Matrix Exponential

Let \(\|\cdot\|\) be a submultiplicative norm on \(M_n(\mathbb{C})\). By induction, \(\|A^k\| \leq \|A\|^k\) for all \(k \geq 1\). Therefore: \[ \begin{align*} \sum_{k=0}^{\infty} \left\|\frac{A^k}{k!}\right\| &\leq \|I\| + \sum_{k=1}^{\infty} \frac{\|A\|^k}{k!} \\\\ &= \|I\| - 1 + e^{\|A\|} \\\\ &\lt \infty. \end{align*} \]

The bound is finite because the scalar exponential series converges at every real number, in particular at \(\|A\|\). By the comparison test, the matrix series converges absolutely. Since \(M_n(\mathbb{C})\) is a finite-dimensional (hence complete) normed vector space, absolute convergence implies convergence.

(a) Immediate: \(\exp(0) = I + 0 + 0 + \cdots = I\).

(b) Assume \(AB = BA\). Then \(A\) and \(B\) generate a commutative subalgebra of \(M_n(\mathbb{C})\), so the binomial theorem applies: \[ (A + B)^k = \sum_{j=0}^{k} \binom{k}{j} A^j B^{k-j}. \] Therefore: \[ \begin{align*} \exp(A + B) &= \sum_{k=0}^{\infty} \frac{(A+B)^k}{k!} \\\\ &= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{A^j}{j!} \cdot \frac{B^{k-j}}{(k-j)!} \\\\ &= \left(\sum_{j=0}^{\infty} \frac{A^j}{j!}\right)\left(\sum_{m=0}^{\infty} \frac{B^m}{m!}\right) \\\\ &= \exp(A)\exp(B). \end{align*} \] The interchange of summation order is justified by absolute convergence.

(c) Since \(A\) and \(-A\) commute, property (b) gives \(\exp(A)\exp(-A) = \exp(A + (-A)) = \exp(0) = I\). Similarly, \(\exp(-A)\exp(A) = I\). Therefore \(\exp(A)\) is invertible with inverse \(\exp(-A)\).

(d) When \(A\) is diagonalizable, say \(A = P \,\mathrm{diag}(\lambda_1, \ldots, \lambda_n)\, P^{-1}\), property (e) gives \(\exp(A) = P\,\mathrm{diag}(e^{\lambda_1}, \ldots, e^{\lambda_n})\,P^{-1}\), so \[ \det(\exp(A)) = \prod_{i=1}^{n} e^{\lambda_i} = e^{\sum_i \lambda_i} = e^{\mathrm{tr}(A)}. \] The general case follows by a density argument. Diagonalizable matrices are dense in \(M_n(\mathbb{C})\) (any matrix can be approximated arbitrarily well by one with distinct eigenvalues, which is diagonalizable), and both \(\det(\exp(\cdot))\) and \(\exp(\mathrm{tr}(\cdot))\) are continuous functions of \(A\).

(e) By induction, \((PAP^{-1})^k = PA^kP^{-1}\) for all \(k \geq 0\). Therefore: \[ \begin{align*} \exp(PAP^{-1}) &= \sum_{k=0}^{\infty} \frac{(PAP^{-1})^k}{k!} \\\\ &= P\left(\sum_{k=0}^{\infty} \frac{A^k}{k!}\right)P^{-1} \\\\ &= P\exp(A)P^{-1}. \end{align*} \]

In parts (a) and (b) the matrix \(A\) is real, and then so is \(\exp(A)\), because every partial sum of the exponential series is real and \(M_n(\mathbb{R})\) is closed in \(M_n(\mathbb{C})\).

(a) By property (d) of the matrix exponential: \(\det(\exp(A)) = \exp(\mathrm{tr}(A)) = \exp(0) = 1\). For complex \(A\) the same computation gives \(\exp(A) \in SL(n, \mathbb{C})\).

(b) If \(A^\top = -A\), then: \[ \exp(A)^\top = \exp(A^\top) = \exp(-A) = \exp(A)^{-1}. \] Hence \(\exp(A)^\top \exp(A) = I\), so \(\exp(A) \in O(n)\). Moreover, \(\det(\exp(A)) = \exp(\mathrm{tr}(A)) = \exp(0) = 1\) (the diagonal entries of a skew-symmetric matrix are all zero, so \(\mathrm{tr}(A) = 0\)), and therefore \(\exp(A) \in SO(n)\).

(c) If \(A^* = -A\), then \(\exp(A)^* = \exp(A^*) = \exp(-A) = \exp(A)^{-1}\), so \(\exp(A) \in U(n)\).

(d) Combine (c) with the computation in (a), which gives \(\det(\exp(A)) = 1\) for complex \(A\) as well.

The key observation is that \(\hat{\mathbf{n}}_\times\) satisfies the identity \(\hat{\mathbf{n}}_\times^{\,3} = -\hat{\mathbf{n}}_\times\). To see why, recall that \(\hat{\mathbf{n}}_\times \mathbf{v} = \hat{\mathbf{n}} \times \mathbf{v}\) extracts the component of \(\mathbf{v}\) perpendicular to \(\hat{\mathbf{n}}\) and rotates it by \(90^\circ\) about \(\hat{\mathbf{n}}\). Applying this twice reverses the perpendicular component (a \(180^\circ\) rotation), and applying it a third time yields a \(270^\circ\) rotation, the negative of the original \(90^\circ\) rotation. Hence \(\hat{\mathbf{n}}_\times^{\,3} = -\hat{\mathbf{n}}_\times\).

Algebraically, this is the Cayley-Hamilton theorem applied to \(\hat{\mathbf{n}}_\times\). As a \(3 \times 3\) real skew-symmetric matrix with eigenvalues \(0, +i, -i\), it has characteristic polynomial \(\lambda^3 + \lambda\). Substituting the matrix into its own characteristic polynomial gives \(\hat{\mathbf{n}}_\times^{\,3} + \hat{\mathbf{n}}_\times = 0\). The geometric and algebraic viewpoints are two faces of the same identity. This gives a cyclic pattern: \[ \begin{align*} \hat{\mathbf{n}}_\times^{\,3} &= -\hat{\mathbf{n}}_\times, \\\\ \hat{\mathbf{n}}_\times^{\,4} &= -\hat{\mathbf{n}}_\times^{\,2}, \\\\ \hat{\mathbf{n}}_\times^{\,5} &= \hat{\mathbf{n}}_\times, \quad \ldots \end{align*} \]

Substituting into the exponential series and grouping by powers of \(\hat{\mathbf{n}}_\times\): \[ \begin{align*} \exp(\theta\,\hat{\mathbf{n}}_\times) &= I + \theta\,\hat{\mathbf{n}}_\times + \frac{\theta^2}{2!}\hat{\mathbf{n}}_\times^{\,2} + \frac{\theta^3}{3!}\hat{\mathbf{n}}_\times^{\,3} + \frac{\theta^4}{4!}\hat{\mathbf{n}}_\times^{\,4} + \cdots \\\\ &= I + \left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)\hat{\mathbf{n}}_\times + \left(\frac{\theta^2}{2!} - \frac{\theta^4}{4!} + \frac{\theta^6}{6!} - \cdots\right)\hat{\mathbf{n}}_\times^{\,2} \\\\ &= I + \sin\theta\hat{\mathbf{n}}_\times + (1 - \cos\theta)\hat{\mathbf{n}}_\times^{\,2}. \end{align*} \]

Let \(\gamma : \mathbb{R} \to G\) be a one-parameter subgroup. A continuous one-parameter subgroup of a matrix Lie group is automatically smooth. For matrix groups this can be established directly by an analytic argument on the exponential series. The same conclusion follows from Cartan's Closed Subgroup Theorem, which makes \(G\) a smooth embedded submanifold of \(GL(n, \mathbb{C})\) on which every continuous group homomorphism is smooth.

Differentiating the identity \(\gamma(t + s) = \gamma(t)\gamma(s)\) with respect to \(s\) at \(s = 0\) gives: \[ \gamma'(t) = \gamma(t)\,\gamma'(0) = \gamma(t)\,A \] where \(A = \gamma'(0)\). This is the matrix ODE \(\gamma'(t) = \gamma(t)\,A\) with initial condition \(\gamma(0) = I\). Since \(\exp(tA)\) also satisfies this ODE (as \(\frac{d}{dt}\exp(tA) = \exp(tA)\,A\)), uniqueness of solutions gives \(\gamma(t) = \exp(tA)\).

The three curves \[ \begin{align*} \gamma_1(t) &= \exp(t\,E_1), \\\\ \gamma_2(t) &= \exp(t\,E_2), \\\\ \gamma_3(t) &= \exp(t\,E_3) \end{align*} \] are one-parameter subgroups of \(SO(3)\), where \(E_1, E_2, E_3\) are the infinitesimal generators defined in the previous section. Each \(\gamma_i\) traces out a great circle in \(SO(3)\), the family of all rotations about a single coordinate axis.

An arbitrary one-parameter subgroup has the form \(\gamma(t) = \exp(t(a_1 E_1 + a_2 E_2 + a_3 E_3))\), and for \((a_1, a_2, a_3) \neq \mathbf{0}\) it corresponds to rotation about the axis \((a_1, a_2, a_3)\) at rate \(\|(a_1, a_2, a_3)\|\).

If \(G\) is a finite group (such as the dihedral group \(D_n\) embedded as matrices in \(O(2)\)), then any continuous homomorphism \(\gamma : \mathbb{R} \to G\) must be constant. Indeed, \(\gamma(\mathbb{R})\) is a connected subset of \(G\) (as the continuous image of the connected space \(\mathbb{R}\)), but a finite group with the discrete topology has no connected subsets other than singletons. Since \(\gamma(0) = I\), we conclude \(\gamma(t) = I\) for all \(t\).

This argument is the rigorous version of an observation from Geometry of Symmetry. The dihedral group \(D_n\) has "no in-between states." One cannot continuously interpolate between a rotation by \(0^\circ\) and a rotation by \(360^\circ/n\) while staying inside \(D_n\). The group \(SO(2)\), by contrast, has a one-parameter subgroup for every real number \(\theta\), reflecting the fact that rotation by any angle is a valid symmetry.