Inner Product formulation of a dual basis? I was reading this post A basis for the dual space of $V$ where given a basis $\mathcal{B} =\{\alpha_{1},\alpha_{2},\dots,\alpha_{n}\}$ for a vector space $V$, and $\mathcal{F} = \{f_{1}, f_{2},\dots, f_{n}\}$ defined by $f_{i}:= \langle v, \alpha_{i}\rangle$.  Then $\mathcal{F}$ is a basis for $V^{*}$.
This piqued my curiosity. Instead of finding the set of linear maps $f_{i}(\alpha_{j}) = \delta_{i,j}$, can we just define an inner product and use that to construct a basis instead? If so, what is the impact of the choice of inner product? Given the basis constructed the usual way a dual basis is calculated for a finite dimensional space, can we figure out what the corresponding inner product is that would create it? 
As an experiment, I took $\mathcal{B} = \{1, 1-x, x-x^{2}\}$ as a basis of $\mathbb{P}_{2}$ and tried to find a basis for the dual space by taking an arbitrary $p(x)\in \mathbb{P}_{2}$ and defining
$$\langle p(x),q(x)\rangle := \int_{-1}^{1} p(x)q(x)dx$$
to be my inner product, but this doesn't seem to produce anything that looks correct because I would have thought that $f_{1}(1-x) = \langle 1-x, 1\rangle$ should equal $0$.
I have a suspicion that I am misunderstanding something in a very big way because the proof by Noah Stein in the above link doesn't seem to depend on a specific inner product.
 A: You've asked several questions and hopefully this will answer some.
Any inner product will do.  Given an inner product space $(V, \langle \cdot, \cdot \rangle)$, if $\{v_i\}$ is a basis for $V$, then $\{\langle v_i, \cdot \rangle\}$ is a basis for $V^*$. I think you see this from the linked proof above, but the point is that $\langle v_i, \cdot \rangle$ picks out the $v_i$-coordinate of a vector.  The family $\{\langle v_i, \cdot \rangle\}$ is linearly independent precisely because $\{v_i\}$ is linearly independent.
In fact, having chosen a basis for $V$, every inner product on $V$ can be written $\langle x, y \rangle = x^TAy$ for some (nice) matrix $A$.
Given a basis $\{v_i\}$ for $V$, you present two different ways of finding a basis for $V^*$.  One is to search for linear maps $\lambda_i$ satisfying $\lambda_i(v_j) = \delta_{ij}$.  The other is to define an inner product $\langle \cdot, \cdot \rangle$ and use the basis $\{\langle v_i, \cdot \rangle \}$.  You ask if, given the results of the first process, we can find an inner product which will make the second process give identical results.  The answer is yes: define $\langle v, w \rangle = \langle \sum a_jv_j, w\rangle = \sum a_j \lambda_j(w)$.
A good, theory-focused linear algebra textbook will address these issues and more.  I learned from Hoffman and Kunze but there are lots of others.
A: Let me check if I am addressing your question correctly. 
Suppose you have two vector spaces $V, W$ of the same dimension $n$ over a field $F$. 
Fix a basis $v_{1}, \dots, v_{n}$ of $V$. 
Now there is a one-to-one correspondence between 


*

*non-degenerate bilinear forms $\langle \cdot, \cdot \rangle : V \times W \to F$, and 

*bases $w_{1}, \dots, w_{n}$ of $W$, 


the correspondence being provided by $\langle v_{i}, w_{j} \rangle = \delta_{ij}$.
