 Solution

For the basis $e_1,\dots ,e_n$, define the dual basis byUniqueness of the coordinates makes each $e_i^{\prime }$ well-defined and linear. If ${\sum }_i{a}_i{e}_i^{\prime }=0$, evaluation at $e_j$ gives $a_j=0$, so these linear functionals are linearly independent. For any $\lambda \in V^{\prime }$, evaluation on each basis vector givesBoth sides agree on a basis, hence on every linear combination. Thus they also span the dual space and are a basis of it.