Proof about linear functionals without reference to a basis. Recall the following: Let $V$ be a vector space, and let $v$ be in $V$.  If $\lambda(v) = 0$ for all $\lambda \in V^*$, then $v = 0$.
Does anyone know of a proof that makes no reference to a basis?  And if not, is there reasonable intuition behind why a choice of basis is necessary in proving the result?
Thanks in advance.
 A: It is known that the axiom of choice (AC) is equivalent to the statement that all vector spaces have a basis, and that these bases cannot be "written down" in general. If $V$ has a basis, then your statement, which says that the canonical map $e : V \to V^{**}$ is injective, is clearly true; but actually something weaker will suffice, namely that all $1$-dimensional subspaces are direct summands. This statement follows from AC, but is probably weaker than AC, and cannot be proven in ZF. That is, there are models in ZF where there a vector spaces $V$ such that $e : V \to V^{**}$ is not injective.
My intuition is that infinite-dimensional vector spaces are no 'reasonable' spaces to study because their properties often depend on set theory. Notice that $V \to V^{**}$ is an isomorphism for finitely generated vector spaces $V$, and that you can prove this in ZF. Also, Banach spaces and Hilbert spaces are often far more well-behaved than infinite-dimensional vector spaces, basically because we can make sense of infinite sums.
A: By the Riesz Representation Theorem: corresponding to each $\lambda \in V^*$ is an element $u \in V^{**} \supseteq V$ such that for any $x \in V$ one has $\langle x,u\rangle = \lambda(x)$.
So, we choose $\lambda_v \in V^*$ such that for any $x \in V$ one has $\lambda_v(x) = \langle x,v\rangle$. By the statement in your question, it follows that
$$0 = \lambda_v(v) = \langle v,v \rangle = \|v\|^2$$
implying that $v \equiv 0$ by norm positivity. 
(For a proof of the Riesz Representation Theorem, please see Folland's Real Analysis pg.212 or 223 for a version which does not rely on a basis for the space)
