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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.

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 $\mathrm{Tr}(A-B{)}^2$.

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.

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 $\lambda$.

For the standard complex inner product, $\Vert y{\Vert }^2=y^{†}y$. Using the conjugate transpose and the defining equality for a normal matrix,Both norms are nonnegative, so $\Vert Ax\Vert =\Vert A^{†}x\Vert$ for every $x$.

The adjoint eigenvector identity for a normal matrix gives $A^{†}{x}_{\lambda }=\overline{\lambda }{x}_{\lambda }$. HenceAs $\mu \ne \lambda$, $x_{\lambda }^{†}{x}_{\mu }=0$. Thus eigenvectors in distinct eigenspaces of a normal matrix are orthogonal in the complex inner product.