# Finite dimensional symplectic vector space over field of arbitrary characteristic has even dimension

Let $V$ be a finite dimensional over an arbitrary field $F$ and $f: V \times V \to F$ be a symplectic form, meaning

• f is bilinear,
• f is alternating, i.e. $f(v,v)=0$ for all $v \in V$, and
• f is non-degenerate, i.e. if $f(v,w)=0$ for all $w \in V$ then $v=0$.

Problem to solve: Now I want to show that every finite dimensional symplectic vector space $(V,f)$ over an arbitrary field $F$ has even dimension.

My partial solution: I am able to prove the claim if $\operatorname{char}(F) \neq 2$:

Since $f$ is alternating, the matrix representation $A$ of $f$ is skew-symmetric, i.e. $A = -A^t$.
If $n = \operatorname{dim} V$, then \begin{align} \operatorname{det}(A) = \det(-A^t) = (-1)^n \det(A^t) = (-1)^n \det(A). \tag{1} \label{1} \end{align}

By non-degeneracy of $f$, the matrix $A$ is invertible, especially $\det(A)\neq 0$. If we divide both sides of (\ref{1}) by $\det(A)$, we obtain $1 = (-1)^n$. Because $\operatorname{char}(F)\neq 2$, the dimension $n$ must be even.

Obstacle: If the characteristic of $F$ is $2$, I can not use the argument above anymore, since $(-1)^n = 1$ for any $n$.

Could you please explain what one can do for this remaining case? Thank you!

• See Corollary $1.3$ here, which also treats the case $p=2$. – Dietrich Burde May 12 '18 at 10:39

Take a nonzero $v$, then a $w$ with $f(v,w)\ne0$. Then prove that $W$, the subspace of $V$ spanned by $v$ and $w$ is non-degenerate. Deduce that $V=W\oplus W^\perp$, where $W^\perp$ is the orthogonal complement of $W$ under $f$. Now induct on dimension.
• A question about the induction step: If $\operatorname{dim}(V)=n$, can I say that $W^\perp$ has even dimension because the induction hypothesis holds for each symplectic subspace with dimension strictly smaller than $n$ (here: $\operatorname{dim}(W^\perp) = n-2$)? – Diglett May 12 '18 at 19:37