Problem
Show that every linear map from a 1-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim V = 1$ and $T \in \mathcal{L}(V,V)$, then there exists $\lambda \in \mathbb{F}$ such that $Tv = \lambda v$ for all $v \in V$
Solution
Let $u \in V$ be non-zero, then the list $u$ is a basis for $V$ as per Example 2.18 and theorem 2.39. Therefore any $v \in V$ is in $\textrm{span}(u)$ because a basis is a spanning list by definition 2.27 or equivalentely, for all $v \in V$ there exist an $\alpha \in \mathbb{F}$ such that $v = \alpha u$.
As $Tu \in V$ we must therefore have $Tu = \lambda u$ for some $\lambda \in \mathbb{F}$ and therefore $$Tv = T(\alpha u) \stackrel{D3.2}{=} \alpha (Tu) = \alpha (\lambda u) \stackrel{D1.19}{=} (\alpha \lambda) u$$ $$\stackrel{E1.4}{=} (\lambda \alpha) u \stackrel{D1.19}{=} \lambda (\alpha u) = \lambda v$$ for all $v \in V$.