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For the homogeneous solution, expand in the independent eigenvectors of $A$. Each coefficient obeys $c_j^{\prime }={\lambda }_j{c}_j$, so the complementary function isIf ${\mathbf{x}}_p$ is any particular integral, subtracting it from any other solution gives a homogeneous solution. Hence $\mathbf{x}={\mathbf{x}}_p+{\mathbf{x}}_c$ is the general solution. Complex-conjugate modes can be combined to produce real solutions when the forcing is real.

For a real two-by-two matrix, the characteristic polynomial is ${\lambda }^2-(\mathrm{tr}A)\lambda +\det A$. Under the stated distinct-eigenvalue assumption, all nonzero homogeneous modes are purely oscillatory exactly when the eigenvalues are $\pm i\omega$ with $\omega >0$. Their sum and product imply $\mathrm{tr}A=0$, $\det A={\omega }^2>0$. Conversely these trace and determinant conditions force those eigenvalues, so there is no exponential growth, decay or Jordan-block secular growth.

For the final specified initial-value problem, direct multiplication gives $A^2=-9I$. The forcing vector $\mathbf{b}=(2,3i-1{)}^T$ satisfies $A\mathbf{b}=3i\mathbf{b}$, so the forcing is resonant and $t\mathbf{b}{e}^{3it}$ is a particular integral. The matrix exponential isSince the resonant particular integral vanishes initially, the initial-value solution isThis complex solution is appropriate to the complex forcing as written. Taking real parts gives the corresponding solution for the real part of that forcing.

Differentiate the proposed first integral along the vector field:At $\alpha =0$, $K$ is conserved and the system is an inverted quartic oscillator phase portrait. Its equilibria are $O=(0,0)$ and $S_{\pm }=(0,\pm 1/\sqrt{2})$. The Jacobian matrix isAt the origin its eigenvalues are $\pm i$. Moreover $K$ has a strict local minimum there, and its nearby positive contours are closed, so the origin is a genuine center equilibrium, not merely an inconclusive linear centre. The arrows circulate clockwise: on the positive $x$-axis, $\dot{y}<0$. At either $S_{\pm }$, the eigenvalues are $\pm \sqrt{2}$, giving saddle equilibria. The stable eigendirection is $\delta x=\sqrt{2}\delta y$, and the unstable one is $\delta x=-\sqrt{2}\delta y$.

The special energy level factors asBetween the two saddles, its left arc has $x<0$ and travels upward from $S_{-}$ to $S_{+}$; its right arc has $x>0$ and travels downward from $S_{+}$ to $S_{-}$. These two heteroclinic orbits bound the central eye. Trajectories inside the eye are closed periodic orbits and do not tend to the centre as $t\to \infty$.

More explicitly, a level $K=C$ hasFor $0<C<1/4$, its bounded central component is periodic, while its exterior components turn at $|y|=\sqrt{(1+\sqrt{1-4C})/2}$ and then escape. For $C>1/4$, $x$ never vanishes and the two branches run from one vertical infinity to the other. Levels $C\le 0$ have only exterior nonconstant components, apart from the origin at $C=0$.

On the upper outer separatrix, the $x>0$ branch arrives from positive infinity and approaches $S_{+}$ in forward infinite time; the $x<0$ branch leaves $S_{+}$ towards positive infinity. On the lower outer separatrix, the $x<0$ branch arrives from negative infinity and approaches $S_{-}$, while the $x>0$ branch leaves towards negative infinity. Approaching a saddle along its stable manifold is exponential at rate $\sqrt{2}$. A trajectory close to but off a separatrix can spend a long time near the saddle before completing a periodic orbit or escaping.

The escape is finite-time blowup, not an asymptote at $t=\infty$. On any escaping level, $|x|=\sqrt{y^4-y^2+C}\sim y^2$, so the remaining travel time is bounded by a convergent integral proportional to ${\int }^{\infty }dy/y^2$. Thus an outward upper branch has $y\sim (T-t{)}^{-1}$, $x\sim -(T-t{)}^{-2}$; an outward lower branch has $y\sim -(T-t{)}^{-1}$, $x\sim (T-t{)}^{-2}$. The statement about approaching from infinity likewise refers to a finite past endpoint. Only equilibria, periodic trajectories and trajectories tending to saddles have a future defined for all time in this conservative portrait.

For small positive $\alpha$, the equilibria remain in place, but the origin's eigenvalues become $\alpha /4\pm i\sqrt{1-{\alpha }^2/16}$, so it becomes a repelling focus. The saddles retain one eigenvalue of each sign, $\alpha /4\pm \sqrt{2+{\alpha }^2/16}$. Since $\dot{K}\ge 0$, and its time integral is strictly positive on every nonconstant trajectory, trajectories cross $K$-contours towards larger values. Near the centre this means outward spiralling. No nonconstant periodic orbit can remain, and the former saddle-to-saddle connections at equal $K=1/4$ are broken. Stable saddle trajectories approach from $K<1/4$; unstable ones depart into $K>1/4$. Generic central trajectories eventually escape, while the exceptional stable saddle manifolds still approach their saddles.

Using the monic convention for the characteristic polynomial, expand the determinant along the last column:For the eigenvalue $1$, the equations $(A-I)v=0$ give $v_2=v_1$ and $v_3=0$. Thus every corresponding nonzero eigenvector is a nonzero multiple of $(1,1,0{)}^T$. For the eigenvalue $2$, the first two equations give $v_2=2v_1$ and $v_1=0$, so every corresponding nonzero eigenvector is a nonzero multiple of $(0,0,1{)}^T$. ThereforeThese eigenspaces together have dimension $2$, so they cannot provide an eigenvector basis of ${\mathbb{R}}^3$. The matrix is not diagonalizable; the eigenvalue $1$ has algebraic multiplicity $2$ but geometric multiplicity $1$.

The characteristic polynomial has only the eigenvalue $\mu$. If $A$ is diagonalizable, there is an invertible matrix $P$ and a diagonal matrix $D$ with $A=PD{P}^{-1}$. Every diagonal entry of $D$ is an eigenvalue, hence equals $\mu$. Thus $D=\mu I$ and $A=P(\mu I)P^{-1}=\mu I$. Conversely, $\mu I$ is already diagonal in every basis. HenceThis argument uses the existence of an eigenvector basis rather than merely the repeated root of the polynomial; repeated eigenvalues alone do not imply diagonalizability.

Let $M_{ij}={\int }_S{x}_i{x}_{j}dS$. A rotation matrix $Q$ maps the unit sphere and its surface measure to themselves. Changing variables therefore gives $Q_{ip}{Q}_{jq}{M}_{pq}=M_{ij}$, so $M$ is an isotropic second-rank tensor. Reflection symmetry also directly makes its off-diagonal entries zero, while interchange of coordinates makes its three diagonal entries equal. Its trace ishence $M_{ij}=(4\pi /3){\delta }_{ij}$. Oddness gives ${\int }_S{x}_{i}dS=0$. Expanding the defining product for $T$ now givesThe action on a vector is $Tv=\lambda v+\mu y(y\cdot v)$. For $y\ne 0$, its eigenvalues and eigenspaces are thereforeThe latter has multiplicity two. If $y=0$, the single eigenvalue $4\pi /3$ has multiplicity three and every vector is an eigenvector apart from the excluded zero vector.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$, with attractive Coulomb potential $V(r)=-\kappa /r$, where $\kappa =e^2/(4\pi {\epsilon }_0)$. Separating $\psi =R(r)Y_{\ell }^m(\theta ,\phi )$ uses the spherical harmonic angular eigenvalue $-\ell (\ell +1)$ of the unit-sphere Laplacian. The radial kinetic contribution is $-{\hslash }^2(R^{\prime \prime }+2R^{\prime }/r)/(2m)$, the angular kinetic contribution is the centrifugal term ${\hslash }^2\ell (\ell +1)R/(2m{r}^2)$, and the remaining term is the attractive potential energy. The right side is the energy eigenvalue times the radial wavefunction. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the reduced mass.

For $\ell =0$, substitution of the ground-state exponential givesIts $1/r$ coefficient must vanish and its constant coefficient determines the energy, soHere $a_0={\hslash }^2/(m\kappa )$ is the Bohr radius.

For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yieldsCancel the nonzero exponential in the radial Schrödinger equation and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,ConsequentlyThe positive decay constants ensure normalizability; the second radial wavefunction has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.

If the entire energy gap is carried by one photon, neglecting atomic recoil, conservation of energy gives the requested frequencyThere is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the electric-dipole selection rules for hydrogen. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0{|}^2\hat{\mathbf{r}}d\Omega =0$, so every component of $⟨1s|\mathbf{r}|2s⟩$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which ${\nu }_1+{\nu }_2={\nu }_{\mathrm{gap}}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in NIST's hydrogen spectral-data compilation. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.

For $x\in \ker (\alpha -\lambda I)$, commutation givesThus each eigenspace of $\alpha$ is an invariant subspace for $\beta$.

If $\alpha ={\beta }^2$, the two endomorphisms commute. The $n$ distinct real eigenvalues of $\alpha$ have one-dimensional real eigenspaces, since their direct sum is the $n$-dimensional vector space. Take a nonzero real eigenvector $x_i$. Its eigenspace is invariant under $\beta$, so $\beta x_i=c_i{x}_i$ for a real scalar $c_i$. Squaring givesThe simple-spectrum hypothesis is essential: the real $90$-degree rotation matrix squares to $-I$ in dimension two, whose negative eigenvalue has a two-dimensional eigenspace.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Use the intended trace bound $\mathrm{Ric}\ge (n-1)kg$ and write $D=\pi /\sqrt{k}$. We state precisely the allowed Bishop volume comparison with a positive-curvature model. Setwhere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere. For $V_p(r)=\mathrm{Vol}B(p,r)$, the ratio $V_p(r)/v_k(r)$ is nonincreasing on $0<r<D$, tends to one at zero, and is at most one. Before the cut time, the radial volume density $J(r,\theta )$ in geodesic polar coordinates is at most $s_k(r{)}^{n-1}$.

By the Bonnet-Myers theorem, $\mathrm{diam}M\le D$. Therefore $V_p(D)=\mathrm{Vol}M$. If distance-$D$ endpoints exist, they do not affect this volume equality: the distance sphere is contained in the smooth image ${\exp }_p(D{S}_p^{n-1})$, which has zero $n$-dimensional Riemannian volume. Taking limits in the comparison ratio at $D$ is legitimate. The assumed equality of total volume gives $V_p(D)/v_k(D)=1$, so monotonicity forcesfor every $p$.

We now extract the equality geometry, rather than merely naming a rigidity theorem. On a small normal neighborhood of $p$, the continuous nonnegative difference $s_k(r{)}^{n-1}-J(r,\theta )$ has integral zero. It is therefore identically zero. Let $A(X)={\nabla }_X{\partial }_r$ on a distance sphere and $h=\mathrm{tr}A={\partial }_r\log J$. Equality of the density gives $h=(n-1)s_k^{\prime }/s_k$. The traced radial Riccati equation for distance spheres readsSince $s_k^{\prime \prime }=-k{s}_k$, substituting $h$ makes the sum of the following two nonnegative quantities zero:Equality in the Cauchy-Schwarz inequality for the eigenvalues of the self-adjoint operator $A$ forces $A=(s_k^{\prime }/s_k)I$. The angular metric $g_r$ in geodesic polar coordinates consequently satisfies ${\partial }_r{g}_r=2(s_k^{\prime }/s_k)g_r$. Smoothness at the center fixes its integration constant, since $r^{-2}{g}_r\to g_{S^{n-1}}$. Henceon the normal ball: the Riemannian metric is locally the round curvature-$k$ metric. As $p$ was arbitrary, $M$ has constant sectional curvature $k$.

The classification of complete positive constant-curvature manifolds now gives $M=S_k^n/\Gamma$ with $\Gamma$ a finite freely acting group of isometries. Covering degree gives $\mathrm{Vol}M=\mathrm{Vol}{S}_k^n/|\Gamma |$. Volume equality forces $|\Gamma |=1$, provingThis establishes spherical rigidity of maximal total volume under the correct normalization.

Under the literal trace bound, the assertion is false in dimension four. Take the Riemannian product $M=S^2(r)\times S^2(r)$ with $r^2=1/(\sqrt{6}k)$. Its Ricci curvature is $r^{-2}g=\sqrt{6}kg\ge kg$, andBut a mixed tangent two-plane has sectional curvature zero, whereas a factor plane has curvature $\sqrt{6}k$. Thus this manifold is not the round sphere despite having the literal hypotheses.

Fix an even weight $k$ and write $\Gamma =S{L}_2(\mathbb{Z})$. Extend the slash operator for modular forms to positive-determinant rational matrices by the determinant-normalized slash operatorThe automorphy factor identity makes this a right action. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. Their left $\Gamma$-cosets have unique determinant-n matrix representatives for Hecke operatorsAn integral row operation obtained from Bézout's identity makes the first column $(a,0{)}^T$, and a row shear reduces the other top entry modulo $d$, proving this description. The usual Hecke operator isThis normalization has $T_1=1$. Right multiplication by $\Gamma$ permutes the left cosets in ${\mathcal{D}}_n$, so $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic in $\mathbb{H}$. Holomorphy at the cusp and preservation of cusp forms follow directly from the coefficient formula below; the full modular group has only one cusp orbit. Thus the operators act on both $M_k$ and $S_k$.

If $f={\sum }_{r\ge 0}{a}_r{q}^r$, summing over $b$ kills terms unless $d\mid r$. Writing $r=d\ell$ gives output exponent $a\ell$, so a term of exponent $m$ requires $a\mid m$ and has original index $r=dm/a=nm/a^2$. Its multiplier is $n^{k-1}{d}^{1-k}=a^{k-1}$. The Fourier coefficients of a composite-index Hecke operator are consequentlyFor $m=0$, this means $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$. For a prime number $p$,with a nonexistent fractional-index coefficient interpreted as zero.

The Hecke multiplication relations follow from these coefficients. For coprime indices, the two divisor choices combine uniquely and give $T_m{T}_n=T_{mn}$. For prime powers put $A=p^{k-1}$, and, on formal Fourier series, let $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_p(\sum a_m{q}^m)=\sum a_m{q}^{pm}$. These individual maps need not preserve level one, but $U_p{V}_p=1$, and the coefficient formula givesMultiplying on the left by $T_p=U_p+A{V}_p$ and cancelling $U_p{V}_p$ yieldsInduction gives $T_{p^u}{T}_{p^v}={\sum }_{j=0}^{\min (u,v)}{p}^{j(k-1)}{T}_{p^{u+v-2j}}$; combining distinct prime factors provesThe modular Hecke algebra is therefore commutative and generated by its prime-index operators. The identity $U_p{V}_p=1$, rather than the generally false reversed identity, is what proves the prime-power recurrence.

On $S_k$ use the positive definite Petersson inner productExponential cusp decay makes this integral finite. The Hecke operators are self-adjoint for the Petersson inner product. Here is the correspondence argument. For a positive-determinant matrix $\alpha$, set $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. The finite sum over the double coset $\Gamma \alpha \Gamma$ unfolds its pairing intoChanging variables $z=\alpha \tau$ preserves hyperbolic area, and $\mathrm{Im}(\alpha \tau )=\det (\alpha )y/|c\tau +d{|}^2$ shows that the determinant-normalized slash factors transfer $\alpha$ from $f$ to ${\alpha }^{-1}$ on $g$. The new quotient is by $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$; folding it back gives the operator for the reversed double coset. For $\alpha \in {\mathcal{D}}_n$, its adjugate $n{\alpha }^{-1}$ again lies in ${\mathcal{D}}_n$. Positive scalar matrices act trivially in the determinant normalization, so ${\alpha }^{-1}$ and its adjugate give the same action. The adjugate involution preserves the entire determinant-$n$ correspondence, and the external normalization $n^{k/2-1}$ is real. Thus

Finite-dimensional spectral theory now gives a simultaneous orthonormal eigenbasis on $S_k$, since the operators commute and are self-adjoint. For a common Hecke eigenform $f$ with eigenvalues ${\lambda }_n$, the first-coefficient formula gives $a_n(f)={\lambda }_n{a}_1(f)$. If $a_1(f)=0$, every coefficient vanishes and $f=0$, a contradiction. Rescale each eigenform to have $a_1=1$. The resulting basis is orthogonal, andAll these eigenvalues and coefficients are real. A fixed simultaneous eigenspace is one-dimensional, because its normalized Fourier coefficients are uniquely determined by the eigenvalues.

Applying the multiplication relations to an eigenform givesThe recurrence yields ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$. Multiplicativity then gives the Euler product of a Hecke eigenforminitially in a right half-plane, for example $\mathrm{Re}s>k/2+1$ using the Fourier coefficient bound for a cusp form. Modular inversion and the Mellin transform of a cusp-form L-function continue $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)$ to an entire function and give $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

There is also an integral structure. If $d=\dim S_k$, choose $F_j={\Delta }^j{R}_{k-12j}$, $1\le j\le d$, with $R_w$ a monomial in $E_4,E_6$ of weight $w$ and constant term one. These forms have integral Fourier coefficients and start with $q^j$; their first $d$ coefficients form a unitriangular matrix. They are a basis, and every cusp form with integral coefficients is an integral combination of them, by successive elimination of its first $d$ coefficients. The coefficient formula shows that $T_n$ preserves this lattice. Its characteristic polynomial is consequently monic with integer coefficients, proving that the Hecke eigenvalues are algebraic integers. Together with self-adjointness, this gives real algebraic integers without requiring any unproved bound for their sizes.

For even $k\ge 4$, the full space decomposes as $M_k=\mathbb{C}{E}_k\oplus S_k$. The normalized Eisenstein series is a simultaneous eigenform withIts coefficients are $-2k{\sigma }_{k-1}(m)/B_k$, and the elementary divisor identity ${\sigma }_{k-1}(m){\sigma }_{k-1}(n)={\sum }_{d\mid \gcd (m,n)}{d}^{k-1}{\sigma }_{k-1}(mn/d^2)$ verifies the eigenvalue formula, including the constant coefficient. Thus the cusp eigenbasis plus $E_k$ diagonalizes the operators on $M_k$. Weight two and odd weights have no forms; in weight zero the constant function has eigenvalue ${\sigma }_{-1}(n)$. As a concrete cuspidal example, $S_{12}=\mathbb{C}\Delta$, so $T_n\Delta =\tau (n)\Delta$ for its Fourier coefficients, and the recurrence gives $\tau (p^2)=\tau (p{)}^2-p^{11}$.