Change of basis and inner product in non-orthogonal basis I have a vector, originally expressed in the standard coordinates system, and want to perform a change of basis and find coordinates in another basis, this basis being non-orthogonal.

*

*Let $B = \{e_1, e_2\}$ be the standard basis for $\Bbb R^2$.

*Let $B' = \{e_1', e_2'\}$ be a non-orthogonal basis for $\Bbb R^2$.

*Let $v$ be some vector in $\Bbb R^2$.

The standard inner product is
$\langle a, b \rangle = \sum_{i=0}^n a_i b_i.$
I want to define an inner product in the non-orthogonal basis $B'$ so that $\langle e_1', e_2' \rangle_{B'} = 0$ since  $\sum_{i=0}^n e_{1i}' e_{2i}' \neq 0$.
Basically, I want to use this new inner product to get the component/coordinates of the vector $v$ on the basis $B'$.

 A: What you want to do is change the basis of the vectors you are working with, then take the inner product on that. since $e_1', e_2'$ are a basis for $\mathbb{R}^2$, every vector $v$ = $v_1 e_1' + v_2 e_2'$ for unique values $v_1, v_2$. Then you set $\langle v, w \rangle = v_1 w_1 + v_2 w_2$
To find an explicit method of converting a vector into this new coordinate system, look into change-of-basis matrices. if $e_1' = a_1 e_1 + a_2 e_2, e_2' = b_1 e_1 + b_2 e_2$, take the matrix with 
$$ \left( \begin{array}{cc}
a_1 & b_1 \\
a_2 & b_2  \end{array} \right)
$$
and take its inverse. Then if $\langle v, w \rangle'$ is your new inner product, you have $\langle v, w \rangle'= \langle M^{-1}v, M^{-1}w \rangle$.
A: Let u,v be any lin ind vectors in the plane. Define u.u=a>0,u.v=0=v.u,v.v=b>0. Extend this linearly for all pairs of vectors. It is easily verified that this is an inner product.  
A: If you write down a $2\times 2$ matrix $P$ whose columns are formed by the coordinates of $e'_1,e'_2$ in the standard basis, then multiplication (on the left) by $P$ does the following in terms of coordinates. If $x,y\in\Bbb R$ these can be interpreted as coordinates of a vector $v$ with respect to the new basis $e'_1,e'_2$, in other words one takes $v=xe'_1+ye'_2$; then $\binom rs=P\binom xy$ gives the coordinates of$~v$ in the standard basis $e_1,e_2$, that is, $v=re_1+se_2$. To see why this is true, note that by definition of$~P$ it is true when $(x,y)=(1,0)$ and also when $(x,y)=(0,1)$ (one gets $v=e'_1$ respectively $v=e'_2$); the general case now follwos by taking linear combinations of these special cases.
But what you want is the opposite: you want to find the coordinates in the new basis of a vector given by its coordinates $\binom rs$ in the standard basis. Therefore you need to compute the inverse matrix $P^{-1}$, and the desired coordinate change is given by (left) mluitplication by$~P^{-1}$.
Note that no inner product plays a role in any way when doing change of basis.
