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A Runge-Kutta method with stage matrix $A$, weights $b$ and step size $h$ isIts order measures accuracy as $h\to 0$, while stability concerns how errors or stiff components propagate. Linear A-stability, strong stiff damping through L-stability, and nonlinear B-stability address different questions.

For the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is $\{z:|R(z)|\le 1\}$. The method is A-stable when it contains the closed left half-plane, and L-stable when it is also true that $R(z)\to 0$ as $|z|\to \infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.

The Forward Euler method has $R(z)=1+z$, so it fails A-stability, for example at $z=-3$. In fact no nontrivial explicit Runge-Kutta method is A-stable: its stability function is a nonconstant polynomial, which is unbounded along the negative real axis. The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z|\ge 1$, and the function tends to zero at infinity, proving both A-stability and L-stability. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability, but its limit $-1$ rules out L-stability.

For nonlinear equations, a dissipative vector field satisfies $⟨f(x)-f(y),x-y⟩\le 0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that inner product. A B-stable method preserves this contraction for every $h>0$, whenever its stages are well defined.

Let $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the update's squared norm, substituting $d=D_i-h{\sum }_j{a}_{ij}{F}_j$, and symmetrizing the double sum proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and the matrix $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the quadratic form of $M$ over each coordinate of the $F_i$. Thus algebraic stability implies B-stability. Applied to complex scalar linear equations with the real part of the Hermitian inner product, the same argument also implies A-stability, where the stages are defined.

The Backward Euler method has $b_1=a_{11}=1$, hence $M=(1)$, and is algebraically stable. One can prove its nonlinear contraction directly: $d_{n+1}=d+h(f(y_{n+1})-f({\tilde{y}}_{n+1}))$ gives $\Vert d_{n+1}{\Vert }^2\le ⟨d,d_{n+1}⟩\le \Vert d\Vert \Vert d_{n+1}\Vert$. The implicit midpoint rule has $a_{11}=1/2$, $b_1=1$ and $M=0$, so it is B-stable even though it is not L-stable.

Linear A-stability does not by itself imply nonlinear B-stability. The ODE trapezoidal rule has the same stability function as the implicit midpoint rule, but take the dissipative scalar field $f(y)=-\max (y,0)$ and $h=6$. For any positive starting value, its trapezoidal update is $y_{n+1}=y_n+3(f(y_n)+f(y_{n+1}))=-2y_n<0$. Thus two positive starting values have their distance doubled, violating B-stability. Its algebraic stability matrix also has a negative first diagonal entry. The field is globally Lipschitz continuous, so this is a well-defined nonlinear counterexample.

These examples distinguish accuracy, scalar stiff-mode stability, nonlinear contraction, and rapid stiff-mode damping. For implicit methods, solvability of the stage equations remains an additional requirement; a formal stability function alone does not establish it for arbitrary nonlinear problems.

The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.

The Hermitian conjugate is $A^{†}={\overline{A}}^{,T}$. A matrix is unitary when $A^{†}A=A{A}^{†}=I$, and Hermitian when $A^{†}=A$.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.

For $x\ne 0$,so the Hermitian matrix $G$ has positive real eigenvalues. If $G{e}_i={\lambda }_i{e}_i$, thenMoreover $|A{e}_i{|}^2={\lambda }_i|e_i{|}^2$, so $|f_i|/|e_i|=\sqrt{{\lambda }_i}$.

A real matrix is orthogonal when $Q^{T}Q=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda |=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian inner product gives $(1-\overline{\lambda }\mu )v^{†}w=0$; distinct unit-modulus eigenvalues therefore have orthogonal eigenvectors.

The nonreal eigenvalues of a real $3\times 3$ matrix occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real eigenvalue is $-1$. If $x\cdot n=0$, then $(Qx)\cdot (Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose determinant is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal basis adapted to $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

The system matrix has eigenvalues $2$ and $3$, with eigenvectors in the directions $y=-x$ and $y=-x/2$, respectively. Both eigenvalues are positive, so the origin is an unstable node. The two eigendirections are straight trajectories directed away from the origin. Every other trajectory also moves outward, tangent near the origin in backward time to the faster eigendirection and asymptotic in forward time to the eigenvalue-$3$ direction.

Choose bases of $V$ and $W$, of dimensions $v$ and $w$. A linear map is uniquely determined by the arbitrary images of the $v$ basis vectors, each with $w$ coordinates, so

Put $a=\dim A$ and $b=\dim B$, and extend bases of $A,B$ to bases of $V,W$. The condition $\varphi (A)\subseteq B$ forces the $(w-b)a$ matrix entries from $A$ to a complement of $B$ to vanish and imposes no other restriction. Hence

For the last part let $r=\dim (S\cap T)=s+t-v$. ChooseA map in $Y$ sends $S\cap T$ into itself, $S_0$ into $S$, and $T_0$ into $T$, with these choices independent. Therefore

False. The real matrixhas the two distinct complex eigenvalues $i$ and $-i$, so it is diagonalizable over $\mathbb{C}$. It has no real eigenvalue and hence no real eigenbasis, so it is not diagonalizable over $\mathbb{R}$.

True. Suppose $A,B$ are real and $AP=PB$ for an invertible complex matrix $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts givesThe polynomial $q(t)=\det (X+tY)$ is not identically zero, since $q(i)=\det P\ne 0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb{R}$, as stated by complex similarity of real matrices implies real similarity.

For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.