Let $M$ be an invertible symmetric $2n \times 2n$ matrix with entries in the finite field $\mathbb{F}\_2$. Is $\mathrm{Ker}\ (M^2 - I\_{2n})$ necessarily even dimensional? 



Fixed space of the square of a symmetric matrix over $\mathbb{F}_2$
The matrix $M = \left(\begin{array}{clcr} 1&1&0&0\\1&1&1&0\\0&1&1&0\\0&0&0&1\end{array} \right)$ is an example where the dimension of the space in question is odd.


(Later edit: However if $M = M^{t}$ and also $M \in {\rm Sp}(2n,2),$ then we in fact have $M^{2} = I\_{2n}$ as Darij Grinberg and Noam Elkies implicitly noted in comments).


