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If we have a vector space $V$ this has a dual $V^*$. When a learned about dual spaces we were told it has the same dimension as the vector space and I think the lecturer said they were unique (if not the notation of just using * seems a bit vague).

Then I noticed something. Say $V$ is a function space, with a set of basis functions $e_i (x)$. Then I can create a range of dual vector spaces:

1) $$ \int e_i(x) w_1 (x) ● $$

2) $$ \int e_i(x) w_2(x) ● $$

Where $w_1\ne w_2$. We now either have 2 dual spaces or the dual space is twice as big as the origional space.

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  • $\begingroup$ They have the same dimension only if they are finitely-dimensional (in fact it is iff, but whatever). $\endgroup$ – Kenny Lau Oct 13 '17 at 9:47
  • $\begingroup$ In your examples, are the two duals not isomorphic? $\endgroup$ – Kenny Lau Oct 13 '17 at 9:48
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    $\begingroup$ By definition the dual space of a $k$-vector space $V$ is the set of linear maps $V\to k$, equipped with the obvious structure of $k$-vector space. How could this definition leave space for non-uniqueness? $\endgroup$ – Hagen von Eitzen Oct 13 '17 at 9:52
  • $\begingroup$ What's a range of dual vector spaces? What are $w_1$ and $w_2$, and how are you integrating? $\endgroup$ – B. Pasternak Oct 13 '17 at 9:59
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The dual of a vector space $V$ is defined as $$V^*:=\hom(V,k)\ ,$$ the vector space of linear maps from $V$ to the base field $k$. It is not an existence statement or something similar, it is a definition, and as such it is unique.

For the dimension statement, if $V$ is finite dimensional, then $V^*$ has the same dimension as $V$. Even then, $V$ and $V^*$ are not canonically isomorphic. In general, you have that $$\dim V^*\ge\dim V\ .$$ It is a good exercise (if possibly a bit hard for you, if I gauge your level correctly) to prove that the dual of any vector space cannot have countable dimension.

If you have more structure (e.g. topological vector spaces, Banach spaces and Hilbert spaces), then there is a notion of continuous dual, again denoted by $V^*$. There is an important theorem (called the Riesz representation theorem) saying that if $V$ is a Hilbert space, then $V$ is canonically isomorphic to its continuous dual $V^*$.

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  • $\begingroup$ "prove that the dual of any vector space cannot have countable dimension". Do you mean the dual of any infinite-dimensional vector space cannot have countable dimension? $\endgroup$ – Joppy Oct 13 '17 at 11:15
  • $\begingroup$ @Joppy By countable dimension, I mean a base of cardinality $\aleph_0:=|\mathbb{N}|$. So the dual of any vector space cannot have countable dimension. $\endgroup$ – Daniel Robert-Nicoud Oct 13 '17 at 11:32
  • $\begingroup$ Ah, ok. The definition of "countable" I know is a set that has an injection into $\mathbb{N}$, which is either finite or has cardinality $|\mathbb{N}|$. $\endgroup$ – Joppy Oct 13 '17 at 13:04
  • $\begingroup$ @Joppy I usually say "at most countable" for that notion. $\endgroup$ – Daniel Robert-Nicoud Oct 13 '17 at 13:33
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The dual space is unique. To demonstrate this, let us formally define some linear maps from the original vector space $V$ with elements $\{u, v,\ldots\}$ to the real line $\mathscr{R}$:

$f: V \to \mathscr{R}$ with $u \mapsto f(u):=\int_0^1 u(x)w_1(x)dx$.

$g: V \to \mathscr{R}$ with $u \mapsto g(u):=\int_2^\pi u(x)w_2(x)dx$.

$h: V \to \mathscr{R}$ with $u \mapsto h(u):=\frac{\mathrm{d}u(x)}{\mathrm{d}x}$ evaluated at $x=5$. … etc.

All these maps are linear in their arguments and are "well-defined" in the sense that if $u=v$ implies $f(u)=f(v)$, $g(u)=g(v)$, $h(u)=h(v)$ etc. Each map leads from one element $v$ to one element in $\mathscr{R}$.
The SET of maps $\{f, g, h,\ldots\}$ (and all kind of linear maps you can think of...) are the element of the dual space. Moreover, the fact that you collect ALL POSSIBLE maps guarantees that this collection is unique! In other words: you do not make any choice at all. Nobody can do this in a different way.

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    $\begingroup$ Please use $Latex$ to format math. See here please. $\endgroup$ – tarit goswami Apr 2 at 14:52

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