Let $F:\mathbf{R}^n\rightarrow\mathbf{R}$ be a non-degenerate quadratic forms. Let
$$
O(F):=\{g\in GL\_n(\mathbf{R}):F(gv)=F(v),\forall v\in \mathbf{R}^n\}
$$
be the isotropy group of $F$.


Q: So how does one prove **in the simplest way possible** that if $O(F)=O(G)$ then
there exists $\lambda\in\mathbf{R}^{\times}$ such that $F=\lambda G$?


P.S. I would like to have a proof that could be explained to undergraduate students who take an advanced linear algebra class.



On the determination of a quadratic form from its isotropy group 
A relatively easy proof also follows from using the reflection identity: First, define the inner product associated to $F$, namely $v\ \cdot\_F\ w = {\frac12}\bigl(F(v{+}w)-F(v)-F(w)\bigr)$, and then, for any $v$ with $F(v)\not=0$, define the *reflection in $v$* by 
$$
\rho^F\_v(w) = w - 2\ \frac{v\ \cdot\_F\ w}{v\ \cdot\_F\ v}\ v
$$
This map $\rho^F\_v$ belongs to $O(F)$, as is easy to verify. By your assumption, it belongs to $O(G)$ as well. This works out to imply 
$$
\bigl(v\ \cdot\_F\ v\bigr)\bigl(v\ \cdot\_G\ w\bigr) = 
\bigl(v\ \cdot\_G\ v\bigr)\bigl(v\ \cdot\_F\ w\bigr).
$$
Now if you replace $v$ by $v+tw$ in this equation and compare $t$-linear terms, you'll get
$$
\bigl(v\ \cdot\_F\ v\bigr)\bigl(w\ \cdot\_G\ w\bigr) = 
\bigl(v\ \cdot\_G\ v\bigr)\bigl(w\ \cdot\_F\ w\bigr),
$$
i.e., $F(v)G(w) = F(w)G(v)$, from which the result is obvious.


