Operator theory

Linear operator theory studies linear maps, their spectra, invariant subspaces, and normal forms such as Jordan normal form. In a finite-dimensional vector space, eigenvalues identify parameters for which a shifted linear operator cannot be inverted. In an infinite-dimensional vector space, inversion can fail even without an eigenvector. The spectrum of a bounded operator captures this larger obstruction and connects linear algebra with the approximation questions of functional analysis.

An annihilating polynomial of a linear operator is a nonzero polynomial $p$ for which the displayed operator is zero. The minimal polynomial is its unique monic annihilating polynomial of least degree and divides every other annihilating polynomial. For a diagonalizable operator with distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_s$, the product ${\prod }_j(t-{\lambda }_j)$ annihilates it. For an individual vector it suffices to annihilate the eigenvalues contributing to that vector; this gives exact finite-step termination bounds for a Krylov subspace method.

For distinct scalars ${\mu }_j$, the Lagrange interpolation polynomials ${\ell }_j(t)={\prod }_{k\ne j}(t-{\mu }_k)/({\mu }_j-{\mu }_k)$ sum to one. Their values at $T$ have images in the respective eigenspaces, since $(t-{\mu }_j){\ell }_j$ is a multiple of the annihilating polynomial. Hence every vector is the sum of its ${\ell }_j(T)$ images. On the $i$th eigenspace, ${\ell }_j(T)$ is multiplication by ${\delta }_{ij}$, which proves directness. This gives the decomposition without invoking the primary decomposition theorem. Under this decomposition a linear operator commutes with $T$ exactly when it preserves every eigenspace.

The differentiation operator sends a differentiable function to its derivative; its domain and the norms must be specified. On $C^1([-1,1])$ with values in $C([-1,1])$, using the supremum norm on both, it is an unbounded linear operator: $f_n(x)=\sin (nx)$ satisfies $\Vert f_n{\Vert }_{\infty }\le 1$ and $\Vert D{f}_n{\Vert }_{\infty }=n$. Its graph norm is $\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, which is complete. Indeed, the kernel consists of constants and is complete, while $Sg(x)={\int }_0^{x}g(t)dt$ is a bounded right inverse by the fundamental theorem of calculus; apply completeness of a graph norm from a bounded right inverse.

For a bounded linear operator, the series $e^A={\sum }_{n=0}^{\infty }{A}^n/n!$ converges in operator norm. For an unbounded self-adjoint operator $H$, the spectral theorem instead defines the bounded unitary operator $e^{-iHt/\hslash }$ by applying the scalar function $e^{-iEt/\hslash }$ to its projection-valued measure. This yields unitary time evolution without assuming that every vector lies in the domain of every power of $H$. In finite dimensions it is the matrix exponential.

For a bounded skew-adjoint operator on a complex Hilbert space, the operator exponential converges in norm because its term norms are bounded by $\Vert T{\Vert }^n/n!$. Continuity of the adjoint operator gives $(e^T{)}^{\ast }=e^{T^{\ast }}=e^{-T}$. Absolute convergence permits the Cauchy product of $e^T$ and $e^{-T}$: its degree-$n$ coefficient is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]$, zero for $n>0$ and the identity for $n=0$. Both multiplication orders therefore give the identity, proving unitarity. The boundedness hypothesis makes this elementary power-series proof valid without domain issues for unbounded generators.

For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.

For a bounded operator $A$ on a Hilbert space, set $|A|=(A^{\ast }A{)}^{1/2}$. There is a unique partial isometry $U$, zero on $\ker A$, satisfying $A=U|A|$; on the range of $|A|$, define $U(|A|x)=Ax$ and extend by continuity. This is well-defined and isometric because $\Vert Ax{\Vert }^2=⟨A^{\ast }Ax,x⟩=\Vert |A|x{\Vert }^2$. For an invertible square matrix, $U$ is unitary.

A shift operator translates the components of a function or sequence by a fixed displacement. Bilateral shifts act on doubly infinite sequences; the unilateral shift operator acts on a one-sided sequence space and has distinct forward and backward versions.

The unilateral shifts on sequences move every component one place left or right. Their spectra illustrate behavior absent in finite dimensions.

On the absolutely summable sequence space, this weighted shift has $T^m{e}_j=e_{j+m}/[j(j+1)\cdots (j+m-1)]$, so $\Vert T^m\Vert =1/m!$. The spectral radius formula gives radius zero. Its spectrum is exactly $\{0\}$ because it is not onto, while it has no eigenvalues: zero is excluded by injectivity, and any nonzero eigenvalue equation forces all coordinates to vanish starting at the first. Thus it is a nonzero quasinilpotent element of the operator algebra.

On ${\ell }_1$, the unilateral shift operator $R$ and left shift operator $L$ satisfy $LR=I$ and $\Vert L\Vert =1$, while $L$ has nontrivial kernel. If $\Vert A\Vert <1$, the Neumann series makes $L(R+A)=I+LA$ invertible. Were $R+A$ invertible, this identity would express $L$ as a product of invertible operators, a contradiction. Thus the open unit ball about $R$ contains no invertible operator.

On the Hilbert space ${\ell }^2(\mathbb{N})$, the left shift deletes the first coordinate. It is the adjoint operator of the unilateral shift operator $S$. The identities $LS=I$ and $SLf=f-f_1{e}_1$ show that it is also the Moore–Penrose inverse of an operator for $S$; both maps have operator norm one.

An operator $T$ is a unilateral shift exactly when it is an isometry, $(\mathrm{Im}T{)}^{\perp }$ is one-dimensional, and ${\bigcap }_{n\ge 1}\mathrm{Im}(T^n)=\{0\}$. A unit vector $e_1$ in the first orthogonal complement is wandering, and $e_n=T^{n-1}{e}_1$ is a complete orthonormal basis.

A solvability condition is a compatibility requirement on data for an equation to admit a solution in a specified function space or with specified boundary behavior. For a bounded linear operator between Hilbert spaces, the equation $Lu=f$ requires $f$ to be orthogonal to the kernel of the adjoint. A closed range makes that requirement sufficient; the Fredholm alternative for a compact operator applies this principle to $I-K$ with compact $K$. in an asymptotic inner problem, exclusion of a growing homogeneous mode can impose an analogous weighted-integral condition.

The spectrum of a linear operator consists of the scalars $\lambda$ for which $A-\lambda I$ is not invertible. In finite dimensions it is the set of eigenvalues.

Let $P$ be a bounded linear operator on a complex Banach space with $P^2=P$, $P\ne 0$ and $P\ne I$. Its image of a linear map and kernel contain nonzero vectors, giving eigenvalues one and zero. For $\lambda \notin \{0,1\}$,Direct multiplication proves the formula, so the spectrum is exactly $\{0,1\}$ and its spectral radius is one. Since $P^j=P$ and $\Vert P\Vert \ge 1$, the spectral radius formula is explicit: ${\inf }_{j\ge 1}\Vert P^j{\Vert }^{1/j}={\lim }_{j\to \infty }\Vert P{\Vert }^{1/j}=1$.

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]

An approximate eigenvalue of a bounded operator $T$ is a scalar $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$. Every approximate eigenvalue belongs to the spectrum; for a normal operator, all spectral points are approximate eigenvalues.

The unit vectors realizing an approximate eigenvalue are approximate eigenvectors: their residual tends to zero, although they need not converge to any genuine eigenvector.

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.

For a self-adjoint operator $T$, a real number $\lambda$ lies in the continuous spectrum when $T-\lambda I$ is injective with dense range but is not surjective. Such spectral values need not be eigenvalues of normalizable states. The free-particle Hamiltonian on the real line has a continuous energy spectrum.

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.

If $V$ is a subspace of a space with an inner product and $P_V$ is its orthogonal projection, the compression of an operator $A$ to $V$ isIf the columns of $Z$ are an orthonormal basis of $V$, its matrix in that basis is $Z^{\ast }AZ$.

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

Let $\ddot{q}=Fq$ on a Hilbert space, with time-independent self-adjoint operator $F$, and conserved quadratic energy $E=\Vert \dot{q}{\Vert }^2/2-⟨q,Fq⟩/2$. If $⟨q_0,F{q}_0⟩>0$, set $\dot{q}(0)=\alpha q_0$ with ${\alpha }^2=⟨q_0,F{q}_0⟩/\Vert q_0{\Vert }^2$ and $\alpha >0$. This gives $E=0$. For $N=\Vert q{\Vert }^2$, $N^{\prime }=2\mathrm{Re}⟨q,\dot{q}⟩$ and $N^{\prime \prime }=4\Vert \dot{q}{\Vert }^2$, soby the Cauchy-Schwarz inequality. Since $(\ln N{)}^{\prime }(0)=2\alpha$, $N(t)\ge N(0)e^{2\alpha t}$. Positivity of $N$ follows from $N^{\prime }(0)>0$ and $N^{\prime \prime }\ge 0$. Thus a negative direction of the restoring energy $W=-⟨q,Fq⟩/2$ yields an exponentially growing solution without a normal-mode expansion or a variational spectral theorem. If $W$ is positive definite, conserved $E$ instead controls the energy norm; a bound in the original Hilbert space norm additionally follows from a coercive lower bound on $W$.

A densely defined semibounded symmetric operator on a Hilbert space has a canonical semibounded extension that is a self-adjoint operator obtained by closing its quadratic form. After shifting the lower bound to zero, complete the initial form domain in the norm $(\Vert u{\Vert }^2+q[u]{)}^{1/2}$; the associated closed form determines the Friedrichs extension. For the positive Laplace-Beltrami operator on a polygonal surface, this specifies the finite-energy realization at corners. Different choices of the initial form domain impose Dirichlet boundary conditions, Neumann boundary conditions, or mixed boundary conditions; the extension does not choose the boundary condition independently of that domain.

For a bounded self-adjoint operator $L$, a spectral Weyl sequence at $\lambda$ consists of unit vectors $f_n$ with $\Vert (L-\lambda I)f_n\Vert \to 0$. Such a sequence exists exactly at points of the spectrum of a bounded operator. If no sequence exists, the shifted operator is bounded below, with closed range and zero kernel. Self-adjointness then makes its range dense, so it is invertible. This spectral criterion is different from the equidistribution Weyl criterion.

A singular Weyl sequence for a bounded self-adjoint operator $L$ at $\lambda$ is a spectral Weyl sequence that also converges weakly to zero. Such a sequence exists exactly when $\lambda$ is in the essential spectrum of a bounded self-adjoint operator. An infinite-dimensional shifted kernel gives a weakly null orthonormal sequence. If that kernel is finite-dimensional but the shifted range is not closed, choose approximate null vectors in the kernel complement and extract a weakly convergent subsequence; its limit belongs to both kernel and complement, so is zero.

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.

Positivity makes $A-z$ bounded below with dense closed range for every $z<0$, excluding negative spectral values. Conversely if the spectrum is nonnegative, $R=(I+A{)}^{-1}$ has spectrum in $[0,1]$ by the resolvent spectral mapping identity. The numerical-range endpoint approximate-eigenvalue theorem gives $0\le R\le I$, and positive-form Cauchy-Schwarz inequality gives $\Vert Rf{\Vert }^2\le ⟨Rf,f⟩$. For $u=Rf$, $⟨Au,u⟩=⟨Rf,f⟩-\Vert Rf{\Vert }^2\ge 0$.

Symmetry gives $\Vert (A-z)u\Vert \ge |\mathrm{Im}z|\Vert u\Vert$ when $z$ is nonreal. Closedness makes the range closed; its orthogonal complement is $\ker (A-\overline{z})=0$, so it is also dense and hence surjective. This proves the resolvent estimate and reality of the spectrum. Equality of the adjoint domains is needed in the range argument.

For a self-adjoint operator $L$, solving $Lu=f$ requires $f$ to be orthogonal to every $v\in \ker L$, because $⟨v,Lu⟩=⟨Lv,u⟩=0$. When the range is closed, this condition is also sufficient: $(\mathrm{ran}L{)}^{\perp }=\ker L$ and closedness gives $\mathrm{ran}L=(\ker L{)}^{\perp }$. In finite dimensions the range is automatically closed. This Fredholm solvability condition determines amplitude equations by projecting perturbative forcing onto critical eigenfunctions.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

On ${\ell }^2$, the bounded self-adjoint operator $(Lv{)}_j=v_j/j$ is strictly positive. Yet $f_j=1/j$ lies in ${\ell }^2$ and $Lu=f$ would require the non-square-summable vector $u_j=1$. A positive uniform lower bound, rather than strict positivity alone, guarantees a bounded inverse for a bounded self-adjoint operator.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

A Hermitian operator on a $2^n$-dimensional Hilbert space, with $2^n$ distinct eigenvalues all belonging to $\{c/2^n:0\le c<2^n\}$, must occupy this entire grid. This follows from the pigeonhole principle: the set of possible values and the set of distinct eigenvalues have the same cardinality. Its minimum eigenvalue is zero, so the operator is necessarily singular. Distinctness therefore does not imply invertibility; it forces a one-dimensional kernel in this specific complete-grid setting.

An orthonormal eigenbasis of a linear operator is an orthonormal basis consisting of eigenvectors. Each vector can be reconstructed from its scalar components along those mutually perpendicular unit directions. With the convention that the inner product is linear in its first argument,The finite-dimensional spectral theorem supplies such a basis for a self-adjoint operator. In its two-dimensional example, the coordinates of $x=(1,0{)}^T$ are $1/\sqrt{2}$ along $e_{+}$ and $-1/\sqrt{2}$ along $e_{-}$. The operator multiplies those coordinates by $3$ and $1$, and reconstruction gives $Ax=(2,1{)}^T$.

Repeated eigenvalues permit different orthonormal choices within one eigenspace, so the basis need not be unique. The spectral projection of a self-adjoint matrix onto that entire eigenspace is unique: it keeps the geometrically meaningful component without choosing individual axes inside it.

For a finite-dimensional Hermitian matrix $A$, its spectral projection $P_{\lambda }$ is the orthogonal projection onto the eigenspace for $\lambda$. In an orthonormal eigenbasis, the displayed sum is over every basis vector with that eigenvalue. The projection is independent of the choice of basis within a repeated eigenspace. For distinct eigenvalues $\lambda ,\mu$,This is the spectral decomposition grouped by distinct eigenvalues: decompose a vector into orthogonal pieces, then scale each piece. For the matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$,Thus $P_3(1,0{)}^T=(1/2,1/2{)}^T$ and $P_1(1,0{)}^T=(1/2,-1/2{)}^T$: the two pieces add back to $(1,0{)}^T$. Squaring the matrix simply squares each scalar, $A^2=9P_3+P_1$, and similarly the matrix exponential is $e^{tA}=e^{3t}{P}_3+e^t{P}_1$. Spectral projections of a self-adjoint matrix turn repeated matrix calculations into scalar calculations on fixed vector subspaces.

On polynomials with weighted inner productthe operatoris self-adjoint because $e^{-x}Lf=(x{e}^{-x}{f}^{\prime }{)}^{\prime }$. Its eigenvalues on polynomials of degree at most $n$ are $0,-1,\dots ,-n$; its eigenvectors are scalar multiples of the Laguerre polynomials.

At eigenvalue zero the Laguerre differential operator on polynomials gives $x{y}^{\prime \prime }+(1-x)y^{\prime }=0$, equivalently $(x{e}^{-x}{y}^{\prime }{)}^{\prime }=0$. Thus $y^{\prime }=D{e}^x/x$ and the displayed solutions follow from the exponential integral. For any $x_0>0$, all solutions with $y(x_0)=1$ can be written $y(x)=1+D{\int }_{x_0}^x{e}^{t}dt/t$. The choices $D=0$ and $D=1$ are linearly independent despite sharing that one prescribed value.

The second branch grows as $e^x/x$ at infinity and has a logarithmic singularity at zero. It is consequently excluded from the regular weighted square-integrable eigenproblem with weight $e^{-x}$. Only the constant zero mode remains in that self-adjoint realization. A normalization at one interior point alone is not enough to select a unique second-order solution.

For polynomial solutions of the Laguerre differential equation, put $D_n=L_n^{\prime }-L_n$. The equation gives $D_n+t{D}_n^{\prime }=-(n+1)L_n$. Integrating $\int t{D}_n(L_m^{\prime }-L_m)e^{-t}$ by parts gives the displayed identity; the factor $t$ kills the endpoint at zero and exponential decay kills the endpoint at infinity. Its left side is symmetric in $n,m$, so subtracting the swapped identities proves orthogonality for distinct degrees. This provides a direct energy proof of orthogonality for the Laguerre polynomials without assuming their norm formula.