I want to solve this problem.
$V$ a given vector space of finite dimension, with a bilinear form as a scalar product, which is positive definite ( $ \forall u \in V, \langle u,u\rangle \geq 0 $ ). I have to show that :
$$ \exists d \,\in \mathbb N^*, \ F : V \rightarrow \mathbb R^d \, \ \text{such that} \\ \forall u,v \in V : \ \langle u, v\rangle \ = \ F(u) \cdot F(v) $$ where $ \cdot $ is the classical scalar product (standard).
I tried to find something for the case where $ V = \mathbb R ^2 $ but in vain... I think that $F$ has to be a square matrix of the same dimension as $V$. But how to show this? I know how to prove the Riesz decumposition on scalar product and it seems almost like the same...
Let $\{f_1, \ldots, f_n\}$ be an orthonormal basis for $V$ with respect to $\langle\cdot, \cdot\rangle$ and let $\{e_1, \ldots, e_n\}$ be the canonical basis for $\mathbb{R}^n$. Define a linear map $F : V \to \mathbb{R}^n$ as $Fx = \sum_{i=1}^n \langle x, f_i\rangle e_i$.
Let $u,v \in V$ we have
\begin{align} \langle u, v\rangle &= \left\langle \sum_{i=1}^n\langle u, f_i\rangle f_i, \sum_{j=1}^n\langle v, f_j\rangle f_j\right\rangle \\ &=\sum_{i=1}^n\sum_{j=1}^n \langle u, f_i\rangle \langle v, f_j\rangle \overbrace{\langle f_i, f_j\rangle}^{=\delta_{ij}} \\ &= \sum_{i=1}^n\langle u,f_i\rangle \langle v, f_i\rangle \\ &= \left(\sum_{i=1}^n \langle u,f_i\rangle e_i\right)\cdot\left(\sum_{i=1}^n\langle v,f_i\rangle e_i \right)\\ &= Fu \cdot Fv \end{align}
so $F$ is the desired map.