Is every quadratic form in characteristic $2$ induced by a bilinear form?

Question

Let $F$ be a field of characteristic $2$, and $q$ a quadratic form on $F^n$ for some positive integer $n$, i.e., $q:F^n\rightarrow F$ is a function such that $g(x,y):=q(x+y)-q(x)-q(y)$ is a bilinear form on $F^n$ and $q(cx)=c^2q(x)$ for all $c\in F$ and $x\in F^n$. Then does there exist a bilinear form $f$ on $F^n$ such that $q(x)=f(x,x)$ for all $x\in F^n$? If the answer is yes, how to find such an $f$?

Answer 1

Yes, $f$ does exist. Keep your notation: If $(e_1,\ldots,e_n)$ is a basis of $F^n$, then let $f$ be the unique bilinear form such that $f(e_i,e_i)=q(e_i), f(e_i,e_j)=0$ if $i>j$ and $f(e_i,e_j)=g(e_i,e_j) $ if $i<j$.

Then $$q(\sum_i x_i e_i)=\sum_iq(e_i)x_i^2+\sum_{i<j}g(e_i,e_j)x_ix_j=\sum_iq(e_i)x_i^2+\sum_{i<j}g(e_i,e_j)x_ix_j+\sum_{i>j}0\cdot x_ix_j=\sum_i f (e_i,e_i)x_i^2+\sum_{i<j}f(e_i,e_j)x_ix_j+\sum_{i>j}f(e_i,e_j)\cdot x_ix_j=f(\sum_i x_ie_i, \sum_i x_i e_i).$$