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A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

A generalized eigenfunction solves the eigenvalue equation for a linear operator in an enlarged space, often a space of distributions, although it may fail to belong to the underlying Hilbert space. A plane wave $e^{ikx}$ on the whole real line is a generalized momentum eigenstate, with momentum $\hslash k$, and is not square-integrable. This usage differs from a generalized eigenvector in a Jordan chain, where a higher power of $L-\lambda$ vanishes rather than $L-\lambda$ itself.

A Jordan basis is a basis consisting of Jordan chains of generalized eigenvectors for a linear operator. Its matrix is in Jordan normal form. Over the complex numbers every finite-dimensional linear operator has such a basis; a real matrix with nonreal eigenvalues requires complexification for this form.

If $(M-\lambda I)e=0$ and $(M-\lambda I)v=e$, the vector $v$ is a generalized eigenvector and $e^{\lambda t}(v+te)$ solves $\dot{x}=Mx$. The polynomial factor distinguishes a nontrivial Jordan block from an ordinary repeated eigenvalue with a full eigenvector basis.

For the homogeneous linear differential equation $\dot{Y}=K(t)Y$ with continuous coefficient matrix of period $P>0$, the monodromy matrix of a periodic linear system is the one-period map from $Y(0)$ to $Y(P)$. If $X$ is any fundamental matrix,Its eigenvalues are the Floquet multipliers. Multipliers are nonzero, since a fundamental matrix is invertible. Changing the starting phase conjugates $N$, so the Floquet multipliers and Jordan blocks are independent of that choice.^[ref]

This turns a continuous-time problem into repeated matrix multiplication:If $Nv=mv$, the associated solution of a differential equation obeys $Y(t+P)=mY(t)$ at every time, not merely at the sampled times. In particular, $Nv=v$ gives a nonzero $P$-periodic solution, and every nonzero $P$-periodic solution arises this way. If $Nv=-v$, the state changes sign after one period and repeats after two. A unit-modulus Floquet multiplier different from $1$ need not give repetition after one period, and a generalized eigenvector in a nontrivial Jordan block need not even give bounded motion.

For the scalar example under periodic solution, $N$ is the one-by-one matrix $(e^{2\pi a})$. Numerically solving a matrix problem uses the same idea: integrate one initial value problem for each column of $X(0)=I$ through one period, then compute the eigenvalues of $X(P)$. This is useful when an explicit solution of a differential equation is unavailable: one period determines the Floquet growth rate over arbitrarily many periods.

For piecewise constant coefficients, which are also covered by the one-period construction, $N$ is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. For example, take period $P=2$ and alternateSince $K_1^2=K_2^2=0$,Replacing the coefficient by its mean would instead give $e^{K_1+K_2}=\left(\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right)$ and the incorrect rate $0.5$. The matrices do not commute. The scalar mean-coefficient formula therefore cannot simply be transferred to matrix systems. Nevertheless the Liouville formula for a fundamental matrix gives the correct determinant even in the noncommuting case:In the example, this is $1=m_{+}{m}_{-}$; it fixes the product of the Floquet multipliers, rather than either multiplier separately.

In Sur les équations différentielles linéaires à coefficients périodiques (1883), Gaston Floquet calls a function satisfying $f(t+P)=mf(t)$ periodic of the second kind, and distinguishes the genuinely periodic function obtained when $m=1$. This historical terminology captures precisely the difference between repetition of the coefficient and repetition of the state.^[ref]

Translate the equilibrium to the origin by $z=(x-a,y-b{)}^T$. If the vector field is continuously differentiable, its first-order Taylor expansion isThe Jacobian matrix $M$ gives the linearization $\dot{z}=Mz$, whose general solution is $z(t)=e^{tM}z(0)$ using the matrix exponential. Its eigenvalues determine the exponential factors, while its eigenvectors and, when necessary, generalized eigenvectors determine their directions.

For distinct eigenvalues ${\lambda }_1,{\lambda }_2$ and corresponding eigenvectors $v_1,v_2$,For a real matrix with nonreal conjugate eigenvalues, choose conjugate coefficients to obtain real solutions. Equivalently, if $\lambda =a+ib$ and $v=p+iq$, a real pair of independent solutions isFor a repeated eigenvalue $\lambda$, there are two cases:

The factor $t$ in the second case is essential. These are exact solutions of the linearization; the remainder describes the error for the nonlinear differential equation. When no eigenvalue has zero real part, the hyperbolic equilibrium theory justifies the corresponding local qualitative classification. Zero real parts require additional nonlinear analysis.

Eigenfrequencies and mode geometry. Write $d=A_{11}-A_{22}$ and $D=\sqrt{d^2+4A_{12}{A}_{21}}>0$. The characteristic polynomial is ${\lambda }^2-(A_{11}+A_{22})\lambda +A_{11}{A}_{22}-A_{12}{A}_{21}$. Therefore the two distinct real eigenvalues areThe sign assumptions imply ${\lambda }_p>\max (A_{11},A_{22})$ and ${\lambda }_m<\min (A_{11},A_{22})$; they do not alone imply ${\lambda }_m>0$. A positive lower frequency additionally requires $\det A>0$, as in the usual nondegenerate physical Laplace-Lagrange secular matrix.

Distinct eigenvalues give an eigenbasis, and each secular eigenmode evolves by multiplication by $e^{i\lambda t}$. Hencewith nonnegative amplitudes and initial phases. The second row of each eigenvalue equation gives $z_1/z_2=(\lambda -A_{22})/A_{21}$. For the upper eigenvalue this ratio is negative, whereas for the lower one it is positive. ThusThe upper secular eigenmode has anti-aligned apsides, and the lower has aligned apsides. If one amplitude vanishes, its phase is immaterial.

The initial vector sums. The planet-1 modal vectors point at $30^{\circ }$ and $120^{\circ }$. Their resultant at $60^{\circ }$ requires $e_{1p}=\sqrt{3}{e}_{1m}$. The planet-2 vectors point at $210^{\circ }$ and $120^{\circ }$; their resultant at $150^{\circ }$ requires $e_{2m}=\sqrt{3}{e}_{2p}$. These follow by resolving components perpendicular to each resultant. In particular the specified phases constrain the matrix: $r_p/r_m=3$. They cannot be imposed on an arbitrary matrix satisfying only the sign assumptions.

Put $s=r_p{r}_m=A_{12}/A_{21}$ and let $B=e_{2p}>0$. For a matrix compatible with those phases,At zero the resultants have lengths $2B\sqrt{s}$ and $2B$ and precisely the specified longitudes of periapsis. The diagram uses $B=1$, $s=1$ as a drawing normalization, not as an additional assumption on the physical planets.

Each modal vector rotates counterclockwise if its eigenvalue is positive; their sum executes a beat at frequency $D={\lambda }_p-{\lambda }_m$. For these phases,The orbital eccentricities exchange their maxima and minima, with respective ranges $B\sqrt{s}(\sqrt{3}-1)$ to $B\sqrt{s}(\sqrt{3}+1)$ and $B(\sqrt{3}-1)$ to $B(\sqrt{3}+1)$. The longitude of periapsis is the argument of each displayed vector sum; it does not in general advance uniformly at either individual eigenfrequency. The complex eccentricity trajectories are generally quasiperiodic; they close only for commensurate modal frequencies. This example has a dominating upper mode in planet 1 and a dominating lower mode in planet 2, so their apsidal phases have different winding rates.

Amplitude product and the angular-momentum convention. Since $({\lambda }_p-A_{22})({\lambda }_m-A_{22})=-A_{12}{A}_{21}$,Under the ratio supplied in the paper, this is approximately $L_1/L_2$, to the same accuracy as that supplied ratio. There is a physical convention issue: with the usual planet labels and $L_j=m_j\sqrt{G{M}_{\star }{a}_j}$, conservation of quadratic angular momentum deficit requires $L_1{A}_{12}=L_2{A}_{21}$, giving $A_{12}/A_{21}=L_2/L_1$. This follows directly by differentiating $\frac{1}{2}(L_1|z_1{|}^2+L_2|z_2{|}^2)$. Thus the paper's stated angular-momentum ratio is reversed for the conventional column-vector equation $\dot{z}=iAz$. The exact algebraic product above is independent of that naming discrepancy.

Tidal dissipation and damping. Assume the tidal dissipation contributes only the specified linear eccentricity damping, without changing the conservative matrix or adding apsidal precession. ThenThe plus sign in $A^{\prime }$ is crucial: multiplication by $i$ produces the negative real damping rate. Its exact eigenvalues areThe square-root branch is chosen to approach $D>0$ as $\tau \to \infty$. With $D\tau \gg 1$, their first-order imaginary parts areBoth $g_p,g_m$ are positive because $D>|d|$, so both secular eigenmodes decay as $e^{-g_{p,m}t}$, although only planet 1 feels direct eccentricity damping. The coupling transmits that damping to planet 2. Their precession frequencies change only at second order; their eigenvectors acquire small complex corrections, so exact alignment or anti-alignment becomes a small apsidal phase lag. The more slowly damped mode eventually dominates, unless its initial coefficient vanishes: it is the lower aligned mode for $d>0$ and the upper anti-aligned mode for $d<0$. At $d=0$ both leading damping rates equal $1/(2\tau )$.

The stated slow-secular condition presumes a nonzero positive ${\lambda }_m$. It need not imply $D\tau \gg 1$ near a nearly degenerate pair, so the exact square-root formula is the appropriate answer there. Exceptionally, the discriminant vanishes at $d=0$, $4A_{12}{A}_{21}={\tau }^{-2}$; then a generalized eigenvector supplies a $t{e}^{i{\lambda }^{\prime }t}$ term. There is still decay. Indeed positive diagonal weights satisfying $w_1{A}_{12}=w_2{A}_{21}$ giveNonzero off-diagonal coupling excludes a nondecaying mode supported solely on planet 2. This checks the damping signs without relying on the weak-damping expansion.

Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.

A Jordan block is $J_m(\alpha )=\alpha I+N$, where $N$ has ones on the superdiagonal and zeros elsewhere. Since $N^2=0$ for $m=2$, Taylor expansion givesThe Jordan normal form is the block-diagonal matrix, unique up to block order, to which $A$ is similar and whose blocks are Jordan blocks.

A block of size at least two for eigenvalue $\lambda$ contributes a generalized eigenvector in $\ker (A-\lambda I{)}^2\setminus \ker (A-\lambda I)$. Thus the displayed equality for every $\lambda$ is equivalent to every block having size one, which is diagonalizability.

The transpose of each Jordan block is similar to it by reversing its basis. Hence $A^T$ has the same Jordan form as $A$.

Write $A=PJ{P}^{-1}$. Let $R$ be block diagonal with the reversal matrix on each Jordan block. Both $R$ and $RJ$ are symmetric and $R$ is invertible. Thereforewhere $C,D$ are symmetric and $C$ is nonsingular.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.