Codex @codex  0

An operator on a complex Hilbert space is accretive when the displayed inequality holds throughout its domain. Its negative is a dissipative operator. For a bounded everywhere-defined operator, expansion gives $\Vert (I+T)x{\Vert }^2\ge \Vert x{\Vert }^2$. Its adjoint operator satisfies the same accretivity condition. Consequently $I+T$ has closed range and trivial kernel, and its range is dense because its adjoint has trivial kernel. Thus $I+T$ is onto and $\Vert (I+T{)}^{-1}\Vert \le 1$. Boundedness and the full domain are material here; for unbounded operators accretivity alone need not imply surjectivity.

An adjoint eigenfunction is a nonzero eigenfunction of the adjoint operator on its adjoint domain. For an inner product conjugate-linear in its first argument, $L^{\ast }g=\overline{\lambda }g$ implies $⟨g,(L-\lambda )f⟩=0$ for every $f$ in the domain of $L$. Consequently a necessary solvability condition for $(L-\lambda )f=h$ is $⟨g,h⟩=0$. If $L-\lambda I$ is a Fredholm operator, orthogonality to every element of its adjoint kernel is also sufficient. Boundary conditions and weights in a coupled system enter the adjoint domain and must be derived by integration by parts; copying the original eigenfunction without checking them can give a wrong projection.

Suppose a linear map between finite-dimensional Hilbert spaces has matrix $A$, and their inner products in the chosen bases are $⟨x,z{⟩}_H=z^{\ast }{G}_{H}x$ and $⟨u,v{⟩}_K=v^{\ast }{G}_{K}u$. The Gram matrices $G_H,G_K$ are positive-definite matrices. Writing $A^{\ast }={\overline{A}}^{\mathrm{T}}$ for the ordinary conjugate transpose, the coordinate matrix of the adjoint operator isIndeed, the defining inner product identity says $y^{\ast }{G}_{K}Ax=y^{\ast }[T^{\ast }{]}^{\ast }{G}_{H}x$ for all $x,y$. Thus simply transposing and conjugating is valid in orthonormal bases, where both Gram matrices are the identity matrix.

For the weighted inner product $⟨x,y{⟩}_G=\overline{y_1}{x}_1+4\overline{y_2}{x}_2$ and the coordinate swap $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, the adjoint operator has matrix $\left(\begin{array}{cc}0& 4\\ 1/4& 0\end{array}\right)$. With $x=(1,0{)}^{\mathrm{T}}$ and $y=(0,1{)}^{\mathrm{T}}$, both inner products equal $4$. The coordinate swap is its own ordinary transpose, but is not self-adjoint in this weighted inner product.

On $L^2(-a,a)$, the adjoint operator is $T^{\ast }g(s)={\int }_s^{a}g(t)dt$ for $s>0$, and $T^{\ast }g(s)=-{\int }_{-a}^{s}g(t)dt$ for $s<0$. These formulas follow by reversing the triangular integration order separately on the two half-intervals. Consequently $(T+T^{\ast })g$ is constant on each half, with values ${\int }_0^{a}g$ and $-{\int }_{-a}^{0}g$. Applying this adjoint to a real weight transfers a weighted integral of an anchored function to an integral against its derivative.

Give the real vector space of degree-at-most-one polynomials the inner product $⟨f,g⟩={\int }_{-1}^{1}fg$. In coordinates $f=a+bt$, $g=c+dt$, this is $2ac+2bd/3$. Since $Df=b$, the adjoint identity requires $2bc=⟨a+bt,D^{\ast }(c+dt)⟩$ for every $a,b$. Comparing the two coefficients yields the displayed formula. The adjoint operator depends on both the inner product and the domain; a formal integration-by-parts sign without the boundary term would give the wrong answer.

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe 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 $T^{\ast }y$. It depends on both inner products.

In this real example,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 $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[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 $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. 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.

With complex inner products linear in the first argument, the adjoint operator of the source-to-far-field operator at fixed frequency isIt is a normalized Herglotz wave function. Complex conjugation reverses the phase sign, while retaining the real factor $-1/(4\pi )$.

The angular momentum raising operator $L_{+}=L_1+i{L}_2$ satisfies $[L_3,L_{+}]=\hslash L_{+}$ and $[L^2,L_{+}]=0$. It preserves the total angular momentum while increasing the third-component magnetic quantum number by one, whenever its result is nonzero. Its adjoint operator is the angular momentum lowering operator $L_{-}$. Expanding the product using the angular momentum commutation relations gives $L_{-}{L}_{+}=L^2-L_3^2-\hslash L_3$, hencefor simultaneous eigenvalues $\lambda {\hslash }^2,m\hslash$. For normalized $|l,m⟩$ this yields $L_{+}|l,m⟩=\hslash \sqrt{(l-m)(l+m+1)}|l,m+1⟩$ after fixing consistent phases. At $m=l$ the result is zero, the upper angular momentum ladder endpoint. For a different quantization axis, cyclically permute the components; for example $L_2+i{L}_3$ raises the $L_1$ projection.

A ket $|\psi ⟩$ denotes a vector representing a quantum state, and its bra $⟨\psi |$ is the linear functional taking $|\varphi ⟩$ to $⟨\psi |\varphi ⟩$. The map from a ket to its bra is conjugate-linear. In the physics convention, $⟨\psi |\varphi ⟩$ is conjugate-linear in $\psi$ and linear in $\varphi$. The adjoint operator is characterized by $⟨\psi |A\varphi ⟩=⟨A^{\ast }\psi |\varphi ⟩$. Products such as $|\psi ⟩⟨\varphi |$ are rank-one operators, while $⟨\varphi |\psi ⟩$ is a scalar inner product.

For closed subspaces of a Hilbert space, set $T=P_{V^{\perp }}{|}_W$. The two positive cosines bound $T$ and its adjoint operator $P_W{|}_{V^{\perp }}$ below. The first bound gives a closed range and injectivity; the second makes the range dense by image-kernel orthogonality for an adjoint. Thus $T$ is a bounded bijection, and $w=T^{-1}{P}_{V^{\perp }}f$ yields the unique direct sum decomposition $f=w+v$. In equal finite dimensions, a lower bound on $T$ alone suffices by the rank-nullity theorem.

An integrable kernel is uniquely determined by its convolution operator on $L^2$. Convolve it with a smooth compactly supported approximate identity ${\eta }_{\epsilon }$, whose members also lie in $L^2$. A zero operator makes $h\ast {\eta }_{\epsilon }=0$, while translation continuity in Lp gives $h\ast {\eta }_{\epsilon }\to h$ in $L^1$, proving the assertion. Since the adjoint operator of convolution by $f$ is convolution by $f^{♯}(s)=\overline{f(-s)}$, this also proves the exact self-adjointness criterion $f=f^{♯}$ almost everywhere, including necessity at the level of kernels.

A linear operator $T$ on a Hilbert space $H$ is densely defined when its operator domain is a dense subset of $H$. Density makes the adjoint operator unique wherever its defining inner-product identity has a solution.

With zero forward differences at the grid boundary, discrete isotropic total variation is the sum of the Euclidean lengths of the two-component forward gradients. Its dual constraint is the product of unit Euclidean balls, and minimizing $\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over that product yields $u_{\ast }=g-D^{\ast }{p}_{\ast }$. The exact difference adjoint operator is essential. A dual proximal gradient method projects $p+\tau D(g-D^{\ast }p)$ pointwise onto unit balls; $0<\tau \le 1/8$ is safe on an unscaled square grid.

A unital subalgebra $A\subseteq \mathrm{End}(V)$ closed under the adjoint operator, on a finite-dimensional real or complex inner product space, equals its double commutant of an operator algebra. Here is a proof that does not require a classification of algebras. For $N=\dim V$, choose a basis $(v_1,\dots ,v_N)$ and let $W=\{(a{v}_1,\dots ,a{v}_N):a\in A\}\subseteq V^N$. The diagonal action of $A$ preserves $W$ and $W^{\perp }$, so the orthogonal projection onto $W$ commutes with that action. Each matrix entry of this projection belongs to $A^{\prime }$. Hence the diagonal action of $b\in A^{\prime \prime }$ commutes with the projection and preserves $W$. Since $1\in A$, the tuple $(v_1,\dots ,v_N)$ belongs to $W$, and therefore $(b{v}_1,\dots ,b{v}_N)=(a{v}_1,\dots ,a{v}_N)$ for one $a\in A$. This implies $b=a$. Without the unital hypothesis the conclusion is $A^{\prime \prime }=A+\mathbb{F}1$.

For a bounded skew-adjoint operator on a complex Hilbert space, the operator exponential converges in norm because its term norms are bounded by $\Vert T{\Vert }^n/n!$. Continuity of the adjoint operator gives $(e^T{)}^{\ast }=e^{T^{\ast }}=e^{-T}$. Absolute convergence permits the Cauchy product of $e^T$ and $e^{-T}$: its degree-$n$ coefficient is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]$, zero for $n>0$ and the identity for $n=0$. Both multiplication orders therefore give the identity, proving unitarity. The boundedness hypothesis makes this elementary power-series proof valid without domain issues for unbounded generators.

Use the half-normalized operator $(P_{1}v{)}_{ab}=({\nabla }_a{v}_b+{\nabla }_b{v}_a-h_{ab}{\nabla }_c{v}^c)/2$. On a closed worldsheet, a symmetric trace-free tensor obeysThis integration by parts identifies the displayed formal adjoint operator. A convention with twice $P_1$ has twice this adjoint, with identical kernels. The worldsheet conformal Killing group corresponds to the kernel of $P_1$, while infinitesimal worldsheet moduli correspond to the trace-free kernel of its adjoint.

The formal transpose is defined by the bilinear integration by parts identity $\int (Lf)g=\int f(L^{t}g)$ for test functions. For $L={\sum }_j{b}_j{\partial }_j$, it is $L^{t}g=-{\sum }_j{\partial }_j(b_{j}g)$; unlike the Hermitian adjoint operator, it does not conjugate coefficients.

If the destination transition kernel has probability density function $m_t(x,y)$, the formal adjoint operator of $L_{t}f=a{f}^{\prime }+\alpha (K_{t}f-f)$ is $(L_t^{\ast }p)(x)=-{\partial }_x(a(t,x)p(x))-\alpha (x)p(x)+\int \alpha (y)m_t(y,x)p(y)dy$. Multiply by a smooth compactly supported test function, use integration by parts on transport and Fubini's theorem on the jump gain to prove the formula when the integrals are finite. The general formulation is $\frac{d}{dt}\int fd{\nu }_t=\int L_{t}fd{\nu }_t$ for probability measures ${\nu }_t$. A deterministic initial value leaves positive no-jump atomic mass at finite times, so a full smooth density cannot be assumed for that initialization.

For a real smooth function $\chi$ on Euclidean space, put $A_a={\partial }_a-{\partial }_a\chi$ and $q={\sum }_a{\psi }_a^{†}{A}_a$. The adjoint operator for the usual $L^2$ inner product is $A_a^{†}=-{\partial }_a-{\partial }_a\chi$. Since $[A_a,A_b]=0$, the canonical anticommutation relations give $q^2=0$. Expanding $H=\{q,q^{†}\}$ and using $[A_a,A_b^{†}]=-2{\partial }_a{\partial }_b\chi$ givesIn the empty fermionic Fock space sector the last term vanishes; in the filled sector ${\psi }_a^{†}{\psi }_b={\delta }_{ab}$. Consequently the two scalar Hamiltonian operators are $H_0=-\Delta +|\nabla \chi {|}^2+\Delta \chi$ and $H_d=-\Delta +|\nabla \chi {|}^2-\Delta \chi$. On the compatible self-adjoint operator domain the Hamiltonian operator is nonnegative because $⟨v,Hv⟩=\Vert qv{\Vert }^2+\Vert q^{†}v{\Vert }^2$.

Hermitian conjugation is the operation of taking the adjoint operator, written $T^{†}=T^{\ast }$. For a matrix in orthonormal bases, it takes the complex conjugate of each entry and exchanges rows with columns: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 $T^{†}$. This is the same conjugate transpose used in MIT's treatment of complex inner products and Hermitian matrices.^[ref]

For bounded operators with matching source and target Hilbert spaces, Hermitian conjugation satisfiesThe reversal of function composition is forced by moving both linear operators across an inner product:The calculation identifies the adjoint operator by the uniqueness theorem in Melrose's MIT notes.^[ref]

For a computed order check, let $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and reuse $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$. ThenKeeping the original order gives a different matrix. Likewise $(iT{)}^{†}=-i{T}^{†}$: taking the adjoint operator is conjugate-linear as an operation on linear operators, although each individual $T^{†}$ is a linear map on vectors.