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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.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

For the Volterra integration operator, Cauchy-Schwarz inequality givesIntegrating and reversing the order yieldsThus $J$ is bounded, with $\Vert J\Vert \le 1/\sqrt{2}$. Since $f\in L^2(0,1)\subseteq L^1(0,1)$, $Jf$ is absolutely continuous and $(Jf{)}^{\prime }=f$ almost everywhere. If $Jf=0$ as an $L^2$ element, its continuous representative is zero everywhere, so its derivative is zero almost everywhere. Hence $J$ is one-to-one.

Its range is exactly $\{u\in H^1(0,1):u(0)=0\}$, and $Au=i{u}^{\prime }$. This contains the smooth compactly supported functions in $(0,1)$, so $A$ is a densely defined operator. Closedness can also be seen directly: if $u_n\to u$ and $A{u}_n\to v$ in $L^2$, then $u_n=J(-iA{u}_n)\to J(-iv)$, giving $u=J(-iv)\in D(A)$ and $Au=v$.

For every $\lambda \in \mathbb{C}$,If $\lambda \ne 0$, the allowed fact $\sigma (J)=\{0\}$ makes $J-(i/\lambda )I$ invertible. At $\lambda =0$, $\lambda J-iI=-iI$ is invertible too. Thus the bounded operatorhas range in $D(A)$ and is a two-sided inverse of $\lambda I-A$. Equivalently, solving the initial-value differential equation givesThis is bounded for every fixed $\lambda$ on the finite interval. Therefore Volterra inverse differentiation has empty spectrum:The familiar nonempty-spectrum theorem for bounded operators does not apply to this unbounded operator.

The subspace $H_0$ is the kernel of the bounded functional $f\mapsto Jf(1)={\int }_0^{1}f$, so it is closed. Its image under $J$ isthe zero-trace Sobolev space. If $u_n\in D(A_0)$, $u_n\to u$ and $A_0{u}_n\to v$, then $-i{A}_0{u}_n\in H_0$ converges to $-iv\in H_0$, and $u=J(-iv)\in D(A_0)$. Thus $A_0$ is closed. It is densely defined because it also contains the smooth compactly supported functions.

For $u,v\in D(A_0)$, integration by parts and their zero endpoint traces giveHence $A_0$ is closed and symmetric.

For any $v\in C^1([0,1])$, the same formula gives $⟨A_{0}u,v⟩=⟨u,i{v}^{\prime }⟩$, with no endpoint term because $u$ vanishes there. Consequently $v\in D(A_0^{\ast })$ and $A_0^{\ast }v=i{v}^{\prime }$, proving the requested inclusion. In fact, testing against compactly supported smooth $u$ shows that every element of $D(A_0^{\ast })$ has weak derivative in $L^2$; conversely the integration-by-parts formula applies to every $v\in H^1(0,1)$. Thuswith no boundary restrictions.

For any $\lambda \in \mathbb{C}$, the nonzero function $v_{\lambda }(x)=e^{-i\lambda x}$ belongs to this domain and satisfies $i{v}_{\lambda }^{\prime }=\lambda v_{\lambda }$. Conversely this first-order equation has only scalar multiples of that exponential. ThereforeFor any $\lambda$, the adjoint eigenfunction with eigenvalue $\overline{\lambda }$ annihilates $\mathrm{Ran}(\lambda I-A_0)$, so this range is not all of $H$. Explicitly the resolvent equation with zero initial value has the solution $R_{\lambda }h$ above, and its endpoint value is zero exactly whenThus the range is a proper closed hyperplane for every $\lambda$. This proves that the two-endpoint symmetric derivative has whole complex spectrum:There are no eigenvalues of $A_0$ itself: the eigenfunction equation together with $u(0)=0$ forces the exponential coefficient to vanish. The operator is symmetric but not self-adjoint, since its adjoint has a strictly larger domain; symmetry alone does not force a real spectrum.

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

A densely defined operator $T$ on a Hilbert space is symmetric whenEquivalently, $T$ agrees with its adjoint operator on $D(T)$, while $D(T^{\ast })$ may be larger. A self-adjoint operator additionally requires equality of these operator domains.