Exercise $7$, Section $3.A$ - Linear Algebra Done Right Exercise : 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 $T \in L(V, V)$, then there exists $\lambda \in F$ such that $Tv = \lambda v$ for all
$v \in V$.
$L(V,V)$ denotes the set of all linear maps $T : V \to V$. $F$ denotes the field of real or complex numbers.
I might be misinterpreting this question, but the following was my attempt at proving that. I am assuming that $V=\{\lambda v : \lambda \in F,v\in W\}$ where $W$ is another vector space. In other words, $V$ is a subspace of $W$.
Proof: Let $v\in V$. Then $T(v)\in V$ by definition of $T$. Then by definition of $V$, we have that $T(v)=\lambda v$. Thus, there exists a scalar $\lambda \in F$ such that $Tv=\lambda v$.
Am I misunderstanding the question or is this proof correct?
Edit: I will update this post once I find a correct proof.
Edit 2: The following is the correct proof.
Proof: Let $u\in V$ where $u=\lambda v$. Then, $Tu=T(\lambda v)=\lambda T(v)=\lambda \lambda 'v=\lambda'\lambda v=\lambda'u$
Hence, ther exists a scalar $\lambda ' \in F$ such that for any $u\in V$ and every $T \in L(V, V)$ we have that $Tu=\lambda 'u$.
Edit 3: As pointed out in the comments, it is not necessary to use $W$. I was just confusing myself.
 A: Let $w\in V$ be a nonzero vector then $\{w\}$ is the basis of $V$ (because $\dim V=1$).
Since $T$ is a linear operator it follows that $T(w) \in V$ then there is $\lambda \in F$ such that $T(w) = \lambda w$.
Now, let $v \in V$ be arbitrary then there exists $c\in F$ such that $v=cw$. On other hand,
\begin{equation*}
T(v) = T(cw)=cT(w)=c(\lambda w)=\lambda(cw)=\lambda v.
\end{equation*}
A: There is no need nor justification for introducing a vector space $W$ that contains $V$. An important hypothesis (which must be used somewhere) is $\dim(V)=1$ which means that $V$ admits a basis (at least one, but it will not be unique) that has exactly one vector. And the statement to be proved is that there exists a scalar $\lambda$ such that for all vectors $v\in V$ one has $T(v)=\lambda v$. This is not the same as showing (as you appear to do) that for all vectors $v\in V$ there exists a scalar $\lambda$ such that $T(v)=\lambda v$ (which is a weaker statement).
So here is an outline of how your proof could proceed. Since a basis of $V$ with exactly one vector exists, let $[u]$ be such a basis. This means that $u\neq\vec0$ (since a zero vector cannot be member of a basis), and every vector in $V$ can be written (uniquely) as a scalar multiple of $u$, i.e., as $\mu u$ for some scalar$~\mu$. (I avoided the letters $v$ and $\lambda$ since they are used as bound variables in the statement to prove, so using them elsewhere may be confusing.)
Now given a linear $T:V\to V$ find a candidate scalar $\lambda$ for which you claim will hold (there is a bit work to do here, and the chosen basis vector $u$ will be useful to find a candidate.) Now let $v\in V$ be an arbitrary vector; show that $T(v)=\lambda v$ for the proposed candidate; this establishes the claim. (Though not asked, it is also true that $\lambda$ is unique; there is only one candidate that will do. Your argument should make this clear in passing.)
A: Alternative:
Given $\dim V_F=1$ and $T\in\mathcal{L}(V) $ has an eigenvalue, say $\lambda$ (linear polynomial always split!)
Then the eigenspace $E(\lambda, T) =V$
$[$$E(\lambda,T)$ is subspace of $V$ containg a non zero vector$]$
i.e $Tv=\lambda v$ for all $v\in V$ and fixed scalar $\lambda$.
