Let $G$ be a group. An involution is an element $g\in G$ such that $g^2=1$.
Let $F$ be a field, $V$ an $F$-vector space and $b:V\times V \rightarrow F$ a nondegenerate alternating bilinear form. The set $\mathrm{Sp}(V,b)=\{ f:V\rightarrow V \mid f \mathrm{\ is\ bijective\ and\ } b(f(x),f(y))=b(x,y) \forall x,y \in V\}$ with the composition of maps is a group called the symplectic group of $(V,b)$. It is well-known that the symplectic group is generated by transvections. Is it also generated by involutions in $\mathrm{Sp}(V,b)$?



Generation of the symplectic by involutions
Here is an answer based on the many comments by myself and by Nick Gill.


In characteristic 2 the transvections are always involutions, so generation by transvections implies generation by involutions in all dimensions.
Below I will assume that the characteristic is not 2.


In dimensions 2 the determinant is a symplectic form, thus $G$ is conjugated to $\text{SL}\_2(F)$. Then it is easy to check that the only involutions in $G$ are $1$ and $-1$. In particular, $G$ is not generated by involtions.
Below I will assume that the dimension is not 2.


I claim that $G$ is generated by involutions.
Note first that $G$ has at least one non-central involution:
for example one can view the form as a direct sum of lower dimensional ones, take 1 on one and -1 on the other. 
Note also that the group generated by all non-central involution is normal and not central. This group must be $G$, as every proper normal subgroup of $G$ is central. Indeed, this is the case for every symplectic group apart of $\text{Sp}(2,\mathbb{F}\_2)$, $\text{Sp}(2,\mathbb{F}\_3)$ and $\text{Sp}(4,\mathbb{F}\_2)$, and by the assumptions above $G$ is not in this list.


