What are some nice examples to illustrate that a basis for $V^\ast$ induces a basis for $V$ Let $V$ be a $n$-dimensional vector space over $\mathbf{C}$ and let $(v_1^\ast,\ldots,v_n^\ast)$ be a basis for $V^\ast$. Then, there is a unique basis $(v_1,\ldots,v_n)$ for $V$ such that $v_i^\ast(v_j) =\delta_{ij}$.
What are some nice examples to illustrate this fact explicitly?
I know two examples.
Let $V$  be the vector space of polynomials of degree less than $n$. Let $a_1,\ldots,a_n$ be distinct complex numbers. Let $\textrm{ev}_{a_i}$ be the evaluation at $a_i$. Note that $(\textrm{ev}_{a_1},\ldots,\textrm{ev}_{a_n})$ is a basis for $V^\ast$. Thus, there are  unique polynomials $P_1,\ldots,P_n$ of degree less than $n$ such that $P_i(a_j) =\delta_{ij}$. These are the Lagrange interpolation polynomials: $$P_i(x) = \prod_{j\neq i} \frac{x-a_j}{a_j -a_i}.$$
The second example is the standard example mentioned below in the comments section by Henning Makholm.
Are there any other nice examples?
 A: Given an $n$-dimensional  commutative $\mathbb C$-algebra $A $ , consider the set $X=Char(A)$ of characters of $A$ , where a character is a $\mathbb C$-algebra morphism $\chi: A\to \mathbb C$. This set is a linearly independent subset $X\subset A^\ast$ of the dual vector space  $A^\ast$ of $A$ and $X$ has cardinality at most $n$.
A necessary and sufficient condition for $A$ to be a diagonal (=product) algebra, i.e to be isomorphic to the algebra $\mathbb C^n$ with its componentwise product, is that $A$ have $n$ characters $\chi_1, \chi_2,\ldots , \chi_n$, in other words that the characters form a basis of $A^\ast$.
If this is the case the basis $e_1, e_2,\ldots ,   e_n $ of $A$ dual to the basis of characters  diagonalizes $A$, that is $e_i . e_j=\delta_{ij}e_i$ and we have an isomorphism (the Gelfand transform) 
$$Gel:A \stackrel {\sim}{\to } \mathbb C^X: a \mapsto \hat a=(\chi (a))_ {\chi \in X }   $$
transforming the basis $(e_i)_{i=1,...,n}$ of $A$ into the canonical basis $(\delta_x)_{x\in X}$ of $\mathbb C^X$.
An important example The group algebra $A=\mathbb C[\mathbb Z/n\mathbb Z]$ of the group $\mathbb Z/n\mathbb Z$ , with convolution as product [which just means that for the natural basis $(e_k)_{k\in \mathbb Z/n\mathbb Z }$ of $A$ we have $e_k*e_l=e_{k+l} $] is diagonalizable.
Indeed it has the $n$ characters $\chi_\omega$ ( where $\omega \in \mathbb C$ is an $n$-th root of unity : $\omega^n=1$), characterized by  $\chi_{\omega} (e_k)= \omega^k $.
This is at the heart of the discrete Fourier transform.
