# adjoint

Let $\mathscr{H}$ be a Hilbert space^{} and let $A:\mathcal{D}(A)\subset \mathscr{H}\to \mathscr{H}$ be a densely defined linear operator^{}. Suppose that for some $y\in \mathscr{H}$, there exists
$z\in \mathscr{H}$ such that $(Ax,y)=(x,z)$ for all $x\in \mathcal{D}(A)$. Then such $z$ is unique, for if ${z}^{\prime}$ is another element of $\mathscr{H}$ satisfying that condition, we have $(x,z-{z}^{\prime})=0$ for all $x\in \mathcal{D}(A)$, which implies $z-{z}^{\prime}=0$ since $\mathcal{D}(A)$ is dense (http://planetmath.org/Dense). Hence we may define a new operator ${A}^{*}:\mathcal{D}({A}^{*})\subset \mathscr{H}\to \mathscr{H}$ by

$\mathcal{D}({A}^{*})=$ | $\mathrm{\{}y\in \mathscr{H}:\text{there is}z\in \mathscr{H}\text{such that}(Ax,y)=(x,z)\},$ | ||

${A}^{*}(y)=$ | $z.$ |

It is easy to see that ${A}^{*}$ is linear, and it is called the adjoint^{} of $A$.

Remark. The requirement for $A$ to be densely defined is essential, for otherwise we cannot guarantee ${A}^{*}$ to be well defined.

Title | adjoint |
---|---|

Canonical name | Adjoint |

Date of creation | 2013-03-22 13:48:09 |

Last modified on | 2013-03-22 13:48:09 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 10 |

Author | Koro (127) |

Entry type | Definition |

Classification | msc 47A05 |

Synonym | adjoint operator |

Related topic | TransposeOperator |