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Expansion along the first column gives the characteristic polynomialThus the eigenvalue $2$ has algebraic multiplicity two. Solving $(M-2I)(x,y,z{)}^T=0$ gives $y=0$ and then $z=0$, leaving $x$ arbitrary. Its eigenspace is $\mathrm{span}\{(1,0,0{)}^T\}$ and has geometric multiplicity one. A diagonalisable matrix needs an eigenvector basis, so $M$ is not diagonalisable.

If $B=S^{-1}AS$ with $S$ invertible, then $tI-B=S^{-1}(tI-A)S$. Multiplicativity of the determinant givesThe roots of the characteristic polynomial, counted with their multiplicities, are the eigenvalues with their algebraic multiplicities. Thus similar matrices have the same eigenvalues with the same algebraic multiplicities.

The converse is false. The matricesboth have characteristic polynomial $(t-1{)}^2$, but every matrix similar to $I_2$ is $S^{-1}{I}_{2}S=I_2$, whereas $J\ne I_2$. Alternatively, their geometric multiplicities for $1$ differ.

The Cayley-Hamilton theorem says that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a $2\times 2$ diagonalisable matrix, write $A=SD{S}^{-1}$ with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. The characteristic polynomial is $(t-{\lambda }_1)(t-{\lambda }_2)$, including repeated roots. Evaluating it at $D$ gives zero in each diagonal position, and polynomial evaluation respects matrix similarity: ${\chi }_A(A)=S{\chi }_A(D)S^{-1}=0$. This proves the requested case.

For the final assertion, let $\lambda$ be any complex eigenvalue of $B$, with nonzero eigenvector $\mathbf{v}$. From $B^k=0$ we obtain $0=B^k\mathbf{v}={\lambda }^k\mathbf{v}$, so $\lambda =0$. The characteristic polynomial of this nilpotent linear map is therefore $t^n$. Applying the full Cayley-Hamilton theorem yields $\boxed{B^n=0}$. This argument does not assume $B$ is diagonalisable.

The symmetry between the last two coordinates gives the eigenvector $(0,1,-1{)}^T$ with eigenvalue $2$. On vectors $(s,t,t{)}^T$, the remaining eigenvalue equation iswhose characteristic polynomial is ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)$. Corresponding eigenvectors are $(1,-1,-1{)}^T$ and $(2,1,1{)}^T$. Thus all eigenvalues and eigenspaces areNonzero scalar multiples give all eigenvectors in each eigenspace. The three displayed vectors are mutually orthogonal and have squared norms $3,2,6$, respectively.

Use the classical risk model surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with $c>0$, and write $\phi$ for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate $\lambda$. A claim at time $t$ leaves capital $u+ct-x$ if $x\le u+ct$, after which the Markov property restarts the same risk model. ConsequentlyPut $C(v)={\int }_0^v\phi (v-x)f(x)dx$ and change variable $v=u+ct$:The convolution $C$ is continuous since $\phi$ is bounded and $f$ is integrable. Differentiating proves the survival integro-differential equation

For the specified claim law, expand its probability density function as $f(x)=\frac{1}{2}{e}^{-x}+\frac{1}{4}{e}^{-x/2}$. This is an equal-weight mixture distribution of exponential distributions with rates $1$ and $1/2$. In the convolution, set $z=u-x$ to obtainSince $\lambda /c=10/21$, the requested coefficients are

For the exponential-mixture differential equation for survival probability, define $A(u)=e^{-u}{I}_1(u)$, $B(u)=e^{-u/2}{I}_2(u)$ and $r=\lambda /c$. Differentiating the integrals yieldsApply $(D+1)(D+\frac{1}{2})$ to the last equation, where $D=d/du$. The first two equations eliminate $A,B$, givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution $C_0+C_1{e}^{-u/6}+C_2{e}^{-6u/7}$.

The claim expected value is $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 2=3/2$. The relative safety loading is $\rho =2.1/(3/2)-1=2/5$, and the given zero-capital survival probability is $\phi (0)=2/7$. The limit at infinity sets $C_0=1$. A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so ${\phi }^{\prime }(0)=(10/21)(2/7)=20/147$. Thereforewhich gives $C_1=-20/29$ and $C_2=-5/203$. The final survival probability and ultimate ruin probability areBoth exponential coefficients of $\psi$ are positive. Hence $\phi$ increases from $2/7$ to one, and $\psi$ decreases from $5/7$ to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.

Take every finite positive drag coefficient $\gamma >0$ and defineThe coefficients of the characteristic polynomial are $c_3=2\gamma$, $c_2=K+{\gamma }^2$, $c_1=2L\gamma$ and $c_0=P{\gamma }^2$. The nontrivial Routh-Hurwitz criterion reduces exactly toThe printed strict chain $K>L>P>0$ makes every coefficient positive and both bracket terms positive, so it proves attraction for every positive drag coefficient.

There is, however, a boundary error in the asserted necessity. To have the bracket positive for every $\gamma >0$, its constant and slope as a function of ${\gamma }^2$ must be nonnegative, and cannot both vanish. Together with coefficient positivity this gives the exact drag-polynomial stability for all positive stopping rates:Necessity follows by taking arbitrarily small and arbitrarily large positive $\gamma$; if one bracket coefficient is negative, the inequality fails in the corresponding limit. Sufficiency follows because at least one coefficient is strictly positive. Thus $K=L>P$ and $K>L=P$ still give strict decay at each finite positive $\gamma$. They need not give a uniform stability margin in the zero- or infinite-drag limit.

In a Keplerian shearing sheet, $S=3\Omega /2$ andThe differences areHence $r>3$ establishes the requested strict chain and trapping conclusion. At $r=3$, however, $K=L={\Omega }^2$ and $P=9{\Omega }^2/16$, soAll coefficients remain positive. This is an explicit counterexample within the stated vortex model: $r=3$ also attracts particles for every finite positive drag coefficient, so the literal strict “if and only if” is false. The exact Keplerian all-positive-drag condition is $r\ge 3$. If $\gamma =0$ is included, the polynomial has zero and purely imaginary roots and there is no asymptotic attraction; “all drag coefficients” must mean positive drag. The local attraction claim presumes the particle remains in the patch, as is assured for sufficiently small perturbations of its centre.

Write $u=u_x$ and $B=B_z$. In this geometry the Lorentz force density is purely longitudinal and equals $-{\partial }_x(B^2/2{\mu }_0){\mathbf{e}}_x$. The ideal magnetohydrodynamic induction equation becomes $B_t+{\partial }_x(uB)=0$. Together with the continuity equation and adiabatic pressure evolution, the four primitive-variable equations areThey have the required quasilinear form ${\mathbf{U}}_t+A{\mathbf{U}}_x=0$ withHere the notation $u_x$ in derivatives means ${\partial }_{x}u$, whereas $u$ denotes the longitudinal velocity. Spatial dependence is one-dimensional; time dependence is allowed in the wave problem.

Put $c_f^2=\gamma p/\rho +B^2/({\mu }_0\rho )$, and assume $\rho >0$ and $c_f>0$. Expansion of the characteristic polynomial givesTwo independent eigenvectors for ${\lambda }_0$ can be chosen as $(1,0,0,0{)}^T$ and $(0,-B/{\mu }_0,0,1{)}^T$. For the two propagating branches, useful eigenvectors areThe repeated branch advects an entropy disturbance and a disturbance with unchanged total gas-plus-magnetic pressure. The $\pm$ branches are perpendicular fast magnetosonic waves, with speed $c_f$ relative to the fluid.

On a uniform background, a Fourier amplitude $\hat{\mathbf{U}}{e}^{i(kx-\omega t)}$ obeys $(kA-\omega I)\hat{\mathbf{U}}=0$. Thus the eigenvalues are the phase velocities $\omega /k$ for $k\ne 0$, and the four dispersion branches are $\omega =ku$ twice and $\omega =k(u\pm c_f)$. The reduced geometry retains neither an independent transverse velocity nor an additional propagating Alfvén wave.

To construct each finite-amplitude simple wave in magnetohydrodynamics, trace a state-space curve with tangent $r_{\pm }$, using $\rho$ as its parameter. Integrating the pressure and magnetic components givesThe constants $K$ and $b$ are the entropy and transverse magnetic flux freezing invariants along this state curve. HenceSince $A{\mathbf{U}}_{\rho }={\lambda }_{\pm }{\mathbf{U}}_{\rho }$, substituting this curve into the original system reduces it to ${\rho }_t+{\lambda }_{\pm }(\rho ){\rho }_x=0$. By the chain rule,Thus the characteristic speed itself satisfies the Inviscid Burgers equation. For initial speed $L(a)$, its smooth solution is ${\lambda }_{\pm }=L(a)$ along $x=a+tL(a)$, until the characteristic curves intersect. This derivation supplies genuine state curves for simple waves, not merely a linear dispersion relation.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.

By the given matrix similarity, $A^{\prime }=P^{-1}AP$ for an invertible $P$. HenceThe characteristic polynomials are identical. Therefore the characteristic equations and eigenvalues, including algebraic multiplicities, agree. The conclusion is independent of the choice between $\det (tI-A)$ and $\det (A-tI)$ conventions.

Let the repeated eigenvalue be $a$, and choose an eigenvector $v_1\ne 0$. Complete it to a basis $(v_1,v_2)$ of ${\mathbb{C}}^2$. In this basis the matrix is an upper triangular matrix, with first column $(a,0{)}^T$, say $T=\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)$. Its characteristic polynomial is $(t-a)(t-d)$. By part (a)(ii), this must equal $(t-a{)}^2$, so $d=a$.

If $c=0$, the matrix is already $aI$. If $c\ne 0$, rescale the first basis vector by $k=c$ and apply part (a)(iii); the upper-right entry becomes one. ThusThe second possibility is a size-two Jordan block. It has only one independent eigenvector, whereas $aI$ has a two-dimensional eigenspace, so the two types cannot be related by matrix similarity. This repeated-eigenvalue classification in dimension two proves the required result without assuming the general Jordan normal form theorem.

The characteristic polynomial factors as $(r-2)(r+1)$, so the complementary solution is $A{e}^{2t}+B{e}^{-t}$. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses $t{e}^{2t}$ and $t{e}^{-t}$. Substitution gives the particular solution $t{e}^{2t}-t{e}^{-t}-3t$. ThusThe initial conditions give $A+B=0$ and $2A-B-3=0$. HenceThe factors of $t$ in the particular solution are essential: an unmodified exponential trial lies in the kernel of the differential operator.

Write $U^{\ast }={\overline{U}}^T$ for the conjugate transpose. Taking determinants in $U^{\ast }AU=A$ gives $|\det U{|}^2\det A=\det A$. Since the Hermitian matrix $A$ is invertible, every such $U$ is invertible and $|\det U|=1$. The identity belongs to $U_A$; if $U,V\in U_A$, then $(UV{)}^{\ast }A(UV)=V^{\ast }AV=A$; and multiplying $U^{\ast }AU=A$ by $(U^{\ast }{)}^{-1}$ and $U^{-1}$ gives $(U^{-1}{)}^{\ast }A{U}^{-1}=A$. Associativity comes from matrix multiplication, so this is a group.

The Hermitian form $h_A(v,w)=v^{\ast }Aw$ obeys $h_A(Uv,Uw)=h_A(v,w)$ precisely when $U\in U_A$. Thus $U_A$ is the group of linear isometries of this nondegenerate Hermitian form; positivity of $A$ is unnecessary.

For $A=I$, $U$ is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over $\mathbb{C}$, its characteristic polynomial has a root, giving a unit eigenvector $v$ with $Uv=\lambda v$. Preservation of the inner product gives $|\lambda |=1$. If $w\perp v$, then $⟨Uw,Uv⟩=⟨w,v⟩=0$, so $v^{\perp }$ is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, $U$ is diagonal, with diagonal entries on the unit circle.

A matrix is in Jordan normal form if it is block diagonal with Jordan blocks $J_r(\lambda )=\lambda I+N$, where $N$ has ones immediately above the diagonal and zeros elsewhere. For a Jordan normal form matrix $J$, number its diagonal positions $1,\dots ,n$ and putChoose positive ${\epsilon }_m\to 0$ avoiding the finitely many values solving ${\lambda }_i+i\epsilon ={\lambda }_j+j\epsilon$ for $i\ne j$. The matrix remains upper triangular and now has distinct eigenvalues, so it is diagonalizable. Its entries converge to those of $J$. For arbitrary $A=SJ{S}^{-1}$, use $A_m=S{J}_m{S}^{-1}$. This proves density of diagonalizable complex matrices.

If $A$ is diagonalizable with eigenvalues ${\lambda }_1,{\lambda }_2$, conjugating $B$ by the same basis change conjugates $T_A$ to $T_{\mathrm{diag}({\lambda }_1,{\lambda }_2)}$. On the four matrix units $E_{ij}$, this acts by $({\lambda }_i+{\lambda }_j)E_{ij}$. Thus, writing $s=\mathrm{tr}A$ and $d=\det A$, its characteristic polynomial isThe entries of $T_A$ depend linearly on those of $A$, and the coefficients of its characteristic polynomial depend polynomially, hence continuously, on those entries. Apply the formula to the approximating $A_m$ and pass to the limit. The boxed formula therefore holds for every $A$, including non-diagonalizable matrices. This is the two-by-two anticommutator characteristic polynomial.

Expanding the characteristic polynomial givesSolving the three nullspace systems gives the eigenvalues and eigenspacesFor example, at $\lambda =1$ the first two equations force $z=0$ and $x=y$; at $\lambda =2$ they force $x=0$ and $y=-z$; at $\lambda =-1$ they force $x=y=z$.

Putting these eigenvectors in columns proves their linear independence:Hence $A$ is a diagonalizable matrix over the reals.

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.