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Let $d=\Delta x$, impose $u_0=u_{M+1}=0$, and write the method of lines system as $U^{\prime }=LU$, whereThus $D_2$ is a negative definite symmetric matrix and $D_1$ is a skew-symmetric matrix. Use the mesh-weighted Euclidean norm $\Vert U{\Vert }_d^2=d{\sum }_{m=1}^M|u_m{|}^2$. Discrete summation by parts yields the centered Dirichlet drift-diffusion energy identityConsequentlyThe same estimate controls perturbations and is uniform in the number of grid points and in the fixed drift coefficient. Finite-dimensional linear ODE theory guarantees existence, so this proves stability of a numerical method for the semidiscretization.

The factor $d^{1/2}$ simply rescales the vector norm and does not change the induced matrix norm. Equivalently the symmetric part is $(L+L^{\ast })/2=D_2$, whose largest eigenvalue is $-4d^{-2}{\sin }^2[\pi /(2(M+1))]<0$. This is the Euclidean logarithmic norm, rather than generally the spectral abscissa of a nonnormal matrix. No periodic Fourier mode assumption has been made: the zero endpoint terms are part of the proof. In particular positivity of both off-diagonal coefficients is not needed for this $L^2$ stability result.

By orthogonal diagonalization of a real symmetric matrix, write $A=Q\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)Q^T$, with orthogonal $Q$. Its matrix exponential has the same eigenvectors and positive eigenvalues $e^{t{\lambda }_j}$. Orthogonal invariance of the induced Euclidean norm gives the exact identityThis proves the requested inequality with equality. If a real number $\beta$ gave the bound for every $t\ge 0$, evaluating on a unit eigenvector for ${\lambda }_{\max }(A)$ at any $t>0$ would give $e^{t{\lambda }_{\max }}\le e^{t\beta }$, so $\beta \ge {\lambda }_{\max }$. Thus the stated exponent is the smallest possible. For a symmetric matrix the spectral abscissa and Euclidean logarithmic norm coincide, unlike the general nonsymmetric case in Question 1.

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.

In diffusion image processing, the observed grey-value image signal $g$ is the initial condition for an evolution $u(x,t)$; time controls the image smoothing scale. A useful model must suppress image noise while respecting the image signal's geometric boundaries. Unless exterior values are intended, use periodic boundaries or Neumann boundary conditions, so the filter does not lose intensity through the image signal border.

The basic linear model is the heat equation, $u_t=\Delta u$, $u(0)=g$. On the whole plane,It is Gaussian image smoothing with variance $2t$ in each coordinate. In Fourier variables, $\hat{u}(\xi ,t)=e^{-t|\xi {|}^2}\hat{g}(\xi )$: high spatial frequencies are damped most strongly. Under zero-flux or periodic boundaries the mean is conserved, the maximum principle keeps values within the initial range, andThese give stable image noise suppression, but a sharp step also contains high frequencies and is blurred across a width of order $\sqrt{t}$. Constant diffusivity has no mechanism for distinguishing image noise from an image edge.

Linear sharpening by the backward heat equation would multiply Fourier modes by $e^{t|\xi {|}^2}$ and amplify arbitrarily fine image noise without bound. It is an ill-posed initial-value problem. A controlled unsharp masking step instead uses $g+\lambda (g-G_t\ast g)$, with multiplier $1+\lambda (1-e^{-t|\xi {|}^2})$. This amplifies high frequencies by at most $1+\lambda$; it can improve apparent contrast but also amplifies image noise and cannot reliably restore information already lost by image smoothing.

Nonlinear models use image signal structure to select the diffusion. A gradient-based energy and its formal $L^2$ gradient flow areFor a smooth solution with the corresponding zero-flux condition, $dE/dt=-\int u_t^2\le 0$. The fidelity term prevents indefinite drift towards a constant reconstruction; pure diffusion is usually stopped at a selected finite time.

The local principal diffusion coefficients distinguish suppression of diffusion from backward diffusion. Where $s=|\nabla u|>0$, the flux derivative isIts coefficient along an image signal level curve is $c(s)$; across that curve it is $c(s)+s{c}^{\prime }(s)={\Psi }^{\prime \prime }(s)$. Forward parabolic behavior requires both coefficients nonnegative, with strict positive lower bounds giving uniform parabolicity. Merely choosing $c>0$ and decreasing does not establish well-posedness.

For a convex area-type penalty $\Psi (s)=\sqrt{{\epsilon }^2+s^2}$, the coefficients areDiffusion across steep image edges is weak but remains forward. As $\epsilon \to 0$, the total variation flow formally becomes $u_t=\mathrm{div}(\nabla u/|\nabla u|)$. Its convex subgradient formulation handles flat regions and discontinuities. It favors piecewise constant image signals and preserves sharp transitions better than Gaussian image smoothing, but can produce staircasing in total variation denoising, shrink small objects and move boundaries by curvature. image edge preservation does not mean exact preservation of all image edge locations or amplitudes.

The Perona-Malik equation takes a decreasing diffusivity such as $c(s)=1/(1+s^2/{\kappa }^2)$. Small gradients are smoothed strongly, while large gradients have weak flux. More precisely,For $s>\kappa$, diffusion in the gradient direction is backward, so a strong transition can steepen. This gives formal image edge enhancement but also causes instability and continuum ill-posedness. The exponential choice $c(s)=e^{-s^2/{\kappa }^2}$ similarly has a negative normal coefficient for $s>\kappa /\sqrt{2}$. Discrete results depend on the stencil, step size and implicit regularization; appealing visual results are not a proof of a well-posed PDE.

A regularized Perona-Malik diffusion computes the conductance from a smoothed image signal, for exampleThe flux still acts on $u$, but the image edge detector is less sensitive to raw image noise. With a smooth kernel and a positive conductance bounded away from zero on the attained range, the local principal diffusion is forward; appropriate regularity and boundary assumptions give a well-posed nonlocal parabolic model. The exact regularization and fixed image smoothing scale are part of the model.

A genuinely directional filter uses a diffusion tensor for image filtering:Here $e_n$ estimates the image edge normal and $e_t$ its tangent. A structure tensor $J_{\rho }=G_{\rho }\ast (\nabla u_{\sigma }\nabla u_{\sigma }^T)$ provides robust directions at a second averaging scale. Strong tangent diffusion smooths image noise along an image edge, while weak normal diffusion reduces mixing across it. Related coherence-enhancing choices connect elongated features along their dominant orientation. Positive tensor eigenvalues preserve forward parabolicity; this type of enhancement must be distinguished from Perona–Malik's backward normal diffusion.

For more direct sharpening, a shock filter for image enhancement formally evolves $u_t=-\mathrm{sign}(\Delta u)|\nabla u|$. It is a Hamilton–Jacobi-type transport mechanism, not a positive diffusion operator, and steepens transitions around inflection boundaries. In practice it can be combined with regularized forward diffusion to control image noise. Any enhancement method needs a scale or stopping rule and an honest treatment of image noise amplification.

Linear heat flow offers predictable image smoothing but blurs image edges; convex nonlinear diffusion can preserve them; backward or shock mechanisms sharpen them at a greater stability cost. Gradient thresholds, conductance regularization and positive tensor directions determine which of these behaviors a proposed filter actually has.

A singular value system of a compact operator consists of positive numbers ${\sigma }_j$ and orthonormal families $v_j\in X$, $u_j\in Y$ satisfyingThus $v_j$ are positive-eigenvalue eigenvectors of $A^{\ast }A$ and $u_j$ of $A{A}^{\ast }$, with eigenvalue ${\sigma }_j^2$. The families are complete in $(\ker A{)}^{\perp }$ and $\overline{\mathrm{ran}A}$ respectively. The singular values can be listed nonincreasingly with multiplicities, and tend to zero in the infinite-rank case. For finite rank there are only finitely many positive singular values; zero-kernel directions are handled separately.

Use the convention that $⟨f,g⟩$ is conjugate-linear in its first entry. Then the singular value system givesThe latter series converges precisely on the admissible range component specified by the Picard criterion:with $R^{\perp }$ components annihilated by the inverse. Merely writing a formal singular expansion does not imply it converges in $X$.

Keep the prescribed outer slip velocity fixed when perturbing the unsteady Prandtl equation. Cancelling the base equation and retaining terms linear in the disturbance gives the linearized unsteady Prandtl equationThe no-slip boundary condition supplies $\tilde{u}=\tilde{v}=0$ at the wall; the fixed outer velocity supplies $\tilde{u}\to 0$ at infinity. There is no prescribed zero normal-velocity disturbance at infinity: a finite displacement-related value is permitted.

With the frozen parallel profile $U=U(y)$, $V=0$, the coefficients are invariant under translations in $x$ and $t$. Normal modes therefore separate those variables. For the mode convention used here the equations areEliminate $u=-v^{\prime }/(i\alpha )$ to obtain the Prandtl normal-mode equationwith a bounded far-field $v$. Requiring $v(\infty )=0$ would incorrectly eliminate the proposed outer profile. For real $U$ and real $\alpha$, complex conjugation of the equation and its boundary conditions replaces $(\alpha ,c,v)$ by $(-\alpha ,c^{\ast },v^{\ast })$. This proves the stated eigenvalue symmetry. The paired modes have the same temporal growth rate $\alpha \mathrm{Im}c$, so choosing positive wavenumber loses no real physical disturbance.

Here the prescribed parallel profile is a local frozen-coefficient model. A nontrivial arbitrary $U(y)$ with $V=0$ and constant outer slip is not generally an exact steady solution of the unforced unsteady Prandtl equation; its base equation would require $U^{\prime \prime }=0$. The following stability calculation uses the simplifying local model stipulated for the mode analysis.

Set $c_0=U_c=U(y_c)$. Away from the critical point the dominant equation is $(U-c_0)v_0^{\prime }-U^{\prime }{v}_0=0$, whose solutions are multiples of $U-c_0$. Choosing the coefficient to be zero below $y_c$ and one above it satisfies the wall and tangential far-field conditions. The choice $c_0=U_c$ makes $v_0$ continuous across the joining point; $U^{\prime }(y_c)=0$ also makes its first derivative continuous. Its second derivative jumps. Thus it is a valid leading outer solution away from $y_c$, but viscosity must smooth the join in a critical layer in a shear flow.

Near the join, $U-c_0\sim (y-y_c{)}^2/2$. If the layer width is $\ell$, matching gives $v=O({\ell }^2)$. The advection side of the mode equation scales as $\alpha {\ell }^3$ and $v^{\prime \prime \prime }$ scales as ${\ell }^{-1}$. Balance gives ${\ell }^4\sim {\alpha }^{-1}=\epsilon /k$. Equivalently, the eigenvalue correction ${\epsilon }^{1/2}{c}_1$ balances $(y-y_c{)}^2$, so the quarter-power critical layer hasSubstitution, retaining the leading terms and cancelling their common factor, gives

At order ${\epsilon }^{1/2}$ outside the layer,A solution consistent with the chosen lower branch and the wall conditions is $v_1=0$ below the join. Above it a particular solution is $v_1=-c_1$; a multiple of $U-c_0$ represents an arbitrary amplitude renormalization and may be set to zero. Matching consequently requiresThe corresponding derivatives match as $w_Y\sim Y$, $w_{YY}\to 1$ on the positive side and as zero on the negative side. The matching statements require that growing homogeneous corrections be absent.

To symmetrize these different end conditions and remove $k$ and $i$, chooseThe polynomial $A(z)=z^2/2-C$ is itself a solution of the transformed homogeneous equation. Direct substitution, using $a^4=ik$, givesThe derivative is third order, as required by the original mode equation. Initially the two ends are along the rotated rays $z=aY$ with real $Y$. Continuing them to the real $z$ axis is compatible with the asymptotic end conditions: the decreasing homogeneous correction behaves exponentially as $\exp [-z^2/(2\sqrt{2})]$ and still decreases throughout the rotation from angle $\pi /8$ to zero. Thus the supplied real-axis spectral normalization uses the same recessive conditions; a complex coordinate change should not be mistaken for a real stretching alone.

For each supplied real eigenvalue $C_n=(4n+7)/\sqrt{2}$, the phase-speed correction isThe mode factor has temporal magnitude $e^{\alpha \mathrm{Im}ct}$. ThereforeAmong the listed indices, $n=-2$ gives positive growth $\gamma \sim \sqrt{\alpha }/2$; $n=-1,1,2,\dots$ give decay. Hence the frozen non-monotone flow has high-wavenumber instability of a non-monotone Prandtl layer, with arbitrarily rapid linear growth as the positive wavenumber increases. The normal-mode calculation demonstrates linear instability; it is not by itself a claim about the final nonlinear state.

For the second-harmonic feedback in a long-wave convection amplitude equation, at $\mu =2$ the linear steady operator is $L_0=-(1+{\partial }_x^2{)}^2$. Its eigenvalue on $e^{inx}$ is $-(1-n^2{)}^2$, so the critical Fourier modes are $n=\pm 1$. More generally the linear dispersion relation is $\lambda (k)=-1+\mu k^2-k^4$, giving first onset at $\mu =2,k=1$.

At order ${ϵ}^2$, the weakly nonlinear expansion givesWith ${\Theta }_1=A{e}^{ix}+\overline{A}{e}^{-ix}$,The constant part disappears under differentiation. Inverting $L_0$ on the second harmonic, whose eigenvalue is $-9$, givesup to a critical-harmonic correction absorbed into the definition of $A$. At order ${ϵ}^3$,The coefficient of $e^{ix}$ in the last three terms is respectivelyThe Fredholm solvability condition requires their sum to vanish, because $L_0$ annihilates the critical Fourier mode. For a nonzero amplitude,This requires ${\mu }_2/(3-8s^2/9)>0$. The cubic amplitude coefficient is positive for $s^2<27/8$, yielding a small-amplitude branch in a supercritical bifurcation; it is negative for $s^2>27/8$, yielding a subcritical bifurcation branch in this leading approximation. At $s^2=27/8$, the cubic coefficient vanishes and the quoted relation is singular: higher-order nonlinear terms and a different detuning balance are needed. One must not assert a finite $p$ there.

If a term $ϵ{\mu }_1$ were included in $\mu$, order ${ϵ}^2$ would also contain $-{\mu }_1{\Theta }_{1xx}={\mu }_1{\Theta }_1$. The quadratic nonlinearity produces only the zeroth and second harmonics, with the zeroth differentiated away, so it cannot balance a first-harmonic contribution at that order. Solvability would give ${\mu }_{1}A=0$. Hence a nontrivial critical-mode expansion forces ${\mu }_1=0$ and first balances detuning against cubic amplitude effects at order ${ϵ}^3$.

To find the threefold phase-locked equilibria, write $C=R{e}^{i\theta }$ with $R>0$; separating real and imaginary parts givesA steady state satisfiessoWriting $y=R_0^2$ and $D=1+4\mu -4{\omega }^2$, the amplitude branches areThere are two distinct positive roots for $\mu >{\omega }^2-1/4$, except at $(\mu ,\omega )=(0,0)$, where $y_{-}=0$ and only the upper root is nonzero. Indeed their sum is positive in this range and their product is ${\mu }^2+{\omega }^2$. Below the fold condition there are none. At equality, there is one positive repeated root $y={\omega }^2+1/4$, the saddle-node bifurcation limit.

For each positive amplitude, the sine and cosine determine $3{\theta }_0$ modulo $2\pi$, giving three phases separated by $2\pi /3$. Thus the source's “two states” means two amplitude branches modulo the threefold spatial symmetry. Generically there are six nonzero complex equilibria, three on each branch, not literally two.

The polar Jacobian matrix at an equilibrium isOn the lower branch, $\det J=-3y_{-}\sqrt{D}<0$, so it is a saddle equilibrium and unstable. On the upper branch, $\det J=3y_{+}\sqrt{D}>0$ andbecause existence implies $\mu >-1/4$. Therefore every upper-branch equilibrium is asymptotically stable, and every lower-branch equilibrium is a saddle equilibrium, away from the degenerate endpoints. The eigenvalues are unchanged by the smooth polar coordinate transformation at $R_0>0$. The origin, not covered by those coordinates, has eigenvalues $\mu \pm i\omega$ and is stable for $\mu <0$ and unstable for $\mu >0$.

The unforced onset is a stationary pattern-forming instability. Under a horizontal translation, the critical Fourier mode transforms as $A\mapsto e^{i\varphi }A$. A cubic amplitude equation without forcing must have the same phase weight: its leading terms are $\tilde{\mu }A-c|A{|}^{2}A$. Reflection of the unforced spatial pattern conjugates $A$, permitting real coefficients in this stationary problem. They are determined by a weakly nonlinear expansion and projection onto the adjoint eigenfunction; symmetry alone does not calculate their values or guarantee a nonzero coupling.

Represent the third-harmonic forcing by a complex coefficient $F$ multiplying $e^{3i{k}_{c}x}$, whose phase weight is three. The product $F{\overline{A}}^2$ has weight $3-2=1$ and therefore resonates with the critical positive harmonic. Neither a direct third-harmonic term nor $F\overline{A}$ has the required wavenumber balance. A travelling boundary pattern makesA response phase and the sign of its coefficient can be incorporated into $\delta$. The resulting three-to-one spatially forced amplitude equation isThis retains the leading resonant forcing term, linear detuning and cubic saturation, while dropping higher powers and nonresonant harmonics. The forcing is weak, the unforced critical eigenvalue is near zero, and the amplitude varies on a slow time; very high forcing frequency outside that slow scaling would require a different averaging argument. Reduction to the specific saturating canonical form in part (i) additionally assumes $c>0$ and nonzero forcing.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

Let $Dx=\lambda x$ with $x\ne 0$. Commutation givesThus $Cx$ belongs to the eigenspace $E_{\lambda }(D)$, with the zero vector allowed. Since $D$ has $n$ distinct eigenvalues, each eigenspace is one-dimensional. Hence $Cx=\mu x$ for some scalar $\mu$. This is the fact that commuting maps preserve eigenspaces.

The standard linear independence theorem for eigenvectors belonging to distinct eigenvalues supplies a basis $x_1,\dots ,x_n$ of eigenvectors of $D$. The preceding argument makes every one an eigenvector of $C$ as well. Their column matrix therefore simultaneously diagonalizes $C$ and $D$. This is simple spectrum gives simultaneous diagonalization.

For real matrices, this conclusion always holds over $\mathbb{C}$; a real common basis follows when the distinct eigenvalues of $D$ are real. Real matrix entries alone do not ensure a real eigenbasis: a quarter-turn of ${\mathbb{R}}^2$, with $C=D$, has eigenvalues $\pm i$. The matrices in this question have real spectra, so their common basis is real.

The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.