Non-degenerate bilinear form on $\mathbb{Z}_2$ vector space I am confused that if $f$ is a non-degenerate bilinear form on $\mathbb{Z}_2$ vector space $V$ of finite dimension, so do we have $f(x,x)=0$ for every $x\in V$?
Thanks for every indication.
 A: Given a bilinear form $B$ on some vector space $V$, vectors $\mathbf{x}\in V$ with $B(\mathbf{x},\mathbf{x}) = 0$ are called isotropic.
If all vectors are isotropic, $B$ is called alternating.
So your question is: Is a non-degenerate bilinear form on a $\mathbb{F}_2$ vector space necessarily alternating?
The answer is No.
The standard bilinear form $$B : \mathbb{F}_2^n\times\mathbb{F}_2^n\to \mathbb{F}_2,\quad ((x_1,\ldots,x_n),(y_1,\ldots,y_n)) \mapsto x_1y_1 + \ldots + x_n y_n$$
is non-degenerate, but not all vectors are isotropic (consider $\mathbf{x} = (1,0,0,\ldots,0)$).
It is worth mentioning that half of the vectors $\mathbf{x}$ are isotropic, so for $n\geq 2$ there are always non-zero isotropic vectors. This is quite different from our experience with vector spaces over real or complex numbers.
However, non-degenerate alternating bilinear forms on $\mathbb{F}_2^n$ indeed do exist, namely for all even dimensions $n$. An example is given by
$$A : \mathbb{F}_2^n\times\mathbb{F}_2^n\to \mathbb{F}_2,\\\quad ((x_1,\ldots,x_n),(y_1,\ldots,y_n)) \mapsto (x_1y_2 + x_2 y_1) + (x_3 y_4 + x_4 y_3) + \ldots + (x_{n-1} y_n + x_n y_{n-1}).$$
As a side note: There are no other isomorphism types of symmetric non-degenerate bilinear forms on $\mathbb{F}_2^n$. The above list is complete, consisting of all bilinear forms $B$, and for even $n$ of all bilinear forms $A$.
A: No. If $V$ has a basis $\{x_1,x_2,\ldots,x_n\}$, then the bilinear form $B:V\times V\to\mathbb{Z}_2$ given by the formula
$$
B(\sum_ia_ix_i,\sum_i b_ix_i)=\sum_i{a_ib_i}
$$
is non-degenerate. This is because, given $x\in V$, we have $B(x,y)=0$ for all $y\in V$ only, if $x=0$. On the other hand we have $B(x_i,x_i)=1$ for all $i=1,2,\ldots,n$, so the condition $B(x,x)=0$ does not hold for all the vectors in $V$.
