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My naive idea of an expression is that it returns (or denotes) exactly one object (be that may a class or a set), but of course there are multiple vector spaces being isomorphic to one another, thus there are multiple vector spaces being the dual of $V$.

The question is what do we have to say when studying vector spaces to erase ambiguity?

Should I just use the predicate "$W$ being a dual vector space of $V$", or just scratch a "here we only care about the vector space structures" on the margin?

Edit: I realize now how stupid the question is, what I have in my mind is how when we prove theorems concerning only the vector space structure, the result of that theorem can safely be applied to isomorphic vector spaces, thus isomorphism behaving as a kind of equality.

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    $\begingroup$ The dual of $V$ is a specific vector space denoted $V^*$. There are vector spaces isomorphic to $V^*$, but they aren't $V^*$. $\endgroup$
    – Michael Albanese
    Commented Jul 29, 2016 at 14:14
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    $\begingroup$ There are multiple vector spaces isomorphic to $V^{\ast}$ (including $V$ itself!). I think we justify "the dual" by describing the dual map as a contravariant functor on the category of vector spaces, and $V^{\ast}$ being the image of an object under this functor. $\endgroup$
    – Moya
    Commented Jul 29, 2016 at 14:14
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    $\begingroup$ Why downvotes with no comment? This seems to me a reasonable question from someone learning to work with abstract mathematics. $\endgroup$
    – Ethan Bolker
    Commented Jul 29, 2016 at 14:16

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The dual space of a $k$-vector field $V$ is defined as the vector space $$V^* := \hom(V,k).$$ Here, $\hom$ means the space of linear functions (from the first argument to the second one).

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    $\begingroup$ Correct, of course - but perhaps not instructive for someone starting out in linear algebra, who may not have seen "hom". $\endgroup$
    – Ethan Bolker
    Commented Jul 29, 2016 at 14:17
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    $\begingroup$ @EthanBolker But we may certainly assume they have seen the (or: a?) definition of dual vector space ... $\endgroup$
    – Hagen von Eitzen
    Commented Jul 29, 2016 at 16:45
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There is just one dual space of $V$: the space of linear functionals. We name that one $V^*$. There are other spaces isomorphic to $V^*$. If you need one of those other spaces $X$ in a proof you show/say that $X$ is isomorphic to $V^*$.

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  • $\begingroup$ In the presence of a dual pairing $V\otimes W\to k$, we could call $V$ and $W$ dual to each other, but wouldn't necessarily call them each other's dual spaces. $\endgroup$
    – anon
    Commented Jul 29, 2016 at 16:51
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The dual space is the vector space of all linear functions $V\to K$ where $K$ is the field of scalars of $V$ (note that $K$ is part of the definition of the vector space $V$, so there's only one $K$ for the given $V$). There is only one such space. If $W$ is isomorphic to $V$, then $W^*$ is isomorphic to $V^*$, but if $W$ is not $V$, then $W^*$ is not $V^*$.

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