Let $\sigma:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be a bilinear symmetric form which is non-degenerate in the sense that for every $0\neq u\in \mathbb{R}^n$ there is $v\in \mathbb{R}^n$ with $\sigma\left(u,v\right)\ne 0$.

It is well-known and easy to see that there is a basis $e_1,...,e_n$ and $m \in \overline{0,n+1}$ such that $\sigma\left(u,v\right)= \sum_{k=1}^{m} u_kv_k-\sum_{k=m}^{n} u_kv_k$, where $v=(v_1,...,v_n)$ and $u=(u_1,...,u_n)$ with respect to $e_1,...,e_n$.

Endow $\mathbb{R}^n$ with an inner product $\langle u,v\rangle=\sum_{k=1}^{m} u_kv_k$, where $v=(v_1,...,v_n)$ and $u=(u_1,...,u_n)$ with respect to $e_1,...,e_n$. Then $T:\mathbb{R}^n\to \mathbb{R}^n$ defined by $Te_k=e_k$, for $1\le k\le m$ and $Te_k=-e_k$, for $m\le k\le n$ is an isometry, such that $\sigma\left(u,v\right)=\langle Tu,v\rangle$.

I am wondering if the following infinite-dimensional version holds:

Let $V$ be a real vector space, and let $\sigma$ be a symmetric and non-degenerate bilinear form on $V$. Can we find a Hilbert space $H$, a linear injection $S:V\to H$ and a surjective isometry $T:H\to H$ such that $\sigma\left(u,v\right)=\langle TSu,Sv\rangle_{H}$, for every $u\in V$?