So we have our definition of the dual basis:
We call the ordered basis $\beta^*=\{\textbf{f}_1,\dots ,\textbf{f}_n\}$ of $V^*$ that satisfies $\textbf{f}_i(x_j)=\delta_{ij}$ the dual basis of $\beta$.
Take this example:
The space of square $n \times n$ matrices. We have an example of a linear functional $\textbf{f}:V \to \mathbb{R}$ that maps a matrix $A$ to $tr(A)$. (an example functional just to demonstrate my understanding).
What would be the dual basis to the standard basis for this space?
Another example $\mathbb{R}^2$:
Let $\beta =\{(2,1),(1,3)\}$. We wish to find the dual basis $\beta^*=\{\textbf{f}_1,\textbf{f}_2\}$. To explicitly determine a formula for $\textbf{f}_1$ we consider
$1=\textbf{f}_1(2,1)=\textbf{f}_1(2e_1+e_2)=2\textbf{f}_1(e_1)+\textbf{f}_1(e_2)$
$0=\textbf{f}_1(1,3)=\textbf{f}_1(e_1+3e_2)=\textbf{f}_1(e_1)+3\textbf{f}_1(e_2)$
Why do we use the standard basis here instead of the given basis explicitly?
After all this is done, does the dual basis always induce a natural inner product? If so, what does that really mean intuitively and what does this do for us?
Best Answer
If you have a basis $\mathcal B$ of a vector space $V$, the dual basis $\mathcal B^*$ of $V^*$ simply consists in the (parallel) projections on the vectors of $\mathcal B$.
Thus, in the example you mention, the standard basis is the set of matrices $E_{ij}=(e_{kl})$, where $e_{kl}=0$ if $(k,l)\ne (i,j)$, $e_{ij}=1$. Its dual basis is made up of the maps $$A=(a_{kl})\longmapsto a_{ij}.$$
Second question: the standard basis is used just because all cordinates are given in the standard basis… Note that since $f_1=2e_1+e_2$, we have $f_1^*=2e_1^*+e_2^*$.
Last question: no, the dual basis does not induce an inner product, because this notion is valid for vector spaces over abstract fields, while inner products are deined for real (or complex) vector spaces. However, for real vector spaces the notions are linked through a natural pairing: \begin{align*} V\times V^*&\to \mathbf R,\\ (x,u)&\mapsto u(x). \end{align*}