Let $V$ be a vector space over a field $F$, and let $T: V \rightarrow V$ be a linear transformation. Define the following two sets $$ A = \{W \colon W \subseteq V \text{ is a 1-dimensional subspace of $V$ satisfying $T(W) \subseteq W$} \} $$ and $$ B = \{\lambda \in F \colon \text{ $\lambda$ is an eigenvalue of $T$} \}. $$ I would like to define a function $f: A \rightarrow B$ as follows. Given $W \in A$, choose any non-zero vector $w \in W$, and define $f(W) = \lambda$, where $\lambda \in F$ is such that $T(w) = \lambda w$. I have two questions.
Does my definition of $f$ require the Axiom of Choice?
If my description requires it, can the same function be described in a way that does not require the Axiom of Choice?
(Admission 1: I have no hesitation assuming the Axiom of Choice, I just don't feel like it should be required here. Admission 2: I haven't thought much about the Axiom of Choice, so this could be a very misguided question.)
Overall my intuition is that since every choice of non-zero $w \in W$ results in the same output $\lambda$, that should help us, but I get more confused the more I think about it.
No, this does not ultimately require choice. Consider the following slight modification of your definition:
(The "nonzero" there is of course unnecessary, but it's useful to include it to demonstrate the general pattern.)
This doesn't invoke choice at all. The point is that when the outcome of a process is independent of a choice made along the way, we can "de-choicify" by asking about the universal behavior.
(Note that we could also go with "for some" - but then the nonzero-ness requirement would not be redundant. In general I prefer the "for every" approach. That said, sometimes - e.g. in computability theory - we want both simultaneously.)