Advanced Calculus by Loomis & Sternberg Problem 1.13:
Let $V$ be $\mathbb{R^2}$, and let $M$ be the line through the origin with slope $k$. Let $\mathbf{x}$ be any nonzero vector in $M$. Show that $M$ is the subspace $\mathbb{R} \mathbf{x} = \{t \mathbf{x} : t \in \mathbb{R}\}$.
Could you please give me some hints for this problem?
Edit - Managed to solve it using the hints GEdgar gave below. Here is my solution:
Proving $\ \mathbb{R}M \subseteq M$
Because $M$ is the line through the origin with slope $k$, $M$ is the set of all tuples $<a,b>$ that satisfies $b=ka$.
We know that $\mathbf{x}=<x,y>$ is in $M$ so $y=kx$. But then $t\mathbf{x}=<tx, ty>$ $\in M$ since $ty=t(kx)=k(tx)$. Therefore $\mathbb{R}\mathbf{x} \subseteq M $.
Proving $\ M \subseteq \mathbb{R}M $
Since $\mathbf{x}=<x,y>$ is non-zero, either $x \neq 0$ or $y \neq 0$ is true. Also, since $\mathbf{x} \in M$, $y=kx$.
If $x \neq 0$ is true - Take any $\mathbf{u}=<u_1,u_2> \in M$. Then $\ u_2=ku_1$. Since $x \neq 0$, $\ k=y/x$. So $u_2=y * u_1 / x$. Therefore $(u_1/x)*<x,y>$ is equal to $<u_1, u_2>$ so $\mathbf{u} \in \mathbb{R}\mathbf{x}$.
Similarly for $y \neq 0$.
Thus $M \subseteq \mathbb{R}M$. $\square$