adjoint endomorphism AdjointEndomorphism 2002-02-26 08:58:59 2006-03-19 09:35:11 Definition Hermitian adjoint \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{xypic} \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}}} \newcommand{\bu}{\mathbf{u}} \newtheorem{proposition}{Proposition} \paragraph{Definition (the bilinear case).} Let $U$ be a finite-dimensional vector space over a field $\kfield$, and $B:U\times U\to\kfield$ a symmetric, non-degenerate bilinear mapping, for example a real inner product. For an endomorphism $T:U\rightarrow U$ we define the adjoint of $T$ relative to $B$ to be the endomorphism $T\adj:U\rightarrow U$, characterized by $$B(u,Tv) = B(T\adj u,v),\quad u,v\in U.$$ It is convenient to identify $B$ with a linear isomorphism $B:U\rightarrow U\dual$ in the sense that $$B(u,v) = (Bu)(v),\quad u,v\in U.$$ We then have $$T\adj = B^{-1} T\dual B.$$ To put it another way, $B$ gives an isomorphism between $U$ and the dual $U^*$, and the adjoint $T\adj$ is the endomorphism of $U$ that corresponds to the \PMlinkname{dual homomorphism}{DualHomomorphism} $T\dual:U\dual\rightarrow U\dual$. Here is a commutative diagram to illustrate this idea: $$ \xymatrix{% U \ar[r]^{T^\star} \ar[d]^B & U \ar[d]^{B} \\ \;U^* \ar[r]^{T^*} & \;U^{*} } $$ \paragraph{Relation to the matrix transpose.} Let $\bu_1,\ldots,\bu_n$ be a basis of $U$, and let $M\in \Mat_{n,n}(\kfield)$ be the matrix of $T$ relative to this basis, i.e. $$\sum_j M^j_{\,i}\, \bu_j = T(\bu_i).$$ Let $P\in\Mat_{n,n}(\kfield)$ denote the matrix of the inner product relative to the same basis, i.e. $$P_{ij} = B(\bu_i,\bu_j).$$ Then, the representing matrix of $T\adj$ relative to the same basis is given by $ P^{-1} M\supt P.$ Specializing further, suppose that the basis in question is orthonormal, i.e. that $$B(\bu_i,\bu_j) = \delta_{ij}.$$ Then, the matrix of $T\adj$ is simply the transpose $M\supt$. \paragraph{The Hermitian (sesqui-linear) case.} If $T:U\rightarrow U$ is an endomorphism of a unitary space (a complex vector space equipped with a \PMlinkname{Hermitian inner product}{HermitianForm}). In this setting we can define 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.