Riesz representation theorem

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 $(\overline{\det B}{)}^n$.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

For bounded positive operators $S,T$, membership $x\in \ker (S+T)$ gives $0=⟨Sx,x⟩+⟨Tx,x⟩$. Both terms are nonnegative, so both vanish. The Cauchy-Schwarz inequality for a positive operator makes $Sx=Tx=0$. The reverse inclusion is immediate. Commutativity is not required.

For a bounded positive operator $T$, the form $B(x,y)=⟨Tx,y⟩$ is positive semidefinite and Hermitian. Expanding $B(x+ty,x+ty)\ge 0$ for complex $t$ and minimizing its quadratic polynomial gives the displayed inequality. If $B(y,y)=0$, varying $t$ along the negative direction of $B(x,y)$ forces $B(x,y)=0$, covering the degenerate case. In particular $⟨Tx,x⟩=0$ implies $Tx=0$, by testing against arbitrary $y$.

A symmetric operator $L$ on a real Hilbert space is strictly positive definite when $⟨Lv,v⟩>0$ for every nonzero $v$ in its operator domain. Strict positivity proves uniqueness of a solution of $Lu=f$, but in infinite dimension it need not prove existence for every $f$. A uniformly positive definite symmetric operator additionally has a positive lower bound independent of $v$. Symmetry must be stated separately when positivity is defined only by a real quadratic expression: adding a skew-symmetric matrix changes the operator without changing that expression.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

The support of a positive operator is the orthogonal complement of its kernel. In finite dimension it is the span of eigenvectors with positive eigenvalues, equivalently the operator's image. Quantum relative entropy is finite only when the first state's support lies in the reference state's support.

Every bounded positive operator $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2}{)}^2=T$. The spectral theorem for normal operators on a separable Hilbert space constructs it by applying the scalar square-root function to the spectrum.

For bounded positive operators satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Self-adjointness and the Hilbert projection theorem decompose the whole Hilbert space as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.

The spectral theorem for normal operators gives a unique decompositionwhere $A_{+},A_{-}\ge 0$ and their supports are orthogonal. They are called the positive and negative parts of $A$. If $\mathrm{Tr}A=0$, then $\mathrm{Tr}{A}_{+}=\mathrm{Tr}{A}_{-}=\Vert A{\Vert }_1/2$.

The negative part of a Hermitian operator is $X_{-}={\sum }_i\max (-{\lambda }_i,0)P_i=(|X|-X)/2$. In this convention $X_{-}$ is itself a positive operator, and $X=X_{+}-X_{-}$. Its support is orthogonal to that of the positive part of a Hermitian operator.

For a Hermitian operator with spectral decomposition $X={\sum }_i{\lambda }_i{P}_i$, its positive part is $X_{+}={\sum }_i\max ({\lambda }_i,0)P_i=(|X|+X)/2$. It is a positive operator supported on the positive spectral subspace. Together with the negative part of a Hermitian operator it satisfies $X=X_{+}-X_{-}$ and $|X|=X_{+}+X_{-}$.

For Hermitian operators $A$ and $B$, one writes $A\le B$ when $B-A$ is positive semidefinite. This partial order is also called the semidefinite order.

For bounded positive self-adjoint operators, $0\le A\le B$ implies $A^{1/2}\le B^{1/2}$. Thus the nonnegative square-root function is operator monotone.

For a subspace $U$ of a finite-dimensional inner-product space,This follows directly by moving $T$ across the inner product and using $(U^{\perp }{)}^{\perp }=U$.

An operator is normal when $T{T}^{\ast }=T^{\ast }T$. Equivalently,for every $x$; the converse follows by polarizing the quadratic form of the self-adjoint commutator $T^{\ast }T-T{T}^{\ast }$.

For a normal operator $T$ on a real or complex inner product space and a polynomial $f(x)={\sum }_j{a}_j{x}^j$, the adjoint operator identity givesEvery power of $T$ commutes with every power of $T^{\ast }$, since $T{T}^{\ast }=T^{\ast }T$. Expanding finite sums consequently proves that $f(T)$ commutes with $f(T{)}^{\ast }$, so it is a normal operator. The bar on $f$ means coefficientwise complex conjugation, and is essential for complex coefficients.

For a normal operator on a real or complex inner product space, equality of $\Vert Tv\Vert$ and $\Vert T^{\ast }v\Vert$ gives $\ker T=\ker T^{\ast }$. If $T^{2}v=0$, put $w=Tv$. Then $Tw=0$ forces $T^{\ast }w=0$, and the adjoint operator identity givesThus $Tv=0$ and $\ker T^2=\ker T$. Applying this implication successively to $T^{k-2}v,T^{k-3}v,\dots ,v$ provesIn particular a nilpotent normal operator is zero. The argument uses no spectral theorem.

For a normal bounded operator, $\sigma (T)={\sigma }_{\mathrm{ap}}(T)$. If $T-\lambda I$ is bounded below, its range is closed; normality makes its adjoint injective, so the range is also dense and the operator is invertible.

The spectral theorem represents a normal operator by multiplication by its spectral parameter. In a finite-dimensional inner product space, this is diagonalization in an orthonormal basis; for a self-adjoint operator on a Hilbert space, it uses a projection-valued measure. Unbounded operators additionally require their natural square-integrability domain.

Every normal operator on a finite-dimensional complex inner-product space has an orthonormal basis of eigenvectors. An eigenvector is also an eigenvector of the adjoint with conjugate eigenvalue, so its orthogonal complement reduces the operator and induction applies.

A finite family of pairwise commuting normal operators on a finite-dimensional complex inner product space, commuting also with each other's adjoints, generates a commutative unital C-star algebra. Its finite maximal ideal space identifies the algebra with functions on finitely many points. The characteristic functions of those points pull back under the Gelfand transform to mutually orthogonal self-adjoint linear projections summing to the identity. Their ranges are the simultaneous eigenspaces, giving simultaneous diagonalization without choosing separate incompatible eigenbases.

For a Hermitian matrix with unitary diagonalization $A=U\mathrm{diag}({\lambda }_i)U^{\ast }$, define $f(A)=U\mathrm{diag}(f({\lambda }_i))U^{\ast }$ for a function on its finite spectrum. This acts on eigenvalues, unlike an entrywise matrix function $f[A]$. For example, the spectral square of a matrix is its ordinary matrix product with itself, whereas its entrywise square is a Hadamard power.

A normal operator on a separable Hilbert space has a unique projection-valued spectral measure $E$ such that $A=\int zdE(z)$, with the natural square-integrability domain in the unbounded case.

Every self-adjoint operator on a separable complex Hilbert space is diagonal modulo a compact operator. For a bounded operator the compact self-adjoint error can have arbitrarily small Hilbert-Schmidt norm. In a cyclic spectral representation, partition a bounded real interval into $N$ intervals of width at most $\delta$ and successively bisect. Orthogonal step-function differences give basis vectors supported in the parent intervals. A diagonal value chosen in each support interval gives total squared column error at most $3N{\delta }^2$. Since $N=O({\delta }^{-1})$, this tends to zero. Apply the construction to countably many cyclic summands with summable squared error bounds. For an unbounded operator first split into bounded spectral intervals, obtaining a bounded compact error on the same domain.

Bounded self-adjoint operators on separable infinite-dimensional complex Hilbert spaces are unitarily equivalent modulo compact operators exactly when they have the same essential spectrum in the Calkin algebra. The Weyl-von Neumann theorem diagonalizes each modulo a compact error. Their diagonal cluster sets are the essential spectra; cluster-set matching of bounded sequences pairs the two diagonal lists with differences tending to zero, giving a compact diagonal difference. The separability assumption is essential: projections with countable and uncountable range dimensions can have the same essential spectrum but cannot differ by a compact operator after unitary conjugation.

A projection-valued measure $E$ assigns an orthogonal projection $E(B)$ to each measurable set, with $E(\varnothing )=0$, $E(X)=I$, $E(B\cap C)=E(B)E(C)$, and countable additivity in the strong operator topology on pairwise disjoint sets.

The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

For a unital star-representation $\pi :C(K)\to \mathcal{B}(H)$, the Riesz-Markov-Kakutani representation theorem gives regular measures ${\mu }_{x,y}$ representing $f\mapsto ⟨\pi (f)x,y⟩$. Define $P(E)$ by $⟨P(E)x,y⟩={\mu }_{x,y}(E)$. Positivity and ${\mu }_{x,x}(K)=\Vert x{\Vert }^2$ make these positive contractions. The identities ${\mu }_{\pi (f)x,y}=f{\mu }_{x,y}$ imply commutation with $\pi (f)$ and ${\mu }_{P(E)x,y}=1_E{\mu }_{x,y}$; hence $P(F)P(E)=P(F\cap E)$. Thus each $P(E)$ is an orthogonal projection. Scalar countable additivity and orthogonality give countable additivity in the strong operator topology, constructing the normalized regular projection-valued measure.

If $A$ is a commutative C-star algebra of bounded operators on a Hilbert space containing its identity, there is a unique regular projection-valued measure $E$ on the character space ${\Phi }_A$ such that $E({\Phi }_A)=I$ and $a={\int }_{{\Phi }_A}\hat{a}dE$ for every $a\in A$, where $\hat{a}$ is the Gelfand transform.

For a faithful representation of $C(K)$ by integration against a regular projection-valued measure $P$, every nonempty open subset $U$ satisfies $P(U)\ne 0$. Choose a nonzero continuous function supported inside $U$, using compact Hausdorff normality. If $P(U)=0$, its spectral integral vanishes, contradicting faithfulness. Regularity is understood through the scalar spectral measures.

A spectral projector is the value $E(B)$ of the projection-valued measure of a normal operator on a measurable part $B$ of its spectrum. In finite dimension, the spectral projector onto an eigenspace extracts the corresponding eigenvector component.

For vectors $f,g$, the scalar spectral measure is ${\mu }_{f,g}(S)=⟨E(S)f,g⟩$. Functional calculus satisfies $⟨h(A)f,g⟩=\int hd{\mu }_{f,g}$.

For a self-adjoint operator $A$, its projection-valued spectral measure of a normal operator assigns each Borel set $S\subseteq \mathbb{R}$ an orthogonal projection $E_A(S)$. For a normalized quantum state $\psi$, the probability of a measurement result in $S$ is $⟨\psi ,E_A(S)\psi ⟩$. In finite dimensions, $E_A(S)$ is the sum of the orthogonal projections onto eigenspaces whose eigenvalues lie in $S$. This generalizes an individual eigenvector probability to degenerate eigenvalues and continuous spectra. The functional-calculus notation $1_S(A)$ uses the indicator function of the set.

For a bounded normal operator $T$ with spectral measure of a normal operator $P$, any projection $P(E)$ commutes with $T$ and $T^{\ast }$ by the Borel functional calculus for a normal operator. Its range is a closed reducing subspace. If the spectrum has two points, choose disjoint nonempty relative open sets around them. Full support of a faithful spectral measure makes both projections nonzero and their product zero, so either range is nonzero and proper, in particular an invariant subspace.

For a projection-valued measure $E$ and vectors $v,w$, the scalar spectral measure is the complex measure ${\mu }_{v,w}(B)=⟨E(B)v,w⟩$. The positive measure ${\mu }_v={\mu }_{v,v}$ has total mass $\Vert v{\Vert }^2$.

Scalar spectral measures ${\mu }_{v,w}^{(n)}$ converge weakly to ${\mu }_{v,w}$ when $\int fd{\mu }_{v,w}^{(n)}\to \int fd{\mu }_{v,w}$ for every bounded continuous function $f$. Uniform second-moment bounds supply the tightness needed to pass from polynomial tests to all such $f$.

Stone's formula recovers spectral projections of a self-adjoint operator from the jump of its resolvent across the real axis, with half weight at interval endpoints.