Hilbert space ($\mathcal{H}$)

A Hilbert space is a real or complex inner product space that is complete in the norm $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Completeness means that every Cauchy sequence has a limit in the space: if its terms become arbitrarily close in this norm, they converge to an allowed vector. The inner product supplies lengths and orthogonality; completeness ensures that limiting approximation procedures do not leave the space.

Finite-dimensional Euclidean spaces are examples, but the same geometry also works for sequences and functions. In the l2 sequence space,In the L2 space on $[-1,1]$ with Lebesgue measure,These formulas use the convention linear in the first argument. Elements of the L2 space are equivalence classes of square-integrable functions agreeing almost everywhere, so zero distance really identifies the same element. A vector can therefore be a whole function, and its squared length can measure an integrated squared error.

For example, approximating $f(x)=x^2$ by $a+bx$ is a nearest-point problem in this Hilbert space. The allowed approximations form the linear span of $1,x$, and the orthogonal projection is the constant $1/3$. Its residual $x^2-1/3$ is orthogonal to both allowed directions because its integral and its inner product with $x$ vanish. The geometry therefore identifies the best approximation without searching through every pair of coefficients.

This is useful whenever a complicated vector must be represented by a tractable collection of directions. Fourier series approximate functions with orthogonal oscillations; the Parseval identity relates the sum of squared coefficients to the squared norm. Least-squares solutions approximate attainable data using a linear operator. The Hilbert projection theorem explains why a nearest element exists and is unique for a nonempty closed convex set. Distinct application settings share the same geometry of useful components and orthogonal residuals. Casey Rodriguez's MIT lectures develop these connections between Hilbert spaces, Fourier series, and nearest-point problems.^[ref]

A historical mathematical setting is David Hilbert's Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen (1912). Its preface states that it reprints six communications published during 1904–1910 on linear integral equations. This provides a concrete source for the integral-equation setting associated with the name, while the abstract definition above incorporates later developments.^[ref]

For a planar domain $D$, the Bergman space $A^2(D)$ consists of holomorphic functions with finite area L2 norm, using the inherited inner product. It is a closed subspace of $L^2(D)$. Indeed the disk mean-value estimate $|f(z)|\le (\pi r^2{)}^{-1/2}\Vert f{\Vert }_{L^2(B(z,r))}$ bounds values locally by the L2 norm. An $L^2$-Cauchy sequence of holomorphic functions consequently converges uniformly on compact subsets to a holomorphic function, agreeing almost everywhere with its $L^2$ limit.

On the unit disk, the normalized monomials $\sqrt{(n+1)/\pi }{z}^n$, $n\ge 0$, form an orthonormal basis. Angular integration makes them orthogonal and gives $\Vert z^n{\Vert }_2^2=\pi /(n+1)$. Applying the same integration to a Taylor expansion gives $\Vert {\sum }_n{a}_n{z}^n{\Vert }_2^2=\pi {\sum }_n|a_n{|}^2/(n+1)$, proving completeness of this basis.

In the area Hilbert space of the unit disk, conjugates of Bergman space functions vanishing at zero are orthogonal to all holomorphic functions, since ${\int }_{D}hgdA=\pi h(0)g(0)$. This follows from the angular mean-value property and the Cauchy-Schwarz inequality, which makes the product integrable. But the radial function $|z{|}^2-1/2$ is also orthogonal to every holomorphic function: its radial integral against the constant angular mean is zero. It is not conjugate holomorphic, since it is real-valued and nonconstant. Thus the conjugate-holomorphic family is a proper subset of the orthogonal complement.

A closed subspace of a Hilbert space reduces a bounded operator when both the operator and its adjoint leave the subspace invariant. Equivalently, the orthogonal projection onto it commutes with the operator: the adjoint invariance makes its orthogonal complement invariant too. For a star-closed algebra of operators, the closed span of the orbit of any vector is reducing for the whole algebra. This enables an orthogonal direct-sum decomposition into cyclic actions when constructing Borel functional calculus for a normal operator.

The Hilbert tensor product is the completion of the algebraic tensor product of Hilbert spaces with $⟨u\otimes v,u^{\prime }\otimes v^{\prime }⟩=⟨u,u^{\prime }⟩⟨v,v^{\prime }⟩$, extended sesquilinearly. Orthonormal bases identify this inner product with the Euclidean inner product on coefficient arrays, proving positivity. In particular, $⟨u^{\otimes m},v^{\otimes m}⟩=⟨u,v{⟩}^m$ and $\Vert u^{\otimes m}\Vert =\Vert u{\Vert }^m$. These identities turn scalar power series into vector embeddings, as in the proof of the Grothendieck inequality.

The Hardy space of the circle is the closed subspace of L2 space whose negative Fourier coefficients vanish. The orthogonal projection onto it is denoted $P_{+}$. A continuous symbol $f$ defines a Toeplitz operator $T_f=P_{+}{M}_f{|}_{H^2}$. For continuous $f,g$, $T_f{T}_g-T_{fg}$ is compact: it is finite rank for Laurent polynomials, and uniform approximation extends the result.

The Bargmann-Fock space is a Hilbert space of entire holomorphic functions with Gaussian-weighted area inner product. If $f(z)=\sum a_n{z}^n$, angular integration and the Gaussian integral give $\Vert f{\Vert }^2=\sum n!|a_n{|}^2$. Conversely any such square-summable coefficient sequence defines an entire function, since $\sum |a_n{z}^n|\le (\sum n!|a_n{|}^2{)}^{1/2}{e}^{|z{|}^2/2}$. This proves completeness and shows that $z^n/\sqrt{n!}$ is an orthonormal basis. The same bound shows boundedness of evaluation and gives reproducing kernel $e^{z\bar{w}}$.

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.

A vector subspace closed in the norm topology of a Hilbert space is complete in the inherited norm and inner product. Its orthogonal complement gives a decomposition of the ambient Hilbert space into an orthogonal direct sum. This closedness permits orthogonal projection and ensures limits used in range arguments remain in the subspace. Every finite-dimensional vector subspace is closed.

A linear map $V:\mathcal{H}\to \mathcal{K}$ is an isometry when it preserves inner products, equivalently $V^{†}V=I$. It need not be surjective. It maps an orthonormal basis to an orthonormal family, and $\rho \mapsto V\rho V^{†}$ preserves the nonzero eigenvalues and the Von Neumann entropy of a density operator.

A linear isometry from a subspace of a finite-dimensional Hilbert space into the same ambient space extends to a unitary operator. Complete an orthonormal basis of the input subspace and its isometric image to full orthonormal bases, then map the first to the second. The equal complement dimensions make this possible. Orthogonality of the specified image columns, not normalization alone, is essential. In an infinite-dimensional space arbitrary isometries need not be surjective; the finite-dimensional same-space hypothesis cannot be dropped.

An N-term approximation in an orthonormal basis uses at most $N$ basis functions. For a chosen index set $\Lambda$, its optimal coefficients are the inner products with those functions, and its squared error is the sum of the omitted squared coefficients by the Parseval identity.

The best N-term approximation in an orthonormal basis retains $N$ coefficients of greatest absolute value. It minimizes the squared Hilbert space error over all index sets of size at most $N$. The selection makes the approximation nonlinear in the data, even though coefficient extraction is linear.

For a bounded function on an interval with finitely many jumps and piecewise $C^{\alpha }$ bounds, localized orthonormal wavelets with sufficient vanishing moments give squared best N-term approximation error $O(N^{-2\alpha })$. There are $O(2^j)$ smooth-region coefficients of size $O(2^{-j(\alpha +1/2)})$, and only $O(K)$ jump-crossing coefficients of size $O(2^{-j/2})$ per level. Retain all coefficients through level $J$ and the jump-crossing coefficients through $L=\lceil 2\alpha J\rceil$ when $\alpha >1/2$. This costs $O(2^J+KJ)$ and leaves squared tails $O(2^{-2\alpha J}+2^{-L})$. Choosing $2^J$ comparable to $N$ proves the rate by the Parseval identity.

A linear N-term approximation fixes its index set independently of the approximated function. Wavelet approximation normally orders by increasing resolution, whereas Fourier series approximation normally keeps the lowest frequencies. An arbitrary reordering can change or destroy a claimed convergence rate.

The Hilbert space completion of an inner product space $V$ is a Hilbert space containing an isometric dense copy of $V$. It can be constructed from Cauchy sequences in $V$, identifying two sequences when the norm of their difference tends to zero.

The space of finitely supported sequences with the l2 sequence space norm illustrates the need for completion. Let $u^{(n)}=(1,1/2,\dots ,1/n,0,\dots )$. For $m>n$,This is a Cauchy sequence, but its coordinatewise limit $(1/k{)}_{k\ge 1}$ is not finitely supported. It belongs to the l2 sequence space, which is the completion. The same construction permits infinitely many small components while controlling their total squared size. An infinite orthonormal basis of a Hilbert space consequently permits norm-convergent series; a Hamel basis represents each vector by a finite sum, so these two uses of "basis" must be distinguished.

Start with finite real linear combinations of $1$, $\cos (\lambda x)$ and $\sin (\lambda x)$ with arbitrary positive real frequencies, and use the inner product $⟨f,g{⟩}_M={\lim }_{R\to \infty }{R}^{-1}{\int }_{-R}^{R}fg$. Product-to-sum identities give existence of these averages and mutual orthogonality of distinct frequencies. The squared norm is $2a_0^2+{\sum }_{\lambda }(a_{\lambda }^2+b_{\lambda }^2)$. Its Hilbert space completion contains an uncountable orthonormal set, so is not a separable Hilbert space. This construction does not equip all locally square-integrable functions with an inner product: compactly supported nonzero functions have zero mean square, and other functions have divergent or nonexistent averages.

A Hilbert space is separable when it has a countable dense subset, equivalently a countable Hilbertian basis.

For every nonempty closed convex set $C$ in a Hilbert space and every $x$ in that space, there is a unique $p\in C$ satisfyingClosedness and completeness let a limiting approximation attain the infimum; convexity prevents two different minimizers. The proof idea is to choose a sequence approaching the infimum. The parallelogram identity and the admissible midpoint make it a Cauchy sequence; applying the same identity to two putative minimizers proves uniqueness.

For a closed subspace of a Hilbert space, this nearest-point map is an orthogonal projection and is linear. For a general closed convex set it need not be linear: projecting the real line onto $[0,1]$ clips values below zero to zero and values above one to one. The theorem is therefore useful for both linear approximation and constrained nearest-point problems.^[ref]

Closedness is essential even in familiar examples. For the open interval $(0,1)\subseteq \mathbb{R}$ and $x=0$, the distance infimum is zero but no nearest point exists. The same obstruction appears when approximating $(1/k{)}_{k\ge 1}$ by finitely supported sequences: the infimum is zero, but the limit lies outside that subspace until we take its Hilbert space completion.

For a closed subspace $F$ of a Hilbert space,The closest point $P_{F}x$ gives the first component, and differentiating $\Vert x-P_{F}x-tu{\Vert }^2$ at zero shows that the remainder is orthogonal to every $u\in F$.

The orthogonal complement of a subset $F$ of an inner-product space is

For a linear subspace $F$ of a Hilbert space,Thus $F=F^{\perp \perp }$ exactly when $F$ is closed.

The orthogonal projection $P_F$ onto a closed subspace of a Hilbert space $F$ sends $x$ to its component in the orthogonal decompositionThe residual $r$ is orthogonal to every permitted direction. For any $z\in F$, the Pythagorean theorem givesThus $P_{F}x$ uniquely minimizes the distance to $F$. It is a linear map, is idempotent, and is nonexpansive: $\Vert P_{F}x-P_{F}y\Vert \le \Vert x-y\Vert$; taking $y=0$ gives $\Vert P_{F}x\Vert \le \Vert x\Vert$.

In ${\mathbb{R}}^2$, project $v=(1,3)$ onto the line spanned by $u=(2,1)$. The inner product givesThe check $⟨r,u⟩=-2+2=0$ confirms orthogonality, and $\Vert v{\Vert }^2=10=5+5=\Vert P_{F}v{\Vert }^2+\Vert r{\Vert }^2$. Projecting again changes nothing, since the result already lies on the line.

For a finite orthonormal basis $e_1,\dots ,e_m$ of $F$, the same operation is $P_{F}x={\sum }_j⟨x,e_j⟩e_j$, using the convention linear in the first argument. Replacing coordinate vectors by functions gives exactly the same approximation principle. For $f(x)=x^2$ in the real L2 space on $[-1,1]$ and $F=\mathrm{span}\{1,x\}$, the directions are orthogonal, with squared norms $2$ and $2/3$. Their inner products with $f$ are $2/3$ and zero, soDirect integration gives the squared error $\Vert x^2-1/3{\Vert }^2=8/45$. For any competing affine polynomial, the orthogonal residual yieldsEvery deviation from $a=1/3$, $b=0$ adds a nonnegative error. The best function need not pass through any selected data point: it minimizes an integrated error across the whole interval.

If idempotent matrices $P_1,\dots ,P_k$ satisfy $P_i{P}_j=0$ whenever $i\ne j$, then $({\sum }_i{P}_i{)}^2={\sum }_{i,j}{P}_i{P}_j={\sum }_i{P}_i$. Thus the sum is idempotent. For a real orthonormal set $q_i$, the matrices $P_i=q_i{q}_i^T$ meet this condition. Their sum is the orthogonal projector onto their span; the symmetric matrix property follows by transposing the sum. Arbitrary sums of idempotent matrices need not be idempotent.

The foot from $\mathbf{c}$ to the line $\mathbf{a}+t\mathbf{u}$, with $\mathbf{u}\ne 0$, is its orthogonal projection onto that line. Solving $(\mathbf{c}-\mathbf{a}-t\mathbf{u})\cdot \mathbf{u}=0$ gives the displayed formula and proves uniqueness. It belongs to the line segment with endpoints $\mathbf{a}$ and $\mathbf{a}+\mathbf{u}$ exactly when $0\le (\mathbf{c}-\mathbf{a})\cdot \mathbf{u}\le \Vert \mathbf{u}{\Vert }^2$.

For a nonzero closed subspace of a Hilbert space $V$ and a closed subspace of a Hilbert space $W$, this angle measures the worst loss under projection from $V$ to $W$. Its positive cosine is a uniform lower bound for $P_W{|}_V$. The order of the subspaces matters in general. This is distinct from the smallest angle between two subspaces, whose cosine is a supremum. For equal finite dimensions the cosine is the smallest singular value of the cross Gram matrix of orthonormal bases. When $V$ is zero, the infimum is over an empty unit sphere and does not define an angle in $[0,\pi /2]$; use a lower-bound formulation instead.

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.

For two finite-dimensional subspaces with projections $P,Q$, define $\cos \vartheta ={\sup }_{\Vert u\Vert =\Vert v\Vert =1}|⟨u,v⟩|=\Vert PQ\Vert$, with vectors drawn from the respective subspaces. The angle is zero when they intersect nontrivially. To measure separation after removing a common intersection, restrict both subspaces to its orthogonal complement. The resulting angle controls sums of positive operators in the Kitaev geometrical lemma.

Suppose positive operators $A,B$ have positive eigenvalues at least $\gamma$ and their nullspaces meet only at zero. If $\vartheta$ is their smallest angle between two subspaces, then $A+B\ge 2\gamma {\sin }^2(\vartheta /2)I$. Indeed $A+B\ge \gamma (2I-P-Q)$, while $(P+Q{)}^2\le (1+\cos \vartheta )(P+Q)$ gives $P+Q\le (1+\cos \vartheta )I$. With a common nullspace, apply the same proof on its orthogonal complement to bound the positive spectral gap.

For orthogonal projections $P,Q$ with $\Vert PQ\Vert \le q<1$, the identity $(PQ{)}^r=PQ(QPQ{)}^{r-1}$ and $\Vert QPQ\Vert =\Vert PQ{\Vert }^2$ imply $\Vert (PQ{)}^r\Vert \le q^{2r-1}$ for $r\ge 1$. When the projections have a common fixed subspace, first remove its orthogonal projection. This estimate is stronger than generic submultiplicativity and is useful for alternating local projection layers.

Every bounded linear functional on a Hilbert space is inner product with one unique vector, with equal norms.

For a nonzero bounded functional $f$ on a Hilbert space, choose nonzero $z\in (\ker f{)}^{\perp }$. Thenimplies $f(x)=⟨x,f(z)z/\Vert z{\Vert }^2⟩$ in the real case. Cauchy-Schwarz and evaluation on the representing vector give equality of the two norms.

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.

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.

On a complex inner product space $V$, with the inner product conjugate-linear in its first argument, an everywhere-defined linear map $T:V\to V$ admits an adjoint on $V$ if there is a linear map $T^{\ast }:V\to V$ satisfyingThe adjoint, when it exists, is unique: the difference of two candidate values at $y$ is orthogonal to every vector and hence to itself. In a finite-dimensional vector space, it always exists, and its matrix in an orthonormal basis is the conjugate transpose. In a space that is not complete, existence must instead be checked.

The identity $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y\in V$ supplies such an adjoint directly, namely $T^{\ast }=T$. This algebraic statement should be distinguished from self-adjointness of a linear operator on a completed Hilbert space, which additionally compares the domains of an operator and its Hilbert-space adjoint. The Gaussian polynomial operator is a useful example where the identity holds on the entire polynomial inner product space and can be used on every finite-dimensional invariant polynomial subspace.

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.

For a densely defined operator on a Hilbert space, $h$ belongs to $D(T^{\ast })$ exactly when the functional $f\mapsto ⟨Tf,h⟩$ is bounded in the ambient Hilbert norm on $D(T)$. Density makes its representing vector $T^{\ast }h$ unique. Differential expressions alone do not determine this domain.

// Target: analysis.bigb

When compositions and adjoints are defined on the common test domain, reversing products gives $[A,B{]}^{\ast }=[B^{\ast },A^{\ast }]$. In particular a self-adjoint Dolbeault Laplacian and the mutually adjoint Lefschetz operators satisfy $[{\Delta }_{\bar{\partial }},L{]}^{\ast }=[\Lambda ,{\Delta }_{\bar{\partial }}]$.

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.

Its adjoint is $i{f}^{\prime }$ on all of $H^1(0,1)$, with no endpoint conditions. Every complex $\lambda$ is an adjoint eigenvalue, with eigenfunction $e^{-i\lambda x}$. Thus every complex $\lambda$ obstructs surjectivity of $\lambda I-A_0$. The original operator has no eigenvalues because its zero initial value kills any solution of $i{f}^{\prime }=\lambda f$.

// Target: analysis.bigb

For a linear map $T:V\to W$ between finite-dimensional inner product spaces,The first equality follows directly from the defining identity for the adjoint operator; the second follows by taking orthogonal complements. In an infinite-dimensional Hilbert space, the second equality generally requires closure of the image.

A bounded operator $T$ on a Hilbert space is invertible exactly when both $T$ and $T^{\ast }$ are bounded below. A lower bound on $T$ makes it injective with closed range, while a lower bound on $T^{\ast }$ gives $\ker T^{\ast }=0$ and hence dense range throughClosed dense range is the whole space, and the lower bound controls the inverse.

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.

The formal adjoint of a differential operator $L$ is characterized by moving $L$ between factors under an integral by integration by parts, while discarding boundary terms for compactly supported functions.

The formal adjoint of an infinite matrix does not specify the domain of the true adjoint. For the weighted difference generator $(Qf{)}_n=4^n(f_{n-1}/2-f_n)$, its formal positive-eigenvalue solution is $h_n=2^{-n}{h}_0{\prod }_{k<n}(1+\alpha 4^{-k})$. This is square summable but does not belong to the true adjoint domain of the maximal operator: testing against $f_n=2^{-n}$ contradicts the adjoint identity. The nonzero limiting boundary term is the obstruction.

// Target: analysis.bigb

For smooth positive $f$, let $L=f^{-1}\nabla \cdot (f\nabla )+k^2$. Its unweighted bilinear transpose is the displayed operator. If $v$ satisfies $\alpha v+\beta {\partial }_{n}v=0$, its transpose boundary condition is $\alpha u+\beta ({\partial }_{n}u-u{\partial }_n\log f)=0$. The resulting Green kernels satisfy $G_L({\mathbf{r}}_2,{\mathbf{r}}_1)=G_{L^{\mathrm{T}}}({\mathbf{r}}_1,{\mathbf{r}}_2)$. Equivalently $G_{L^{\mathrm{T}}}(\mathbf{r},{\mathbf{r}}_2)=f(\mathbf{r})G_L(\mathbf{r},{\mathbf{r}}_2)/f({\mathbf{r}}_2)$, exhibiting weighted acoustic Green-function reciprocity.

For an order-$m$ operator with smooth coefficients on a bounded piecewise smooth domain, integration by parts has no boundary remainder when at each boundary point either the entire boundary jet of $u$ through order $m-1$ vanishes or the entire such jet of $v$ vanishes. Every boundary summand pairs derivatives of the two functions of order at most $m-1$. It is insufficient to require only that, for each same multi-index $\beta$, one of ${\partial }^{\beta }u$ and ${\partial }^{\beta }v$ vanish: for $P={\partial }_x^2$, $u=x(a-x)\chi (y)$ and $v=1$ on a rectangle, all those same-index products vanish on the boundary but $\int (Pu)v=-2a\int \chi \ne 0$.

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.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

The Hermitian part of a square matrix is half its sum with its conjugate transpose. It is a Hermitian matrix and satisfies $\mathrm{Re}(x^{\ast }Ax)=x^{\ast }Hx$. Its largest eigenvalue is the numerical abscissa, controlling instantaneous growth of the Euclidean norm under $x^{\prime }=Ax$.

For a complex $n\times n$ matrix $B$, the real linear map $A\mapsto BA{B}^{\ast }$ on the $n^2$-dimensional real vector space of Hermitian matrices has determinant $|\det B{|}^{2n}$. A real basis of Hermitian matrices is also a complex basis of all matrices, so the matrix of this restriction equals the matrix of the complex-linear congruence map. Left and right multiplication separately have determinants $(\det B{)}^n$ and $(\overline{\det B}{)}^n$.