# Find a scalar product where the given basis is an orthogonal basis.

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 $\;\;$.