Injective linear map is surjective then vector space is finite

Question

We know such conclusion, Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$, If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.

My question is whether the below propositions are true?

(i)Let $V$ be a vector space and if for any $\alpha \in \operatorname{End}(V)$, $\alpha$ is injective then $\alpha$ is surjective, then $V$ is finite dimensional.

(ii)Let $V$ be a vector space and if for any $\alpha \in \operatorname{End}(V)$, $\alpha$ is surjective then $\alpha$ is injective, then $V$ is finite dimensional.

(iii)Let $V$ be a vector space and if for any $\alpha \in \operatorname{End}(V)$, $\alpha$ is surjective if and only if $\alpha$ is injective, then $V$ is finite dimensional.

Answer 1

Yes, all three statements hold, if you accept the axiom of choice. To prove this, it suffices to find a surjective but not injective map and an injective but not surjective map for each infinite-dimensional vector space $V$. Let $\{v_i\}_{i \in I}$ be a basis of $V$. Pick a countable subset $J \cong \mathbb{N_0}$ of $I$. Then define the maps

$$ \phi \colon V \to V, v_i \mapsto \begin{cases} v_i \quad \ \ \ i \notin J \\ v_{i+1} \quad i \in J \end{cases}$$

$$ \psi \colon V \to V, v_i \mapsto \begin{cases} v_i \ \ \ \quad i \notin J \text{ or } i=0\\ v_{i-1} \quad i \in J\backslash \{0\} \end{cases}$$

Then $\phi$ is injective but not surjective and $\psi$ is surjective but not injective.

Note: the notation is a bit sloppy. $J$ is litterally identified with $\mathbb{N}_0$; also, $\phi$ and $\psi$ are only defined on the basis (but this is enough to have unique, well-defined linear maps).

Answer 2

Yes, they are all true. I would express this as follows: let $V$ be a vector space. Then the following conditions are equivalent:

1. $V$ is finite-dimensional;
2. every injective linear endomorphism of $V$ is surjective;
3. every surjective linear endomorphism of $V$ is injective.