How would you show that $ℝv=\{tv\mid t \in ℝ\}$ is a subspace of $ℝ^n$? These are my steps:


*

*$0$ is certainly in $ℝ^n$

*If $at \in ℝ$, then $atv=t(av)=ℝ(av)=a(ℝv)$ 

*If $t+s \in ℝ$, then $(t+s)v=tv+sv=ℝ(tv+sv)=(ℝt+ℝs)v$.
Since I felt my steps are partly erroneous, could anyone fix it please? 
 A: The vector space axioms are stated in a certain way which you are violating in 2 and 3. 
In general, given a vector space $V$ and a subset $W \subset V$, in defining "$W$ is a subspace of $V$" here is what the abstract versions of 2 and 3 say, together with the way you should write that for your particular example:
Item 2 in the abstract:


*

*If $a \in \mathbb{R}$ and $w \in W$ then $aw \in W$


Item 2 for your example:


*

*If $a \in \mathbb{R}$ and $tv \in \mathbb{R}v$, then $a(tv) \in \mathbb{R}v$, because $a(tv) = (at)v \in \mathbb{R}v$.


Item 3 in the abstract:


*

*If $u,w \in W$ then $u+w \in W$.


Item 3 for your example:


*

*If $tv,sv \in \mathbb{R}v$ then $tv + sv \in \mathbb{R}v$, because $tv + sv = (t+s)v \in \mathbb{R}v$.

A: With a little bit more knowledge about linear mappings one can do it without having to check the vector space axioms:
with $$\varphi_t:\mathbb R^n\rightarrow\mathbb R^n,~v\mapsto t\cdot I_n\cdot v = \begin{pmatrix} t & 0 & \ldots & 0 \\ 0 & t & \ldots & 0 \\ \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & t \end{pmatrix}\cdot v$$ we get that $\mathbb Rv=\operatorname{im}(\varphi_t)$ and thus, as the image of a linear mapping, it is a vector space.
