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The bilateral shift $(Tx{)}_n=x_{n+1}$ is unitary. It has empty point spectrum, while its approximate point spectrum is the unit circle. For $|\lambda |=1$, normalized vectors with entries ${\lambda }^n$ on an interval of length $N$ fail to satisfy $Tx=\lambda x$ only at the two endpoints, giving residual norm $\sqrt{2/N}$.

It is the two-sided sequence case of a shift operator.

An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

The spectrum of a bounded operator consists of $\lambda$ for which $T-\lambda I$ has no bounded two-sided inverse. The point spectrum consists of eigenvalues, so $\ker (T-\lambda I)\ne 0$. The approximate point spectrum consists of $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$; equivalently $T-\lambda I$ is not bounded below.

If $T-\lambda I$ is invertible, its inverse bounds it below and its Banach-space adjoint is invertible. Thus neither alternative on the right of the claimed union occurs. Conversely suppose $\lambda$ belongs to neither term and write $S=T-\lambda I$. There is $c>0$ with $\Vert Sx\Vert \ge c\Vert x\Vert$, so $S$ is injective and has closed range: convergence of $S{x}_n$ makes $x_n$ Cauchy, and completeness supplies a preimage of its limit. Also $\ker S^{\prime }=0$. The allowed annihilator identity makes the range dense, hence equal to $X$. Its inverse has norm at most $1/c$. This proves the spectrum decomposition through the Banach-space adjoint:For the right shift on ${\ell }^1$, $\Vert T\Vert =1$, so the allowed norm bound puts its spectrum in the closed unit disc. Under the linear pairing with ${\ell }^{\infty }$, its Banach-space adjoint is the left shift $T^{\prime }(b_1,b_2,\dots )=(b_2,b_3,\dots )$. For each $|\lambda |\le 1$ the bounded nonzero sequence $(1,\lambda ,{\lambda }^2,\dots )$ is an eigenvector of $T^{\prime }$ with eigenvalue $\lambda$. The union identity gives the reverse inclusion. Hence $\sigma (T)=\{\lambda :|\lambda |\le 1\}$. This Banach-space adjoint uses the complex-linear dual, so its eigenvalue is $\lambda$ itself.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda :\lambda I-T\text{ has no bounded two-sided inverse}\}$. Its point spectrum is the set of eigenvalues, where $\lambda I-T$ is not injective.

Here $T$ is linear and takes continuous functions to continuously differentiable functions. Also $\Vert Tx{\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$; the constant function one attains this bound at $t=1$, so $\Vert T\Vert =2$. Its range is contained in $C^1([0,1])$, a proper subset of $C([0,1])$, so zero is in its spectrum. It is not an eigenvalue: $Tx=0$ gives $x(0)=0$ at zero and $x=0$ upon differentiation.

For a nonzero eigenvalue, $\lambda x=Tx$ implies $x^{\prime }=x/\lambda$ and $(\lambda -1)x(0)=0$. A nonzero solution is therefore possible exactly at $\lambda =1$, with eigenfunctions $x(t)=c{e}^t$. Hence $p\sigma (T)=\{1\}$.

For the proposed resolvent putThe proposed $x$ is $y/\lambda +y(0)e^{t/\lambda }/[\lambda (\lambda -1)]+F/{\lambda }^2$. Consequently $x(0)=y(0)/(\lambda -1)$ andAdding $x(0)$ and subtracting from $\lambda x(t)$ gives exactly $y(t)$. Thus $\lambda I-T$ is onto for $\lambda \ne 0,1$. It is also injective by the eigenvalue calculation. The displayed formula defines a bounded map in the supremum norm, since its exponential kernels are bounded on the compact interval. It is therefore the bounded inverse, and

The weighted right shift $T$ is injective, so zero is not an eigenvalue. If $Ta=\lambda a$ with $\lambda \ne 0$, the first coordinate forces $a_1=0$ and subsequent coordinates force every $a_j=0$. Thus its point spectrum is empty.

For $m\ge 1$,since the largest column sum is attained at $j=1$. Therefore the spectral radius formula gives $\nu (T)={\lim }_m(m!{)}^{-1/m}=0$. The limit follows, for example, from $m!\ge (m/2{)}^{m/2}$ for large $m$. Zero is in the spectrum because the output first coordinate is always zero, so $T$ is not onto. HenceThis is a nonzero quasinilpotent element, specifically the factorial weighted right shift.

First give the general definitions and radius facts. The spectrum of a bounded operator is $\sigma (A)=\{\lambda :A-\lambda I\text{ has no bounded inverse}\}$, and the point spectrum is $p\sigma (A)=\{\lambda :\ker (A-\lambda I)\ne 0\}$. If $|\lambda |>\Vert A\Vert$, the Neumann series inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda (I-A/\lambda )$. Thus $\sigma (A)$ is contained in the closed disc of radius $\Vert A\Vert$ and is bounded.

The spectral radius is $\nu (A)={\sup }_{\lambda \in \sigma (A)}|\lambda |$. The spectral radius formula isFor the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j={\lambda }^{j-1}{a}_1$. A nonzero such sequence is in ${\ell }_1$ exactly when $|\lambda |<1$. Since $\Vert R\Vert =1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as point spectrum and therefore its closure by closedness of the spectrum. Hence

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

A linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

A bounded operator on a complex Hilbert space is self-adjoint when $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$. The spectral theorem for compact self-adjoint operators says its nonzero spectrum consists of real eigenvalues of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their eigenspaces are mutually orthogonal; their closed span is the orthogonal complement of the kernel. On that span the operator has the norm-convergent spectral action $Tx={\sum }_j{\lambda }_j⟨x,e_j⟩e_j$.

Take $H={\ell }^2(\mathbb{N})$ and $T{e}_n=e_n/n$. It is self-adjoint and has infinite-dimensional range. Truncation to the first $N$ coordinates is a finite-rank operator and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.

The resolvent set comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the spectrum, and the point spectrum consists of eigenvalues. In this exampleFor $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda {)}^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has eigenvector $e_n$. Zero is not an eigenvalue, but $\Vert T{e}_n\Vert \to 0$ prevents a bounded inverse; also $(1/n{)}_n\in {\ell }^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.

For a bounded operator on a complex Banach space, the resolvent set is $\rho (T)=\{\lambda :\lambda I-T\text{ is bijective with bounded inverse}\}$, and the resolvent operator is $R_T(\lambda )=(\lambda I-T{)}^{-1}$. The spectrum is its complement $\sigma (T)=\mathbb{C}\setminus \rho (T)$; the point spectrum consists of $\lambda$ with $\ker (\lambda I-T)\ne 0$. For a real Banach space these definitions use its complexification.

If $|\lambda |>\Vert T\Vert$, the norm-convergent Neumann series gives $R_T(\lambda )={\lambda }^{-1}{\sum }_{n\ge 0}(T/\lambda {)}^n$, so the spectrum is bounded. If ${\lambda }_0\in \rho (T)$, factorThe second factor is invertible by a Neumann series whenever $|\lambda -{\lambda }_0|\Vert R_T({\lambda }_0)\Vert <1$. Thus the resolvent set is open and the spectrum is closed and bounded.

The point spectrum need not be closed. On ${\ell }^2$, define $T{e}_n=e_n/n$. Every $1/n$ is an eigenvalue, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other eigenvalues occur, so ${\sigma }_p(T)=\{1/n:n\ge 1\}$ is not closed.

Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne 0$, then $⟨Tv,v⟩$ is real and equals $\lambda \Vert v{\Vert }^2$ under the convention linear in the first argument. Therefore every eigenvalue of a self-adjoint operator is real.

For a complex Banach space, the spectrum of a bounded operator and resolvent set areThe bounded inverse theorem says that a bounded bijection between Banach spaces automatically has a bounded inverse, so failure of bijectivity suffices here. The point spectrum isthe set of eigenvalues. For a real Banach space, these definitions use its complexification.

No. On the complex l2 sequence space, let $L(x_n)=(x_n/n)$. This is a bounded self-adjoint operator. At $\lambda =0$, $\ker L=\{0\}$, so zero is not an eigenvalue (an eigenvector must be nonzero). But $y=(1/n)\in {\ell }^2$ has no preimage in ${\ell }^2$: its only formal preimage is the constant-one sequence. Thus $L$ is not surjective andAlternatively $\Vert L{e}_n\Vert =1/n\to 0$ excludes a bounded inverse. This example separates the spectrum of a bounded operator from its point spectrum.

For a bounded linear operator $T$ on a complex Hilbert space, the spectrum of a bounded operator consists of $\lambda \in \mathbb{C}$ for which $T-\lambda I$ does not have a bounded everywhere-defined inverse. A bijective bounded operator has a bounded inverse by the bounded inverse theorem. Using the disjoint spectral convention,These are respectively the point spectrum, continuous spectrum and residual spectrum, and partition the spectrum. Some texts let residual spectrum include noninjective operators with nondense range; the displayed convention excludes overlap with point spectrum.

For a bounded linear operator $T:H\to H$, the spectrum iswhile the point spectrum is the set of eigenvalues, equivalently the $\lambda$ for which $T-\lambda I$ is not injective.

A compact operator maps the unit ball to a relatively compact set. A self-adjoint operator satisfies $T=T^{\ast }$, or $⟨Tx,y⟩=⟨x,Ty⟩$ for all $x,y\in H$. A finite-rank operator has finite-dimensional image.

For a bounded linear operator $T:X\to X$ on a complex Banach space, the point spectrum is precisely its set of eigenvalues:Here $I$ is the identity operator. Thus membership requires an actual nonzero eigenvector, whereas the spectrum of a bounded operator only requires failure of a bounded inverse. Every eigenvalue belongs to the spectrum, but the inclusion can be strict. The multiplication operator discussed below has an entire interval in its spectrum and empty point spectrum.^[ref]

In the disjoint convention for a bounded linear operator $T$, the residual spectrum consists of $\lambda$ for which $T-\lambda I$ is injective but has nondense range. Together with the point spectrum and continuous spectrum it partitions the spectrum of a bounded operator. Some conventions also include noninjective operators with nondense range, so the convention must be stated.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.