Let $V$ be an $n$-dimensional real vector space. For $\alpha \in \mathbb{R} \setminus \{0\}$, we define $\Omega^\alpha V$, the space of $\alpha$-densities on $V$, as the set of all maps $\omega : V^n \to \mathbb{R}$ which satisfy $$\omega(Tv_1, \cdots, Tv_n) = |\det(T)|^{\alpha} \omega(v_1, \cdots, v_n), $$ for every $v_1, \cdots, v_n \in V$ and for every invertible linear operator $T : V \to V$.
It is clear that this space is $1$-dimensional real vector space, as if we have a basis $e_1, \cdots, e_n$ of $V$, then $\Omega^\alpha V$ is spanned by $|\varepsilon^1 \wedge \cdots \wedge \varepsilon^n|^\alpha$, where $\varepsilon^1, \cdots, \varepsilon^n$ is the dual basis.
My question is the following: how can we prove that $(\Omega^\alpha V)^* \simeq \Omega^{-\alpha}V$ is a "natural" way, i.e. by constructing a map without firstly choosing a basis for $V$. I do not know how to prove this statement, as I cannot think of any linear functional on $\Omega^\alpha V$ besides the constant functional.
This question arises when considering distributions on a manifold. Another "natural" isomorphism is, for instance, $$\Omega^\alpha V \otimes \Omega^\beta V \simeq \Omega^{\alpha + \beta} V$$ via the map $\omega \otimes \eta \mapsto \omega \cdot \eta$.