In characteristic not $2$, the Theorem of Cartan-Dieudonné states:

- [Grove, Theorem 6.6]: Let $q$ be a nondegenerate symmetric quadratic form of dimension $n$ in characteristic not $2$. Then every element in the orthogonal group $O(q)$ can be written as a product of at most $n$ reflections.

**In a nutshell:** I am looking for a similar statement but over a field in **characteristic $2$** and in particular in **odd dimension**. At the end I will make the statement that I would like to have more precise.

As I am new to quadratic forms, I got confused by different notions of non-degeneracy, so I want to fix the terminology; I take the one from [Grove]. To a quadratic form $q$ on $V$ one can associate a bilinear form $b$ on $V$ via $b(v,w)=q(v+w)-q(v)-q(w)$. In characteristic $2$, several quadratic forms can give the same bilinear form (e.g. $y^2+xz$ and $xz$). Let $\mathrm{rad}_b(V)=\{w\in V\mid b(v,w)=0 ~\forall v\in V\}$. If $\mathrm{rad}_b(V)\neq\{0\}$, then $q$ is called

*degenerate*if the characteristic is not $2$,*defective*if the characteristic is $2$.

**In characteristic $2$ and dimension odd, all quadratic forms are defective!**
In characteristic $2$, we say that $q$ is *regular* if $q(v)\neq0$ for all non-zero $v\in\mathrm{rad}(V)$. (In [Connors], regular is called *nondegenerate*).

**Example:** The two quadratic forms $q_1=y^2+xz$ and $q_2=xz$ on $V=k^3$, where $k$ is a field of characteristic $2$, give the same bilinear form and they both have $\mathrm{rad}(V)=\langle(0,1,0)\rangle$. So they are defective. While $q_1$ is regular, $q_2$ is not.

From now on, assume that $k$ is a field in characteristic $2$. Here transvections play the role of reflections. I have found the following statements that go in the right direction:

- [Grove, Theorem 14.16]: Assume that $q$ is regular and
**non-defective**(see the assumption on p. 129 which I believe is still taken in Theorem 14.16). Then $O(q)$ is generated by transvections, except for one example over the field $\mathbb{F}_2$ where $V$ is $4$-dimensional.

The above theorem can not be applied in odd dimension. The next statement is exactly what I want, but it is stated only for dimension $3$:

- [Connors, Proposition 3.2]: Let $q$ be a defective, anisotropic regular quadratic form in
**dimension $3$**. Then every element of $O(q)$ is the product of at most $2$ transvections.

**I am looking for a reference of the following statement** (if true): Let $V$ be a vector space over any field in **characteristic $2$**, and let $q$ be a **regular**, **anisotropic** quadratic form on $V$ of **odd dimension** $n$ (i.e. $q$ **defective**). Then every element in the orthogonal group $O(q)$ can be written as a product of (at most $n$) transvections.

[Grove] : Larry C. Grove, "Classical groups and geometric algebra", 2002

[Connors] : Edward A. Connors, ''The Structure of $O'(V)/DO(V)$ in the Defective Case'', 1973