|
|
|
Viewing Version
3
of
'adjoint endomorphism'
|
[ view 'adjoint endomorphism'
|
back to history
]
| Title of object: |
adjoint endomorphism |
| Canonical Name: |
AdjointEndomorphism |
| Type: |
Definition |
| Created on: |
2002-02-26 08:58:59-05 |
| Modified on: |
2002-07-25 06:50:58.426825-04 |
| Classification: |
msc:15A63, msc:15A04 |
| Defines: |
Hermitian adjoint |
| Synonyms: |
adjoint endomorphism=adjoint |
Preamble:
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcommand{\Hom}{\mathop{\mathrm{Hom}}\nolimits}
\newcommand{\Mat}{\mathop{\mathrm{Mat}}\nolimits}
\newcommand{\kfield}{\mathbb{K}}
\newcommand{\supt}{^t}
\newcommand{\dual}{^*}
\newcommand{\adj}{^{\displaystyle \star}}
\newcommand{\reals}{\mathbb{R}}
\newcommand{\natnums}{\mathbb{N}}
\newcommand{\cnums}{\mathbb{C}}
\newcommand{\lp}{\left(}
\newcommand{\rp}{\right)}
\newcommand{\lb}{\left[}
\newcommand{\rb}{\right]}
\newcommand{\supth}{^{\text{th}}}
\newtheorem{proposition}{Proposition} |
Content:
\paragraph{Definition.} Let $U$ be a finite-dimensional inner product space
over a field $\kfield$. For an endomorphism $T:U\rightarrow U$ we
define the adjoint of $T$ (relative to the inner product) to be the
endomorphism $T\adj:U\rightarrow U$, characterized by
$$\left<u,Tv\right> = \left<T\adj u,v\right>,\quad u,v\in U.$$
Identifying $U$ and $U^*$ by means of the inner product, we can more
succinctly define the adjoint $T\adj$ as the endomorphism of $U$
identified with the dual homomorphism $T\dual:U\dual\rightarrow
U\dual$. To be more precise, the inner-product structure on $U$ comes
from a non-degenerate bilinear form that can be canonically identified
with a linear isomorphism $B:U\rightarrow U\dual$ such that
$$\left<u,v\right> = (Bu)(v),\quad u,v\in U.$$
We then have
$$T\adj = B^{-1} T\dual B.$$
\paragraph{Relation to the matrix transpose.} Let $M\in
\Mat_{n,n}(\kfield)$ be the matrix of $T$ relative to a basis of $U$,
and let $P\in\Mat_{n,n}(\kfield)$ be the matrix of the inner product
relative to the same basis. Then, the representing matrix of $T\adj$
is given by $ P^{-1} M\supt P.$ Specializing further, suppose that the
basis in question is orthonormal. Then, the matrix of $T\adj$ is
simply $M\supt$.
\paragraph{The Hermitian case.}
If $T:U\rightarrow U$ is an endomorphism of a complex vector space
with a Hermitian inner product, then we define the Hermitian adjoint
$T\adj:U\rightarrow U$ by means of the familiar adjointness condition
$$\langle u,Tv\rangle = \langle T\adj u,v\rangle,\quad u,v\in U.$$
However, the analogous operation at the matrix level is the conjugate
transpose. Thus, if $M\in \Mat_{n,n}(\cnums)$ is the matrix of $T$
relative to an orthonormal basis, then $\overline{M\supt}$ is the
matrix of $T\adj$ relative to the same basis. |
|
|
|
|
|
