Today, we were showing the following in class:
Let $V$ be a finite-dimensional vector space over $F$. Prove that the bilinear map $$\beta:V \times V^* \to F, \ \ (v,f) \mapsto f(v)$$ is nondegenerate in the second component.
It's a half-liner:
Let $f\in V^*$ such that $\forall v \in V \ (\beta(v,f)=f(v)=0) \implies f=0$.
Our TA made the following remark:
The implication holds because of the properties of the vector space.
I said that I don't see which properties of the vector space were used and how exactly, and asked him to clarify.
He first said about zero-function possibly being non-unique (i.e. due to the remark, we know that $f$ is the zero function) then about some set possibly being empty. I then asked for a counterexample where we don't have any vector space structure. He responded that he is not well versed in functional analysis. I said that I don't see why would we need anything beyond $$g,h:A\to B \ (\forall x \in A:g(x)=h(x)) \iff g=h.$$ To which he responded that it's not always true in measure theory with quotient spaces. At this point I thought that he must have gotten up on the wrong side of the bed and didn't ask any further.
I suspect that, even though he did a poor job defending the remark, he might have gotten it from the professor and maybe somewhat misinterpreted.
Since these issues are a bit tricky sometimes, I've decided to ask you if we need any special justification for concluding $f=0$?
You are right, in the ordinary dual vector space the zero vector is the zero (linear) function, and a function is zero if and only if it is zero in every point of its domain (here $V$). There aren't any properties of $V$ that enter into the validity of this.