Matrix exponential

For a square matrix $A$, its exponential is the convergent power seriesIt is the fundamental matrix for $y^{\prime }=Ay$ and satisfies $e^{(s+t)A}=e^{sA}{e}^{tA}$.

A matrix exponential defines a positive matrix semigroup if $e^{tA}$ maps nonnegative vectors to nonnegative vectors for all $t\ge 0$. Here the inequality means entrywise nonnegativity, with zero entries allowed. In finite dimensions this holds exactly when the real matrix $A$ has nonnegative off-diagonal entries: if $A+\gamma I\ge 0$, expand $e^{tA}=e^{-\gamma t}{\sum }_{k\ge 0}{t}^k(A+\gamma I{)}^k/k!$; conversely, the off-diagonal entries of $e^{tA}=I+tA+O(t^2)$ force $a_{ij}\ge 0$ for $i\ne j$. Zero column sums additionally preserve $1^{T}c$ under $c^{\prime }=Ac$.

For finite matrices, or formally on a common invariant domain where the expansion is meaningful, exponential conjugation is generated by the commutator map. Differentiating $F(s)=e^{sS}H{e}^{-sS}$ gives $F^{\prime }=[S,F]$, and its Taylor series yields the nested-commutator expansion. An anti-Hermitian $S$ makes this a unitary transformation. For perturbation theory this may be used order by order without asserting convergence for arbitrary unbounded operators.

When a square matrix satisfies $M^2=\kappa I$, its powers reduce to $M^{2n}={\kappa }^{n}I$ and $M^{2n+1}={\kappa }^{n}M$. Absolute convergence of the matrix exponential allows even and odd subseries to be separated. Thus $C(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n)!$ and $S(\kappa )={\sum }_{n\ge 0}{\kappa }^n/(2n+1)!$. For real $\kappa >0$, these are $\cosh \sqrt{\kappa }$ and $\sinh \sqrt{\kappa }/\sqrt{\kappa }$. For $\kappa <0$, they are $\cos \sqrt{-\kappa }$ and $\sin \sqrt{-\kappa }/\sqrt{-\kappa }$. At $\kappa =0$, $e^M=I+M$, including nonzero nilpotent matrices of square zero. This also explains the limiting sine-over-argument coefficients without dividing by zero.

If every eigenvalue of a finite square matrix $A$ has negative real part, its matrix exponential decays and its integral converges. The Jordan normal form shows that polynomial factors in $t$ are dominated by exponential decay. Integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, proving the displayed matrix inverse formula. In particular, for a Hermitian matrix with negative eigenvalues, the spectral theorem reduces the integral to scalar exponentials with integrals $-1/{\lambda }_j$. The minus sign is essential: the integral is positive definite in this Hermitian case, while $A^{-1}$ has negative eigenvalues.

A Hermitian matrix has a unitary spectral decomposition $A=Q\mathrm{diag}({\lambda }_j)Q^{†}$ with real eigenvalues ${\lambda }_j$. The matrix exponential respects similarity, giving $e^A=Q\mathrm{diag}(e^{{\lambda }_j})Q^{†}$. Its eigenvalues are positive and its conjugate transpose equals itself. In particular $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$ for $v\ne 0$, so it is a Hermitian positive-definite matrix.

If $A$ is Skew-Hermitian, then $-iA$ is a Hermitian matrix. The finite-dimensional spectral theorem gives $A=Q\mathrm{diag}(i{\theta }_j)Q^{†}$, with real ${\theta }_j$ and a unitary matrix $Q$. Since the matrix exponential respects similarity, $e^A=Q\mathrm{diag}(e^{i{\theta }_j})Q^{†}$. Each diagonal entry has modulus one, so $e^A$ is a unitary matrix. Thus a skew-Hermitian matrix generates norm-preserving linear evolution.

For an invertible matrix $P$, induction gives $(PA{P}^{-1}{)}^k=P{A}^k{P}^{-1}$, including $k=0$. The power series for the matrix exponential is absolutely convergent in every matrix norm, so multiplication by the fixed matrices $P,P^{-1}$ can be taken through the sum. This proves the displayed identity without any diagonalizability assumption. In particular, if $A=PD{P}^{-1}$ with $D$ diagonal, then $e^A=P\mathrm{diag}(e^{D_{11}},\dots ,e^{D_{nn}})P^{-1}$.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

This formula allows a matrix and its derivative to fail to commute. Differentiate the absolutely convergent power series of the matrix exponential to obtain ${\sum }_{k\ge 1}k{!}^{-1}{\sum }_{j=0}^{k-1}{f}^j{f}^{\prime }{f}^{k-1-j}$. Expanding the integral gives the same coefficients, since ${\int }_0^1(1-u{)}^a{u}^{b}du=a!b!/(a+b+1)!$. The formula reduces to $e^f{f}^{\prime }$ when $[f,f^{\prime }]=0$.

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 $\{i,-i\}$.

The Baker--Campbell--Hausdorff formula expresses the local matrix logarithm of a product of exponentials as

On its half-plane of convergence,Indeed, applying the transform to $y^{\prime }=Ay$, $y(0)=y_0$, gives $(sI-A)Y=y_0$ for every $y_0$.