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Two linear operators $A$ and $B$ commute when $AB=BA$. Their matrix exponentials then satisfy $e^{A+B}=e^A{e}^B=e^B{e}^A$.

For $X=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$, the image of its matrix exponential is compact precisely when $a^2+bc<0$ or $X=0$. If $a^2+bc=-{\omega }^2<0$, then $e^{tX}=\cos (\omega t)I_2+\sin (\omega t)X/\omega$ and its image is conjugate to the circle group. A nonzero nilpotent generator or a generator with real nonzero eigenvalues produces an unbounded image.

Complex inclination packages small orbital inclination $I$ and longitude of ascending node $\Omega$. It differs from complex eccentricity. Linear Laplace-Lagrange secular theory gives $\dot{\mathbf{y}}=iB\mathbf{y}$, whose matrix exponential solution decomposes into nodal secular eigenmodes.

An invertible complex Jordan block $\lambda I+N$, with $N^s=0$, has logarithm $\ell I+{\sum }_{k=1}^{s-1}(-1{)}^{k+1}(N/\lambda {)}^k/k$, where $e^{\ell }=\lambda$. Applying this construction blockwise and conjugating back proves surjectivity of the matrix exponential on the complex general linear group. There is no globally single-valued continuous choice of logarithm.

For a Matrix Lie group $G$ with Lie algebra $\mathfrak{g}$, its exponential map is the restriction of the matrix exponential, $X\mapsto e^X$. It maps $\mathfrak{g}$ into $G$ and is a local diffeomorphism at zero, but it need not be surjective globally.

A real matrix with at least one simple negative eigenvalue cannot be a matrix exponential of a real matrix: a logarithm would commute with it and preserve each one-dimensional real negative-eigenvalue space, where exponentiation can only produce a positive eigenvalue. In particular $\mathrm{diag}(-2,-1/2)$ belongs to the connected special linear group ${\mathrm{SL}}_2(\mathbb{R})$ but has no real logarithm.

With Adjoint representation convention ${\mathrm{ad}}_X(Y)=[X,Y]$, the matrix exponential identity is $e^{-X}Y{e}^X=e^{-{\mathrm{ad}}_X}Y$. Inverse conjugation reverses product order, so it is a right action or antirepresentation, whereas ${\mathrm{Ad}}_g(Y)=gY{g}^{-1}$ is an ordinary left group representation. The first-order terms $Y\mp [X,Y]$ detect a sign mismatch immediately. Negating every adjoint generator without reversing the Lie bracket does not give a new Lie-algebra homomorphism.

For a real quadratic form $Q(v)=v^{T}Av$ with $A$ a symmetric matrix, its stabilizer is $G_Q=\{g\in {\mathrm{GL}}_n(\mathbb{R}):g^{T}Ag=A\}$. Its Lie algebra consists exactly of $X$ satisfying $X^{T}A+AX=0$. Differentiation proves necessity; the matrix exponential proves sufficiency because $e^{t{X}^T}A{e}^{tX}=A$. The result holds even when the quadratic form is degenerate.

For bounded matrices $A_j$, the Lie product formula gives operator norm convergenceFor two summands, expanding their matrix exponentials gives the one-step difference below. For Hermitian matrices multiplied by $-i$, the factors are unitary operators and the first-order unitary product-formula error bound gives an explicit error estimate.

The matrix logarithm near the identity maps $Sp(n)=\{A\in U(2n):AJ{A}^t=J\}$ onto an open set of its real compact symplectic Lie algebra $\{X:X^{\ast }=-X,XJ+J{X}^t=0\}$. Its inverse is the matrix exponential. These statements follow by applying the logarithm to $A^{\ast }=A^{-1}$ and $A^t=J^{-1}{A}^{-1}J$, and differentiating $e^{tX}J{e}^{t{X}^t}$. Left translations provide manifold charts at every group element. In block form $X=\left(\begin{array}{cc}P& Q\\ -\bar{Q}& \bar{P}\end{array}\right)$ with $P^{\ast }=-P$ and $Q^t=Q$, giving real dimension $n^2+n(n+1)=n(2n+1)$.

The determinant is a smooth Lie group homomorphism from the real general linear group to ${\mathbb{R}}^{\times }$. Its derivative at the identity is the trace, and the flow-defined exponential on ${\mathbb{R}}^{\times }$ is the ordinary scalar exponential. Naturality of the Lie group exponential therefore proves the identity without diagonalizability assumptions. The series ${\sum }_{k\ge 0}{t}^k{A}^k/k!$ is the flow-defined matrix exponential because it solves $E^{\prime }=EA$, $E(0)=I$, and has inverse $E(-t)$.

If a real square matrix satisfies $M^2=-I$, split the absolutely convergent matrix exponential into even and odd powers to obtain $e^{tM}=I\cos t+M\sin t$. It is the fundamental matrix for ${\mathbf{x}}^{\prime }=M\mathbf{x}$. Consequently every solution has period $2\pi$, and the eigenvalues over the complex numbers belong to $\{i,-i\}$.

A matrix Lie group is a subgroup of a general linear group that is also a Lie group for its matrix topology. Near the identity, the matrix exponential and matrix logarithm give inverse coordinate charts whenever they are restricted to sufficiently small neighborhoods in the group and its tangent space.

For a positive-definite Hermitian matrix with spectral decomposition $A=U\mathrm{diag}({\lambda }_j)U^{\ast }$, the matrix logarithm is $\log A=U\mathrm{diag}(\log {\lambda }_j)U^{\ast }$. More generally it is a branch of the inverse of the matrix exponential when such a branch exists.

For $\dot{Y}=K(t)Y$ with coefficient period $P$, the monodromy matrix maps $Y(0)$ to $Y(P)$. Its eigenvalues are the Floquet multipliers. For piecewise constant coefficients it is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. Its determinant is $\exp ({\int }_0^P\mathrm{tr}K(t)dt)$.

If a spatial method of lines has a Hermitian matrix $H$, its generator $-iH$ is a skew-Hermitian matrix. The matrix exponential $e^{-itH}$ is a unitary matrix, so it preserves the Euclidean norm and any constant-volume discrete L2 norm. Equivalently, differentiating the squared norm gives $2\mathrm{Re}(-i{U}^{\ast }HU)=0$. A real sampled potential preserves the Hermitian property of a symmetric discrete Laplacian. Subsequent time discretization must be assessed separately.

For the matrix exponential $A(t)=e^{Lt}$, maximizing the energy ratio over nonzero initial conditions givesThe Rayleigh quotient achieves its maximum at an eigenvector of $A^{\ast }A$ with largest eigenvalue, equivalently a right singular vector of $A$. Such an optimal disturbance is generally not an eigenvector of $L$.