## Definition (the bilinear case).

Let $U$ be a finite-dimensional vector space over a field $\mathbb{K}$, and $B:U\times U\to\mathbb{K}$ 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 $T^{\displaystyle\star}:U\rightarrow U$, characterized by

 $B(u,Tv)=B(T^{\displaystyle\star}u,v),\quad u,v\in U.$

It is convenient to identify $B$ with a linear isomorphism $B:U\rightarrow U^{*}$ in the sense that

 $B(u,v)=(Bu)(v),\quad u,v\in U.$

We then have

 $T^{\displaystyle\star}=B^{-1}T^{*}B.$

To put it another way, $B$ gives an isomorphism between $U$ and the dual $U^{*}$, and the adjoint $T^{\displaystyle\star}$ is the endomorphism of $U$ that corresponds to the dual homomorphism (http://planetmath.org/DualHomomorphism) $T^{*}:U^{*}\rightarrow U^{*}$. Here is a commutative diagram to illustrate this idea:

 $\xymatrix{U\ar[r]^{T^{\star}}\ar[d]^{B}&U\ar[d]^{B}\\ \;U^{*}\ar[r]^{T^{*}}&\;U^{*}}$

## Relation to the matrix transpose.

Let $\mathbf{u}_{1},\ldots,\mathbf{u}_{n}$ be a basis of $U$, and let $M\in\mathop{\mathrm{Mat}}\nolimits_{n,n}(\mathbb{K})$ be the matrix of $T$ relative to this basis, i.e.

 $\sum_{j}M^{j}_{\,i}\,\mathbf{u}_{j}=T(\mathbf{u}_{i}).$

Let $P\in\mathop{\mathrm{Mat}}\nolimits_{n,n}(\mathbb{K})$ denote the matrix of the inner product relative to the same basis, i.e.

 $P_{ij}=B(\mathbf{u}_{i},\mathbf{u}_{j}).$

Then, the representing matrix of $T^{\displaystyle\star}$ relative to the same basis is given by $P^{-1}M^{t}P.$ Specializing further, suppose that the basis in question is orthonormal, i.e. that

 $B(\mathbf{u}_{i},\mathbf{u}_{j})=\delta_{ij}.$

Then, the matrix of $T^{\displaystyle\star}$ is simply the transpose $M^{t}$.

## The Hermitian (sesqui-linear) case.

If $T:U\rightarrow U$ is an endomorphism of a unitary space (a complex vector space equipped with a Hermitian inner product (http://planetmath.org/HermitianForm)). In this setting we can define we define the Hermitian adjoint $T^{\displaystyle\star}:U\rightarrow U$ by means of the familiar adjointness condition

 $\langle u,Tv\rangle=\langle T^{\displaystyle\star}u,v\rangle,\quad u,v\in U.$

However, the analogous operation at the matrix level is the conjugate transpose. Thus, if $M\in\mathop{\mathrm{Mat}}\nolimits_{n,n}(\mathbb{C})$ is the matrix of $T$ relative to an orthonormal basis, then $\overline{M^{t}}$ is the matrix of $T^{\displaystyle\star}$ relative to the same basis.

Title adjoint endomorphism AdjointEndomorphism 2013-03-22 12:29:36 2013-03-22 12:29:36 rmilson (146) rmilson (146) 12 rmilson (146) Definition msc 15A04 msc 15A63 adjoint Transpose Hermitian adjoint