Question about dual basis If we have a $k$-bilinear pairing $V \times W \to k$ where $V$ and $W$ are $k$-vectorspaces of same dimension. Whenever having a basis $A$ of $V$ what does it mean to say that $B$ is a basis for $W$ which is dual to $A$?
 A: Let denote $\varphi$ the given $k$-bilinear map. Let $(v_i)$ a basis for $V$ then $(w_i)$ is a basis for $W$ dual to $(v_i)$ if
$$\varphi(v_i,w_j)=\delta_{ij}$$
where $\delta$ is the Kronecker's symbol.
A: If $n=\dim V$
 , if $A=\left(a_{1},\ldots,a_{n}\right)$
  and if $B=\left(b_{1},\ldots,b_{n}\right)$
 , then $B$
  is the dual basis of $A$
  if for all $1\leq i,j\leq n$
  we have$$\phi\left(a_{i},b_{j}\right)=\delta_{ij}=\begin{cases}
1 & \textrm{if }i=j\\
0 & \textrm{else}
\end{cases}$$
 where $\phi$
  is the $k$
 -bilinear form you've mentioned. 
A: Let $(\cdot,\cdot)$ denote the bilinear pairing of two vectors.  For a basis $A = \{v_1,\dots,v_n\}$, the corresponding dual basis is the (unique) basis $B = \{w_1,\dots,w_n\}$ that satisfies
$$
(v_i,w_j) = 
\begin{cases}
1 & i=j\\
0 & i \neq j
\end{cases}
$$
A: Bases $\{v_1,\dots,v_n\}$ of $V$ and $\{w_1,\dots,w_n\}$ of $W$ are dual to each other if 
$$
\varphi(v_i,w_j)=\begin{cases}
1 & \text{if $i=j$,}\\
0 & \text{if $i\ne j$,}
\end{cases}
$$
where $\varphi\colon V\times W\to k$ is the pairing.
A dual basis of $W$ always exists, given a basis of $V$. Indeed, the map
$$
f\colon W\to V^*
$$
defined by
$$
f(w)\colon v\mapsto \varphi(v,w)
$$
is an isomorphism of vector spaces. Building a dual basis on $V^*$ is obvious.
A: Let be $V$
  a vector space on a commutative field $k$
  such that $\dim_{k}V=n<+\infty$. Let be $\mathcal{B}=\left(e_{1},\ldots,e_{n}\right)$
  a basis of $V$. Now let us denote by $V^{*}$
  the dual vector space of $V$, that is $V^{*}$
  is the vector space of all linear forms on $V$ (it always exists). Then we define $e_{i}^{*}$
  an element of $V^{*}$
  by$$e_{i}^{*}\left(e_{j}\right)=\delta_{ij}=\begin{cases}
1 & \textrm{if }i=j\\
0 & \textrm{else}
\end{cases}$$
 for all $1\leq i,j\leq n$
 . Then, $\mathcal{B}^{*}=\left(e_{1}^{*},\ldots,e_{n}^{*}\right)$
  is a basis of $V^{*}$, called the dual basis of $\mathcal{B}$.
To see that it is a basis, let us show that it is formed by $n$ non-zero linearly independant elements : it is ok that $e_{i}^{*}\neq0_{E^{*}}$ for all $1\leq i\leq n$ (because, for example, $e_{i}^{*}\left(e_{i}\right)=1\neq0_{E}$). Now let us suppose that there are $\alpha_{1},\ldots,\alpha_{n}\in k$
  such that$$\sum_{i=1}^{n}\alpha_{i}e_{i}^{*}=0_{E^{*}}.$$
 Then for any $1\leq j\leq n$
 $$0_{k}=\left(\sum_{i=1}^{n}\alpha_{i}e_{i}^{*}\right)\left(e_{j}\right)=\alpha_{j}$$
 and then $\alpha_{1},\ldots,\alpha_{n}$ are all zero.
Now let be $W$
  another vector space on $k$
  such that $\dim_{k}W=n$
 . Let be $\mathcal{W}=\left(w_{1},\ldots,w_{n}\right)$
  a basis of $W$
 . Then the application $\psi:W\longrightarrow E^{*}$
  defined by $\psi\left(\lambda w_{i}+\mu w_{j}\right)=\lambda e_{i}^{*}+\mu e_{j}^{*}$
  for all $1\leq i,j\leq n$ and all $\lambda,\mu\in k
 $ is an isomorphism (I let you check this).
Then we can define a bilinear form$\phi:V\times W\rightarrow k$
  by$$\phi\left(e_{i},w_{j}\right):=\psi\left(w_{j}\right)\left(e_{i}\right)=e_{j}^{*}\left(e_{i}\right)=\delta_{ij}$$
 and thus $\mathcal{W}$
  is the dual basis of $\mathcal{B}$
  with regard to the $k$-bilinear form $\phi$. 
