Let $V,W$ be (not necessarily finite-dimensional) vector spaces over the field $\mathbb F$ and $A\in L(V,W)$ s.t. $\operatorname{rank} A=1$. Prove that $\exists f\in V^*$ and $w\in W$ s.t. $$A(v)=f(v)w\;\forall v\in V$$
My attempt:
$$\operatorname{rank} A=1\iff \dim A(V)=1\implies A(V)=\operatorname{span}\lbrace w\rbrace$$
Case : $\dim V=n\lt\infty$
$V^*=L(V,\mathbb F)\implies \dim V^*=\dim V\dim\mathbb F=n$
We could define linear forms $E_{1j}$, or just $e_j^*$, $$e_j^*(e_k)=\begin{cases}0, &k\ne j\\1,&k=j\end{cases}$$
So, $\left\{e_1^*,e_2^*,\ldots,e_n^*\right\}$ could be a dual basis. Since $e_j^*$ is linear: $e_j^*\left(\lambda e_j\right)=\lambda e_j^*\left(e_j\right)=\lambda\in\mathbb F$
$\implies\exists f\in V^*$ s.t. $f=e_j\left(\lambda e_k\right)$ and $f(V)=\mathbb F$
But this isn't proof and it doesn't cover an infinite-dimensional $V$.
I've seen that for $V,\dim V=\infty,\;\dim V^*=2^{\dim V}$ and considered what I've read on the intuition behind the dual space, transpose of compositions of functions and here, but we haven't reached that.
I don't believe all those threads are relevant for this particular task, but I' m going to leave them here just in case.
May I ask for advice on solving this task? Thank you in advance!
You know that there exists $w \in W$ such that every $A(v)$ is of the form $f(v) w$ for a (necessarily unique) $f(v) \in \mathbb F$.
All you need to show is that $f : V \to \mathbb F$ is a linear map, which follows from the linearity of $A$. (Check this!)