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For circular planetary orbits with angular momenta $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$, the inclination matrix has positive off-diagonal entries, zero row sums, and $L_j{B}_{jk}=L_k{B}_{kj}$. Positive diagonal symmetrization of a matrix with $W=\mathrm{diag}(L_j)$ makes $W^{1/2}B{W}^{-1/2}$ a symmetric matrix. Moreover,All eigenvalues are real and nonpositive. The zero eigenvalue is a common tilt of every orbital plane; nonzero modes describe relative nodal precession.

Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.

Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the spectrum of a three-dimensional matrix with zero row sums. In particular, $B(1,1,1{)}^T=0$. DefineExpanding the characteristic polynomial gives $\det (\lambda I-B)=\lambda ({\lambda }^2+S\lambda +T)$, soFor the physical coefficients, $L_j{B}_{jk}=L_k{B}_{kj}$ with $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$. Thus $W=\mathrm{diag}(L_j)$ gives positive diagonal symmetrization of a matrix, andConsequently the eigenvalues are real and nonpositive, and $B$ has a complete eigenvector basis. This justifies the mode expansion in part (i).

The variables $y_j=I_j{e}^{i{\Omega }_j}$ are complex inclinations, despite the question's description as eccentricities. Diagonalize the Laplace-Lagrange inclination matrix: let real eigenvectors ${\mathbf{v}}_k$ satisfy $B{\mathbf{v}}_k={\lambda }_k{\mathbf{v}}_k$, and form $V=({\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3)$. The physical matrix is diagonalizable by positive diagonal symmetrization of a matrix, as shown below.

The matrix exponential solution isWrite $c_k=I_k{e}^{i{\gamma }_k}$ with $I_k\ge 0$, and define $I_{jk}=I_k{v}_{jk}$. This givesThe eigenvalues determine nodal precession frequencies, the eigenvectors determine relative inclinations within each secular eigenmode, and the initial complex inclinations determine their amplitudes and phases. The real $I_{jk}$ can have either sign, encoding opposite nodal phases. The zero-frequency mode is a common fixed tilt.

Let a real tridiagonal matrix have diagonal entries $a_i$, positive upper entries $b_i$, and positive lower entries $c_i$. Set $d_1=1$, $d_{i+1}=d_i\sqrt{c_i/b_i}$ and $D=\mathrm{diag}(d_i)$. Then $D^{-1}AD$ is a symmetric matrix with diagonal $a_i$ and adjacent entries $\sqrt{b_i{c}_i}$. Indeed its two adjacent entries are $b_i{d}_{i+1}/d_i$ and $c_i{d}_i/d_{i+1}$, which are equal by construction. All other zeros are preserved. This matrix similarity implies real eigenvalues and diagonalizability, by the spectral theorem for real symmetric matrices. No sign assumption on $a_i$ is needed. In the weighted formulation of positive diagonal symmetrization of a matrix, the positive diagonal matrix is $W=D^{-2}$.

ConsiderIts row sums vanish, so $(1,1,1{)}^T$ is an eigenvector for the zero eigenvalue. Its characteristic polynomial iswith sum of principal minors of order twoThe other eigenvalues are $(-S\pm \sqrt{S^2-4T})/2$. No symmetry is required for this formula; positive diagonal symmetrization of a matrix is a sufficient condition for real roots.