When are $V$ and $V^{*}$ canonically isomorphic?

Question

Generally speaking, for a finite dimensional vector space $V$, $V$ and $V^*$ are not canonically isomorphic after we fix a basis for $V$. One of my questions is:

If we do not fix a basis for $V$, can we say that $V$ and $V^*$ are canonical isomorphic?

For example, the following is a special case.

Let $L$ be a finite dimensional vector space over a field $K$, and let $V:=\text{Hom}(L,L)$. We know that $$\text{Hom}(L,L)\cong L^*\otimes L\cong L^*\otimes L^{**}\cong(L\otimes L^*)^*\cong\text{Hom}(L,L)^*,$$ and the above isomorphisms are canonical, so $V\cong V^*$ canonicaly.

The other question is:

Are there other vector spaces $V$ such that $V$ and $V^*$ are canonically isomorphic? Or, is the above the only case such that $V$ and $V^*$ are canonically isomorphic?

Answer 1

An isomorphism $V\cong V^\star$ is the same thing as a non-degenerate bilinear form $(-,-)\colon V\times V\to k$. So a vector space is canonically equivalent to its dual exactly if it comes equipped with a non-degenerate bilinear form.

In your example $\operatorname{Hom}(L,L)$ has the bilinear form $(f,g)=\operatorname{tr}(fg)$.

Answer 2

This is the same as Tashi Walde's answer, but perhaps more concrete.

If $(\cdot,\cdot)$ is an inner product on $V$, then the isomorphism is given by $\phi:V \to V^*$, where $\phi(v) \mapsto (v, -)$, where $(v,-)$ is a linear functional given by $(v,-)(w)=(v,w)$.

Indeed, this is an isomorphism, since if the functional $(x,-)$ is zero, then $(x,x)=0$, which implies that $x=0$. Of course, injectivity suffices for vector spaces of the same dimension.

$\phi$ (as previously mentioned) can be identified with a bilinear map $V \times V \to k$ via a "currying argument."