Is it possible to have a vector space in which $\vec{v}=-\vec{v}$, yet $\vec{v}\neq \vec{0}$?

Silly question. I was pondering whether given a vector space $V$ over $K$ and a vector $\vec{v}$ such that $\vec{v}=-\vec{v}$ then $\vec{v}\neq \vec{0}$. This is certainly true as long as $1/2\in K$ (or, equivalently, if $1+1\neq 0$ in $K$), yet what if this condition is not met?

Is it possible to have a vector space in which $\vec{v}=-\vec{v}$, yet $\vec{v}\neq \vec{0}$?

If $1+1=0$, then $1=-1$ and hence $v=-v$ for all vectors $v$.

From $\vec{v} = -\vec{v}$, we get $(1+1)\vec{v}=2\vec{v}=\vec{0}$.

• If $K$ has characteristic $2$ (that is, if $1+1=0$, than $\vec{v}=-\vec{v}$ for all vectors $\vec{v}$.
• Otherwise, we multiply the equation with $2^{-1}$, and get $\vec{v} = 0$