The following question is taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes
Exercise: Let $X$ be a vector space. For $x,y\in X$ define $x\leq y$ to mean there exists $\lambda \geq 1$ with with $\lambda \cdot x=y.$ Show that $(X,\leq)$ is a poset. Show that, when vector spaces are considered as posets in this way, linear maps are order preserving.
I am trying to show that $(X,\leq)$ is a poset.
For the case of reflexivity, we have $\lambda \cdot x=x.$ iff $\lambda = 1$
For transitivity, suppose $\lambda_1 \cdot x=y.$ and $\lambda_2 \cdot y=z,$ for some $\lambda_1,\lambda_2\geq 1,$ then $\lambda_2\lambda_1 \cdot x=\lambda_2 \cdot y=z.$ Hence $\lambda_2\lambda_1 \cdot x =z.$ where $\lambda_2\lambda_1\geq 1.$
For antisymmetry, suppose $\lambda_1\cdot x=y$ and $\lambda_2\cdot y=x,$ then $\lambda_2\lambda_1 \cdot x=\lambda_2 \cdot y=x$ and $\lambda_1\lambda_2 \cdot y=\lambda_2 \cdot x=y.$ Since $\lambda_1\lambda_2=\lambda_2\lambda_1=1$, and both $\lambda_1, \lambda_2 \geq 1$, then $\lambda_1=\lambda_2=1.$ So let $\lambda=\lambda_1=\lambda_2\geq 1.$ We then have ${\lambda}^2 \cdot x=x,{\lambda}^2 \cdot y=y.$ and we get ${\lambda}^2=1$ and hence $\lambda=1$. But $\lambda_1 \cdot x=\lambda_2 \cdot y$ and so $x=y.$