Origin of the term dual space? Basically, why is a dual vector space called as such? Is the reason for the term "dual" simply because the two vector spaces are related by a one-to-one mapping, or is there something more to it?
Sorry for such a basic question, but I've never been able to find a definitive answer so far.
 A: In one of the comments to the original question, Will says:
"@GiuseppeNegro In Terence Tao's preamble [reference to this blog entry] what does he mean by "A vector space V over a field F can be described either by the set of vectors inside V, or dually by the set of linear functionals λ:V→F from V to the field F (or equivalently, the set of vectors inside the dual space V∗)"? Is it meant that you can either describe a vector in terms of a given basis for V or by linear maps which provide a way of determining it's components with respect to a given basis for V?"
Here's my answer. 
"Of course: if $V$ is finite dimensional then you can interpret that statement in terms of bases, like you did. But the principle is more general, as you can also give a more intrinsic interpretation that does not use bases and coordinates (actually it does, but in a silent way). 
Namely, let $v, w\in V$ be vectors. Assume that $f(v)=f(w)$ for all dual vectors (or linear functionals) $f$. Then (claim) $v=w$. 
Proof. Let $x=v-w$. We need to show that $x=0$. Suppose not. Then there exist a basis of $V$ that contains $x$:
$$\left\{x, e_\alpha\ :\ \alpha\in A\right\}$$
Here $A$ is some index set, possibly infinite (in which case, the existence of such a basis is a consequence of the axiom of choice). So any vector in $V$ is uniquely described by a sum $$\lambda x+\sum_{\alpha\in A}\lambda_\alpha e_\alpha, $$ where $\lambda_\alpha\ne 0$ for at most a finite number of indices $\alpha\in A$. (So that sum over $A$ is actually a finite sum).  Define a linear functional $$f\left(\lambda x+ \sum_{\alpha\in A}\lambda_\alpha e_\alpha\right)=\lambda.$$
This functional has the property that $f(x)=1$. But our assumptions give that $f(x)=0$. This is a contradiction, therefore $x=0$, as claimed. $\square$
This statement says if you know how all linear functionals act on a vector, then you know the vector itself with no possible ambiguity. It is a general principle that is widely used in many fields of mathematics, as far as I know."
