Let $V$ be an $\mathbb{K}$-Vectorspace with basis $B :=(b_i)_{i \in I}$ and $B^* = (b_i^*)_{i \in I}$ corresponding dual basis. Let $\Phi: V \rightarrow V^*$ the isomorphism defined as $\Phi(v_i) = v_i^*$.
The map:
$$ \pi: V \times V^* \rightarrow \mathbb{K}, \,\,\,\,\,\,(v, \,f) \mapsto f(v) $$
is called the dual pairing of $V$. Let $$ \gamma_B: V \times V \rightarrow \mathbb{K}, \,\,\,\,\,\,(v, \,w) \mapsto \pi(v, w^*) $$
Show that $\gamma_B$ is a perfect, symmetrical bilinear form. What is $\gamma_B$ in the case of $V := \mathbb{R}^n$ and $B := (e_1, \dots, e_n)$?
I was able to show that $\gamma_B$ is a perfect, symmetric bilinear form. But I don't quite understand what is meant by
What is $\gamma_B$ in the case of $V := \mathbb{R}^n$ and $B := (e_1, \dots, e_n)$?
Let us consider $v,w \in V$, and let us compute $\gamma_B(v,w)$ step by step.
$\gamma_B(v,w) = \pi(v,w^*) = w^*(v).$
But recall that $w^*$ was defined in the following way : the isomorphism sends $b_i$ to $b_i^*$ (the linear form which sends $b_i$ to $1$ and the other $b_j$ to $0$)
So let us write $v = \sum_{i=1}^n v_ib_i$ and $w = \sum_{i=1}^n w_ib_i$ the decomposition of $v,w$ in the basis $B$. You then have $w^* = \sum_{i=1}^n w_ib_i^*$ so that
$\gamma_B(v,w) = \sum_{i=1}^n w_ib_i^*(v) = \sum_{i=1}^n w_iv_i = (v|w)$
where $( \cdot |\cdot)$ is just the usual inner product on $\mathbb{R}^n$ : which was the special thing to notice in this case.