Linear operator theory

Linear operator theory studies linear maps, their spectra, invariant subspaces, and normal forms.

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.

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.

For a bounded operator on a complex Banach space, the spectrum is a nonempty compact subset of the complex plane contained in the closed disk of radius equal to the operator norm.

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.

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 inner-product space has an orthonormal basis of eigenvectors and only real eigenvalues.

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.

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.

The generalized Laguerre polynomial $L_n^{(\alpha )}$ is orthogonal on $[0,\infty )$ for the weight $x^{\alpha }{e}^{-x}$ when $\alpha >-1$.

A Rodrigues formula constructs an orthogonal polynomial by differentiating a weight multiplied by a power. For generalized Laguerre polynomials,

Every real symmetric matrix is orthogonally diagonalizable. Distinct eigenspaces are orthogonal because symmetry gives $\lambda ⟨u,v⟩=\mu ⟨u,v⟩$.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. With a unique dominant eigenvalue separated by a spectral gap and a starting vector that is nonorthogonal to its eigenvector, the directions converge at the spectral-ratio rate.

For a real symmetric matrix with $|{\lambda }_1|>|{\lambda }_2|$ and a starting vector having nonzero component in the leading eigendirection, power-method direction errors arewhereas the Rayleigh quotient error isThe improvement occurs because orthogonality cancels terms linear in the direction error.

The active spectrum consists of eigenvalues whose eigenspaces have nonzero projection of the starting vector. The power method converges toward the eigenvalue of largest modulus in this active spectrum; an exactly absent dominant component is never created in exact arithmetic.

Inverse iteration repeatedly solves $(A-sI)y_{k+1}=q_k$ and normalizes. It is the power method for $(A-sI{)}^{-1}$ and converges toward an eigenvector whose eigenvalue is closest to the shift $s$.

If $A$ is real symmetric, $s$ is not an eigenvalue, one eigenvalue ${\lambda }_{\ast }$ is uniquely closest to $s$, and the initial vector has a nonzero component in its eigenspace, fixed-shift inverse iteration converges to that eigenspace. Its asymptotic direction-error ratio iswhere ${\lambda }_{\mathrm{next}}$ is second closest to $s$.

If the closest eigenvalue ${\lambda }_{\ast }$ lies above the shift, the dominant eigenvalue $({\lambda }_{\ast }-s{)}^{-1}$ of the inverse is positive and normalized iterates approach one fixed sign of its eigenvector. If ${\lambda }_{\ast }<s$, that inverse eigenvalue is negative, so the direction converges but consecutive normalized vectors alternate signs. At a shift exactly midway between two simple eigenvalues, the two inverse eigenvalues have equal modulus and opposite signs, so generic iterates alternate between two mixtures instead of selecting one eigenvector.

Subspace iteration repeatedly applies a matrix to several independent vectors and orthonormalizes them, converging to a dominant invariant subspace when a spectral gap separates the selected eigenvalue cluster.

A dominant invariant subspace is spanned by eigenvectors whose eigenvalue moduli exceed those outside the subspace.

When several eigenvalues share the largest modulus, normalized power iterates need not converge individually, but every limit lies in their joint invariant subspace.

The characteristic polynomial isIts roots are precisely the eigenvalues of $A$.

For square matrices $A,B$ over a field,If $A$ is invertible, the products are similar matrices. In general, apply that case to $(A-sI)B$ and $B(A-sI)$ for all but finitely many $s$, then use polynomial identity in $s$.

Ifthe Faddeev–LeVerrier algorithm computes the coefficients recursively from traces of powers:

For real matrices $A,B$, $p(t)=\det (A+tB)$ is a real polynomial. If $p(i)\ne 0$, it is not the zero polynomial, so some real $\lambda$ satisfies $p(\lambda )\ne 0$.

Every square matrix over an algebraically closed field is similar to an upper-triangular matrix. Choose an eigenvector, extend it to a basis, and apply induction to the induced lower-right block.

Eigenvalue interlacing bounds the ordered eigenvalues of a compressed or rank-modified operator between those of the original operator.

Let a real symmetric matrix $M$ have ordered eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, with $n\ge 2$. If $A$ is obtained by deleting one matching row and column, its largest eigenvalue ${\mu }_1$ satisfiesThe Rayleigh quotient of $A$ is that of $M$ restricted to vectors whose deleted coordinate is zero. This gives the upper bound. Some nonzero linear combination of the top two orthonormal eigenvectors has that coordinate zero; normalizing it gives a quotient at least ${\lambda }_2$. The argument does not require distinct eigenvalues.

Perturb each diagonal position of a Jordan normal form matrix by a different arbitrarily small scalar, avoiding the finitely many collisions between diagonal entries. The resulting upper triangular matrix has distinct eigenvalues, hence is diagonalizable. Conjugate back to approximate any complex matrix. Continuous identities proved for diagonalizable matrices can therefore extend to all matrices.

A complex two-dimensional linear operator whose two eigenvalues both equal $a$ is similar either to $aI$ or to $\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$. Choose an eigenvector and complete it to a basis. The matrix becomes $\left(\begin{array}{cc}a& c\\ 0& a\end{array}\right)$ because its characteristic polynomial is $(t-a{)}^2$. A nonzero $c$ can be normalized to one by rescaling the first basis vector. The two types are distinguished by eigenspace dimension.

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

The commutator action of $X_s$ is diagonalizable on endomorphisms: on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ it acts by $\mu -\lambda$. The action of $X_n$ is nilpotent by nilpotence of commutation by a nilpotent endomorphism. The two actions commute, so uniqueness of the Additive Jordan decomposition identifies them as the parts of $\mathrm{ad}X$.

For a complex semisimple Lie algebra $\mathfrak{g}\subseteq \mathrm{End}(V)$ and $X\in \mathfrak{g}$, both parts of its Additive Jordan decomposition lie in $\mathfrak{g}$. The polynomial semisimple part of $\mathrm{ad}X$ shows $X_s$ normalizes $\mathfrak{g}$. Split $\mathrm{End}(V)=\mathfrak{g}\oplus M$ by the Weyl complete reducibility theorem; the $M$-component of $X_s$ centralizes $\mathfrak{g}$. On each irreducible summand of $V$ it is scalar by the Schur lemma. Its trace is zero because $\mathfrak{g}$ is a perfect Lie algebra and the nilpotent part has zero trace, so those scalars vanish. Nonsemisimple subalgebras may fail this property: $\mathbb{C}(I+N)$ with $N\ne 0$ nilpotent contains neither Jordan part of $I+N$.

The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial. In Jordan form it is the sum of the sizes of all blocks carrying that eigenvalue.

The geometric multiplicity of an eigenvalue is the dimension of its eigenspace. In Jordan form it equals the number of blocks carrying that eigenvalue.

A Jordan block has one eigenvalue on its diagonal and ones on its superdiagonal.

The model matrix $B=\left(\begin{array}{cc}0& 1\\ ϵ& d\end{array}\right)$ has a defective double eigenvalue when $d=ϵ=0$. Its characteristic polynomial is ${\lambda }^2-d\lambda -ϵ$, giving the displayed roots. For fixed nonzero $d$ the small root is $-ϵ/d+O({ϵ}^2)$ and the other is $d+ϵ/d+O({ϵ}^2)$. These expansions cease to be ordered at $d=O(\sqrt{ϵ})$, the distinguished limit in which both roots split on a square-root scale. This is the generic leading behavior when a perturbation closes a two-dimensional Jordan block through a nonzero lower-left entry; it need not occur for a perturbation that leaves that entry zero.

A nilpotent Jordan block has zeros on its diagonal and ones on its superdiagonal. Its $k$th power has ones on the $k$th superdiagonal, and a block of size $d$ satisfies $J_d(0{)}^d=0$.

If $\beta$ has eigenvalues $\lambda ,\mu \ne 0$ and is diagonalizable on a two-dimensional complex vector space, thenIf $\beta$ has one size-two Jordan block, then

Over an algebraically closed field, the characteristic polynomial gives the total size of the Jordan blocks for each eigenvalue, while the minimal polynomial gives the largest block size. For total dimension at most three, these two numbers determine the partition into block sizes: a multiplicity-three eigenvalue has partition $3$, $2+1$, or $1+1+1$ according as its largest block has size $3$, $2$, or $1$. Hence two matrices of size at most three with the same characteristic and minimal polynomials are similar.

A generalized eigenvector lies in the kernel of a positive power of A minus lambda I.

A Jordan chain of a linear operator $T$ for an eigenvalue $\lambda$ is a list of nonzero vectors satisfying $(T-\lambda I)v_1=0$ and $(T-\lambda I)v_j=v_{j-1}$ for $2\le j\le r$. These vectors are linearly independent: in a vanishing linear combination, applying the largest relevant power of $T-\lambda I$ isolates the last coefficient times $v_1$, then descending induction removes all coefficients. On their span the operator has one Jordan block. This is different from a generalized eigenfunction that solves $(T-\lambda I)u=0$ outside the original function space.

If $(M-\lambda I)e=0$ and $(M-\lambda I)v=e$, the vector $v$ is a generalized eigenvector and $e^{\lambda t}(v+te)$ solves $\dot{x}=Mx$. The polynomial factor distinguishes a nontrivial Jordan block from an ordinary repeated eigenvalue with a full eigenvector basis.

The generalized eigenspace of a linear operator $A$ for an eigenvalue $\lambda$ is the union of the kernels $\ker (A-\lambda I{)}^N$. In finite dimension this union stabilizes for sufficiently large $N$.

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.

For a linear operator $f$, the generalized eigenspacesbelonging to distinct eigenvalues are linearly independent. On the generalized $\alpha$-eigenspace, $f-\beta I=(\alpha -\beta )I+N$ is invertible for $\beta \ne \alpha$ because the nilpotent part $N$ has a finite geometric-series inverse.

A generalized eigenvalue problem asks for a nonzero vector $v$ and scalar $\lambda$ satisfying $Av=\lambda Bv$. When $B$ is invertible, it is equivalent to the ordinary eigenvalue problem $B^{-1}Av=\lambda v$.

The generalized characteristic polynomial of a matrix pair $(A,B)$ is $\det (A-\lambda B)$. Every finite generalized eigenvalue is one of its roots.

For a square matrix $A$, its exponential is the convergent power seriesIt is the fundamental matrix for $y^{\prime }=Ay$ and satisfies $e^{(s+t)A}=e^{sA}{e}^{tA}$.

The determinant is a smooth Lie group homomorphism from the real general linear group to ${\mathbb{R}}^{\times }$. Its derivative at the identity is the trace, and the flow-defined exponential on ${\mathbb{R}}^{\times }$ is the ordinary scalar exponential. Naturality of the Lie group exponential therefore proves the identity without diagonalizability assumptions. The series ${\sum }_{k\ge 0}{t}^k{A}^k/k!$ is the flow-defined matrix exponential because it solves $E^{\prime }=EA$, $E(0)=I$, and has inverse $E(-t)$.

If a real square matrix satisfies $M^2=-I$, split the absolutely convergent matrix exponential into even and odd powers to obtain $e^{tM}=I\cos t+M\sin t$. It is the fundamental matrix for ${\mathbf{x}}^{\prime }=M\mathbf{x}$. Consequently every solution has period $2\pi$, and the eigenvalues over the complex numbers belong to $\{i,-i\}$.

The Baker--Campbell--Hausdorff formula expresses the local matrix logarithm of a product of exponentials as

On its half-plane of convergence,Indeed, applying the transform to $y^{\prime }=Ay$, $y(0)=y_0$, gives $(sI-A)Y=y_0$ for every $y_0$.

The minimal polynomial $m_A$ is the unique monic polynomial of least degree satisfying $m_A(A)=0$. It divides every polynomial that annihilates $A$.

A square matrix $A$ is invertible exactly when its minimal polynomial has nonzero constant term. If $m_A(t)=tq(t)$, then $Aq(A)=0$, and invertibility would contradict the minimality of $m_A$. Conversely, if $m_A(t)=tq(t)+c$ with $c\ne 0$, then $A[-q(A)/c]=I$.

For square matrices $A,B$,Indeed, $(BA{)}^{k+1}=B(AB{)}^{k}A$. Thus the two minimal polynomials have the same nonzero factors with the same multiplicities, while their powers of $t$ differ by at most one.

If $p_1,\dots ,p_r$ are pairwise coprime, thenBezout identities between the factors construct the corresponding projections.

If an upper-triangular $n$ by $n$ matrix has diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, its standard invariant flag satisfiesThe commuting product ${\prod }_k(A-{\lambda }_{k}I)$ therefore vanishes, proving $\mathrm{deg}{m}_A\le n$ without invoking the full Cayley-Hamilton theorem.

Let $C=[B,A]$ and suppose $AC=CA$. Then $C$ is nilpotent. For the derivation $D(X)=[B,X]$, one has $D(p(A))=p^{\prime }(A)C$. If $f(A)=0$, induction givesIndeed, if $u{C}^m=C^{m}u=0$, differentiating $u{C}^m=0$ and multiplying on the left by $C^m$ gives $C^{m}D(u)C^m=0$; for $u=f^{(k)}(A)$ this replaces $m$ by $2m+1$. Taking $k=\mathrm{deg}f$ makes $f^{(k)}$ a nonzero constant and proves nilpotence.

On all real-valued functions on $\mathbb{R}$, define $D_{r}f(x)=f(x+r)-f(x)$. The operators commute, and every real number except $-1$ is an eigenvalue. For eigenvalue $\lambda \ne -1$, values may be chosen independently on the cosets of $r\mathbb{Z}$ and extended by $f(x+r)=(1+\lambda )f(x)$, so every eigenspace is infinite-dimensional.

If $p$ has degree $n$ and leading coefficient $a_n$, then for nonzero steps $r_1,\dots ,r_n$,Consequently a degree-$n$ polynomial cannot be the sum of $n$ functions each periodic with some nonzero period.

A linear map $T:V\to V$ is nilpotent if $T^k=0$ for some positive integer $k$.

For a nilpotent endomorphism $N$ of a vector space, the commutator action $\mathrm{ad}N:T\mapsto NT-TN$ on its endomorphisms is nilpotent. The commuting left and right multiplication maps give $(\mathrm{ad}N{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0ex}{}{r}{j}\right)N^{r-j}T{N}^j$. If $N^q=0$ and $r\ge 2q-1$, every summand vanishes. This works in every characteristic of a field and is useful when applying Engel theorem to an Adjoint representation.

A scalar $\lambda$ is an eigenvalue of a linear operator $A$ when $Av=\lambda v$ for some nonzero vector $v$.

The spectral abscissa is the largest real part of a matrix's eigenvalues. It describes asymptotic exponential rates of its matrix exponential, but a non-normal matrix can also have transient amplification. It is generally distinct from the Euclidean logarithmic norm, which gives an immediate Euclidean energy-growth bound.

Geometric multiplicity is the dimension of an eigenspace; algebraic multiplicity is its multiplicity as a root of the characteristic polynomial. For a real symmetric matrix they coincide because it has an orthonormal eigenbasis.

Zero is an eigenvalue of a linear operator exactly when the operator has a nontrivial kernel. Its eigenvectors are precisely the nonzero vectors in that kernel.

ConsiderIts row sums vanish, so $(1,1,1{)}^T$ is an eigenvector for the zero eigenvalue. Its characteristic polynomial iswith sum of principal minors of order twoThe other eigenvalues are $(-S\pm \sqrt{S^2-4T})/2$. No symmetry is required for this formula; positive diagonal symmetrization of a matrix is a sufficient condition for real roots.

An eigenvalue problem asks for the eigenvalues and corresponding eigenvectors or eigenfunctions of a linear operator. It is commonly written as the eigenvalue equation $Av=\lambda v$.

An eigenfunction is a nonzero function $f$ satisfying $Lf=\lambda f$ for a linear operator $L$ and scalar eigenvalue $\lambda$.

An adjoint eigenfunction is a nonzero eigenfunction of the adjoint operator on its adjoint domain. For an inner product conjugate-linear in its first argument, $L^{\ast }g=\overline{\lambda }g$ implies $⟨g,(L-\lambda )f⟩=0$ for every $f$ in the domain of $L$. Consequently a necessary solvability condition for $(L-\lambda )f=h$ is $⟨g,h⟩=0$. If $L-\lambda I$ is a Fredholm operator, orthogonality to every element of its adjoint kernel is also sufficient. Boundary conditions and weights in a coupled system enter the adjoint domain and must be derived by integration by parts; copying the original eigenfunction without checking them can give a wrong projection.

A generalized eigenfunction solves the eigenvalue equation for a linear operator in an enlarged space, often a space of distributions, although it may fail to belong to the underlying Hilbert space. A plane wave $e^{ikx}$ on the whole real line is a generalized momentum eigenstate, with momentum $\hslash k$, and is not square-integrable. This usage differs from a generalized eigenvector in a Jordan chain, where a higher power of $L-\lambda$ vanishes rather than $L-\lambda$ itself.

A nonzero vector $v$ is an eigenvector of a linear operator $A$ when $Av=\lambda v$ for some eigenvalue $\lambda$.

Applying the product of the other eigenvalue factors to a linear relation isolates each coefficient. Distinct eigenvalues therefore give linearly independent eigenvectors. A complete eigenvector basis diagonalizes the associated linear system of ordinary differential equations.

The eigenvalue equation $Av=\lambda v$ asks for a nonzero eigenvector $v$ and its eigenvalue $\lambda$.

A right eigenvector satisfies $Av=\lambda v$.

A left eigenvector satisfies $u^{T}A=\lambda u^T$, equivalently $A^{T}u=\lambda u$ over the real numbers.

The eigenspace for $\lambda$ is $\ker (A-\lambda I)$.

An eigenbasis is a basis consisting of eigenvectors of one operator.

An eigenvalue is simple when it has algebraic multiplicity one.

For unit left eigenvector $u$ and right eigenvector $v$ of a real matrix at a simple eigenvalue $\lambda$, its normwise sensitivity isIt measures the first-order amplification of a matrix perturbation into an eigenvalue perturbation.

If $A(t)=A+tE$ and $\lambda (t)$ is a differentiable function defining a simple eigenvalue, thenThe Cauchy-Schwarz inequality and the operator norm give

A matrix is diagonalizable exactly when its eigenvectors span the whole vector space.

A diagonalization is a factorization $A=PD{P}^{-1}$ in which $D$ is diagonal and the columns of $P$ form an eigenvector basis.

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for some $m>0$, then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A{|}_E=\lambda I+N$ has $\lambda \ne 0$ andThe first nonzero nilpotent term prevents this power from being diagonalizable.

Suppose real matrices $A,B$ satisfy $AP=PB$ for an invertible complex matrix $P=X+iY$. Then $AX=XB$ and $AY=YB$. The real polynomial $\det (X+tY)$ is not identically zero because its value at $t=i$ is nonzero, so some real $t$ makes $X+tY$ invertible. This real matrix intertwines $A$ and $B$.