Given a basis $B$ there exist a unique positive defined scalar product such that $B$ is ortonormal Given a basis $B=\{v_1,...,v_n\}$ of a $\mathbb{R}$-vector space $V$, how can I prove that there exist and is unique a positive defined scalar product $g$ such that $B$ is ortonormal with respect to $g$?
I have figured out that I should define $g$ in this way: $g(v,u)=X^tY$, where $X$ and $Y$ are the column vectors of the components of $v$ and $u$ with respect to $B$, but I would like some guidance in proving that such $g$ satisfies our assumptions and is unique. 
 A: If $B$ is orthonormal with respect to $g$, then $g(v_i,v_j)=\delta_{ij}$ (here the delta is Kronecker delta). Now, for any two vectors $v=\sum_ix_iv_i$ and $u=\sum_iy_iv_i$, you can calculate $g(v,u)$ using the bilinearity of $g$.
A: Assume $g$ is a scalar product. By definition, $g$ is linear in every variable so for any 2 general vectors $u=\sum_{i=1}^n \alpha_i v_i , w= \sum_{j=1}^n \beta_j v_j $ , the following occurs:
$$g(u,w)=g(\sum_{i=1}^n \alpha_i v_i , w)=\sum_{i=1}^n \alpha_i\cdot  g(v_i,w) = \sum_{i=1}^n \alpha_i\cdot  g(v_i,\sum_{j=1}^n \beta_j v_j)=$$
$$ \sum_{i=1}^n \alpha_i \sum_{j=1}^n \beta_j \cdot g(v_i, v_j) = \sum_{i=1}^n \sum_{j=1}^n \alpha_i \beta_j g(v_i,v_j) $$ 
Our conclusion is that if we know how the scalar product works on a basis $\{v_1,\cdots,v_n\}$ then the scalar product is uniquely defined.
You want an orthonormal basis, so $g(v_i,v_j)=\delta_{ij}$ ($1$ if $i=j$ and $0$ if $i\neq j$). You can check that this satisfies the conditions of a scalar product, and as I proved, it's uniquely defined. 
