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If a normal matrix satisfies $Ax=\lambda x$, then $A^{†}x=\overline{\lambda }x$. Indeed, $M=A-\lambda I$ is normal and $\Vert Mx{\Vert }^2=x^{†}{M}^{†}Mx=x^{†}M{M}^{†}x=\Vert M^{†}x{\Vert }^2$. Since $Mx=0$, also $M^{†}x=0$. This identity proves orthogonality of distinct eigenspaces and gives spectral restrictions for Hermitian matrices, skew-Hermitian matrices and unitary matrices.

The Euclidean logarithmic norm is the largest eigenvalue of a matrix's Hermitian part. Differentiating $\Vert y{\Vert }_2^2$ along $y^{\prime }=Ay$ gives $\Vert e^{tA}{\Vert }_2\le e^{t{\mu }_2(A)}$. It is the least exponent for such a bound with prefactor one, as a first-order expansion at zero shows. It equals the spectral abscissa for a normal matrix, but need not do so otherwise. A scalar rational stability function does not generally inherit a bound by its value at this real number.

Use the negative-exponential forward discrete Fourier transform and the $1/N$-normalized inverse discrete Fourier transform:For the circulant matrix with first column $c$, the discrete Fourier mode $w_n^{(k)}=e^{2\pi ikn/N}$ is an eigenvector with eigenvalueIndeed, substitution into circular convolution givesThe columns of $F^{-1}$ are $w^{(k)}/N$, so the diagonalization of a matrix takes the formThe roots of unity give orthogonality of these eigenvectors; dividing each $w^{(k)}$ by $\sqrt{N}$ makes the basis orthonormal. Consequently, every complex circulant matrix is a normal matrix, including ones that are not Hermitian matrices.

The sign convention matters. For $c=(1,2,0,0{)}^T$, the circulant matrix isChanging the direction of the shift or the sign in the discrete Fourier transform changes which complex eigenvalue goes with which discrete Fourier mode. A real symmetric example alone would conceal that distinction.

For a linear system $Cx=b$, put $\hat{x}=Fx$ and $\hat{b}=Fb$. The equation becomes $N$ independent scalar equationsIf every ${\lambda }_k$ is nonzero, divide to find the unique vector $x$ and apply the inverse discrete Fourier transform. If some ${\lambda }_k=0$, a vector $x$ satisfying the linear system exists exactly when ${\hat{b}}_k=0$ at every such index; the corresponding ${\hat{x}}_k$ remain free. This is a concrete version of the Fredholm alternative in a finite-dimensional setting.

With an appropriate Fast Fourier transform, multiplication or a compatible solve takes $O(N\log N)$ arithmetic operations, instead of forming a dense matrix and applying general Gaussian elimination. The gain comes from the periodic structure; an arbitrary matrix does not have this shared basis of discrete Fourier modes.^[ref]

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

For two normal matrices of the same size, a pairing of their eigenvalues has squared Euclidean discrepancy at most the squared Frobenius norm of their difference. For Hermitian matrices, the increasing real eigenvalue order gives such a pairing, because it minimizes squared discrepancy. For real symmetric matrices, the norm square is $\mathrm{Tr}(A-B{)}^2$.

Every eigenvalue of this nonnormal matrix is $-1$, but $e^{t{L}_h}=e^{-t}\left(\begin{array}{cc}1& t/h\\ 0& 1\end{array}\right)$. Applied to the second coordinate unit vector, its norm at $t=1$ is at least $1/(eh)$. Thus negative eigenvalues give no mesh-uniform semigroup bound by themselves. Normal matrices, uniformly conditioned eigenvector bases, or a direct dissipative energy estimate supply the missing control.

For $A=\left(\begin{array}{cc}-i{\omega }_1& a\\ c& -i({\omega }_1-b)\end{array}\right)$ with real parameters,Thus the displayed conditions are necessary and sufficient for a normal matrix. In the family $a=1$, $b=c=q$, normality requires $q=-1$, not $q=1$. If $q=10/Re$, this is the formal value $Re=-10$ and no positive Reynolds number qualifies. At that value $A$ is a skew-Hermitian matrix and its exponential is unitary, so every initial direction has optimal energy amplification of a linear system equal to one.

For a normal matrix, $N^{†}N=N{N}^{†}$. Consequently $\Vert Nx{\Vert }^2=x^{†}{N}^{†}Nx=x^{†}N{N}^{†}x=\Vert N^{†}x{\Vert }^2$. In particular, $N$ and its Hermitian conjugate have the same kernel of a linear map.

Let the columns of the given unitary matrix be $u_j$, with $M{u}_j={\lambda }_j{u}_j$. Columnwise, $MU=UD$ for $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Multiplication by $U^{\ast }$ and unitarity giveThe premise includes an orthonormal eigenbasis. It is not true that an arbitrary matrix admits such a basis; over $\mathbb{C}$ it exists precisely for a normal matrix. The real symmetric inertia matrix in part (b) does satisfy the required condition.

Take the adjoint matrix of $M^{†}{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i$ to obtain ${\mathbf{v}}_i^{†}M={\lambda }_i^{\ast }{\mathbf{v}}_i^{†}$. Evaluating ${\mathbf{v}}_i^{†}M{\mathbf{u}}_j$ using this equation and $M{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ givesHence $\boxed{{\mathbf{v}}_i^{†}{\mathbf{u}}_j=0\text{ if }{\lambda }_i^{\ast }\ne {\lambda }_j}$. The proof does not assume that $M$ is a normal matrix or diagonalizable.

In general the spectrum of an adjoint matrix is the complex conjugate of the original spectrum, not necessarily the same set. The equality of the two sets in the question is an additional assumption, possible for a conjugation-invariant spectrum; the displayed orthogonality proof only uses the two stated eigenvalue equations.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

Since $A$ is an invertible matrix, so is its Hermitian conjugate, andIf $A$ is a normal matrix, substitute $A^{†}A=A{A}^{†}$ into the second identity to obtain $B{B}^{†}=I$, so $B$ is a unitary matrix. Conversely, if $B$ is a unitary matrix, multiply $B{B}^{†}=I$ on the left by $A$ and on the right by $A^{†}$. This gives $A^{†}A=A{A}^{†}$, the defining normal matrix identity. Thus $B$ is a unitary matrix exactly when $A$ is a normal matrix.

Using the complex Euclidean norm and the normal matrix identity,Both sides are nonnegative. Therefore $\Vert Cx\Vert =0$ if and only if $\Vert C^{†}x\Vert =0$. This proves the stronger reusable fact that normal matrices have equal adjoint norms.

The shifted matrix $N=D-\lambda I$ is still a normal matrix, sinceand $N{N}^{†}$ has the same expansion except that its first term is $D{D}^{†}$, which equals $D^{†}D$. For the given eigenvector, $Ne=0$. Part (b), applied to $N$, therefore gives $N^{†}e=0$. Since $N^{†}=D^{†}-\overline{\lambda }I$,The corresponding eigenvalue of the Hermitian conjugate is the complex conjugate of the original eigenvalue.

The eigenvalue stability analysis of a finite difference method reduces a linear evolution discretization to its modal behavior, but it must retain the norm and the dependence on the mesh. For a homogeneous system obtained by the method of lines, $U^{\prime }=A_{h}U$, the solution is $e^{t{A}_h}U(0)$. A one-step time discretization gives $U^{n+1}=G_{h,k}{U}^n$; for a Runge-Kutta method, where its stage matrices are invertible, $G_{h,k}=R(k{A}_h)$. With a multistep method one instead uses an amplification matrix on the augmented vector of several time levels.

The relevant finite-time stability estimate iswith $C_T$ independent of the refining spatial mesh and allowed time steps. Similarly, semidiscrete stability requires $\Vert e^{t{A}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. The norm should represent the continuum problem, such as the discrete L2 norm $\Vert U{\Vert }_{2,h}^2=h{\sum }_j|U_j{|}^2$ in one dimension. A fixed finite matrix having decaying solutions as $t\to \infty$ is not by itself a mesh-uniform result for the partial differential equation.

If $G=X\Lambda X^{-1}$, thenThus eigenvalues of modulus at most one give a uniform bound when the diagonalizing bases have uniformly bounded condition numbers. For a normal matrix in the chosen inner product, the spectral theorem for normal operators supplies an orthonormal basis, so the condition number is one and the norm of $G^n$ is exactly ${\max }_j|{\lambda }_j{|}^n$. For a self-adjoint or skew-adjoint spatial matrix, applying a scalar stability function preserves this favorable modal structure. This is the setting where eigenvalue calculations provide especially clean, reliable step restrictions.

A precise fixed-matrix statement is also useful: a matrix is power bounded for all nonnegative integers $n$ if and only if all its eigenvalues lie in the closed unit disk and every Jordan block at a unit-modulus eigenvalue has size one. Indeed, powers of an interior Jordan block contain a polynomial in $n$ times a decaying geometric factor and are bounded; a larger block on the unit circle produces unbounded polynomial growth. For a family of matrices, the resulting bounds must still be uniform. Moreover, finite-time stability versus power boundedness distinguishes this all-time condition from the physical-time estimate above. For example $G_k=I+kN$, $N^2=0$, has $G_k^n=I+nkN$, bounded for $nk\le T$ despite a nontrivial Jordan block at one. Even the scalar update $1+k$ is stable on fixed intervals, since $(1+k{)}^n\le e^T$, although it represents a growing equation rather than a contraction.

For a constant-coefficient periodic difference operator, Fourier modes diagonalize the spatial matrix. The discrete Fourier transform is a unitary operator, so Parseval's identity identifies its modal maximum with the operator norm. The eigenvalue stability analysis of a finite difference method then becomes von Neumann stability analysis. For the Dirichlet discrete Laplacian, a discrete sine transform plays the same role. These facts explain its advantages: a large matrix calculation becomes a scalar symbol or a known spectral interval, and the extremal eigenvalue directly determines a safe time step.

As a first example, centered space discretization of the heat equation $u_t=u_{xx}$ has eigenvaluesForward Euler method gives $R(k\lambda )=1-4r{\sin }^2(\theta /2)$ with $r=k/h^2$. Requiring all factors in $[-1,1]$ yields $0\le r\le 1/2$ for all periodic meshes. In two dimensions the corresponding uniform bound is $r\le 1/4$. Because these are normal matrices, these scalar bounds prove contraction rather than merely suggest it.

Backward Euler method has $R(z)=1/(1-z)$, so the centered discretization of the heat equation is contractive for every $k>0$. The Crank-Nicolson method has $R(z)=(1+z/2)/(1-z/2)$ and is likewise contractive for these negative real modes. However, its factor tends to $-1$ for very stiff modes, leaving high-frequency oscillations weakly damped; Backward Euler method tends to zero. The eigenvalues therefore expose the presence of stiff differential equations and damping as well as binary stability. Unconditional stability does not remove accuracy requirements.

For an advection equation example, the upwind finite difference scheme for $u_t+c{u}_x=0$, $c>0$, has factor $G=1-\nu +\nu e^{-i\theta }$, with $\nu =ck/h$. Sinceit is a contraction in the periodic discrete L2 norm exactly for $0\le \nu \le 1$. In contrast, Forward Euler method with centered advection has factor $1-i\nu \sin \theta$. Its modulus exceeds one at nonzero active frequencies, and repeated amplification makes it unstable under fixed nonzero Courant refinement. A much smaller scaling $k=O(h^2)$ can bound its finite-time growth, demonstrating again that exact contraction and the most general stability estimate are different claims.

The major limitation is the possible presence of a nonnormal matrix. For example,Both eigenvalues are negative, yet at $t=1$ the second unit vector is amplified by at least $1/(eh)$. This proves that negative spectra do not imply uniform semidiscrete stability. Ill-conditioned eigenvectors, Jordan normal form and transient amplification can invalidate a spectral-only argument. Direct energy estimates or estimates for the resolvent can supply the missing norm control.

Boundary conditions are another limitation. A periodic symbol proves a periodic or whole-line result, not an arbitrary initial-boundary-value problem. Boundary closure of a difference scheme can introduce growing modes or mesh-dependent amplification. A spatial operator with variable coefficients can still be studied through its assembled matrix, but then whether it is a normal matrix, its condition number and the physical inner product must be checked rather than borrowed from a Fourier stability analysis. For the finite element method the natural mass matrix inner product often turns a generalized eigenvalue problem into a self-adjoint one.

For time-dependent or split updates, separate spectra are insufficient to control products. For instance, $G_1=\left(\begin{array}{cc}0& K\\ 0& 0\end{array}\right)$ and $G_2=\left(\begin{array}{cc}0& 0\\ K& 0\end{array}\right)$ each have only zero eigenvalues, while $G_2{G}_1$ has eigenvalue $K^2$. For $K>1$, alternating the two is unstable. A common contractive norm would control the products, but their individual eigenvalues do not.

Finally, stability and consistency of a numerical method have different roles. The Lax equivalence theorem states that, for a well-posed linear initial-value problem and a consistent finite difference method on compatible grid spaces, stability on every fixed time interval is equivalent to convergence of a numerical method as the admissible meshes refine. Bounded grid representation and the correct initial and boundary setup are part of that framework. An eigenvalue estimate can establish its stability hypothesis in the favorable settings described above; it neither proves consistency of a numerical method nor bypasses its uniformity requirements. The method is therefore a powerful modal tool, provided its spectral calculation is connected to a genuine norm estimate.

Eigenvalue analysis proves mesh-uniform stability when it is accompanied by control of eigenvectors or an energy norm; the signs of eigenvalues alone are insufficient. For an evolution equation, the method of lines first produces $u_h^{\prime }=L_h{u}_h$. A time integrator then gives an amplification matrix $S_{h,\delta }$, for example $S_{h,\delta }=R(\delta L_h)$ for a Runge-Kutta method. The useful stability requirement on a fixed physical time interval iswhere $C_T$ is independent of the refining spatial and temporal meshes. The semidiscrete analogue is $\Vert e^{t{L}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. These are bounds on propagation of initial errors and residuals, not merely statements that each finite matrix has a bounded solution over its own finite time interval.

If $L_h$ is a normal matrix in the chosen discrete inner product, it has a unitary eigenbasis. ThenFor dissipative problems it is sufficient that every $\delta {\lambda }_j$ lie in the integrator's linear stability domain. A diagonalizable family also suffices if its eigenvector matrices have uniformly bounded condition numbers: the same maximum is multiplied by $\Vert V_h\Vert \Vert V_h^{-1}\Vert$. Without that uniform bound the argument does not establish mesh stability. For finite-time estimates, an amplification modulus $1+O(\delta )$ can also be acceptable; strict contraction at every step is a stronger property.

For a constant-coefficient stencil on the full lattice or a periodic grid, von Neumann stability analysis provides the eigenvectors explicitly. A Fourier mode diagonalizes each translation-invariant spatial difference, giving a scalar symbol ${\lambda }_h(\theta )$ or a small matrix symbol for a system. The discrete Fourier transform and Parseval identity turn a uniform modal bound into an actual discrete L2 norm bound. The two-dimensional stencil in question 3 has $\mathrm{Re}{\lambda }_h=-(1-\cos \eta {)}^2/h\le 0$, so its continuous-time amplification is contractive; a later time integrator must still be checked. With boundaries one must analyze the boundary rows as well: a whole-line Fourier symbol cannot detect an unstable boundary closure.

For the heat equation $u_t=\kappa u_{xx}$ on a unit interval with zero Dirichlet data, the centred second difference has a sine eigenbasis and eigenvaluesExplicit Euler method multiplies mode $j$ by $1+\delta {\lambda }_j$. Requiring all factors to have modulus at most one yields $\kappa \delta /h^2\le 1/[2{\sin }^2((N-1)\pi /(2N))]$ on a given grid, and the sharp grid-independent sufficient limit isThese parabolic time-step restrictions express the growth of the largest-magnitude diffusion eigenvalue like $h^{-2}$. Backward Euler method instead has factor $(1-\delta {\lambda }_j{)}^{-1}$ and the Crank-Nicolson method factor $(1+\delta {\lambda }_j/2)/(1-\delta {\lambda }_j/2)$, both bounded by one for every $\delta \ge 0$. Their unconditional stability follows from A-stability, with much stronger damping of the shortest scales for backward Euler.

For $u_t=-a{u}_x$, $a>0$, explicit time stepping with a backward spatial difference givesThus the upwind finite difference scheme is contractive exactly for $0\le \nu \le 1$, a Courant–Friedrichs–Lewy condition matching the direction and speed of information propagation. A centred spatial difference has purely imaginary eigenvalues, and forward Euler gives $|S{|}^2=1+{\nu }^2{\sin }^2\theta >1$. Under a fixed positive hyperbolic Courant number, high-frequency data grow by a fixed factor in each of $O(1/h)$ steps, proving instability as the mesh is refined. This is not a claim that no exceptionally small coupled step can ever give a finite-time bound: if $\delta =O(h^2)$, the excess per step is only $O(\delta )$, but the usual hyperbolic scaling is unstable. Higher-order explicit integrators can possess nontrivial imaginary-axis stability intervals and are then suitable for skew-adjoint advection or the energy-form first-order wave equation.

Variable coefficients usually prevent Fourier diagonalization, but may preserve a self-adjoint spatial operator. The positive-face diffusion discretization in question 5 is an example. Its discrete energy identity makes $A+B$ symmetric negative definite, so a unitary eigenbasis still exists even without a closed eigenvalue formula. More generally, a dissipative operator $L_h$ in a discrete energy inner product obeyswhich directly proves a uniform semigroup estimate. The dissipative Cayley-transform contraction also follows without normality: for $\alpha \ge 0$,Taking $v=(I-\alpha L_h{)}^{-1}u$ proves contraction of the Crank–Nicolson factor. The contractivity of split Crank-Nicolson diffusion then follows from multiplying two contraction bounds; simultaneous eigenvectors of the two directional matrices are unnecessary.

The negative spectra do not imply uniform semidiscrete stability example shows what can go wrong. LetBoth eigenvalues are $-1$ for every mesh, but applying the propagator to the second unit vector at $t=1$ gives norm at least $1/(eh)$. The evolution is therefore not mesh-uniformly stable. For a discrete matrix, a unit-modulus eigenvalue with a nontrivial Jordan block gives powers growing polynomially in the step number. Even strictly interior eigenvalues can allow large transient growth when eigenvectors are badly conditioned. A pseudospectrum or numerical range of an operator, and especially a direct energy estimate, can reveal behavior missed by eigenvalues.

A resolvent estimate provides another way to see the needed uniformity. If $\Vert S_h^n\Vert \le C$ for every $n$, then for $|z|>1$,This shows directly that a uniform power estimate imposes more than the location of the spectrum. Conversely, resolvent-to-power estimates in varying dimension must themselves be uniform; a bound with constants depending on the number of grid points is not enough.

For a linear multistep method, each spatial eigenvalue gives an amplification polynomial, and all roots must satisfy the root condition for a multistep method, including simplicity of any unit-modulus root. Merely following the root approximating $e^{\delta \lambda }$ misses parasitic modes. The temporal and spatial analyses must therefore be combined with the appropriate zero-stability and uniform modal bounds.

Finally stability connects to convergence. If the exact grid samples have normalized residual ${\tau }^n$ and the error obeys $e^{n+1}=S_h{e}^n+\delta {\tau }^n$, iteration givesA residual tending uniformly to zero then gives convergence on $[0,T]$. For a well-posed linear initial-value problem and a consistent linear finite-difference approximation, this is the forward implication of the Lax equivalence theorem. The practical conclusion is to analyze the full operator including boundaries, select a time method whose stability region contains the relevant scaled spectrum, and verify that eigenvectors or an energy estimate make the resulting bounds uniform as the grid is refined.

The Hoffman–Wielandt inequality says that for normal matrices $A,B$ with eigenvalues ${\alpha }_i,{\beta }_i$, there is a permutation $\pi$ such thatwhere $\Vert \cdot {\Vert }_F$ is the Frobenius norm and $\ast$ denotes the conjugate transpose. For Hermitian matrices, the increasingly ordered real eigenvalues may be paired directly: this ordering minimizes the sum of squared distances over all permutations, by removing crossed pairs. In particular, for the real symmetric matrices in this question,The square on the right is justified by symmetry. For arbitrary matrices, the appropriate expression is the conjugate-transpose product, not the ordinary square.

A stiff differential equation can contain rapidly decaying modes as well as a slowly changing solution of interest. An explicit scheme may need a very small step solely to keep those already small fast modes from numerical growth. A-stability addresses this restriction through the Dahlquist test equation $y^{\prime }=\lambda y$, whose exact solution decays when $\mathrm{Re}\lambda <0$.

For a one-step method, write $y_{n+1}=R(z)y_n$, $z=h\lambda$. Its linear stability domain consists of the points where the update is defined and $|R(z)|\le 1$. A-stability means that this domain contains the closed left half-plane. This is a scalar linear stability property, not an order condition, an existence theorem for an implicit solve, or a general nonlinear contractivity assertion.

For a rational approximation to the exponential, consistency requires $R(z)=1+z+O(z^2)$, and order $p$ requires $R(z)-e^z=O(z^{p+1})$. Poles must be absent from the relevant half-plane. The Forward Euler method has $R=1+z$ and the disk $|1+z|\le 1$, so negative real modes require $h|\lambda |\le 2$; it is not A-stable. The Backward Euler method has $R=(1-z{)}^{-1}$, and $|1-z|\ge 1$ on the left half-plane, so it is A-stable. The trapezoidal rule and the implicit midpoint rule both have, on the scalar linear problem,They are second-order and A-stable despite being different nonlinear methods. No nonconstant polynomial approximation is A-stable, because its modulus grows without bound on the negative real axis. In particular every nontrivial explicit Runge-Kutta method has a bounded-step stability restriction.

A-stability prevents growth but need not eliminate extremely stiff modes: the trapezoidal factor tends to minus one as $z\to -\infty$. L-stability additionally requires $R(z)\to 0$ within the left half-plane. Backward Euler is L-stable; trapezoidal and implicit midpoint are not. The distinction explains persistent alternating transients in otherwise stable Crank-Nicolson calculations. Accuracy still constrains useful step sizes even for an A-stable method.

A linear multistep method is characterized by polynomials $\rho ,\sigma$, and its scalar modes satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. There is generally no single scalar amplification factor: every root must lie in the closed unit disk and unit roots must be simple. At $z=0$, this is the zero-stability root condition. Consistency plus zero-stability gives convergence by the Dahlquist equivalence theorem, with suitable starting values. A-stability demands the amplification root condition throughout the left half-plane, including the zero-step condition. The Second Dahlquist barrier says that an irreducible A-stable linear multistep method has order at most two, and an explicit linear multistep method cannot be A-stable.

Examples include backward Euler, trapezoidal, and the second-order backward differentiation formulaIts zero-step roots are $1,1/3$. Its unit-circle boundary quotient has real part $(1-\cos \theta {)}^2\ge 0$, and the leading coefficient is nonzero on the left half-plane, giving A-stability by the same continuation argument as in Question 2. Conversely the third-order member of that question fails A-stability. A formal high order without zero-stability, or after silently cancelling a problematic unit factor, does not evade the barrier.

For an implicit Runge-Kutta method, the stages on the test equation satisfy $Y=e{y}_n+zAY$. Whenever these stages are solvable, its stability function isThis rational function is used to test scalar A-stability; the Butcher order conditions separately establish nonlinear order. The Lobatto IIIA method in Question 3 is an A-stable fourth-order example with $R$ the diagonal degree-two Padé approximant. More generally the $s$-stage Gauss--Legendre Runge-Kutta method has order $2s$ and is A-stable, with the diagonal Padé approximation to $e^z$; its stiff-limit factor is nonzero. The $s$-stage Radau IIA method has order $2s-1$ and is L-stable. Thus Runge-Kutta stages permit arbitrarily high A-stable order, unlike irreducible linear multistep formulas.

For a normal matrix, scalar amplification bounds transfer through unitary diagonalization. For a non-normal matrix, eigenvalue information alone does not assert Euclidean contraction: transient growth and eigenvector conditioning can matter. Direct energy method arguments, such as the dissipative Cayley-transform proof in Question 1, use the symmetric part and are more informative in that setting.

For nonlinear dissipative vector fields, the stronger B-stability property asks that distances between two numerical solutions not increase. A standard sufficient condition for Runge-Kutta methods is algebraic stability: $b_i\ge 0$ andis positive semidefinite. It follows from the Runge-Kutta contractivity identity when the implicit stages exist. The three-stage Lobatto IIIA method is A-stable but has the first diagonal entry of that matrix equal to $-1/36$, so scalar A-stability should not be confused with that sufficient nonlinear condition. Stage solvability of an implicit Runge-Kutta method remains a separate issue and computational cost of the implicit equations must be considered. A-stability removes a scalar decay-mode stability restriction; order, stiff damping, nonlinear stability and solvability remain distinct properties.

If $Ax=\lambda x$, then $M=A-\lambda I$ is a normal matrix by part (ii), and $Mx=0$. Part (i), applied to $M$, yields $\Vert M^{†}x\Vert =\Vert Mx\Vert =0$. ThusThe nonzero vector $x$ is consequently an eigenvector of the conjugate transpose with eigenvalue $\overline{\lambda }$. This adjoint eigenvector identity for a normal matrix will also determine the special spectral restrictions below.

Set $M=A-\lambda I$. Its conjugate transpose is $M^{†}=A^{†}-\overline{\lambda }I$, soSince $A$ is a normal matrix, the two products coincide. Therefore $A-\lambda I$ is normal for every complex $\lambda$.