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Factor the first characteristic polynomial asThe other two roots of a polynomial have product one. If $-1<\alpha <1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$: distinct unit-modulus roots, both different from one. At $\alpha =-1$, the root $-1$ is double; at $\alpha =1$, the root one is triple. Outside $[-1,1]$, one of the reciprocal real roots has modulus greater than one. Thus the root condition for a multistep method holds exactly for $-1<\alpha <1$.

For the order conditions for a linear multistep method, expand the exponential defect:The constant, linear and quadratic coefficients vanish for every $\alpha$. The cubic coefficient vanishes only at $\alpha =11/5$, where the quartic coefficient is $-3/5\ne 0$. Consequently the formal order isFor all parameters satisfying the root condition for a multistep method, ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )\ne 0$, so this formal calculation gives ordinary second-order consistency of a numerical method. The Dahlquist equivalence theorem therefore yieldsThe exceptional third-order formula is not zero-stable. At $\alpha =1$, $\sigma \equiv 0$ and $\rho =(w-1{)}^3$: the recurrence contains no vector-field evaluations. It satisfies the displayed Taylor identities only formally and is a degenerate, nonconvergent formula, not a usable second-order ODE solver. This distinction avoids interpreting formal cancellation as convergence.

Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.

The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At $b=0$ the eigenvalues coalesce and this diagonalizability is lost; when $b<0$ the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.

Since $S\mathbf{w}=c{\mathbf{e}}^{(1)}$ and $A\mathbf{w}={\lambda }_1\mathbf{w}$,Thus the first column of $\hat{A}$ is $({\lambda }_1,0,\dots ,0{)}^T$, and $\hat{A}$ is a block upper triangular matrix:where $B$ is its bottom-right $(n-1)\times (n-1)$ submatrix. ThereforeA similarity transformation preserves the characteristic polynomial, sowith algebraic multiplicities included.

To construct $S$, use a Householder transformation. With $\alpha =\Vert \mathbf{w}{\Vert }_2$, choose the sign of $\alpha$ to avoid cancellation and setThen $S$ is an orthogonal matrix and maps $\mathbf{w}$ to $\alpha {\mathbf{e}}^{(1)}$ up to the chosen sign. If $\mathbf{w}$ already lies on the first coordinate axis, take a suitable diagonal sign matrix. This is the one-vector case of orthogonal coordinate reduction of a subspace.

The characteristic polynomial of the homogeneous linear ordinary differential equation isso its complementary solution is $A{e}^{3x}+B{e}^{-x}$. Because the forcing contains the resonant factor $e^{-x}$, set $y=e^{-x}v$. Substitution givesWriting $w=v^{\prime }$, a polynomial particular solution of $w^{\prime }-4w=-16x$ is $w=4x+1$, and hence $v=2x^2+x$. ThusBoundedness as $x\to \infty$ forces $A=0$, while the initial condition $y(0)=1$ gives $B=1$. Therefore

Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

The adjugate matrix is the transpose of the cofactor matrix, and it satisfiesPut $M=tI-A$ and substitute the two given expansions into this identity:Comparing coefficients of powers of $t$ givesHere $c_0=1$ because the characteristic polynomial is monic.

Taking the matrix trace of the adjugate expansion and using part (b) givesHence $\mathrm{Tr}(B_i)=(n-i)c_i$. Taking traces in $B_i=A{B}_{i-1}+c_{i}I$ gives, for $1\le i<n$,For $i=n$, the same formula follows by taking the trace of $-A{B}_{n-1}=c_{n}I$. Iterating the recursion from part (a) also givesConsequently the Faddeev–LeVerrier algorithm isStarting with $c_0=1$, this recursively expresses every $c_i$ using only $\mathrm{Tr}(A),\dots ,\mathrm{Tr}(A^i)$; these are the Newton identities for the characteristic polynomial.

If $A$ is an invertible matrix, thenThus $AB$ and $BA$ are similar matrices, and similar matrices have the same characteristic polynomial.

Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the spectrum of a three-dimensional matrix with zero row sums. In particular, $B(1,1,1{)}^T=0$. DefineExpanding the characteristic polynomial gives $\det (\lambda I-B)=\lambda ({\lambda }^2+S\lambda +T)$, soFor the physical coefficients, $L_j{B}_{jk}=L_k{B}_{kj}$ with $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$. Thus $W=\mathrm{diag}(L_j)$ gives positive diagonal symmetrization of a matrix, andConsequently the eigenvalues are real and nonpositive, and $B$ has a complete eigenvector basis. This justifies the mode expansion in part (i).

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

Two square matrices are similar when $B=P^{-1}AP$ for some invertible matrix $P$. A Jordan normal form is a block-diagonal matrix whose blocks have one eigenvalue on the diagonal, ones on the superdiagonal, and zeros elsewhere; over an algebraically closed field every matrix is similar to such a form, unique up to reordering its Jordan blocks.

Both displayed matrices have characteristic polynomialHowever,so their eigenspace dimensions are respectively two and one. Thus $A$ has Jordan form $J_2(1)\oplus J_1(1)$, whereas $B$ has Jordan form $J_3(1)$. By uniqueness of Jordan normal form,

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. It says that a finitely generated $R$-module is isomorphic towhere the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the invariant factors.

For $V_{\alpha }$, multiplication by $X$ is the linear map $\alpha$. Since $V$ is finite-dimensional over $F$, $V_{\alpha }$ is a finitely generated torsion module over the polynomial ring $F[X]$, so it has no free summand. Choosing each invariant factor $a_i$ to be monic givesThe basis $1,X,\dots ,X^{\mathrm{deg}{a}_i-1}$ of each cyclic summand makes multiplication by $X$ a companion matrix. Concatenating these bases therefore puts $\alpha$ in rational canonical form.

On $F[X]/(a_i)$, a polynomial annihilates multiplication by $X$ exactly when it is divisible by $a_i$. It follows that the minimal polynomial and characteristic polynomial areThe second identity follows blockwise from the characteristic polynomial of a companion matrix. Since every $a_i$ divides $a_k$, the product ${\chi }_{\alpha }$ annihilates every cyclic summand. Thus ${\chi }_{\alpha }(\alpha )=0$, which is the Cayley-Hamilton theorem.

For the displayed matrix, the given generators of $\ker \theta$ are the columns ofElementary row and column operations over $\mathbb{R}[X]$ give the Smith normal formConsequently the nonunit invariant factors areand, as a check,

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an Isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.

Using the convention of the question, the characteristic polynomial and characteristic equation areForone hasDirect multiplication givesentry by entry. Thus ${\chi }_A(A)=0$, verifying the Cayley-Hamilton theorem for $2\times 2$ matrices.

PutUsing ${\alpha }^3=5\alpha -8$, multiplication by $\beta$ on the ordered basis $(1,\alpha ,\beta )$ is determined byIts matrix, with these coordinate vectors as columns, isThe characteristic polynomial isBy the Cayley-Hamilton theorem, this monic integer polynomial annihilates $\beta$, so $\beta$ is an algebraic integer.

LetThe element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq {\mathcal{O}}_K$. Sincethe lattice $\mathbb{Z}[\alpha ]$ has index two in $A$. The given power-basis discriminant isso the discriminant-index formula for an integral lattice gives

If $m=[{\mathcal{O}}_K:A]$, thenBecause $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thusso $(1,\alpha ,\beta )$ is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation ${\beta }^2=-4-\alpha +3\beta$ givesConsequently ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$, which will allow direct use of Dedekind's theorem.

The Euler--Lagrange equations for the quadratic Lagrangian areFor a normal mode $x(t)=a{e}^{i\omega t}$, they become the generalized eigenvalue problem for small oscillationsPut $\lambda =m{\omega }^2/k$. With the stated spring constants,The final matrix has characteristic polynomialso its eigenvalue are ${\nu }_{\pm }=\pm \sqrt{1+{\delta }^2}$. HenceThe two angular frequencies therefore satisfyThis is the spectrum recorded by normal modes of two equal masses between three springs.

The characteristic polynomial isFor ${\lambda }_1=0$ and ${\lambda }_2=1$, corresponding eigenvectors areThey are linearly independent, so the general homogeneous solution is

Suppose for a contradiction that $m>n$. Compose the given injective module homomorphism with the standard injection $B^n\hookrightarrow B^m$ that appends $m-n$ zero coordinates. This gives an injective endomorphism $T$ of the finite free module $B^m$ whose matrix has a zero final row.

Its characteristic polynomial has zero constant term, so the Cayley-Hamilton theorem givesInjectivity lets us cancel $T$. Repeating this argument eventually gives the identity endomorphism equal to zero. That would imply $B^m=0$, contrary to $B\ne 0$. Hence $m\le n$. This proves the rank inequality for an injection of finite free modules.

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.