A finite dimensional vector space that is not naturally isomorphic to its dual.

Question

I need an example of a finite dimensional vector space $V$ that is not naturally isomorphic to $V^\ast$.

I know that, at least in finite dimensional case, there is a one-to-one correspondence between natural isomorphisms of $V$ to $V^*$ and nondegenerate bilinear forms on $V$; so for the above example it is enough to show a vector space without nondegenerate bilinear forms.

Thanks in advance.

Edit: Clearly $V$ and $V^\ast$ are always isomorphic, but I need an isomorphism independent from a choice of a basis.

Answer 1

Hint: give up your search: $\dim(V)=\dim(V^{\ast})$.

Answer 2

People may well disagree with this, but let me essay an answer. Consider a line $L$ in the Euclidean plane. It makes sense to say whether a function $f\colon L\to\mathbb R$ is a polynomial function, because once you choose a coordinatization of $L$, such a function will be a polynomial in the single coordinate on $L$. Equally, it makes sense to speak of the polynomial functions of degree $<n$ on $L$, and we know that this is an $n$-dimensional real vector space. Since our space doesn’t have a natural basis, it’s not clear to me that there is any natural isomorphism with its dual space.

Answer 3

Maybe I don't understand your question, but a finite dimensional vector space always has nondegenerate bilinear forms, hasn't it?

I mean: chose a basis of your vector space and you have an isomorphism $V \cong \mathbf{k}^n$, right?

Then, consider the bilinear form on $\mathbf{k}^n$ with associated matrix the identity -or whatever symmetric matrix $A$ you prefer with non-zero determinant:

$$ \varphi(X,Y) = X^tAY \ . $$

And you have defined a nondegenerate bilinear form on $V$.

Am I missing something?