Same quadratic forms on $\mathbb R^n$

Question

Let $q$ be an inner product on $\mathbb R^n$ and $Q$ be its matrix expressed in the canonical basis of $\mathbb R^n$.

Assume that the group $$SO(q)=\{A\in M_n(\mathbb R) \ | \ A^TQA=Q\}$$ of matrices with positive determinant preserving the quadratic form $q$ is equal to $SO(n,\mathbb R)$.

Is it true that $q$ is the canonical scalar product on $\mathbb R^n$ ? Is it possible to recover the orthogonal group of $q$ just from $SO(q)$ ?

Answer 1

Since $SO(q)=SO(n,\mathbb{R})$ then every $A\in SO(q)$ satisfy $A^tA=AA^t=Id$. Thus, $QA=AQ$ for every $A\in SO(n)$.

Let $v$ be a normalized eigenvector of the symmetric matrix $Q$ associated to the eigenvalue $a$.

Let $w$ be any normalized vector of $\mathbb{R}^n$. It is always possible to find a matrix $A\in SO(n)$ such that $Av=w$.

Notice that $Qw=QAv=AQv=aAv=aw$. Thus, any $w\in\mathbb{R}^n$ is an eigenvector of $Q$ associated to $a$. Thus, $Q=aId$.