Let $q$ be a quadratic form on $E$. Let $E'$ be a second $\mathbb{K}$-vector space of dimension $n$, and let $q'$ be a quadratic form on $E'$. We call an isometry from $(E,q)$ to $(E',q')$ any isomorphism $f$ from $E$ to $E'$ satisfying: for all $x \in E$, $q'(f(x)) = q(x)$. We will say that $(E,q)$ and $(E',q')$ are isometric if and only if there exists an isometry from $(E,q)$ to $(E',q')$. Show that $(E,q)$ and $(E',q')$ are isometric if and only if there exists a basis $e$ of $E$ and a basis $e'$ of $E'$ such that $\operatorname{mat}(q,e) = \operatorname{mat}(q',e')$.
Let $q$ be a quadratic form on $E$. Let $E'$ be a second $\mathbb{K}$-vector space of dimension $n$, and let $q'$ be a quadratic form on $E'$.
We call an isometry from $(E,q)$ to $(E',q')$ any isomorphism $f$ from $E$ to $E'$ satisfying: for all $x \in E$, $q'(f(x)) = q(x)$. We will say that $(E,q)$ and $(E',q')$ are isometric if and only if there exists an isometry from $(E,q)$ to $(E',q')$.
Show that $(E,q)$ and $(E',q')$ are isometric if and only if there exists a basis $e$ of $E$ and a basis $e'$ of $E'$ such that $\operatorname{mat}(q,e) = \operatorname{mat}(q',e')$.