Linear Maps on a $1$ Dimensional Vector Space . Is the the following Proof Correct?
Given a vector space $V$ such that $\dim V=1$ and $T\in\mathcal{L}(V,V)$ where $\mathcal{L}=\{H|H:V\to V\}$ then there exists $\lambda\in\mathbf{F}$ such that $Tv=\lambda v$ $\forall v\in V$.
Proof. Assume that $T\in\mathcal{L}(V,V)$ let $w$ be some vector in $V$ such that $w\neq 0$, evidently $w$ is basis for $V$ consequently 
$$\forall u\in V\exists\tag{1}\lambda\in \mathbf{F}(u=\lambda w)$$
Now let $v$ be an arbitrary vector in $V$ since $T:V\to V$ it follows that $Tv\in V$ thus by $(1)$ it follows that for some $\beta\in \mathbf{F}$. $Tv=\beta w$
$\blacksquare$
 A: No, you haven't shown what was asked. You were asked to show that there exists $\lambda \in \mathbb{F}$ such that for all $v \in V$ we have $Tv = \lambda v$ (note the order of quantifiers). In particular, $\lambda$ shouldn't depend on $v$.
You have shown instead that for every $v \in V$ there exists $\beta \in \mathbb{F}$ (which possibly depends on $v$) such that $Tv = \beta w$. This is not what was asked. Note that your proof doesn't use the linearity of $T$ and without the linearity of $T$, the result is not true.
A: Yep, looks fine to me. Alternatively you could have written the $1 \times 1$ matrix representing the linear transformation and concluded directly from there.
A: You have to be careful since in (1) the scalar $\lambda$ depends on $v$, but in your statement, it doesn't.
The ansatz to choose a basis $w\neq 0$ of $V$ is very good. Now use that $T$ is linear!

Define $u:=Tw$. There exists $\lambda\in\mathbb F$ such that $u=\lambda w$. (Consider that $\lambda$ depends on $u$ and $w$!)We get $Tw=\lambda w$. Now let be $v\in V$ arbitrary and $\mu\in\mathbb F$ such that $v=\mu w$. (Here $\mu$ depends on $v$ and $w$)Since $T$ is linear, we conclude $$Tv=T(\mu w)=\mu Tw=\mu\lambda w=\lambda\mu w=\lambda v.$$Since $\lambda$ depends on $w$ and $u=Tw$ but not on $v$ we are done.

