Proving properties of linear maps on one-dimensional vectors An exercise from the book "Linear Algebra Done Right" asks to prove the following: 
'Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\textrm{dim}\ V = 1$ and $T \in L(V, V)$ then there exists $a \in F$ such that $Tv = av$ for all $v \in V$.'
I'm a bit stuck here. My thought is: surely $V$ contains within it a basis for $V$, meaning there's a linear combination of vectors in $V$ that can express each element of the vector space, i.e. $a_1v_1,...,a_nv_n = v$ where $v_1,...,v_n \in V$ and $v \in V$. Now it seems that the key thing is to show that this linear combination only involves one element, i.e. is of form $a_kv_k$ and this must somehow fall out from the constraint that $V$ is one dimensional, but I don't really see it?
The dimension of $V$, $\textrm{dim} V$ is defined as $n$ if the vector space is $R^{n}$ or $F^{n}$ basically.

Without words (almost):
$$\dim_{\Bbb F}V=1\iff V=\text{Span}_{\Bbb F}\{v\}\;,\;\;v\neq 0\iff \forall\,x\in V\;\exists\,\alpha\in\Bbb F\;\;s.t.\;\;x=\alpha v\implies$$
$$Tx=T(\alpha v)=\alpha Tv$$
Now, since $\;Tv=\beta v\;,\;\;\beta\in\Bbb F\;$, substitute above and we're done....
