# left / right perpendicular

Given a sesquilinear form $b:V\times V\rightarrow k$ 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 $v\bot u$. Likewise $u$ is left perpendicular to $v$ and can be denoted by $u\top v$.

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

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 $(0,1)\bot(1,0)$, or equivalently, $(1,0)\top(0,1)$. However $b((1,0),(0,1))=1\neq 0$ so $(1,0)$ is not right perpendicular to $(0,1)$ and $(0,1)$ is not left perpendicular to $(1,0)$.

Title left / right perpendicular LeftRightPerpendicular 2013-03-22 16:13:05 2013-03-22 16:13:05 Algeboy (12884) Algeboy (12884) 7 Algeboy (12884) Derivation msc 15A63 left perpendicular right perpendicular