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Since $A{A}^T=I$ and $\det A=1$, $A$ is an orthogonal matrix in the special orthogonal group. It also satisfies $A^{T}A=I$ and thus preserves the dot product. Normalize the fixed eigenvector to $e_3=v/\Vert v\Vert$. For $u\cdot e_3=0$,so the orthogonal complement of the fixed axis is invariant. Choose an oriented orthonormal basis $e_1,e_2,e_3$. In this basis the matrix has form $\mathrm{diag}(B,1)$, where $B$ is a real $2\times 2$ orthogonal matrix of determinant one.

Write the first column of $B$ as $(\cos \theta ,\sin \theta {)}^T$. Its unit second column must be perpendicular to the first; positivity of the determinant selects $(-\sin \theta ,\cos \theta {)}^T$. Thus $B=R(\theta )$, and $A$ fixes the axis through $v$ while rotating its perpendicular plane by $\theta$. Since trace is invariant under change of basis,This proves that $A$ is the required axial rotation. The trace determines the unsigned angle; the action in the oriented perpendicular plane determines its sign. The identity case has angle zero and any axis.

Try a particular integral along each forcing eigenvector. Since $A{\mathbf{a}}_j={\lambda }_j{\mathbf{a}}_j$,With both denominators nonzero, add the two responses:Only the forcing component's own eigenvalue enters its denominator; a frequency matching a different eigenmode does not create resonance in this already diagonal forcing component.

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.

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.

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.

For every $u\in U$, write $u=(u-\alpha u)+\alpha u$. The first term lies in $\ker \alpha$, since ${\alpha }^2=\alpha$, and the second lies in $\mathrm{im}\alpha$. If $v$ belongs to both spaces, write $v=\alpha w$; then $\alpha v={\alpha }^{2}w=\alpha w=v$, whereas membership in the kernel gives $\alpha v=0$. Thus the intersection is zero andIn a basis assembled from bases of these two spaces, the idempotent linear map has matrix $\mathrm{diag}(0_{d-r},I_r)$, where $d=\dim U$ and $r=\mathrm{rank}\alpha$. ThereforeThis basis consists entirely of eigenvectors, so $\alpha$ is diagonalizable over the reals, including the extreme cases $r=0$ and $r=d$.

The Pauli group includes scalar phases $\mathcal{Z}=\{\pm I,\pm iI\}$. For a nonzero stabilizer code, its Abelian subgroup must satisfy $-I\notin S$. Otherwise a common positive eigenvector would satisfy $-|\psi ⟩=|\psi ⟩$ and vanish; for example $S=\{I,-I\}$ is Abelian but yields no nonzero stabilizer code. We impose this standard admissibility condition.

Every element of such a stabilizer group is Hermitian and squares to $I$: an imaginary-phase Pauli with square $-I$ is excluded. The associated stabilizer code and its orthogonal projector areThe group multiplication proves $P^2=P$, Hermiticity gives $P^{†}=P$, and its range is exactly the common positive eigenspace. If $S$ has $r$ independent generators, $|S|=2^r$. Every nonidentity element has trace zero, while $\mathrm{tr}I=2^n$, soThus it encodes $k=n-r$ logical qubits.

For the symplectic representation of a stabilizer code, label $i^t{X}^a{Z}^b$ by $(a,b)\in {\mathbb{F}}_2^{2n}$, ignoring scalar phase. Products add labels, andThis is the nondegenerate alternating form $\omega ((a,b),(a^{\prime },b^{\prime }))=a\cdot b^{\prime }+b\cdot a^{\prime }$ on the binary symplectic space of Pauli labels. The image $L$ of $S$ is an r-dimensional isotropic subspace of a binary symplectic space, since the generators commute and no nontrivial scalar lies in $S$. Consequently $L\subseteq L^{\perp }$ and $r\le n$. A binary generator matrix $(A\mid B)$ satisfies $A{B}^T+B{A}^T=0$. Conversely an isotropic subspace can be lifted by choosing commuting Hermitian generators with independent labels; their products cannot equal $-I$, and the orthogonal projector formula supplies the stabilizer code. The labels of the centralizer of a stabilizer group are $L^{\perp }$, with nontrivial logical actions represented by $L^{\perp }\setminus L$.

For an example, the three-qubit bit-flip repetition code has $S=⟨Z_1{Z}_2,Z_2{Z}_3⟩$ and ${\mathcal{C}}_S=\mathrm{span}\{|000⟩,|111⟩\}$. Its two binary rows have zero X part and Z parts $110,011$. The errors $I,X_1,X_2,X_3$ have respective error syndromes $00,10,11,01$, so all single bit flips can be corrected. This is not a stabilizer code correcting every single-qubit error: $Z_1$ acts as a nontrivial logical phase error.

Let $C(S)$ denote the Pauli centralizer. The exact Pauli error criterion for a stabilizer code isEquivalently, no difference of error labels lies in $L^{\perp }\setminus L$. The scalar phases in $\mathcal{Z}S$ are essential: an error product $-I$ is harmless even though it is not in $S$.

We prove the criterion. For $T=E_a^{†}{E}_b$ outside $C(S)$, some $s\in S$ anticommutes with $T$. Since $sP=Ps=P$, we have $PTP=PsTsP=-PTP$, hence $PTP=0$. If $T=\lambda s\in \mathcal{Z}S$, then $PTP=\lambda P$. If $T\in C(S)\setminus \mathcal{Z}S$, it restricts to a unitary operator on the stabilizer code, butno $sT$ is scalar. A scalar restriction would have scalar coefficient zero by this trace equation, contradicting unitarity. Thus this third case is not scalar.

To justify why scalar compressions are exactly what correction requires, suppose one recovery corrects every $E_a$. Dilate that recovery to an isometry $W$ including its auxiliary system. Correct recovery of every pure stabilizer code state gives $W{E}_a|\psi ⟩=|\psi ⟩\otimes |{\eta }_a⟩$. The auxiliary state is independent of $\psi$: linearity on basis states and their superpositions forces their auxiliary vectors to coincide. Preservation of inner products then giveswhich is the Knill--Laflamme condition $P{E}_a^{†}{E}_{b}P=c_{ab}P$. The three cases above prove necessity of the boxed criterion.

For sufficiency, put errors in the same class when $E_a^{†}{E}_b\in \mathcal{Z}S$. They act identically on the stabilizer code up to scalar phase, so give the same error-image subspace. Under the criterion, different classes give mutually orthogonal subspaces $E_a{\mathcal{C}}_S$, because their cross-compressions vanish. Measure which of these subspaces is occupied and apply the inverse of a representative error. This returns every stabilizer code state unchanged; extend the recovery on the orthogonal complement by preparing any fixed stabilizer code state. Linearity also corrects any valid noise channel whose Kraus operators are linear combinations of the specified errors: within each syndrome all amplitudes multiply the same logical state, and trace preservation normalizes the recovered state. This proves sufficiency, including degenerate error classes.

The concavity inequality is true, but its printed equality condition is false. At $b=1$ equality always holds, even for distinct states; for example take ${\rho }_1=|0⟩⟨0|$ and ${\rho }_2=|1⟩⟨1|$. The correct statement on $0\le b\le 1$ isFor $0<b<1$, entropy is therefore strictly concave.

We give the matrix inequality needed for the proof. Define the quantum relative entropy $D(\tau \Vert \sigma )=\mathrm{tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. Each row and column of $q$ sums to one. The elementary scalar inequalityfollows from $x\ln x-x+1\ge 0$, with equality exactly at $x=1$; use limits at zero. ConsequentlyEquality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every eigenvector $|i⟩$ of $\tau$ is also a $\sigma$-eigenvector with eigenvalue $r_i$, so $\tau =\sigma$. Conversely identical states give zero relative entropy. This proves nonnegativity of quantum relative entropy and its equality condition, including states with zero eigenvalues under the support convention.

Now set $\rho =b{\rho }_1+(1-b){\rho }_2$. For $0<b<1$ its support contains both component supports. Direct expansion givesSince both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently ${\rho }_1={\rho }_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected strict concavity of Von Neumann entropy.

The command centres the log measurements and performs principal component analysis of their covariance matrix. With cor=F, it does not first standardize the variables to variance one. For a unit principal component loading vector $q$, the score $q^T(z_i-\bar{z})$ has sample variance $q^T\hat{\Sigma }q$. By the spectral theorem for real symmetric matrices, expand $q$ in an orthonormal eigenbasis; that variance is a weighted average of the eigenvalues. Its maximum is the largest eigenvalue, reached at the corresponding eigenvector. Repeating subject to orthogonality to previous directions gives the other principal components.

The software uses the divisor-$31$ sample covariance matrix for its component standard deviations, whereas var used divisor $30$. Diagonalizing the displayed matrix gives approximatelyTaking square roots of the latter gives $(0.5671603,0.07312405,0.03064834)$, agreeing with the output to the rounding of the displayed sample covariance matrix. The normalization changes these numerical standard deviations but leaves the eigenvectors and proportions of explained variance of a principal component unchanged.

The first principal component explains about $98.08\%$ of the total log-variable variance, the second another $1.63\%$, and the last about $0.286\%$. Keeping two preserves about $99.714\%$. An original scree plot displays the sharp drop after the first component:

Consequently the centered cloud is very elongated along one direction: a one-dimensional approximation represents almost all of its squared variation. From the displayed covariance matrix, a unit first eigenvector, with its arbitrary sign chosen positive, is approximatelyIts principal component score is therefore $0.3980(\log G_i-\overline{\log G})+0.09514(\log H_i-\overline{\log H})+0.9124(\log V_i-\overline{\log V})$. All coefficients have the same sign, so this direction describes overall tree size, with log volume dominating the unstandardized variation. The summary alone supplies proportions, not the loading directions; the displayed sample covariance matrix supplies the latter. This is a geometric approximation, not evidence of an exact deterministic relation or independent original variables.

One further use is principal component analysis on a correlation matrix, obtained with cor=T, which equalizes marginal variances and answers a different scaling-sensitive question. Another is principal component regression: fit a response to selected principal component scores rather than highly correlated original predictors, choosing the number using prediction assessment. The fitted transformation also projects new measurements into the same low-dimensional coordinates, provided the training centering and scaling are retained.

Write ${\beta }_T=(k_{B}T{)}^{-1}$ to distinguish inverse temperature from an order-parameter critical exponent. For the Ising model, define the full two-point correlation function $G(r)=⟨{\sigma }_0{\sigma }_r⟩$ and its connected correlation function $G_c(r)=G(r)-M^2$ in a translation-invariant phase. Away from a thermodynamic critical point, the large-distance envelope decays exponentially, possibly multiplied by a power of distance:This defines the correlation length in physical distance units when the limit exists. At a continuous phase transition, $\xi \to \infty$ and the critical connected correlation function typically has algebraic decay $G_c(r)\sim |r{|}^{-(D-2+\eta )}$, where $\eta$ is the anomalous dimension. Below the transition the full $G(r)$ tends to $M^2$, so the decay definition must use $G_c$.

Let $S={\sum }_r{\sigma }_r$ and $F=-{\beta }_T^{-1}\log Z$ be the extensive Helmholtz free energy. Holding the field-independent constant $C$ and other physical couplings fixed,Because the Hamiltonian contains $-hS$, ${\partial }_h\log Z={\beta }_T⟨S⟩$ and ${\partial }_h^2\log Z={\beta }_T^2(⟨S^2⟩-⟨S{⟩}^2)$. Translation invariance then proves the correlation-function susceptibility sum rule:In the ordered zero-field phase these statements refer to a selected pure thermodynamic phase; averaging equally over opposite magnetizations introduces the separate spin-mixture contribution to zero-field susceptibility.

A normalized blocking kernel $K_b({\sigma }^{\prime },\sigma )\ge 0$ satisfies ${\sum }_{{\sigma }^{\prime }}{K}_b({\sigma }^{\prime },\sigma )=1$. For example, assign one block spin to each block of $b^D$ microscopic spins using majority sign, with equal probabilities for a tie. Define the blocked Hamiltonian, retaining the entire generated operator space, byThe field-independent contribution $N{g}_E$ belongs to the identity operator. With $N^{\prime }=N{b}^{-D}$ andsumming over ${\sigma }^{\prime }$ proves $Z(u,C,N)=Z(u^{\prime },C^{\prime },N^{\prime })$ exactly. Iterating gives $N_p=N{b}^{-pD}$ and the same partition function with $(u_p,C_p,N_p)$. Observables at large scales are reproduced by appropriately blocked observables and sources. Truncating the generated interactions makes this an approximation; dropping the additive constant already destroys the equality of free energies even when normalized spin expectations remain correct.

The extensive Helmholtz free energy is unchanged: $F(u_0,C_0,N)=F(u_p,C_p,N_p)$. In contrast, its value per site changes because the number of sites changes. Define the reduced zero-constant free-energy density $f_0(u)={\beta }_{T}F(u,0,N)/N$ in the thermodynamic limit. The preceding equality yieldsand iteration proves the additive free-energy recursion under blocking:Equivalently $C_p=b^{pD}[C_0+{\sum }_{j=0}^{p-1}{b}^{-jD}{g}_E(u_j)]$. Thus the inhomogeneous source records the identity-operator contribution to renormalization-group free energy from eliminated short-distance degrees of freedom.

To express this recursion for the singular part, write $f_0=a+f_s$, where $a$ is the regular analytic background. Then the same displayed recursion holds with $f_s$ andThis is the analytic subtraction of an inhomogeneous renormalization recursion. A purely analytic source can often be absorbed into $a$, after which it does not control the leading singular powers. It is not legitimate simply to discard an analytic source in the full free energy, where it can dominate the regular background. Moreover an analytic monomial $c{t}^m{h}^n$ contributesIf $m{l}_t+n{l}_h=D$, the sum is $p$ and hence generates a logarithm when $p\sim -\log |t|/(l_t\log b)$: a renormalization-group free-energy resonance. Such a term cannot be discarded when describing that logarithmic singularity. Neglect of the inhomogeneous term in the leading power-law scaling therefore assumes that its regular part has been subtracted, that any remaining source only changes finite scaling amplitudes, and that there are no relevant resonances or marginal logarithms. Omitting an irrelevant operator further requires absence of a dangerously irrelevant coupling, whose vanishing limit can make a scaling function singular.

A renormalization-group fixed point obeys $R_b{u}_{\ast }=u_{\ast }$. Linearization gives $\delta u^{\prime }=A\delta u$; in eigenvector coordinates $w_i^{\prime }=b^{l_i}{w}_i$. Positive $l_i$ label relevant operators, negative $l_i$ label irrelevant operators, and zero $l_i$ labels a marginal operator requiring nonlinear analysis. The critical surface is the stable manifold obtained by tuning all relevant scaling fields to zero. A repulsive renormalization-group trajectory leaves the fixed point along a relevant direction as the observation length increases. Locally a typical zero-field section has $dw/d\ell =-\omega w$, $dt/d\ell =l_{t}t$, $\ell =\log b$, so the critical surface $t=0$ attracts along $w$, while trajectories with either sign of $t$ depart from it. A magnetic scaling field adds another relevant direction.

Under the stated homogeneous-scaling assumptions and after fixing irrelevant fields at their limiting values, two relevant scaling fields obey $t^{\prime }=b^{l_t}t$, $h^{\prime }=b^{l_h}h$. These fields are analytic coordinates near the renormalization-group fixed point, proportional to the physical reduced temperature and conjugate field to leading order. The singular free-energy density consequently obeysHere $F_s$ denotes the singular free energy per original site or unit volume, with regular dimensional factors absorbed; the extensive singular part is $N{F}_s$. Iterate and choose $b^p=|t{|}^{-1/l_t}$ to obtain the scaling hypothesis for critical phenomena:The sign labels the two sides of the transition: $f_{+}$ describes $t>0$, the disordered side, while $f_{-}$ describes $t<0$, the ordered side. The two functions need not coincide; $f_{-}$ has the zero-field one-sided derivative appropriate to spontaneous magnetization. The numbers $l_t,l_h$ are scaling eigenvalues in logarithmic form: the corresponding discrete RG multipliers are $b^{l_t},b^{l_h}$.

The correlation length obeys $\xi (t,h)=b\xi (b^{l_t}t,b^{l_h}h)$. The same scale choice gives $\nu =1/l_t$. Differentiating $F_s$ once and twice with respect to the conjugate field, and twice with respect to temperature, givesAt $t=0$, instead choose the scale $b^p=|h{|}^{-1/l_h}$. Then $F_s(0,h)\sim |h{|}^{D/l_h}$ and $M\sim \mathrm{sgn}(h)|h{|}^{D/l_h-1}$, so $\delta =l_h/(D-l_h)$. Thus differentiation and scale matching establishThe first is the Widom scaling relation and the second the hyperscaling relation. The hyperscaling relation requires precisely the regular homogeneous scaling used here. Above the ordinary upper critical dimension, the quartic coupling is dangerously irrelevant in the ordered phase and this derivation cannot discard it: the mean-field critical exponents $\nu =1/2$, $\alpha =0$ do not obey $\alpha =2-D\nu$ for $D>4$. At marginal dimensions logarithmic factors likewise require care beyond the displayed pure powers.

Use local coordinates $q=\theta -\pi /2$, $p=\varphi$ at $P_1$. The Jacobian matrix, its trace and its determinant areConsequently $P_1$ is attracting when $T<0,D>0$, repelling when $T>0,D>0$, and a saddle equilibrium when $D<0$. A transition between a node and a focus at $T^2=4D$ changes the linear geometry but is not a bifurcation of the local topological phase portrait.

The stationary bifurcation curve is $\lambda +\kappa =2$. To identify it, write $C=\cos 2\varphi$, $S=\sin 2\varphi$. At a non-equatorial equilibrium point, eliminating $S$ givesNear $(3,-1)$, if $d=2-\lambda -\kappa >0$, these yield $p=\pm \sqrt{d}/2+O(d^{3/2})$ and $q=p+o(\sqrt{d})$. The two branches are related by the local reflection $(q,p)\mapsto (-q,-p)$. Their determinants are $-4d+o(d)$, so they are saddle equilibria. This is a pitchfork bifurcation; on the portion with $T<0$ the saddle branches lie on the side where $P_1$ is stable, giving a subcritical pitchfork bifurcation.

A Hopf bifurcation requires $T=0,D>0$, henceThe oscillation frequency is $\omega =\sqrt{-2(1+\kappa )}$. At the intersection $\boxed{(\lambda ,\kappa )=(3,-1)}$, $J\ne 0$ but $J^2=0$: there is a double zero eigenvalue with one eigenvector. The map from $(\lambda ,\kappa )$ to $(T,D)$ has nonzero Jacobian determinant, so two independent parameter conditions produce this codimension-two bifurcation. Reflection removes quadratic terms; the appropriate initial comparison is with a reflection-symmetric cubic double-zero normal form, subject to the cubic degeneracy checked below.

For completeness, the printed vector field has a subcritical, rather than supercritical, Hopf bifurcation. In the mechanical coordinates $x=p$, $y=\dot{p}$, at $\lambda =3$ its cubic acceleration has $x^{2}y$ coefficient $b=6(\kappa -3)(\kappa +1)/(\kappa -1{)}^2$ and $y^3$ coefficient $d_3=-2/(\kappa -1{)}^2$. For a small harmonic oscillation, the averaged cubic change of $H=(y^2+{\omega }^2{x}^2)/2$ is ${\omega }^2{A}^4(b+3d_3{\omega }^2)/8$, whereThus a small unstable periodic orbit lies on the attracting side of the Hopf bifurcation. This calculation matters when comparing the original equations with the later assumed normal form.

Let $q=r_{\Omega }^2$ and $h=\varpi$. The $c,e$ eigenvalues at the conductive equilibrium point are $-h$ and $-(4-h)\sigma$, both negative since $0<h<4$ and $\sigma >0$. The coupled $(a,b,d)$ Jacobian matrix has characteristic polynomialwhereA simple zero eigenvalue occurs on $\boxed{r_P=1+q}$. The reflection $(a,b,d)\mapsto (-a,-b,-d)$, with $c,e$ fixed, makes its generic steady-state bifurcation a pitchfork bifurcation.

For a nonzero imaginary pair, substitute $s=i\omega$ into $P$: the real and imaginary parts require $A_3=A_1{A}_2$ and ${\omega }^2=A_2>0$. HenceWith $0<\sigma <1$, this Hopf bifurcation curve is present only for $q>q_{TB}=(1+\sigma )/(1-\sigma )$. At its intersection with the pitchfork bifurcation curve,and $P(s)=s^2(s+1+2\sigma )$. The zero eigenvalue has geometric multiplicity one: its eigenvector has $b=a$, $d=a/\sigma$. Thus this is a reflection-symmetric Bogdanov–Takens bifurcation, rather than two independent zero modes.

The Routh-Hurwitz stability criterion gives the stable trivial state below $r_P$ for $q<q_{TB}$ and below $r_H$ for $q>q_{TB}$. The rest of the line $r_P$, beyond $q_{TB}$, is a stationary bifurcation of an already unstable trivial state. Likewise the extension of $r_H$ below $q_{TB}$ is only an algebraic line, since its putative frequency is not real. The middle panel of the bifurcation diagram in part 1(b) distinguishes the actual curves from this extension.

Write the linear differential equation as $\dot{\mathbf{x}}=M\mathbf{x}$, withIts characteristic polynomial is $(\lambda +4{)}^2+9$. Thus a choice of eigenvalues and eigenvectors isThe distinct eigenvalues give two linearly independent solutions, and the general complex solution isReal solutions have $a_2=\overline{a_1}$. Equivalently,They spiral counterclockwise towards the origin, which is a stable spiral.

In a forced linear differential equation, resonance in a differential equation occurs when exponential forcing excites a mode with the same exponent as its eigenvalue, producing a polynomial factor such as $t$ multiplying that exponential. Merely matching an exponent is not enough: its projection onto that mode must be nonzero. Here $t{e}^{{\lambda }_{j}t}$ still decays in absolute magnitude because the real part of ${\lambda }_j$ is negative; resonance refers to this extra factor, not necessarily to unbounded growth.

For each forcing vector, resolve it in the eigenvector basis:If $u_k$ is the coefficient of the $k$th eigenvector, its scalar equation is ${\dot{u}}_k={\lambda }_k{u}_k+{\sum }_j{c}_{kj}{e}^{{\lambda }_{j}t}$. Multiplying by $e^{-{\lambda }_{k}t}$ and integrating shows that the $j=k$ contribution is $c_{kk}t{e}^{{\lambda }_{k}t}$, whereas a $j\ne k$ contribution is $c_{kj}{e}^{{\lambda }_{j}t}/({\lambda }_j-{\lambda }_k)$. Therefore the exact conditions for no resonant response areUnder these conditions one particular solution iswhich contains no factor of $t$. For a real forcing, $p_2=\overline{p_1}$ and $q_2=\overline{q_1}$, and the two boxed conditions are conjugate to each other.