Proving that □A → A is valid iff R is reflexive?

Question

How to prove, in modal logic, that \square $A\to A$ is valid iff $R$ is reflexive? (shouldn't this be the definition of $T$ axiom in modal logic)?

(NOTE: The question was edited because I made a mistake by misplacing the necessity sign. Former formulation does not make sense at all)

Answer 1

$T$ axiom is not defined as such, instead it's defined in most modal logics the othere way around as below:

T, Reflexivity Axiom: □p → p (If p is necessary, then p is the case.)

T holds in most but not all modal logics. Zeman (1973) describes a few exceptions, such as $S1^0$.

$A \to \square A$ is an inference rule (not an axiom) in the weakest normal modal logic $K$ only when $A$ is a theorem:

Necessitation rule: $\vdash A$ implies $\vdash \square A$.

Obviously in general if A is a mere contingent proposition, we cannot assert A must be necessarily true in our world even if our world is reflexive normal modal system $T$ accessible to itself.