= Solution
{c}
For the <basis> $e_1,\ldots,e_n$, define the <dual basis> by
$$
\boxed{e_i'(e_j)=\delta_{ij},\qquad e_i'\Bigl(\sum_j x_je_j\Bigr)=x_i.}
$$
Uniqueness of the <coordinates> makes each $e_i'$ well-defined and linear. If $\sum_i a_ie_i'=0$, evaluation at $e_j$ gives $a_j=0$, so these <linear functionals> are <linearly independent>. For any $\lambda\in V'$, evaluation on each <basis vector> gives
$$
\lambda=\sum_i\lambda(e_i)e_i'.
$$
Both sides agree on a <basis>, hence on every <linear combination>. Thus they also <span> the <dual space> and are a <basis> of it.
Back to article page
