Let $ u = (1, −1, 0), v = (0, 1, 1), w = (1, 1, 0) $ be vectors of $\mathbb{R}^ 3$.
Prove that exists only one scalar product $\langle \ , \ \rangle$ in $\mathbb{R}^3$ that has $u, v,w$ as orthonormal basis.
Any idea to begin with?
I've think about the idea that the matrix $M(\langle \ , \ \rangle, \mathscr{S} ) = I_3 $, with $\mathscr{S}$ a Sylvester basis. Because of the fact that an euclidian geometry is an orthogonal geometry with Sylvester invvariant $r_0=0, r_+=n, r_-=0$. But I'm not sure at all...
Any help? Thanks!!
We can be more general than that. Let $V$ be a vector space and let $u_1,\dots, u_n$ be any basis on $V$. Let $\langle\cdot, \cdot \rangle_0$ and $\langle\cdot, \cdot \rangle_1$ be two inner products on $V$ such that $$ \langle u_i,u_j\rangle_0 = \langle u_i,u_j\rangle_1 $$ for all $i,j$. We want to show that $$ \langle \cdot, \cdot\rangle_0=\langle \cdot, \cdot\rangle_1 $$ Let $v,w\in V$. Then we can express it as a linear combination of $u_1,\dots, u_n$: $$ v=\sum_i v^i u_i,\hspace{0.2in} w=\sum_i w^iu_i $$ Then $$ \begin{align} \langle v,w\rangle_0&=\sum_{i,j}v^iw^j\langle u_i,u_j\rangle_0\\ &=\sum_{i,j}v^iw^j\langle u_i,u_j\rangle_1\\ &=\sum_{i,j}\langle v^iu_i,w^ju_j\rangle_1\\ &=\langle v,w\rangle_1, \end{align} $$ which proves the claim.