# 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.

• I would guess that most names in math don't come from a committee sitting down for years discussing what the most insightful name would be -- some bloke just chooses a name which seems appropriate. So I'd guess it's called the dual space because it forms a pair with the space it's the dual of -- and "dual" means "$2$". – user137731 May 23 '15 at 18:13
• Typically, the term dual is applied when the dual of the dual is the original object. Wikipedia has a host of examples. – Rahul May 23 '15 at 18:13
• one-to-one but no natural choice of such a mapping. If you put a positive definite inner product on the original space, this provides a specific mapping, given $v$ we define $v^ast$ by $v^\ast(w) = \langle v,w \rangle$ – Will Jagy May 23 '15 at 18:14
• Usually, there is no one-to-one mapping between a vector space and it's dual... And if you're talking about algebrical dual (and not topological dual), the cardinal of the dual is strictly greater than the cardinal of the "base" vector space for infinite vector spaces. – Tryss May 23 '15 at 18:29
• The preamble to this blog entry of T.Tao contains his view on the term "duality". HTH – Giuseppe Negro May 23 '15 at 18:55

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?"

"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."

• So is the concept of duality in this sense is that they are connected to one another in that if you know the dual space then you know the original vector space and vice versa?! – Will May 23 '15 at 21:45
• Maybe that's true, I don't know. However, I hear that, in the case of topological vector spaces, there exist spaces that are not isomorphic yet they have the same dual space. Those are technicalities for the specialists, that's not the point of the present discussion: I just want to emphasize that dual spaces are not there to classify vector spaces. It is the pairing $V^\star\times V\to K$ that is interesting. You can choose to work on vectors $v\in V$ directly, or to work on the action of $V^\star$ through this pairing. – Giuseppe Negro May 23 '15 at 21:57
• "You can choose to work on vectors v∈V directly, or to work on the action of V⋆ through this pairing. " this is the part I'm confused about. Is it that their pairing maps to the field $K$ over which the vector field $V$ is defined? – Will May 23 '15 at 22:24
• @Will: Yes, $K$ is the scalar field. – Giuseppe Negro May 23 '15 at 22:36
• Does the duality follow from the fact that we can consider dual vectors as linear functionals acting on vectors, but we can equally consider vectors as linear functionals acting on dual vectors (as the double dual is naturally isomorphic to the "original" vector space)? – Will May 25 '15 at 9:48