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The oscillator operators satisfy $[a,a^{†}]=1$ and $[a,a]=[a^{†},a^{†}]=0$. Substitution into the kinetic and potential energies givesThe number operator $N=a^{†}a$ is nonnegative because $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2$. The commutators $[N,a]=-a$ and $[N,a^{†}]=a^{†}$ lower and raise an eigenvalue by one. If an eigenvalue were a noninteger $\nu \ge 0$, repeated lowering would produce a nonzero eigenvector with eigenvalue between zero and one, and one further lowering would have negative eigenvalue; its norm is nonzero because $\Vert a\psi {\Vert }^2=\nu \Vert \psi {\Vert }^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0⟩=0$, and raising it supplies every level, withFor the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3⟩=|n_1⟩\otimes |n_2⟩\otimes |n_3⟩$, with energy $\hslash \omega (n_1+n_2+n_3+3/2)$.

Write the perturbation as $\lambda W$, where $W=(\hslash \omega /2){\sum }_{i<j}(a_i+a_i^{†})(a_j+a_j^{†})$. Its ground-state expectation is zero. Acting on the ground state, it produces $|110⟩,|101⟩,|011⟩$, each with coefficient $\hslash \omega /2$. They lie $2\hslash \omega$ above the ground state, so second-order perturbation theory givesIn the first excited subspace with basis $|100⟩,|010⟩,|001⟩$, the first-order perturbation matrix is $(\lambda \hslash \omega /2)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$. Its symmetric eigenvector has eigenvalue $\lambda \hslash \omega$, while the perpendicular two-dimensional space has eigenvalue $-\lambda \hslash \omega /2$. Thus the lowest excited triplet splits intoAs an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega \sqrt{1+2\lambda }$ and $\omega \sqrt{1-\lambda }$ twice; their small-$\lambda$ expansions reproduce both results.

For ${\theta }_k=k\pi /(M+1)$, extend the proposed vector by $v_0=v_{M+1}=0$. The sine addition identity gives $v_{i-1}+v_{i+1}=2\cos {\theta }_k{v}_i$. Thus row-by-row multiplication yieldsThese vectors do not depend on $a,b$. To prove they form a basis even when eigenvalues of $A$ coincide, apply the calculation first to the symmetric matrix with $a=0,b=1$, whose eigenvalues $2\cos {\theta }_k$ are distinct. Its $M$ nonzero eigenvectors are orthogonal and form a basis. Their squared lengths are $(M+1)/2$, so multiplying each by $\sqrt{2/(M+1)}$ gives a common orthonormal eigenbasis for every matrix in the family.

With the endpoint values zero, move the new-level differences to the left and keep the old-level differences on the right. Then $B$ is tridiagonal with diagonal $1+\mu$ and off-diagonal entries $-\mu /2$, while $C$ has diagonal $1-\mu$ and off-diagonal entries $\mu /2$. Thus $B{u}^{n+1}=C{u}^n$.

Part (a) supplies their common orthonormal eigenbasis. Their eigenvalues are $1+\mu (1-\cos {\theta }_k)$ and $1-\mu (1-\cos {\theta }_k)$. For $\mu \ge 0$, $B$ is invertible, and the step matrix $B^{-1}C$ has eigenvaluesSince the eigenbasis is orthonormal, $\Vert (B^{-1}C{)}^n{\Vert }_2={\max }_k|{\eta }_k{|}^n\le 1$, uniformly in the number of steps and mesh. Hence the method is unconditionally stable for every $\mu \ge 0$, which includes all physical positive time steps. For negative $\mu$, every nonzero modal parameter has modulus larger than one whenever its denominator is nonzero; if a denominator vanishes the step is undefined. Thus the algebraic stable range is exactly $[0,\infty )$. This proof uses simultaneous matrix diagonalization, not Fourier stability analysis.

For a bounded operator on a complex Banach space, the resolvent set is $\rho (T)=\{\lambda :\lambda I-T\text{ is bijective with bounded inverse}\}$, and the resolvent operator is $R_T(\lambda )=(\lambda I-T{)}^{-1}$. The spectrum is its complement $\sigma (T)=\mathbb{C}\setminus \rho (T)$; the point spectrum consists of $\lambda$ with $\ker (\lambda I-T)\ne 0$. For a real Banach space these definitions use its complexification.

If $|\lambda |>\Vert T\Vert$, the norm-convergent Neumann series gives $R_T(\lambda )={\lambda }^{-1}{\sum }_{n\ge 0}(T/\lambda {)}^n$, so the spectrum is bounded. If ${\lambda }_0\in \rho (T)$, factorThe second factor is invertible by a Neumann series whenever $|\lambda -{\lambda }_0|\Vert R_T({\lambda }_0)\Vert <1$. Thus the resolvent set is open and the spectrum is closed and bounded.

The point spectrum need not be closed. On ${\ell }^2$, define $T{e}_n=e_n/n$. Every $1/n$ is an eigenvalue, but zero is not: $Tx=0$ forces every coordinate of $x$ to vanish. No other eigenvalues occur, so ${\sigma }_p(T)=\{1/n:n\ge 1\}$ is not closed.

Finally, if $T$ is self-adjoint and $Tv=\lambda v$, $v\ne 0$, then $⟨Tv,v⟩$ is real and equals $\lambda \Vert v{\Vert }^2$ under the convention linear in the first argument. Therefore every eigenvalue of a self-adjoint operator is real.

Label the two displacements in cell $n$ by $u_n,v_n$, with cell length $2a$. Choose the $K$ spring within a cell and the $G$ spring between cells. The equations areA Bloch wave $(u_n,v_n)=(u,v)e^{i(2aqn-\omega t)}$ gives a two-by-two eigenvalue problem, whose determinant isThus the alternating-spring chain dispersion relation isPeriodic boundaries require $e^{2iaqN}=1$, so $q=\pi j/(Na)$ modulo $\pi /a$. There are $N$ distinct wavenumbers in the Brillouin zone $-\pi /(2a)\le q<\pi /(2a)$ and two branches, accounting for $2N$ modes. The minus branch is acoustic, with zero frequency and in-phase motion at the centre; the plus branch is optical, with opposite cell displacements.

Expanding the square root near zero givesAt the zone boundary the square root is $K-G$, so the acoustic maximum is $\sqrt{2G/m}$ and the optical minimum is $\sqrt{2K/m}$. The frequency gap is thereforeQuantizing each harmonic mode gives phonons, bosonic excitations of energy $\hslash {\omega }_{\pm }(q)$ and crystal momentum $\hslash q$ defined modulo a reciprocal lattice vector. Their group velocity is $d\omega /dq$; acoustic phonons carry long-wavelength sound, while optical phonons have a nonzero centre frequency. There is one longitudinal polarization per branch here, and phonon number is not conserved in thermal equilibrium.

Because $Sw=c{e}_1$ with $c\ne 0$, the similarity transform satisfies $\hat{A}{e}_1={\lambda }_1{e}_1$. Thus its first column is $({\lambda }_1,0,\dots ,0{)}^T$ and it has block form $\left(\begin{array}{cc}{\lambda }_1& \ast \\ 0& B\end{array}\right)$. Taking the determinant gives $\det (zI-\hat{A})=(z-{\lambda }_1)\det (zI-B)$. Similarity preserves the characteristic polynomial. Hence the eigenvalues of $A$, with algebraic multiplicities, are ${\lambda }_1$ and those of $B$.

The two independent columns of $V$ remain independent after multiplication by invertible $S$. Since $R=SV$ has zeros below the second row, its leading two-by-two triangular block is invertible. Hence the image under $S$ of the invariant subspace is exactly $\mathrm{span}(e_1,e_2)$. This coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$, forcing the bottom-left block to vanish:The block determinant factors $\det (zI-\hat{A})=\det (zI-B)\det (zI-D)$. Therefore the eigenvalues of $A$ are exactly those of the indicated two-by-two and $(n-2)$-by-$(n-2)$ blocks, counted with algebraic multiplicity. No diagonalizability assumption is needed.

The terms $u$ and $\alpha v$ represent intrinsic exponential population growth. The losses $-uv$ and $-\alpha uv$ describe competition between the species; $-{ϵ}_1{u}^2$ and $-\alpha {ϵ}_2{v}^2$ describe crowding within each species. The positive constant $\alpha$ sets the relative time scale of the second population.

The four equilibrium points in the nonnegative quadrant areThe last follows by solving ${ϵ}_{1}u+v=1$ and $u+{ϵ}_{2}v=1$ and is strictly positive under the assumptions. The Jacobian matrix isAt the origin its eigenvalues are $1,\alpha >0$, so it is unstable. At the first single-species equilibrium they are $-1$ and $\alpha (1-1/{ϵ}_1)<0$; at the second they are $1-1/{ϵ}_2<0$ and $-\alpha$. Both are asymptotically stable. At coexistence,Thus coexistence is a saddle; either species-only equilibrium is stable, and extinction is unstable. The saddle's stable manifold separates the competing basins of attraction.

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

Let $H$ be a Hermitian matrix and $He=\lambda e$ with $e\ne 0$. The inner product $e^{†}He$ is real because its conjugate equals $e^{†}{H}^{†}e=e^{†}He$. But it also equals $\lambda e^{†}e$, with $e^{†}e>0$. Thus every eigenvalue is real. For two eigenvectors,When ${\lambda }_i\ne {\lambda }_j$, this proves $e_i^{†}{e}_j=0$, so the eigenvectors are orthogonal.

For a real antisymmetric matrix, $(iA{)}^{†}=-i{A}^T=iA$, so $iA$ is a Hermitian matrix. Its real eigenvalues imply that the eigenvalues of $A$ are purely imaginary. To ensure a nonzero one without assuming it, writeExpansion of the characteristic polynomial gives $\det (tI-A)=t(t^2+|w{|}^2)$. Therefore the nonzero eigenvalue $-i\theta$ exists with $\theta =|w|>0$, and a nonzero complex eigenvector $e_1=u+iv$ satisfiesSeparating real and imaginary parts givesNeither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ andThus $u,v$ are orthogonal and have equal length.

In odd dimension $\det A=\det A^T=\det (-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real eigenvector $e_3$ with $A{e}_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(A{e}_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an orthonormal basis.

The matrix exponential series converges absolutely, allowing even and odd terms to be grouped. Since $A^{2}u=-{\theta }^{2}u$ and $A^{2}v=-{\theta }^{2}v$, it acts asConsequently the skew-symmetric exponential as an axial rotation has, in the displayed orthonormal basis, the matrixIt is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence $R=e^A$ is a rotation matrix, with angle $\theta$ about the chosen oriented axis.

For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt the origin it is $2I$, so there is a strict local minimum with value zero. At an axis point its eigenvalues are $-4$ and $2(1-b)$, giving a saddle point of a scalar function if $b<1$ and a strict local maximum if $b>1$. All four axis values are $1/2$.

At a diagonal point the Hessian has eigenvaluesHence these four points are strict local maxima for $b<1$ and saddle points of a scalar function for $b>1$. Their value is $1/(1+b)$. None of these critical points is degenerate when $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the level curves. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the local maxima identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.

For negative $c$ only the plus sign gives a positive radius, producing one outer closed level curve. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four saddle points of a scalar function. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no level curves above it.

For $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted level curve is the saddle level.

Finally, at $b=1$ the expression depends only on radius:Consequently $\boxed{\max f=1/2\text{, attained exactly on }{x}^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.