term being the product of a scalar and an element of S)
For simplicity, we will assume that our given vector space is over some subfield F of ${\displaystyle \mathbb{C}}$. Then, applying the above definition:
${\displaystyle span\left(\{\overrightarrow{v_1},\overrightarrow{v_2}\}\right)=\{\sum _{i=1}^k{\lambda }_i{v}_i\mid {\lambda }_i\in F\}}$
${\displaystyle =\{{\lambda }_1\overrightarrow{v_1}+{\lambda }_2\overrightarrow{v_2}\mid {\lambda }_1,{\lambda }_2\in F\}}$
But note that ${\displaystyle \overrightarrow{v_2}=2\overrightarrow{v_1}}$, and so, for any ${\displaystyle {\lambda }_1,{\lambda }_2\in F}$
${\displaystyle {\lambda }_1\overrightarrow{v_1}+{\lambda }_2\overrightarrow{v_2}={\lambda }_1\overrightarrow{v_1}+{\lambda }_2\left(2\overrightarrow{v_1}\right)=({\lambda }_1+2{\lambda }_2)\overrightarrow{v_1}}$
Then, as any linear combination of ${\displaystyle \left\{\overrightarrow{v_1}\text{and}\overrightarrow{v_2}\right\}}$ can be expressed as a scalar multiple of ${\displaystyle \{\overrightarrow{v_1}}$, and any scalar multiple of ${\displaystyle \{\overrightarrow{v_1}}$ can be expressed as a linear combination of ${\displaystyle \left\{\overrightarrow{v_1}\text{and}\overrightarrow{v_2}\right\}}$ by setting ${\displaystyle {\lambda }_2=0}$ we have
${\displaystyle \left(\{\overrightarrow{v_1},\overrightarrow{v_2}\}\right)=\{\lambda \overrightarrow{v_1}\mid \lambda \in F\}}$