I am reading about the orthogonal group $O(V)$ of a real finite dimensional quadratic vector space $(V,Q)$ with $Q$ nondegenerate. By definition $$O(V)=\{f:V\mapsto V |\quad Q(f(v))=Q(v) \quad \forall v\in V\}.$$ I don't know if this definition is enough to derive that $O(V)\subset GL(V)$. Also can we imply from definition that $|\det(f)|=1$ for every $f\in V$ ? (For the case $V$ is positive definite, i.e. the bilinear form associated to $Q$ is an inner product, it's true, but I don't know in general case.)
Thanks for any help!
A quadratic form $Q$ can be given by a symmetric matrix $A$ as $$Q(x)=x^TAx\,.$$ This being nondegenerate means that $A$ is invertible, i.e. $\det A\ne 0$.
As you commented, $Q$ determines a nondegenerate (but not necessarily positive definite) bilinear form $B$: $$B(x,y):=\frac12(Q(x+y)-Q(x)-Q(y))\,.$$ Thus, $B(x,y)=x^TAy$.
If $f:V\to V$ is linear and preserves $Q$ (i.e. $f\in O(V,Q)$), then it also preserves $B$, so, also writing $f$ for its matrix (w.r.t. a fixed basis of $V$), $$x^Tf^TAfy=(fx)^TA(fy)=x^TAy$$ for all $x,y$. Applying it to the basis vectors, they imply that $f^TAf=A$. So, we also have $$(\det f)^2\cdot\det A=\det A$$ and hence $|\det f|=1$ as $\det A\ne 0$.