# 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 $v\u27c2u$ if and only if $u\top v$. However, $v\u27c2u$ need not imply $u\u27c2v$.

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 $(1,0)$.

Title | left / right perpendicular |
---|---|

Canonical name | LeftRightPerpendicular |

Date of creation | 2013-03-22 16:13:05 |

Last modified on | 2013-03-22 16:13:05 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 7 |

Author | Algeboy (12884) |

Entry type | Derivation |

Classification | msc 15A63 |

Defines | left perpendicular |

Defines | right perpendicular |