 Solution

The dual map is defined by ${\alpha }^{\ast }(\lambda )=\lambda \circ \alpha$. Suppose ${\alpha }^{\ast }=0$. Then $\lambda (\alpha u)=0$ for every $u\in U$ and every $\lambda \in V^{\ast }$. The dual space separates vectors: a nonzero vector can be included in a basis and a coordinate functional takes value one on it. Thus $\alpha u=0$ for every $u$, and $\alpha =0$. This proves injectivity of $\Phi$.

Its source and target have the same finite dimension:An injective linear map between equal finite dimensions is surjective by the rank-nullity theorem. Therefore $\Phi$ is an isomorphism. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.