Does existence of nonzero linear functional depends on axiom of choice?

Question

Given an arbitrary nonzero vector space $V$, is there a nonzero linear functional on $V$, without assuming axiom of choice? I know that by assuming existence of a basis for $V$, we can consider the dual basis for a subspace of $V^*$, which justifies the existence of nonzero linear functional, but this argument fails without axiom of choice. I guess there may not always be a nonzero linear functional, but my knowledge on axiom of choice and infinite-dimensional vector spaces is lacking. A quick search on Google fails to give an answer.

Note that I am not talking about normed spaces or continuous linear functionals, just plain vector spaces with no additional structure.

Answer 1

The answer is negative. In other words, there are models of set theory (without the axiom of choice) for which there are vector spaces $V\neq\{0\}$ such that $V^*=\{0\}$. See here, for instance.