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In the mathematical discipline offunctional analysis, the concept of acompact operator on Hilbert space is an extension of the concept of a matrix acting on a finite-dimensional vector space; inHilbert space,compact operators are precisely the closure offinite-rank operators (representable by finite-dimensional matrices) in thetopology induced by theoperator norm. As such, results from matrix theory can sometimes be extended to compact operators using similar arguments. By contrast, the study of general operators on infinite-dimensional spaces often requires a genuinely different approach.

For example, thespectral theory of compact operators onBanach spaces takes a form that is very similar to theJordan canonical form of matrices. In the context of Hilbert spaces, a square matrix is unitarily diagonalizable if and only if it isnormal. A corresponding result holds for normal compact operators on Hilbert spaces. More generally, the compactness assumption can be dropped. As stated above, the techniques used to prove results, e.g., thespectral theorem, in the non-compact case are typically different, involving operator-valuedmeasures on thespectrum.

Some results for compact operators on Hilbert space will be discussed, starting with general properties before considering subclasses of compact operators.

Let${\displaystyle H}$ be aHilbert space and${\displaystyle L(H)}$ be the set ofbounded operators on${\displaystyle H}$. Then, an operator${\displaystyle T\in L(H)}$ is said to be acompact operator if the image of each bounded set under${\displaystyle T}$ isrelatively compact.

We list in this section some general properties of compact operators.

IfX andY are separable Hilbert spaces (in fact,X Banach andY normed will suffice), thenT :X →Y is compact if and only if it issequentially continuous when viewed as a map fromX with theweak topology toY (with the norm topology). (See (Zhu 2007, Theorem 1.14, p.11), and note in this reference that the uniform boundedness will apply in the situation whereF ⊆X satisfies (∀φ ∈ Hom(X,K)) sup{x**(φ) = φ(x) :x} < ∞, whereK is the underlying field. The uniform boundedness principle applies since Hom(X,K) with the norm topology will be a Banach space, and the mapsx** : Hom(X,K) →K are continuous homomorphisms with respect to this topology.)

The family of compact operators is a norm-closed, two-sided, *-ideal inL(H). Consequently, a compact operatorT cannot have a bounded inverse ifH is infinite-dimensional. IfST =TS =I, then the identity operator would be compact, a contradiction.

If sequences of bounded operatorsB_n →B,C_n →C in thestrong operator topology andT is compact, then${\displaystyle B_{n}T{C}_n^{\ast }}$ converges to${\displaystyle BT{C}^{\ast }}$ in norm.^[1] For example, consider the Hilbert space${\displaystyle {\ell }^2(\mathbf{N}),}$ with standard basis {e_n}. LetP_m be the orthogonal projection on the linear span of {e₁, ...,e_m}. The sequence {P_m} converges to the identity operatorI strongly but not uniformly. DefineT by${\displaystyle T{e}_n=\frac{1}{n^2}{e}_n.}$T is compact, and, as claimed above,P_mT →IT =T in the uniform operator topology: for allx,$${\displaystyle \Vert P_{m}Tx-Tx\Vert \le {\left(\frac{1}{m+1}\right)}^2\Vert x\Vert .}$$

Notice eachP_m is a finite-rank operator. Similar reasoning shows that ifT is compact, thenT is the uniform limit of some sequence of finite-rank operators.

By the norm-closedness of the ideal of compact operators, the converse is also true.

The quotient C*-algebra ofL(H) modulo the compact operators is called theCalkin algebra, in which one can consider properties of an operator up to compact perturbation.

A bounded operatorT on a Hilbert spaceH is said to beself-adjoint ifT =T*, or equivalently,

$${\displaystyle ⟨Tx,y⟩=⟨x,Ty⟩,x,y\in H.}$$

It follows that ⟨Tx,x⟩ is real for everyx ∈H, thus eigenvalues ofT, when they exist, are real. When a closed linear subspaceL ofH is invariant underT, then the restriction ofT toL is a self-adjoint operator onL, and furthermore, theorthogonal complementL^⊥ ofL is also invariant underT. For example, the spaceH can be decomposed as the orthogonaldirect sum of twoT–invariant closed linear subspaces: thekernel ofT, and the orthogonal complement(kerT)^⊥ of the kernel (which is equal to the closure of the range ofT, for any bounded self-adjoint operator). These basic facts play an important role in the proof of the spectral theorem below.

The classification result for Hermitiann ×n matrices is thespectral theorem: IfM =M*, thenM is unitarily diagonalizable, and the diagonalization ofM has real entries. LetT be a compact self-adjoint operator on a Hilbert spaceH. We will prove the same statement forT: the operatorT can be diagonalized by an orthonormal set of eigenvectors, each of which corresponds to a real eigenvalue.

Theorem For every compact self-adjoint operatorT on a real or complex Hilbert spaceH, there exists anorthonormal basis ofH consisting of eigenvectors ofT. More specifically, the orthogonal complement of the kernel ofT admits either a finite orthonormal basis of eigenvectors ofT, or acountably infinite orthonormal basis {e_n} of eigenvectors ofT, with corresponding eigenvalues{λ_n} ⊂R, such thatλ_n → 0.

In other words, a compact self-adjoint operator can be unitarily diagonalized. This is the spectral theorem.

WhenH isseparable, one can mix the basis {e_n} with acountable orthonormal basis for the kernel ofT, and obtain an orthonormal basis {f_n} forH, consisting of eigenvectors ofT with real eigenvalues {μ_n} such thatμ_n → 0.

Corollary For every compact self-adjoint operatorT on a real or complex separable infinite-dimensional Hilbert spaceH, there exists a countably infinite orthonormal basis {f_n} ofH consisting of eigenvectors ofT, with corresponding eigenvalues{μ_n} ⊂R, such thatμ_n → 0.

Let us discuss first the finite-dimensional proof. Proving the spectral theorem for a Hermitiann ×n matrixT hinges on showing the existence of one eigenvectorx. Once this is done, Hermiticity implies that both the linear span and orthogonal complement ofx (of dimensionn − 1) are invariant subspaces ofT. The desired result is then obtained by induction for${\displaystyle T_{x^{\perp }}}$.

The existence of an eigenvector can be shown in (at least) two alternative ways:

1. One can argue algebraically: The characteristic polynomial ofT has a complex root, thereforeT has an eigenvalue with a corresponding eigenvector.
2. The eigenvalues can be characterized variationally: The largest eigenvalue is the maximum on the closed unitsphere of the functionf:R²ⁿ →R defined byf(x) =x*Tx = ⟨Tx,x⟩.

Note. In the finite-dimensional case, part of the first approach works in much greater generality; any square matrix, not necessarily Hermitian, has an eigenvector. This is simply not true for general operators on Hilbert spaces. In infinite dimensions, it is also not immediate how to generalize the concept of the characteristic polynomial.

The spectral theorem for the compact self-adjoint case can be obtained analogously: one finds an eigenvector by extending the second finite-dimensional argument above, then apply induction. We first sketch the argument for matrices.

Since the closed unit sphereS inR²ⁿ is compact, andf is continuous,f(S) is compact on the real line, thereforef attains a maximum onS, at some unit vectory. ByLagrange's multiplier theorem,y satisfies$${\displaystyle \mathrm{\nabla }f=\mathrm{\nabla }{y}^{\ast }Ty=\lambda \cdot \mathrm{\nabla }{y}^{\ast }y}$$for some λ. By Hermiticity,Ty = λy.

Alternatively, letz ∈Cⁿ be any vector. Notice that if a unit vectory maximizes ⟨Tx,x⟩ on the unit sphere (or on the unit ball), it also maximizes theRayleigh quotient:$${\displaystyle g(x)=\frac{⟨Tx,x⟩}{\Vert x{\Vert }^2},0\ne x\in {\mathbf{C}}^n.}$$

Consider the function:$${\displaystyle \{\begin{array}{l}h:\mathbf{R}\to \mathbf{R}\\ h(t)=g(y+tz)\end{array}}$$

By calculus,h′(0) = 0, i.e.,

Define:$${\displaystyle m=\frac{⟨Ty,y⟩}{⟨y,y⟩}}$$

After some algebra the above expression becomes (Re denotes the real part of a complex number)$${\displaystyle \mathrm{Re}(⟨Ty-my,z⟩)=0.}$$

Butz is arbitrary, thereforeTy −my = 0. This is the crux of proof for spectral theorem in the matricial case.

Note that while the Lagrange multipliers generalize to the infinite-dimensional case, the compactness of the unit sphere is lost. This is where the assumption that the operatorT be compact is useful.

Claim  IfT is a compact self-adjoint operator on a non-zero Hilbert spaceH and$${\displaystyle m(T):=sup\{|⟨Tx,x⟩|:x\in H,\Vert x\Vert \le 1\},}$$thenm(T) or −m(T) is an eigenvalue ofT.

Ifm(T) = 0, thenT = 0 by thepolarization identity, and this case is clear. Consider the function$${\displaystyle \{\begin{array}{l}f:H\to \mathbf{R}\\ f(x)=⟨Tx,x⟩\end{array}}$$

ReplacingT by −T if necessary, one may assume that the supremum off on the closed unit ballB ⊂H is equal tom(T) > 0. Iff attains its maximumm(T) onB at some unit vectory, then, by the same argument used for matrices,y is an eigenvector ofT, with corresponding eigenvalueλ = ⟨λy,y⟩ =⟨Ty,y⟩ =f(y) =m(T).

By theBanach–Alaoglu theorem and the reflexivity ofH, the closed unit ballB is weakly compact. Also, the compactness ofT means (see above) thatT :X with the weak topology →X with the norm topology is continuous^{[disputed –discuss]}. These two facts imply thatf is continuous onB equipped with the weak topology, andf attains therefore its maximumm onB at somey ∈B. By maximality,${\displaystyle \Vert y\Vert =1,}$ which in turn implies thaty also maximizes the Rayleigh quotientg(x) (see above). This shows thaty is an eigenvector ofT, and ends the proof of the claim.

Note. The compactness ofT is crucial. In general,f need not be continuous for the weak topology on the unit ballB. For example, letT be the identity operator, which is not compact whenH is infinite-dimensional. Take any orthonormal sequence {y_n}. Theny_n converges to 0 weakly, but limf(y_n) = 1 ≠ 0 =f(0).

LetT be a compact operator on a Hilbert spaceH. A finite (possibly empty) or countably infinite orthonormal sequence {e_n} of eigenvectors ofT, with corresponding non-zero eigenvalues, is constructed by induction as follows. LetH₀ =H andT₀ =T. Ifm(T₀) = 0, thenT = 0 and the construction stops without producing any eigenvectore_n. Suppose that orthonormal eigenvectorse₀, ...,e_{n − 1} ofT have been found. ThenE_n := span(e₀, ...,e_{n − 1}) is invariant underT, and by self-adjointness, the orthogonal complementH_n ofE_n is an invariant subspace ofT. LetT_n denote the restriction ofT toH_n. Ifm(T_n) = 0, thenT_n = 0, and the construction stops. Otherwise, by theclaim applied toT_n, there is a norm one eigenvectore_n ofT inH_n, with corresponding non-zero eigenvalue λ_n =±m(T_n).

LetF = (span{e_n})^⊥, where {e_n} is the finite or infinite sequence constructed by the inductive process; by self-adjointness,F is invariant underT. LetS denote the restriction ofT toF. If the process was stopped after finitely many steps, with the last vectore_m−1, thenF=H_m andS =T_m = 0 by construction. In the infinite case, compactness ofT and the weak-convergence ofe_n to 0 imply thatTe_n =λ_ne_n → 0, thereforeλ_n → 0. SinceF is contained inH_n for everyn, it follows thatm(S) ≤m({T_n}) = |λ_n| for everyn, hencem(S) = 0. This implies again thatS = 0.

The fact thatS = 0 means thatF is contained in the kernel ofT. Conversely, ifx ∈ ker(T) then by self-adjointness,x is orthogonal to every eigenvector {e_n} with non-zero eigenvalue. It follows thatF = ker(T), and that {e_n} is an orthonormal basis for the orthogonal complement of the kernel ofT. One can complete the diagonalization ofT by selecting an orthonormal basis of the kernel. This proves the spectral theorem.

A shorter but more abstract proof goes as follows: byZorn's lemma, selectU to be a maximal subset ofH with the following three properties: all elements ofU are eigenvectors ofT, they have norm one, and any two distinct elements ofU are orthogonal. LetF be the orthogonal complement of the linear span ofU. IfF ≠ {0}, it is a non-trivial invariant subspace ofT, and by the initial claim, there must exist a norm one eigenvectory ofT inF. But thenU ∪ {y} contradicts the maximality ofU. It follows thatF = {0}, hence span(U) is dense inH. This shows thatU is an orthonormal basis ofH consisting of eigenvectors ofT.

IfT is compact on an infinite-dimensional Hilbert spaceH, thenT is not invertible, hence σ(T), the spectrum ofT, always contains 0. The spectral theorem shows that σ(T) consists of the eigenvalues {λ_n} ofT and of 0 (if 0 is not already an eigenvalue). The set σ(T) is a compact subset of the complex numbers, and the eigenvalues are dense in σ(T).

Any spectral theorem can be reformulated in terms of afunctional calculus. In the present context, we have:

Theorem. LetC(σ(T)) denote theC*-algebra of continuous functions on σ(T). There exists a unique isometric homomorphismΦ :C(σ(T)) →L(H) such that Φ(1) =I and, iff is the identity functionf(λ) =λ, thenΦ(f) =T. Now we may define${\displaystyle g(T):=\mathrm{\Phi }(g)}$ (clearly this would hold when${\displaystyle g}$ is polynomial). Then it also holds, thatσ(g(T)) =g(σ(T)).

The functional calculus map Φ is defined in a natural way: let {e_n} be an orthonormal basis of eigenvectors forH, with corresponding eigenvalues {λ_n}; forf ∈C(σ(T)), the operator Φ(f), diagonal with respect to the orthonormal basis {e_n}, is defined by setting$${\displaystyle \mathrm{\Phi }(f)(e_n)=f({\lambda }_n)e_n}$$for everyn. Since Φ(f) is diagonal with respect to an orthonormal basis, its norm is equal to the supremum of the modulus of diagonal coefficients,$${\displaystyle \Vert \mathrm{\Phi }(f)\Vert =\underset{{\lambda }_n\in \sigma (T)}{sup}|f({\lambda }_n)|=\Vert f{\Vert }_{C(\sigma (T))}.}$$

The other properties of Φ can be readily verified. Conversely, any homomorphism Ψ satisfying the requirements of the theorem must coincide with Φ whenf is a polynomial. By theWeierstrass approximation theorem, polynomial functions are dense inC(σ(T)), and it follows thatΨ = Φ. This shows that Φ is unique.

The more generalcontinuous functional calculus can be defined for any self-adjoint (or even normal, in the complex case) bounded linear operator on a Hilbert space. The compact case described here is a particularly simple instance of this functional calculus.

Consider an Hilbert spaceH (e.g. the finite-dimensionalCⁿ), and a commuting set${\displaystyle \mathcal{F}\subseteq \mathrm{Hom}(H,H)}$ of self-adjoint operators. Then under suitable conditions, it can be simultaneously (unitarily) diagonalized.Viz., there exists an orthonormal basisQ consisting of common eigenvectors for the operators — i.e.,$${\displaystyle (\mathrm{\forall }q\in Q,T\in \mathcal{F})(\mathrm{\exists }\sigma \in \mathbf{C})(T-\sigma )q=0}$$

Notice we did not have to directly use the machinery of matrices at all in this proof. There are other versions which do.

We can strengthen the above to the case where all the operators merely commute with their adjoint; in this case we remove the term "orthogonal" from the diagonalisation. There are weaker results for operators arising from representations due to Weyl–Peter. LetG be a fixed locally compact hausdorff group, and${\displaystyle H=L^2(G)}$ (the space of square integrable measurable functions with respect to the unique-up-to-scale Haar measure onG). Consider the continuous shift action:$${\displaystyle \{\begin{array}{l}G\times H\to H\\ (gf)(x)=f(g^{-1}x)\end{array}}$$

Then ifG were compact then there is a unique decomposition ofH into a countable direct sum of finite-dimensional, irreducible, invariant subspaces (this is essentially diagonalisation of the family of operators${\displaystyle G\subseteq U(H)}$). IfG were not compact, but were abelian, then diagonalisation is not achieved, but we get a uniquecontinuous decomposition ofH into 1-dimensional invariant subspaces.

The family of Hermitian matrices is a proper subset of matrices that are unitarily diagonalizable. A matrixM is unitarily diagonalizable if and only if it is normal, i.e.,M*M =MM*. Similar statements hold for compact normal operators.

LetT be compact andT*T =TT*. Apply theCartesian decomposition toT: define$${\displaystyle R=\frac{T+T^{\ast }}{2},J=\frac{T-T^{\ast }}{2i}.}$$

The self-adjoint compact operatorsR andJ are called the real and imaginary parts ofT, respectively. ThatT is compact implies thatT* and, consequently,R andJ are compact. Furthermore, the normality ofT implies thatR andJ commute. Therefore they can be simultaneously diagonalized, from which follows the claim.

Ahyponormal compact operator (in particular, asubnormal operator) is normal.

The spectrum of aunitary operatorU lies on the unit circle in the complex plane; it could be the entire unit circle. However, ifU is identity plus a compact perturbation,U has only a countable spectrum, containing 1 and possibly, a finite set or a sequence tending to 1 on the unit circle. More precisely, supposeU =I +C whereC is compact. The equationsUU* =U*U =I andC =U −I show thatC is normal. The spectrum ofC contains 0, and possibly, a finite set or a sequence tending to 0. SinceU =I +C, the spectrum ofU is obtained by shifting the spectrum ofC by 1.

• LetH =L²([0, 1]). The multiplication operatorM defined by$${\displaystyle (Mf)(x)=xf(x),f\in H,x\in [0,1]}$$ is a bounded self-adjoint operator onH that has no eigenvector and hence, by the spectral theorem, cannot be compact.
• An example of a compact operator on a Hilbert space that is not self-adjoint is theVolterra operator, defined for a function${\displaystyle f\in L^2([0,1])}$ and a value${\displaystyle t\in [0,1]}$ as$${\displaystyle V(f)(t)={\int }_0^{t}f(s)ds.}$$ It is the operator corresponding to theVolterra integral equations.
• Define a Hilbert-Schmidt kernel${\displaystyle K:\mathrm{\Omega }\times \mathrm{\Omega }\to \mathbb{C}}$ on${\displaystyle \mathrm{\Omega }=[0,1]}$ and its associatedHilbert-Schmidt integral operator${\displaystyle T_K:L^2(\mathrm{\Omega })\to L^2(\mathrm{\Omega })}$ as$${\displaystyle (T_{K}f)(x)={\int }_0^{1}K(x,y)f(y)\mathrm{d}y.}$$ Then${\displaystyle T_K}$ is a compact operator; it is aHilbert–Schmidt operator with Hilbert-Schmidt norm${\displaystyle \Vert T_k{\Vert }_{\mathrm{H}\mathrm{S}}=\Vert K{\Vert }_{L^2}}$.
• ${\displaystyle T_K}$ is a compact self-adjoint operator if and only if${\displaystyle K(x,y)}$ is ahermitian kernel which, according toMercer's theorem, can be represented as$${\displaystyle K(x,y)=\sum {\lambda }_n{\phi }_n(x)\overline{{\phi }_n(y)},}$$ where${\displaystyle \{{\phi }_n\}}$ is an orthonormal basis of eigenvectors of${\displaystyle T_K}$, with eigenvalues${\displaystyle \{{\lambda }_n\}}$ and the sum converges absolutely and uniformly on${\displaystyle [0,1]}$.

• Calkin algebra
• Compact operator – Type of continuous linear operator
• Decomposition of spectrum (functional analysis) – Construction in functional analysis, useful to solve differential equations − If the compactness assumption is removed, operators need not have countable spectrum in general.
• Fredholm operator – Part of Fredholm theories in integral equations
• Singular value decomposition#Bounded operators on Hilbert spaces – Matrix decomposition − The notion of singular values can be extended from matrices to compact operators.
• Spectral theory of compact operators
• Strictly singular operator

• J. Blank, P. Exner, and M. Havlicek,Hilbert Space Operators in Quantum Physics, American Institute of Physics, 1994.
• M. Reed and B. Simon,Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1972.
• Zhu, Kehe (2007),Operator Theory in Function Spaces, Mathematical surveys and monographs, vol. 138, American Mathematical Society,ISBN 978-0-8218-3965-2

• Hilbert spaces
• Operator theory
• Linear operators

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