Let $q$ be quadratic form on $V=K^n$ where $K$ is a field in characteristic $2$.
- Let $b(x,y)=q(x+y)-q(x)-q(y)$ be the bilinear form associated to $q$ and set $V^\perp=\{x\in V\mid b(x,y)=0 \text{ for all }y\in V\}$.
- Assume that $q$ is non-singular, i.e. either $V^\perp=\{0\}$, or $\dim(V^\perp)=1$ and there exists $x\in V^\perp$ such that $q(x)\neq 0$.
- Consider $O(K^n,q)$ the orthogonal group, i.e. $O(K^n,q)=\{A\in\mathrm{GL}_n(K)\mid q\circ A(x)=q(x)\text{ for all }x\in GL_n(K)\}$.
Question: Why is the determinant of $A\in O(K^n,q)$ always $1$?
What I know: There is a matrix $M\in\mathrm{Mat}_{n\times n}(K)$ such that $q(x)=x^tMx$, and so $x^tMx=q(x)=q\circ A(x)=x^t A^tMA x$. Since $M$ is symmetric, $M=A^tMA$. In characteristic different from $2$ the matrix $M$ is invertible, implying that the determinant is $\pm1$. In characteristic $2$, however, $M$ is not invertible and I don't know what to do.
Beware that the representative matrix of $q$ in a fixed basis is NOT symmetric (it is upper triangular). However, the representative matrix of $b$ is.
Note that an element of $O(q)$ induces an automorphism of $b$ (easy).
The assumptions on $(V,q)$ imply that $b$ is non degenerate (in the classical sense for symmetric bilinear forms) .This is clear if $V^\perp=0$. If $V^\perp=Kv$ has dim. 1, this is because $q(v)\neq 0$ by assumption. Hence $b$ will be the orthogonal sum of $(x,y)\mapsto axy$ and of a non degenerated symmetric bilinear form. Hence you can conclude as in char $\neq 2$ that any automorphism $u$ of $b$ (in particular, any automorphism $u$ of $q$) satisfies $\det(u)^2=1$.