TITLE: Why is not a vector space isomorphic to its dual space?
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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.