In the book of Geometric Algebra, Artin says (p. 108)
If $V$ is a non-singular bilinear space then any isometry $T:V\rightarrow V$ has determinant $1$ or $-1$. The map $$\mbox{Isom}(V)\rightarrow \{1,-1\}, \,\,\,\,\,\,T\mapsto \det(T)$$ is a homorphism. The image of this homomorphism is $\{1,-1\}$ or just $\{1\}$ (in case there are no reflections or if the characteristic of k is 2).
Question 1. Can there be a case of non-survectivity of above map where field is of characteristic $\neq 2$ for some non-singular space?
I mean, I want to know example (if exists) of a non-singular space with all isometries of determinant $1$ but characteristic of field is not $2$.
On the page 108, perhaps, he do not considers whether bilinear form is symmetric/skew-symmetric etc; he just consider's non-singular.
Question 2. If a bilinear space $V$ is symmetric or skew symmetric over a field of characteristic $\neq 2$, then is there always an isometry with determinant $-1$?