 Solution

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$, with pointwise operations. For the given basis define coordinate forms $e^i$ by $e^i({\sum }_j{a}_j{e}_j)=a_i$. These are linear and satisfy $e^i(e_j)={\delta }_{ij}$. If ${\sum }_i{c}_i{e}^i=0$, evaluation at $e_j$ gives $c_j=0$. For every linear form $\varphi$,so the coordinate forms span the dual and form its dual basis.

In the polynomial example write $p(X)=a+bX+c{X}^2$. ThenThese recover all three coefficients, so they span $V^{\ast }$ and, since the dual has dimension three, form a basis. Explicitly the basis dual to $1,X,X^2$ isEvaluation on the three monomials verifies the dual-basis identities directly.