= Solution
The <dual map> is defined by $\alpha^*(\lambda)=\lambda\circ\alpha$. Suppose $\alpha^*=0$. Then $\lambda(\alpha u)=0$ for every $u\in U$ and every $\lambda\in V^*$. 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:
$$
\dim L(U,V)=(\dim U)(\dim V)=\dim L(V^*,U^*).
$$
An injective <linear map> between equal finite dimensions is <surjective> by the <rank-nullity theorem>. Therefore \b[$\Phi$ is an <isomorphism>]. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.
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