If $B=((1,0,1),(2,1,2),(3,3,0))$ is a basis of $\mathbb{R}^3$, find a scalar product where $B$ is an orthogonal basis.
Given the standard scalar product of $\mathbb{R}^3$, to find an orthogonal basis I use the Gram-Schmidt process. However, I don't know what to do in this case where given the orthogonal basis I have to find his scalar product.
How can I start?
Write your ordered basis $B$ as $$ B=(b_1,b_2,b_3). $$ You want to define a scalar product $()_B$ on $\mathbb{R}^n$ so that $$ (b_i,b_j)_B=\delta_{ij}\tag{*} $$ where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i\neq j$. Given $(*)$, use the properties of scalar products to define $ (x,y)_B $ for arbitrary $x,y\in\mathbb{R}^3$.
Note that any $x\in\mathbb{R}^3$ can be written as a linear combination of elements of the basis $B$.
You need a positive definite matrix for the inner product:
$$A:=\begin{pmatrix}a&x&y\\x&b&z\\y&z&c\end{pmatrix}\;$$
with the principal minors positive, and it has to fulfill all the usual stuff, for example:
$$(1,0,1)A\begin{pmatrix}1\\0\\1\end{pmatrix}\stackrel{\text{We want}}=1\iff a+2y+c=1$$
$$(2,1,2)A\begin{pmatrix}2\\1\\2\end{pmatrix}\stackrel{\text{We want}}=1\iff 4a+b+4c+4x+8y+2z=1$$
and etc. Observe that you also have the orthogonality relations: $\;r^tAs=0\;$ , with $\;r,\,t\;$ two different elements of $\;\;$.
