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I have seen similar questions on s.e. but I really can't seem to understand the proofs given.

So the question is the following:

Let $V$ be a finite-dimensional vector space. show that there is an isomorphism between $V$ and $V^{**}$ where $V^{*}$ is the dual space of $V$.

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  • $\begingroup$ Assume I do not allow you to pick a base in either of th espaces involved. Can you write down any nonzero linear map from $V$ to $V^{**}$? $\endgroup$ – Hagen von Eitzen Nov 18 '15 at 18:03
  • $\begingroup$ Can you show that the dimension of $V^*$ should be the same as the dimension of $V$? $\endgroup$ – Omnomnomnom Nov 18 '15 at 18:04
  • $\begingroup$ Surely a multicate. $\endgroup$ – Martin Brandenburg Nov 18 '15 at 20:38
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Hint. Elements of $V^*$ are linear maps $V\to k$. Elements of $V^{**}$ are linear maps $V^*\to k$. We want a nice linear map from $V\to V^{**}$. So assume we are given $v\in V$. We need to assign to each $f\in V^*$ a value in $k$. Recall that $f$ assigns a value in $k$ to each vector in $V$. So given $v\in V$, what is the only natural choice to assign a value in $k$ to any $f$?

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  • $\begingroup$ Would the value be f(k)? $\endgroup$ – Mathphilo Nov 18 '15 at 18:24
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    $\begingroup$ @Mathphilo Indeed. Now go ahead and show that this map (that can be somewhat confusingly written as $v \mapsto (f\mapsto f(v))$) is injective. Then use that $\dim V^{**}=\dim V^*=\dim V$ to show that it is in fact an isomorphism for finite-dimensional $V$. $\endgroup$ – Hagen von Eitzen Nov 18 '15 at 18:34
  • $\begingroup$ Why does the fact that dimV**=dimV*=dimV imply that the map is surjective? $\endgroup$ – Mathphilo Nov 18 '15 at 19:46
  • $\begingroup$ @Mathphilo I would assume that, it would mean that the basis $B$ of $V$ would have to map one to one with the basis $B^{**}$ of $V^{**}$. Which would give you the subjection. Since any $v^{**}\in V^{**}$ can be written as a sum of elements in $B^{**}$ which in turn can be written as a set of mapped elements from $B$. $\endgroup$ – user160110 Nov 18 '15 at 20:50
  • $\begingroup$ @Mathphilo The image under an injective linear map has the same dimension as the domain. $\endgroup$ – Hagen von Eitzen Nov 18 '15 at 22:34
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$V \simeq V^{**} $ by $ v \mapsto e_v$ where $e_v (f) = f(v)$ $\forall f \in V^*$.

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Show the isomorphism of a vector space and its dual then, by transitity, show what you want. Hint: the vector space and its dual will have same dimension

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  • $\begingroup$ Well, it is true, that you can reason by dimension, but not specifying the existence of a canonical isomorphism between $V$ and $V^{**}$ might be reductive. $\endgroup$ – user228113 Nov 18 '15 at 18:29
  • $\begingroup$ Well, when he finds the canonical isomorphism between the vector space and its dual, using transitivity he finds the explicit isomorphism wanted. The hint is to give an idea on what the first isomorphism could be. $\endgroup$ – Shoutre Nov 18 '15 at 18:53
  • $\begingroup$ There exists no canonical isomorphism between $V$ and $V^*$. $\endgroup$ – user228113 Nov 18 '15 at 20:57

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