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The kinetic energy of the three unit masses is $\frac{1}{2}{\sum }_i{\dot{x}}_i^2$. Expanding the potential givesThus the small-oscillation mass and stiffness matrices arewhere $J$ is the all-ones matrix.

The vector $(1,1,1{)}^T$ is an eigenvector of the stiffness matrix with eigenvalue $1$. Every vector orthogonal to it has coordinate sum zero and is an eigenvector with eigenvalue $4$. Since $T=I$, the generalized eigenvalue problem for small oscillations gives ${\omega }^2$ equal to these eigenvalues. One orthonormal choice of normal modes isThe last two may be replaced by any orthonormal basis of the sum-zero plane because the frequency is degenerate. Hence the normal frequencies are $\boxed{1,2,2}$.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.

A simple Lie algebra is a nonabelian Lie algebra whose only ideals of a Lie algebra are $0$ and the whole algebra.

Use the standard basis $e,f,h$ of the sl2 Lie algebra, withLet $I\ne 0$ be an ideal and choose $0\ne x=ae+bf+ch\in I$. Since $I$ is invariant under the Adjoint representation, it is invariant under the linear operator $\mathrm{ad}h$. The three basis vectors are eigenvectors of $\mathrm{ad}h$ with distinct eigenvalues $2,-2,0$. Applying the corresponding polynomial spectral projections to $x$ shows that $I$ contains at least one nonzero multiple of $e$, $f$, or $h$.

If $e\in I$, then $[f,e]=-h\in I$ and $[f,h]=2f\in I$; the cases $f\in I$ and $h\in I$ are identical after taking brackets with the other basis vectors. Hence $e,f,h\in I$, so $I={\mathfrak{sl}}_2(\mathbb{C})$. Therefore ${\mathfrak{sl}}_2(\mathbb{C})$ is simple.

A Cartan subalgebra $\mathfrak{h}$ of a finite-dimensional complex semisimple Lie algebra is a maximal abelian subalgebra consisting of semisimple elements. The root-space decomposition isand the nonzero functionals $\alpha$ are the roots. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$. Its brackets have the formand $[E_{\alpha },E_{\beta }]=N_{\alpha \beta }{E}_{\alpha +\beta }$ when $\alpha +\beta$ is a root, and zero when $\alpha +\beta$ is neither a root nor zero.

For the complexified so4 Lie algebra, take $H^1=T^{(12)}$ and $H^2=T^{(34)}$. WriteDirect use of the stated commutation relations givesThe simultaneous eigenvectors, hence the step generators, may be chosen asThus the roots relative to $(H^1,H^2)$ are $(\pm i,\pm i)$. Replacing $H^a$ by $-i{H}^a$ gives the usual real coordinates $(\pm 1,\pm 1)$. The only nonzero brackets between step generators, apart from those obtained by antisymmetry, are

An isomorphism of Lie algebras is a bijective linear map preserving the Lie bracket. DefineThenThe two spans are commuting copies of the complexified ${\mathfrak{su}}_2$, and together contain all six basis elements of ${\mathfrak{so}}_4$. Chiral decomposition of the complexified so4 Lie algebra therefore givesUnder this isomorphism the Adjoint representation is the direct sum of the adjoint representations of the two factors:

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).

The variables $y_j=I_j{e}^{i{\Omega }_j}$ are complex inclinations, despite the question's description as eccentricities. Diagonalize the Laplace-Lagrange inclination matrix: let real eigenvectors ${\mathbf{v}}_k$ satisfy $B{\mathbf{v}}_k={\lambda }_k{\mathbf{v}}_k$, and form $V=({\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3)$. The physical matrix is diagonalizable by positive diagonal symmetrization of a matrix, as shown below.

The matrix exponential solution isWrite $c_k=I_k{e}^{i{\gamma }_k}$ with $I_k\ge 0$, and define $I_{jk}=I_k{v}_{jk}$. This givesThe eigenvalues determine nodal precession frequencies, the eigenvectors determine relative inclinations within each secular eigenmode, and the initial complex inclinations determine their amplitudes and phases. The real $I_{jk}$ can have either sign, encoding opposite nodal phases. The zero-frequency mode is a common fixed tilt.

For real ${\lambda }_1,{\lambda }_2$, computeThus $L$ is a non-normal matrix. For distinct eigenvalues, takeThese satisfy $L{v}_j={\lambda }_j{v}_j$ but have inner product $v_1^T{v}_2=1$. The eigenvectors are therefore nonorthogonal. When the eigenvalues coincide, only the direction $v_2$ remains as an eigenvector; $L$ is a defective Jordan block.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

Replace coordinate inequalities by the Loewner order and coordinate sums by matrix trace. The maximizing semidefinite program isIts feasible set is the fantope. In an orthonormal eigenbasis of $X$, its objective is ${\sum }_i{\lambda }_i(X)Y_{ii}$, and the diagonal entries of $Y$ belong to the capped simplex. Choosing $Y$ as the orthogonal projection matrix onto the largest $k$ eigenvectors attains the sum of the largest eigenvalues, as in the Ky Fan maximum principle.

The minimizing semidefinite program isFor example, its Lagrangian arises by assigning $S⪰0$ to $I-Y⪰0$, keeping $Y⪰0$ in the domain, and using free $t$ for $\mathrm{tr}Y=k$. Finiteness of the supremum over $Y$ requires $S+tI-X⪰0$.

Equality follows directly by choosing $S$ to have eigenvalues $({\lambda }_i(X)-t{)}_{+}$ in the same orthonormal eigenbasis, with $t$ between the $k$th and $(k+1)$st eigenvalues, or below the smallest one for $k=n$. This is the threshold semidefinite program for the largest eigenvalues, and it also verifies the endpoint case $k=n$.

For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.

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.

Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.

Here $D=I$, so the Jacobi iteration matrix isThe all-ones vector is an eigenvector with eigenvalue $-2\alpha$. On its two-dimensional orthogonal complement, the coordinates sum to zero and $Hv=\alpha v$, so the other eigenvalue is $\alpha$ with multiplicity two. ThereforeBy Jacobi convergence for a three-by-three equicorrelation matrix, convergence occurs exactly forAt either endpoint the spectral radius is one, so convergence for arbitrary initial data fails.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

If $\varphi v=\mu v$ for a nonzero eigenvector $v$, thenHence every eigenvalue satisfies $\mu =0$. Since a complex endomorphism has an eigenvalue, zero is the only possible eigenvalue.

Choose the least $k\ge 1$ such thatThenis nonzero by minimality, andTherefore $fw=\alpha w$, so $w$ is the required nonzero eigenvector.

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.