= Hermitian conjugation
{c}
{title2=$A^\dagger=\overline A^{\mathrm T}$}
= Hermitian conjugate
{c}
{synonym}
Hermitian conjugation is the operation of taking the <adjoint operator>, written $T^\dagger=T^*$. For a <matrix> in <orthonormal bases>, it takes the <complex conjugate> of each entry and exchanges rows with columns:
$$
(A^\dagger)_{jk}=\overline{A_{kj}},\qquad A^\dagger=\overline A^{\mathrm T}.
$$
For a real <matrix> it is the ordinary <transpose>. For the complex <matrix> $T$ in the <adjoint operator> example, its second row $(2,1-i)$ becomes the second column $(2,1+i)^{\mathrm T}$ of $T^\dagger$. This is the same <conjugate transpose> used in MIT's treatment of complex <inner products> and <Hermitian matrices>.https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/53f17d47bf99d26fdc472b3177c256a8_MIT18_06SCF11_Ses3.2sum.pdf[Strang, MIT 18.06SC, lecture 26 summary, Complex matrices, pp. 1–2, 2011]{ref}
For <bounded operators> with matching source and target <Hilbert spaces>, <Hermitian conjugation> satisfies
$$
(A+B)^\dagger=A^\dagger+B^\dagger,\quad (\alpha A)^\dagger=\overline\alpha A^\dagger,\quad (BA)^\dagger=A^\dagger B^\dagger,\quad (A^\dagger)^\dagger=A.
$$
The reversal of <function composition> is forced by moving both <linear operators> across an <inner product>:
$$
\langle BAx,y\rangle=\langle Ax,B^\dagger y\rangle=\langle x,A^\dagger B^\dagger y\rangle.
$$
The calculation identifies the <adjoint operator> by the uniqueness theorem in Melrose's MIT notes.https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2021/3d4cc88026d44a01f936cd6a0aa995cb_MIT18_102s20_lec_FA.pdf[Melrose, MIT 18.102, Functional Analysis, Chapter 3, §11, 2020]{ref}
For a computed order check, let $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and reuse $T=\begin{pmatrix}1&i\\2&1-i\end{pmatrix}$. Then
$$
(ST)^\dagger=T^\dagger S=\begin{pmatrix}2&1\\1+i&-i\end{pmatrix},\qquad ST^\dagger=\begin{pmatrix}-i&1+i\\1&2\end{pmatrix}.
$$
Keeping the original order gives a different <matrix>. Likewise $(iT)^\dagger=-iT^\dagger$: taking the <adjoint operator> is conjugate-linear as an operation on <linear operators>, although each individual $T^\dagger$ is a <linear map> on <vectors>.
Back to article page
