Let $V$ be a one-dimensional vector space over $\Bbb R$ and $\beta:V \times V \rightarrow \Bbb R$ a positive defined bi-linear pairing on $V$. Show that $(V,\beta)$ is isometrically isomorphic to $(\Bbb R,(\lambda,\mu)\rightarrow\lambda \mu)$
I know that since V is one dimension, then it must be isomorphic to $\Bbb R$ , let $f:V\rightarrow \Bbb R$. Then lets take $\vec v$ be the base of $V$, so $f(v)$ is base of $\Bbb R $. Now if $ \vec u,\vec w \in V $ then, $\beta (\vec u,\vec w) = \lambda_u \mu_w \beta (\vec v,\vec v) = \lambda_u \mu_w \delta_{\vec v \vec v}= \lambda_u \mu_w$. Taking $f(\vec u)=\lambda_u f(\vec v)$, and $f(\vec w)=\mu_w f(\vec v)$ $(f(\vec u),f(\vec w) \rightarrow f(\vec u)f(\vec w) = \lambda_u \mu_w)$.
But I'm not sure this is enough to say that these are Isometrically Isomorphic.
If $v\neq 0$ in $V$, then the map $t\mapsto tv$ is an isomorphism from $\mathbb{R}$ to $V$, but not necessarily isometric. Fix this by using instead $t\mapsto tv/\beta(v,v)^{1/2}$