Prove that there exists $\lambda \in F$ such that for any $u\in U$, $T(u)=\lambda u$ Let $U$ be a one-dimensional vector space over a field $F$ and let $T : U \to U$ be a linear map.
Prove that there exists $\lambda\in F$ such that for any $u\in U$, $T(u)=\lambda u$.
How do I go about proving this?
 A: Since $U$ is one-dimensional, it has a basis consisting of just one vector, say $e$. Every vector in $U$ is a scalar multiple of $e$, so in particular $T(e)$ is a scalar multiple of $e$. Now you can use linearity of $T$.
A: As $U$ is one-dimensional, any two vectors are linearly dependent. So are $v$ and $T(v)$. $T(v)$ is a scaled version of $v$. Lets say it is scaled by $\lambda$. Now take any vector $w\in U$. For the same reason $v$ and $w$ are scaled versions of each other, say $w=\alpha v$. Then
$$T(w)=T(\alpha v)=\alpha T(v)=\alpha\cdot\lambda v=\lambda \cdot\alpha v=\lambda w.$$
This works for any vector $w$, and we are done.
A: Let $e$ be a base of the 1-dim. vector space $U$, which exists by definition of a 1-dimensional vector space. Then, $\exists\lambda$ such that $T(e)=\lambda e$, again from the definition of base of a 1-dimensional vector space. Now let $v$ be any other vector of the space. Then, $v=ce$ for some $c\in F$. From linearity you have $T(v)=T(ce)=cT(e)=c\lambda e=\lambda ce=\lambda v$. Done.
