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Let $V$ be a finitely generated vector space with a basis $\mathcal{B}=\{\alpha_1,\cdots,\alpha_n\}$ and let $\mathcal{B}^*= \{f_1,\cdots,f_n\}$ be the dual basis of $\mathcal{B}$.

In this situation, I defined a function $T:V\to V^*$ with $T(\alpha_i)=f_i$. I think this function is linear and bijective, thus an isomorphism.

Please disprove my claim.

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    $\begingroup$ $V$ and $V^*$ have the same dimension, they're therefore isomorphic. $\endgroup$ – Gabriel Romon Oct 18 '14 at 16:01
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    $\begingroup$ For finite dimensional vector spaces the space is isomorphic to its dual. However, there is no "canonical" isomorphism (one that does not depend on the choice of a base); while there is a "canonical" isomorphism from the space to its second dual. You might have mixed up something. $\endgroup$ – quid Oct 18 '14 at 16:05
  • $\begingroup$ Essentially the dot product is an isomorphism from a vector space to its dual space - i.e. we can characterize all dot products as, for some map T, being T(a)b where the function T(a) is applied to the vector b. So, there's a natural isomorphism only in inner product spaces; not necessarily in vector spaces. $\endgroup$ – Milo Brandt Oct 18 '14 at 16:16
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Any two vector spaces with the same dimension are isomorphic as vector spaces, but there are many isomorphisms between them (choose a basis of one, and a basis of the other, and map the first basis to the second anyway you like). However, these isomorphisms all depend on a choice of basis.

So, if $V$ is finite dimensional, it is has the same dimension as $V^*$, so as vector spaces they are isomorphic.

As quid mentions in the comments, a finite-dimensional vector space $V$ is canonically isomorphic to its double-dual via $v\mapsto \hat{v}$ where $\hat{v}(f)=f(v)$ for all $f\in V^*$. Notice this is a vector space isomorphism independent of any choice of basis.

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