Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 3 / a / i / Solution

The dual basis is the collection of linear functionals characterized byExplicitly, $e_i^{\prime }({\sum }_j{\alpha }_j{e}_j)={\alpha }_i$. Every linear functional $\ell \in V^{\prime }$ satisfies $\ell ={\sum }_i\ell (e_i)e_i^{\prime }$, proving that the displayed functionals span the dual space. Evaluating a zero linear combination at each $e_j$ makes all its coefficients zero, so they are also linearly independent. Thus they form a basis of $V^{\prime }$.