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For an $n\times n$ matrix here, $D=I$. The sine vectors $v_j(i)=\sin (ij\pi /(n+1))$ give eigenvaluesThus every eigenvalue lies strictly between $1/2$ and $3/2$. For $0<\omega \le 4/3$, $0<\omega {\lambda }_j<2$, so $|1-\omega {\lambda }_j|<1$, including the upper endpoint $\omega =4/3$. Hence the iteration converges for the entire stated range.

The extreme eigenvalues are $1-a$ and $1+a$, with $a=\frac{1}{2}\cos (\pi /(n+1))$. The spectral radius is $\max (|1-\omega (1-a)|,|1-\omega (1+a)|)=|1-\omega |+a\omega$ for $\omega >0$. It decreases up to one and increases after one, since $0\le a<1$. ThereforeFor $n=1$ the same formula gives zero spectral radius and exact solution in one ordinary Jacobi step.

Solving the two factors in each equation gives $(0,0),(3,0),(0,3),(1,1)$. The Jacobian isAt the three boundary equilibria its eigenvalues are $3,-3$. These equilibria are hyperbolic fixed points; the Hartman-Grobman theorem gives local topological conjugacy to their saddle linearizations. At $(1,1)$ the matrix is $\left(\begin{array}{cc}1& 2\\ -2& -1\end{array}\right)$, with eigenvalues $\pm i\sqrt{3}$. The linearization is a centre, but the hyperbolic theorem does not apply there.

The nonlinear classification follows from the actual HamiltonianAlong solutions $\dot{H}=H_x{H}_y-H_y{H}_x=0$. At $(1,1)$, $H=-1$ and its Hessian is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, which is positive definite. Nearby regular level curves are closed ovals around this strict minimum. Their vector field is nonzero and tangent, so each is a periodic orbit. The nonlinear equilibrium is a centre.

The separatrix level $H=0$ consists of $x=0$, $y=0$, and $x+y=3$. Within the triangle they connect the saddles as $(3,0)\to (0,0)\to (0,3)\to (3,0)$. The interior levels $-1<H<0$ are nested clockwise periodic orbits; outside the triangle the remaining level branches are unbounded. On $y=0$, flow is left for $0<x<3$ and right for $x<0$ or $x>3$; on $x=0$, it is up for $0<y<3$ and down otherwise. These directions and the invariant levels determine the full phase portrait.

The origin has Jacobian $\left(\begin{array}{cc}r& 0\\ 1-b& 0\end{array}\right)$ and is unstable because $r>1$. At a positive equilibrium, $r{e}^{-p_c}=1$, givingThe linearization isIts trace is $\tau =1+b{n}_c/r$ and determinant $n_c$. Elimination gives $n_{t+2}^{\prime }-\tau n_{t+1}^{\prime }+n_c{n}_t^{\prime }=0$. For distinct roots,At a repeated root replace the first expression by $(A+Bt){\lambda }^t$ and use $p_t^{\prime }=(n_t^{\prime }-n_{t+1}^{\prime })/n_c$.

As $r\downarrow 1$, $n_c\to 0$, so the roots approach one and zero. The characteristic polynomial at one is $n_c(1-b/r)>0$, putting the larger real root just below one and the other positive root below it. No real root can ever equal one. Moreoversince its numerator exceeds $r-1-\log r>0$. At $n_c=1$, $\tau =1+b/r<2$, so the roots are conjugate and nonreal. Just above this point their modulus is $\sqrt{n_c}>1$.

The unit-disc conditions are $1-n_c>0$, $1-\tau +n_c>0$, and $1+\tau +n_c>0$. The last two hold automatically here, so stability is equivalent to $n_c<1$. The unique crossing and its eigenvalue argument satisfyIndeed $\log r_{\ast }=1-b/r_{\ast }<1$. This is the stability threshold of a discrete exponential predator-prey model. The sketch uses $b=1/2$: the real moduli merge and then coincide, increasing through one at the threshold. Linearization decides stability below and instability above; it does not determine nonlinear stability exactly at the crossing.

With $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$, the matrix of $S_z\cos \theta +S_x\sin \theta$ is $(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Multiplication by $(c,s{)}^T$ and $(-s,c{)}^T$ gives eigenvalues $+\hslash /2$ and $-\hslash /2$, using the double-angle identities.

The one-particle transformation is the matrix $R=\left(\begin{array}{cc}c& -s\\ s& c\end{array}\right)$, with determinant one. The antisymmetric tensor $|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩$ transforms by this determinant under $R\otimes R$, so the singlet is unchanged. The total $S_z$ kills it. Its invariance under these rotations implies that total $S_z\cos \theta +S_x\sin \theta$ also kills it for every $\theta$, so total $S_x$ kills it. The spin commutator then gives total $S_y$ acting as zero. Therefore $S^2|\chi ⟩=0$ and it has total spin zero.

Rotating only particle B producesThe measurement probabilities are thereforeEach arrow represents the corresponding value $\pm \hslash /2$. The sequential measurements commute because they act on different particles. The two parallel outcomes leave triplet states with total spin one and $S^2=2{\hslash }^2$. Each antiparallel product state is a superposition of the spin-zero singlet and the spin-one zero-projection triplet, and is not an $S^2$ eigenstate. Outcomes of zero probability at a special angle do not actually occur.

For a periodic potential, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting unitary operators, so within each energy eigenspace one can choose simultaneous translation eigenstates. Their eigenvalues have modulus one and can be written $e^{i\mathbf{k}\cdot {\mathbf{a}}_i}$. Translation by any lattice vector then givesThis proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.

The reciprocal lattice consists of vectors $\mathbf{g}$ with $\mathbf{g}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$. Changing $\mathbf{k}$ to $\mathbf{k}+\mathbf{g}$ and $u$ to $e^{-i\mathbf{g}\cdot \mathbf{r}}u$ gives the same physical wavefunction with a periodic new $u$. The eigenvalue problem on one cell has a discrete set of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\mathbf{k}+\mathbf{g})=E_n(\mathbf{k})$. A reciprocal-lattice fundamental domain, conventionally a Brillouin zone, therefore labels inequivalent wavevectors.

Periodic boundary conditions in physical volume $\mathcal{V}$ give one allowed wavevector per reciprocal-space volume $(2\pi {)}^3/\mathcal{V}$. Including the two spin states, there are $2\mathcal{V}{d}^{3}k/(2\pi {)}^3$ states; per unit physical volume the count is the printed $2d^{3}k/(2\pi {)}^3$. Each occupied state carries charge $-e$ and velocity ${\hslash }^{-1}{\nabla }_k{E}_n$, soFor a full band, the integral of ${\nabla }_k{E}_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence a full band carries zero electric current.

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.

For a nonidentity element of ${\mathrm{PSL}}_2(\mathbb{R})$ represented by a determinant-one matrix, the classification is elliptic, parabolic or hyperbolic according as the squared trace is less than, equal to or greater than four. This follows either from the eigenvalues or from the discriminant of its fixed-point equation on the boundary of the upper half-plane.

The first matrix has determinant one and trace two, and is not scalar. It is therefore parabolic. Its fixed-point equation isThus its unique fixed point is $z=-2$ on the real boundary, of multiplicity two; infinity is not fixed and there is no fixed point in the upper half-plane. The second matrix has determinant one and trace $-4$, so it is hyperbolic. Its two boundary fixed points are $(-1\pm \sqrt{3})/2$, consistently with this classification.

A homogeneous equilibrium has $v_0=u_0^2$. Besides $u_0=0$, a nonzero equilibrium satisfies $r=u_0^2-u_0$, giving $u_{\pm }=(1\pm \sqrt{1+4r})/2$. Concentrations are nonnegative, so only nonnegative roots are physically admissible. At a positive equilibrium, the reaction Jacobian matrix isNegative trace and positive determinant are equivalent toThus a positive homogeneous stable branch exists precisely when $s>1/2$ and $-1/4<r<s^2-s$, on the upper root $u_{+}$. The lower positive branch has negative determinant and is unstable. At $r=-1/4$ the branches meet and have a zero eigenvalue; at $u_0=s$ the strict linear stability condition also fails.

The trivial equilibrium $(0,0)$ has eigenvalues $r,-s$, so it is asymptotically stable for $r<0$. It exists for all $r$, independently of the nonzero branches. If negative $u_0$ were allowed mathematically, the negative branch for $r>0$ would also be homogeneously stable, but it does not represent a chemical concentration.

A Fourier mode with $q=k^2\ge 0$ evolves under $J-q\mathrm{diag}({\kappa }_1,{\kappa }_2)$. Its trace remains negative for a homogeneously stable positive branch. Its determinant isIf $u_0{\kappa }_2\le s{\kappa }_1$, this is positive for every $q\ge 0$. Otherwise its minimum is attained at $q_{\ast }=(u_0{\kappa }_2-s{\kappa }_1)/(2{\kappa }_1{\kappa }_2)$. Stability requires $D(q_{\ast })>0$, orIn terms of $d=\sqrt{{\kappa }_2/{\kappa }_1}$, the boundary is the positive root of $u_0{d}^2-s=2d\sqrt{s(2u_0^2-u_0)}$:Within $1/2<u_0<s$, all modes decay for $d<d_c$; a finite band grows for $d>d_c$, giving a Turing instability. Equality is neutral at one nonzero wavenumber.

For $s=2$, the positive lower branch $0<u_0<1/2$ is unstable for every $d$. The upper branch is homogeneously stable for $1/2<u_0<2$: it is stable to all modes to the left of $d_c=2+\sqrt{4-2/u_0}$ and unstable to spatial modes to the right. For $u_0>2$ it is already homogeneously unstable. The zero branch is stable to every spatial mode when $r<0$, because its mode eigenvalues are $r-{\kappa }_1{k}^2$ and $-s-{\kappa }_2{k}^2$.

For $rN>a$ and $p>0$, a nonzero infected equilibrium requires $S_{\ast }=a/r$. Using $a{I}_{\ast }=p{R}_{\ast }$ and population conservation givesEliminate $R=N-S-I$ before assessing stability, since perturbations must preserve the total population. The resulting Jacobian matrix at this endemic equilibrium isIts trace is negative and its determinant positive, so both eigenvalues have negative real part: the endemic equilibrium is locally asymptotically stable on the fixed-population plane. The full three-variable system has a neutral direction corresponding to changing the conserved population.

Writing $\kappa =rN-a>0$, as $p\to 0$ the trace is $-prN/a+O(p^2)$ and the determinant is $p\kappa$. Hence the discriminant $(\mathrm{tr}J{)}^2-4\det J=-4p\kappa +O(p^2)<0$. As $p\to \infty$, the trace is $-p-\kappa +O(p^{-1})$ and the determinant is again $p\kappa$, giving a positive discriminant $p^2+O(p)$. Both quantities are $O(p)$ in both limits, but the eigenvalues are complex for sufficiently small $p$ and real for sufficiently large $p$.

Representative normal forms are $\dot{x}=\mu +x^2$ for a saddle-node bifurcation, $\dot{x}=\mu x-x^2$ for a transcritical bifurcation, and $\dot{x}=\mu x-x^3$ for a supercritical pitchfork bifurcation. Replacing the last cubic term by $+x^3$ gives the subcritical pitchfork bifurcation.

Treat $\mu$ as a variable with $\dot{\mu }=0$. The extended center manifold is tangent to $y=0$, so write $y=h(x,\mu )=A{x}^2+Bx\mu +C{\mu }^2+O((|x|+|\mu |{)}^3)$. Its invariance equation is $h_x\dot{x}=\dot{y}$. At quadratic order this readsMatching coefficients gives $A=2$, $B=-4$, $C=4$. Substitution into the first equation therefore givesHere $O(3)$ denotes total degree at least three in $(x,\mu )$. With $T=t$, $X=x-2\mu$ and $\tilde{\mu }=\mu$, this becomes $dX/dT=\tilde{\mu }+X^2+O(3)$. Thus $\alpha =1$, $\beta =2$, $\gamma (\mu )=\mu$, and the bifurcation is a saddle-node. For negative parameter, the negative-$X$ branch is stable and the positive-$X$ branch unstable along the center manifold; the transverse eigenvalue is near $-1$.

The Lie group $PSL(2,\mathbb{C})$ is a second-countable space. For every point of its discrete subgroup, choose the first basic neighbourhood isolating that point; different elements require different such neighbourhoods. This injects $G$ into a countable set, proving countability.

A discrete subgroup is closed and meets compact sets in finitely many points. Indeed, distinct subgroup elements accumulating anywhere would give quotients converging to identity, contradicting an identity neighbourhood containing no other subgroup element. To apply this, use the positive Hermitian determinant-one model of hyperbolic space. At the base point $I$, the image under a determinant-one representative $g$ is $g{g}^{\ast }$ andThis distance formula follows by unitary diagonalization: eigenvalues are $e^t,e^{-t}$ and the axial hyperbolic displacement is $|t|$. Thus bounded displacement bounds all matrix entries. The determinant-one condition is closed, so the resulting set is compact; the quotient by $\{I,-I\}$ is compact too. Conjugation gives the same conclusion at any base point. These are the bounded displacement sets of hyperbolic isometries are compact facts.

Hence, for each $R$, only finitely many $g\in G$ satisfy $d(p,gp)\le R$. Any enumeration of an infinite group without repetitions therefore has $d(p,g_{n}p)\to \infty$. A finite discrete group is a harmless exception to the wording: it has no infinite enumeration of distinct elements, and proper discontinuity is immediate.

For a compact set $C$, put $R={\max }_{x\in C}d(p,x)$. If $gC\cap C\ne \varnothing$, choose $x,gx\in C$. The triangle inequality gives $d(p,gp)\le d(p,gx)+d(x,p)\le 2R$. Only finitely many such $g$ exist, proving properly discontinuous group action. Conversely, if the subgroup is not discrete, there are distinct $g_n\to e$. A closed ball about $p$ then meets its images under infinitely many $g_n$, violating this compact-set definition. Thus discreteness is equivalent to proper discontinuity. Here proper discontinuity allows finite point stabilizers; a stronger convention requiring freeness would exclude discrete elliptic subgroups and is not the equivalence intended.

In the usual nondegenerate convention, a strongly regular graph of order $n$ and parameters $(d,a,b)$ is a finite simple $d$-regular graph with $0<d<n-1$ in which adjacent vertices have $a$ common neighbours and distinct nonadjacent vertices have $b$. Its adjacency matrix of a graph $A$ obeysFor the usual nondegenerate parameters with $b>0$ the graph is connected, so eigenvalue $d$ has multiplicity one. On the perpendicular complement of the all-ones vector, $J=0$ and the possible eigenvalues areIf $r$ has multiplicity $m$, then zero trace gives $d+mr+(n-1-m)s=0$. Solving provesThe denominator is positive in this nondegenerate convention. Otherwise all eigenvalues on the all-ones complement would coincide, forcing $A=rI+(d-r)J/n$. All off-diagonal entries would then be the same, making the graph complete or edgeless. If complete graphs are admitted with arbitrary unused $b$, the printed claim needs qualification: $K_2$ with $(d,a,b)=(1,0,2)$ satisfies the common-neighbour conditions vacuously but makes the fraction $0/0$. For complete graphs with nonzero denominator, the spectral calculation still works with an unused root of multiplicity zero.

For the triangle-free graph claim, suppose first there are at least two vertices. Nonadjacent vertices have a common neighbour, so the graph is connected. If $u,v$ are adjacent, put $U=N(u)\setminus \{v\}$ and $V=N(v)\setminus \{u\}$. Every member of $U$ is nonadjacent to $v$, and exactly two of its three common neighbours with $v$ lie in $V$, the third being $u$. By symmetry every member of $V$ has exactly two neighbours in $U$. Counting these edges gives $2|U|=2|V|$, hence $d(u)=d(v)$. Connectivity makes the graph $d$-regular. Counting length-two paths from a vertex to non-neighbours givesHere $a=0$, $b=3$, and the larger-root multiplicity becomes $[n-1+d^2/\sqrt{4d-3}]/2$. Thus $t=\sqrt{4d-3}$ is rational and therefore an integer. It is odd, $d=(t^2+3)/4$, and integrality also gives $t\mid d^2$. Since $16d^2\equiv 9(\mathrm{mod}t)$, one obtains $t\mid 9$, henceThe complete bipartite graph $K_{3,3}$ realizes $d=3$: vertices in the same part have all three vertices of the other part as common neighbours, and vertices in opposite parts are adjacent. There are no triangles.

As literally worded, the one-vertex graph is a vacuous exception: it has no triangles or nonadjacent distinct pairs, but has $d=0$. The displayed positive-degree classification therefore requires at least two vertices. This qualification completes the triangle-free graphs with three common neighbours argument rather than silently excluding a counterexample.

A bounded operator on a complex Hilbert space is self-adjoint when $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$. The spectral theorem for compact self-adjoint operators says its nonzero spectrum consists of real eigenvalues of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their eigenspaces are mutually orthogonal; their closed span is the orthogonal complement of the kernel. On that span the operator has the norm-convergent spectral action $Tx={\sum }_j{\lambda }_j⟨x,e_j⟩e_j$.

Take $H={\ell }^2(\mathbb{N})$ and $T{e}_n=e_n/n$. It is self-adjoint and has infinite-dimensional range. Truncation to the first $N$ coordinates is a finite-rank operator and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.

The resolvent set comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the spectrum, and the point spectrum consists of eigenvalues. In this exampleFor $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda {)}^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has eigenvector $e_n$. Zero is not an eigenvalue, but $\Vert T{e}_n\Vert \to 0$ prevents a bounded inverse; also $(1/n{)}_n\in {\ell }^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.