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Put $u=\epsilon v+O({\epsilon }^2)$. The linearization at the zero solution isThe appropriate Fourier sine series isOrthogonality of the modes givesThus each mode grows or decays exponentially with its own eigenvalue, and the largest growth rate is the $n=1$ value. Strict linear stability analysis, meaning decay of every mode, is equivalent toAt equality the first mode is neutral in the linearization; the nonlinear terms decide its subsequent behavior. Above this length the first mode grows, giving linear instability.

Expanding the characteristic polynomial givesThus the eigenvalues are $1,2,3$. Solving the corresponding homogeneous linear systems givesEvery nonzero vector in the displayed eigenspace is an eigenvector. Direct multiplication verifies the three eigenvalue equations. The distinct eigenvalues imply these eigenvectors are linearly independent and form a basis of ${\mathbb{R}}^3$.

Choose real eigenvectors $u,v$ with distinct eigenvalues $\lambda ,\mu$. Symmetry givesTherefore $(\lambda -\mu )u^{T}v=0$, and $\boxed{u\cdot v=0}$. For complex choices, replace transpose by conjugate transpose; the reality proved in part (i) gives the same conclusion for the Hermitian inner product. Thus different eigenspaces are orthogonal.

Write a complex eigenvector as $v=u+iw$, where $u,w\in {\mathbb{R}}^n$. Since $M$ and its eigenvalue $\lambda$ are real,At least one of $u,w$ is nonzero, and it is a real eigenvector for the same eigenvalue. More generally every complex eigenvector is a complex linear combination of real ones: its real and imaginary parts lie in the real eigenspace. Thus each eigenspace has a basis chosen over the reals. The conclusion does not assert that every arbitrary complex eigenvector is itself real.

Let $Mv=\lambda v$ with $v\in {\mathbb{C}}^n$ nonzero. For a real symmetric matrix, $M^{\ast }=M$, where the star denotes the conjugate transpose. Consequently $v^{\ast }Mv$ equals its complex conjugate and is real. The eigenvalue equation givessince $v^{\ast }v>0$. Allowing a complex eigenvector in this argument is essential: reality of an eigenvalue must be proved, rather than assumed in advance.

For a real antisymmetric matrix, $A^T=-A$, so $(A^2{)}^T=A^2$. The previous result makes every eigenvalue $\eta$ of $A^2$ real, with a real nonzero eigenvector $v$. Moreover,Thus every eigenvalue of $A^2$ is real and nonpositive.

Take real unit eigenvectors $u,w$ of $A^2$ for the respective distinct nonzero eigenvalues $-{\lambda }^2,-{\mu }^2$, where the real parameters $\lambda ,\mu$ are nonzero. DefineThe displayed quadratic identity gives $|u^{\prime }|=|w^{\prime }|=1$. Also $u\cdot Au=0$ and $w\cdot Aw=0$, because a real antisymmetric matrix has zero quadratic form. Thus $u\perp u^{\prime }$ and $w\perp w^{\prime }$. Since $A$ commutes with $A^2$, $u^{\prime }$ lies in the same eigenspace of $A^2$ as $u$, and $w^{\prime }$ in the same one as $w$. Distinct eigenspaces of the real symmetric matrix $A^2$ are orthogonal, so all four vectors are orthonormal. Finally,Thus on each of the two perpendicular planes, $A$ acts as a scaled quarter-turn. This constructs the required vectors, rather than merely counting dimensions.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

Identify the Möbius group with ${\mathrm{PSL}}_2(\mathbb{C})$, the orientation-preserving isometry group of three-dimensional hyperbolic space. Here a properly discontinuous group action permits finite stabilizers: for each compact set $K$, only finitely many $g$ satisfy $gK\cap K\ne \varnothing$.

Fix $o\in {\mathbb{H}}^3$. The set of isometries with $d(o,go)\le R$ is compact. To see this, an orientation-preserving isometry is uniquely determined by the image of an oriented orthonormal frame at $o$. The image point lies in a compact closed hyperbolic ball and its frame lies in a compact rotation-group fiber. These choices parametrize the indicated set of isometries continuously.

A discrete subgroup of a topological group is closed: otherwise distinct subgroup elements accumulating at the same point would have quotients tending to the identity, contradicting isolation of the identity. Its intersection with a compact set is therefore finite. If $K\subseteq B(o,R)$ and $gK\cap K\ne \varnothing$, choosing $x,gx\in K$ gives $d(o,go)\le d(o,gx)+d(gx,go)\le 2R$. Hence discreteness implies a properly discontinuous group action. Conversely, if the subgroup is not discrete, distinct $g_n$ tend to the identity. For a compact ball $K$ with $o$ in its interior, eventually $g_{n}o\in K$, so infinitely many $g_{n}K$ meet $K$, contradicting proper discontinuity. This proves the equivalence.

Matrices of determinant one over the Gaussian integers are closed under multiplication and inversion, with inverse $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Quotienting by $\{\pm I\}$ gives the specified group of Möbius transformations. The Gaussian integers form a discrete set in the complex plane, so these matrices form a discrete subgroup of ${\mathrm{SL}}_2(\mathbb{C})$, and the finite quotient remains discrete. Thus this group acts properly discontinuously on ${\mathbb{H}}^3$.

Explicit representatives of all four types areThe first is $z\mapsto -1/z$ and has order two; the second is the nonidentity translation $z\mapsto z+1$. The third has positive real reciprocal eigenvalues $(3\pm \sqrt{5})/2$, giving translation along an axis without rotation. For the last, a nonreal trace excludes both unit-modulus and real eigenvalues. Its multiplier therefore has modulus different from one and a nonzero rotation angle: it is a loxodromic Möbius transformation.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. The Jacobian matrix isAt the origin its eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle equilibrium for every allowed $\mu$. At either nonzero fixed point the characteristic polynomial is ${\lambda }^2+\lambda +2{\mu }^2$, givingThus both are stable nodes for $0<|\mu |<1/(2\sqrt{2})$ and stable focuses for $|\mu |>1/(2\sqrt{2})$. At equality the repeated eigenvalue is $-1/2$ and the matrix is not scalar, so there is only one independent eigenvector: the fixed point is a stable improper node.

The stable manifold of the origin is $x=0$, exactly: there $\dot{y}=-y$. Conversely, a trajectory converging to the origin with $x\ne 0$ would eventually have $\dot{x}/x={\mu }^2-y>{\mu }^2/2$, which precludes convergence of $x$ to zero.

For the local unstable manifold, write $y=h(x)=a{x}^2+b{x}^4+O(x^6)$. Reflection symmetry makes $h$ even, and the invariance equation for a graph requiresComparing quadratic and quartic terms gives $2a{\mu }^2=1-a$ and $4b{\mu }^2-2a^2=-b$. Therefore

For a spatially homogeneous positive steady state, $\alpha n(n_0^2-n^2)=0$ and $\beta n-\gamma a=0$, givingWrite $n=n_0+\eta$, $a=a_0+\xi$, and define $A=2\alpha n_0^2>0$. Homogeneous linearization givesThe eigenvalues are $-A,-\gamma$, both negative. Hence this steady state is stable to homogeneous disturbances, including the repeated-eigenvalue case, where a possible $t{e}^{-At}$ factor still decays.

A steady-state bifurcation is a change in the local equilibrium branches or their stability as a parameter varies. At such a bifurcation the Jacobian matrix at the equilibrium has a zero eigenvalue, so the implicit function argument for a unique persistent local equilibrium fails.

For this system,At $(0,\mu )$ its eigenvalues are $1-{\mu }^2,-\mu$; at $(1,0)$ they are $-2,\mu -1$. Both have one simple zero eigenvalue at $\mu =1$, with the other eigenvalue negative. Interior equilibrium branches, obtained below, meet these boundary equilibria there, establishing the actual bifurcations rather than merely a necessary eigenvalue test.

In the original eigenbasis, $Q^{\prime }=\left(\begin{array}{cc}-1& iϵ\\ -iϵ& 1\end{array}\right)$. Insert the proposed first-order state for eigenvalue near minus one. The first component gives $c_1=0$; the second gives $-i{a}_1+a_2=-a_2$, hence $a_2=i{a}_1/2$. Similarly for the eigenvalue near one, $c_2=0$ and $b_2=i{b}_1/2$. Normalization gives $|a_1|=|b_1|=1$, with their phases freely selectable. Choosing the original-state phase conventions,The states are orthogonal through first order because the two cross terms cancel.

Exactly, $(Q^{\prime }{)}^2=(1+{ϵ}^2)I$, so putting $R=\sqrt{1+{ϵ}^2}$ givesFor an eigenvalue $q^{\prime }$, the first row requires the component ratio $v_2/v_1=-i(1+q^{\prime })/ϵ$, so the coefficient in the printed parametrization is $\boxed{B=1}$. With $D=\sqrt{(1+R{)}^2+{ϵ}^2}$, smooth normalized choices areThey have unit norm and zero inner product, and direct multiplication gives the two eigenvalue equations. For $ϵ\ne 0$, the printed scalar factors can be chosen $A_1=(1+R)/D$ and $A_2=iϵ/D$; the second is allowed a complex phase. The displayed ratio form is singular at zero, but the normalized vectors have the correct continuous limits there.

Expanding $R=1+{ϵ}^2/2+O({ϵ}^4)$ and $D=2+O({ϵ}^2)$ gives $|1^{\prime }⟩=|1⟩+iϵ|2⟩/2+O({ϵ}^2)$ and $|2^{\prime }⟩=|2⟩+iϵ|1⟩/2+O({ϵ}^2)$, while the eigenvalues have no first-order shift. This recovers all six perturbative coefficients with the same phase convention.