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.

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*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 $B'$.


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.
