= Adjoint operator
{title2=$T^*$}
{wiki}
= Hermitian adjoint
{c}
{synonym}
For a <bounded linear operator> $T:H\to K$ between real or complex <Hilbert spaces>, its adjoint is the unique <bounded linear operator> $T^*:K\to H$ satisfying
$$
\langle Tx,y\rangle_K=\langle x,T^*y\rangle_H\qquad(x\in H,\ y\in K).
$$
The <inner product> here is linear in its first argument. The <adjoint operator> transfers an output measurement back to the input: measuring $Tx$ against the <vector> $y$ gives exactly the same scalar as measuring $x$ against $T^*y$. It depends on both <inner products>.
\Image[/adjoint-operator-pairing.png]
{title=The adjoint transfers a measurement direction while preserving its measured value}
In this real example,
$$
A=\begin{pmatrix}2&1\\0&1\end{pmatrix},\quad x=\begin{pmatrix}1\\1\end{pmatrix},\quad y=\begin{pmatrix}1\\0\end{pmatrix},\quad Ax=\begin{pmatrix}3\\1\end{pmatrix},\quad A^*y=\begin{pmatrix}2\\1\end{pmatrix}.
$$
Both <inner products> equal $3$. The dashed lines collect all <vectors> giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The <adjoint operator> therefore moves the direction that describes the measurement. It generally differs from the <inverse matrix>: here $A^{-1}y=(1/2,0)^{\mathrm T}$.
For an explicitly complex example, use the standard <inner product> $\langle u,v\rangle=\sum_j u_j\overline{v_j}$ on $\mathbb C^2$. In the standard <orthonormal basis>, the <adjoint operator> is represented by the <conjugate transpose>:
$$
T=\begin{pmatrix}1&i\\2&1-i\end{pmatrix},\qquad T^*=\begin{pmatrix}1&2\\-i&1+i\end{pmatrix}.
$$
For $x=(1+i,2)^{\mathrm T}$ and $y=(1,i)^{\mathrm T}$, direct <matrix multiplication> gives
$$
Tx=\begin{pmatrix}1+3i\\4\end{pmatrix},\qquad T^*y=\begin{pmatrix}1+2i\\-1\end{pmatrix},
$$
and hence
$$
\langle Tx,y\rangle=(1+3i)+4(-i)=1-i,\qquad \langle x,T^*y\rangle=(1+i)(1-2i)-2=1-i.
$$
The <complex conjugate> is essential to this agreement.
Existence follows from the <Riesz representation theorem>: for fixed $y$, the bounded <linear functional> $x\mapsto\langle Tx,y\rangle$ has a unique representing <vector> $T^*y$. The resulting <adjoint operator> has the same <operator norm> as $T$, and
$$
\|T^*\|=\|T\|,\qquad \|T^*T\|=\|T\|^2.
$$
MIT's lecture 18 develops the representation argument and the <conjugate transpose> example.https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2021/7f23bfc43ee2e1efd238a5384e5b9d51_MIT18_102s21_lec18.pdf[Rodriguez, notes by Andrew Lin, MIT 18.102, lecture 18, Theorem 185 and Example 186, 2021]{ref}
This transfer explains the <normal equation for a linear inverse problem>: a <least-squares solution> has residual $Tx-b$ <orthogonal> to every attainable direction $Th$, so $\langle h,T^*(Tx-b)\rangle=0$ for every $h$, equivalently $T^*Tx=T^*b$. An output residual becomes an input condition through the same <adjoint operator>. All identities in this section concern <bounded operators> defined on the whole source <Hilbert space>; for a <densely defined operator> that is not bounded, use the <adjoint of a densely defined operator> and check its <operator domain>.
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