# Why is not a vector space isomorphic to its dual space?

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.

• $V$ and $V^*$ have the same dimension, they're therefore isomorphic. – Gabriel Romon Oct 18 '14 at 16:01
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.