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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.

For distinct eigenvalues, the formula holds for $n=0,1$. If the lower-left entry of $L^n$ is $b_n=({\lambda }_1^n-{\lambda }_2^n)/({\lambda }_1-{\lambda }_2)$, multiplication by $L$ gives $b_{n+1}={\lambda }_1{b}_n+{\lambda }_2^n=({\lambda }_1^{n+1}-{\lambda }_2^{n+1})/({\lambda }_1-{\lambda }_2)$. This proves by mathematical induction thatSumming the absolutely convergent matrix exponential series givesFor the repeated eigenvalue ${\lambda }_1={\lambda }_2=\lambda$, the continuous limit is $A(t)=e^{\lambda t}\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. This is also the exponential of a two-dimensional Jordan block.

Termwise differentiation of the matrix exponential gives $A^{\prime }=LA$ and $A(0)=I$. Thus $x(t)=A(t)x(0)$ satisfies $x^{\prime }=Lx$ and the prescribed initial condition. Uniqueness for a linear differential equation makes it the solution.

Termwise differentiation of the matrix exponential givesThus $y(t)=e^{tA}{y}_0$ satisfies $y^{\prime }=Ay$ and $y(0)=y_0$. If two solutions existed, multiplying their difference by $e^{-tA}$ would give a vector with zero derivative and zero initial value, proving uniqueness.

For the inhomogeneous equation, multiply by the integrating factor $e^{-tA}$:Integration and multiplication by $e^{tA}$ give the variation-of-constants formula

For the stated matrix,soWriting $f(t)=v\sin t$ with $v=(1,3,0{)}^T$, one findsThereforeor explicitly

For $\epsilon =0$, $L^2=-(1-{\beta }^2)I=-{\Gamma }^{2}I$. The matrix exponential therefore gives

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.

Near $t=0$, take $\Omega (t)=\log F(t)$ to be the continuous local matrix logarithm with $\Omega (0)=0$. The symmetry identity gives $F(-t)=F(t{)}^{-1}$, and uniqueness of this logarithm yieldsThus $\Omega$ is an odd function. The local-logarithm qualification is necessary because the matrix exponential is not globally injective; the statement is naturally understood either near $t=0$ or as an identity of formal power series.

For the Strang splittingreversing $t$ reverses all three factors, soIt is therefore a time-symmetric numerical method. Multiplication of the three exponential series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the Baker--Campbell--Hausdorff formula gives an odd modified generatorfor a matrix $D$ made from nested commutators. Exponentiating givesfor a matrix $C$ depending on $A$ and $B$.

Direct multiplication gives $Q^2=-sQ$, hence for $n\ge 1$,Therefore the matrix exponential iswhich expands to exactly the transition matrix found in part (i).

If $X$ belongs to the Special orthogonal Lie algebra, which is also the Lie algebra of $O(3)$, then $e^{tX}\in O(3)$ near $t=0$. Differentiatingat zero gives $X^{\mathsf{T}}+X=0$. Conversely, the matrix exponential of every real skew-symmetric matrix is orthogonal. Hence

The first component obeys ${\dot{x}}_1={\lambda }_1{x}_1$, so $x_1(t)=e^{{\lambda }_{1}t}{x}_1(0)$. Variation of constants inthen givesHence the matrix exponential is

The hypotheses imply $a=1/5=0.2$ and $b=1/1.25=0.8$, so $a+b=1$. Conditional on the negative result at month 1, the likelihood is the probability of remaining in $S$ at month 2 and moving to $I$ by month 3. The supplied matrix exponential givesThe Markov property therefore gives

Diagonalize the real Laplace-Lagrange secular matrix $\mathbf{A}$. Its eigenvalues areand choose corresponding real eigenvectors ${\mathbf{v}}_k$. If $V=({\mathbf{v}}_1{\mathbf{v}}_2)$, the initial complex eccentricity vector determines complex mode coefficientsThe matrix exponential solution isThuswhere a negative eigenvector component may equivalently be made positive by adding $\pi$ to its phase. The $g_k$ are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and ${\beta }_k$ fixes the phase selected by the initial conditions.

Now $B^2={ϵ}^{2}I$, so splitting its matrix exponential into even and odd powers yields hyperbolic functions:The limiting coefficient at $ϵ=0$ is again one, and $T=I$. The eigenvectors $(1,1)$ and $(1,-1)$ have factors $e^{ϵ}$ and $e^{-ϵ}$, respectively, showing how this hyperbolic transformation differs from a rotation.

Since $A^2=-{ϵ}^{2}I$, its even powers are $A^{2k}=(-1{)}^k{ϵ}^{2k}I$ and its odd powers are $A^{2k+1}=(-1{)}^k{ϵ}^{2k}A$. Splitting the matrix exponential into these two absolutely convergent subseries gives the matrix exponential for a scalar square:At $ϵ=0$, the coefficient $\sin ϵ/ϵ$ has limiting value one and $A=0$, so $S=I$. The result is a planar rotation through angle $-ϵ$.

The displayed power series is the matrix exponential $G=e^H$. From $H^3=-{\alpha }^{2}H$, its odd powers and nonconstant even powers sum separately:The scalar power series are $\sin \alpha /\alpha$ and $(1-\cos \alpha )/{\alpha }^2$, with continuous values $1$ and $1/2$ at zero. Substituting $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$ and the preceding formula for $H^2\mathbf{x}$ yieldsThis is the Rodrigues rotation formula: the component parallel to $\mathbf{n}$ is fixed, while the perpendicular component undergoes a rotation through $\alpha$. For $\alpha =0$ the linear operator is the identity operator.

Induction on $p$ gives $S^p{\mathbf{u}}_i={\lambda }_i^p{\mathbf{u}}_i$, including $p=0$. The absolutely convergent matrix exponential series may therefore be applied term by term to each eigenvector:ThusFor a repeated eigenvalue, every nonzero vector in its eigenspace remains an eigenvector. Because the real exponential function is injective, distinct real eigenvalues of $S$ do not merge under this transformation.

Let $J=A/a$, so $J^2=-I$. The even and odd powers in the matrix exponential giveHere the two scalar sums are the Taylor series for cosine and sine. ConsequentlyWith the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by $-a$; the parameter in the form printed in the paper is $\theta =a$.

An eigenvalue $\lambda$ and associated eigenvector $e\ne 0$ satisfy $Ae=\lambda e$. The characteristic polynomial $\det (A-\lambda I)$ has degree equal to the positive matrix size. The fundamental theorem of algebra gives a complex root, at which $A-\lambda I$ is singular and therefore has a nontrivial kernel. Hence every nonempty square complex matrix has an eigenvector.

A nondiagonalizable example is the Jordan block $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: its only eigenvalue is zero, and its eigenspace is one-dimensional. A Hermitian matrix satisfies $A^{†}=A$ and has an orthonormal eigenbasis by the spectral theorem. One proof finds an eigenvector, notes that its orthogonal complement is invariant, and repeats. Its eigenvalues are real becausewhere $e^{†}e>0$.

With all eigenvalues negative, $A$ is invertible. Its same complete eigenbasis givesSome powered eigenvalues can coincide, but the original eigenvectors remain a complete basis. Writing $A=U\mathrm{diag}({\lambda }_i)U^{†}$, the convergent matrix exponential series givesThus the matrix inverse from a decaying matrix exponential isThe identity printed in the original PDF is missing a minus sign. Already for the one-by-one matrix $A=(-1)$ its integral is $1$, whereas its inverse is $-1$. Equivalently, integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, confirming the correction without diagonalization.

The convergent power series defining the matrix exponential gives $e^A=U{e}^D{U}^{†}$, with $e^D=\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})$. Hence the matrix exponential determinant identity isThe diagonalization proof uses the given Hermitian hypothesis; the identity itself holds for every square real or complex matrix.