left / right perpendicular

Given a sesquilinear form $b:V\times V\to k$ over the field $k$, if $v,u\in V$ such that $b(v,u)=0$ then we say $v$ is right perpendicular to $u$ and denote it $v\perp u$. Likewise $u$ is left perpendicular to $v$ and can be denoted by $u\top v$.

By definition $v⟂u$ if and only if $u\top v$. However, $v⟂u$ need not imply $u⟂v$.

For example, let $V=k\oplus k$ and $b((v_1,v_2),(u_1,u_2))=v_1{u}_2$. Then$b((0,1),(1,0))=0$ so $(0,1)\perp (1,0)$, or equivalently, $(1,0)\top (0,1)$.However $b((1,0),(0,1))=1\ne 0$ so $(1,0)$ is not right perpendicular to $(0,1)$ and $(0,1)$ is not left perpendicular to $(1,0)$.