Bielefeld, June 1997

Reihenpower.

Many applications involve differential equations like dX/dt = AX, where A is a given matrix and X = X(t) is a variable vector having a given initial value X(0) = V. In analogy to the well-known scalar case, it is tempting to write down something like X(t) = e^{At}V, where

(∗)  e^{At} = I + At + A^{2} \frac{t^{2}}{2!} + A^{3} \frac{t^{3}}{3!} + ⋯,

but does that make sense? Differentiated formally, the expression E(t) on the right of (∗) certainly seems to have the requisite properties, namely

(∗∗)  \frac{d}{dt}E(t) = AE(t) and E(0) = I,

but what does it mean? To compute e^{At} you are summing a bunch of matrices, adding A^{k}t^{k}/k! at the k-th stage of the calculation. The entry in row i and column j of your sum therefore accumulates the terms of the power series

a_{ij}^{(0)} + a_{ij}^{(1)}t + \frac{a_{ij}^{(2)}}{2!} t^{2} + \frac{a_{ij}^{(3)}}{3!} t^{3} + ⋯,

where a_{ij}^{(k)} denotes the ij-entry of A^{k}. Nothing wrong, so far. If α is a number greater than 2|a_{ij}| for all ij, an easy induction shows that |a_{ij}^{(k)}| < α^{k} for all k (see?). Hence all these power series converge wherever the one for e^{αt} does — i.e., everwhere — and uniformly on any finite closed interval −T ≤ t ≤ T. Each entry of e^{At} is therefore a continuous function of t.

Suppose someone hands you a variable matrix E(t) and a constant matrix A, for instance

E(t) = e^{t} [ \begin{matrix}1 + t & t \\ −t & 1 − t \end{matrix} ] , A = [ \begin{matrix}2 & 1 \\ −1 & 0 \end{matrix} ] ,

and asks whether E(t) equals e^{At}. To answer the question in the affirmative, all you have to do is check the conditions (∗∗) — please try this! In other words, e^{At} is the only E(t) satisfying those conditions. To see why this works in general, just differentiate the product e^{−At}E(t), getting −Ae^{−At}E(t) + e^{−At}AE(t) = 0, and conclude that e^{−At}E(t) must be constant = I. Note that this argument depends on the fact — obvious from (∗) — that A commutes with e^{−At}.

The same kind of reasoning can be applied to the product e^{Bt}E(t), where E(t) is as above and A commutes with B (and hence with e^{Bt}). Differentiation yields

\frac{d}{dt}[e^{Bt}E(t)] = Be^{Bt}E(t) + e^{Bt}AE(t) = (B + A)e^{Bt}E(t).

Since e^{Bt}E(t) satisfies the analogue of (∗∗) for B + A, it must be equal to e^{(B+A)t}. In symbols:

(†)  e^{Bt}e^{At} = e^{Bt+At}

as long as (remember!) AB = BA. As it stands, (†) can be seen as an equation between formal power series, but nothing stops us from substituting actual numbers for the indeterminate t. Given two numbers u and v (say, u ≠ 0), let us take B = (v/u)A and get

(‡)  e^{Au}e^{Av} = e^{A(u+v)} ,

by setting t = u in (†). In spite of its similar appearance, the new equation (‡) lives in a different world: it is not about power series but about the continuous matrix-valued function F (u) = e^{Au} of the real variable u. To see that this function is even differentiable, consider the power series

\frac{e^{At} − I}{t} = A + A^{2} \frac{t}{2!} + A^{3} \frac{t^{2}}{3!} + A^{4} \frac{t^{3}}{4!} + ⋯,

which converges about as nicely as e^{A}t itself (see?), and hence defines a continuous matrix-valued function G(u) on the real line satisfying G(0) = A. Applying (‡), we now have

lim_{h→0} \frac{F (u + h) − F (u)}{h} = lim_{h→0} \frac{F (h) − I}{h} · F (u) = G(0)F (u) ,

which says that Ḟ (u) = AF (u). Since this operation can be repeated indefinitely, F (u) is a smooth function. Conclusion:

For any A, the matrix-valued function F_{A}(u) = e^{Au} defines a smooth homomorphism from the additive group of real numbers into the group of invertible 2 × 2 real matrices.

As an example, consider A = J. Then, since J^{2m} = ±I and J^{2m+1} = ±J, according to whether m is even or odd, we get

e^{Jt} = (1 − \frac{t^{2}}{2!} + \frac{t^{4}}{4!} − \frac{t^{6}}{6!} + ⋯) I + (t − \frac{t^{3}}{3!} + \frac{t^{5}}{5!} − \frac{t^{7}}{7!} + ⋯) J .

This is a new and apparently effortless solution of the equation Q̇(t) = JQ(t) with which we struggled in “Kreisbewegung”. Because of the uniqueness of the latter, we can conclude that e^{Jt} = Q(t) parametrizes the whole 2 × 2 rotation group. A similar conclusion could be based on the continuous parametrization R(μ) obtained in “Bogenmessung”.

But what if we had not done that previous work? Could we make the present approach stand on its own feet? It worked so smoothly up to here, but now we are struck with blindness: the power series do not easily tell us what kind of functions they represent.

To avoid any confusion, let us give our new solution a new name: e^{Jt} = ξ(t)I +η(t)J = P (t). The differential equation Ṗ (t) = JP (t), quickly resolves into ξ̇(t) = −η(t) and η̇(t) = ξ(t), which — multiplied crosswise — again yield the message that ξ(t)^{2} + η(t)^{2} is constant = 1. Good, at least we are landing in the rotation group! But how do we know that we get all rotations? To obtain some numerical information, note that

η(t) = t − \frac{t^{3}}{3!} + (\frac{t^{5}}{5!} − \frac{t^{7}}{7!} ) + ⋯ > t(1 − \frac{t^{2}}{6} ) > 0 for 0 < t < \sqrt{6} ,

because in this range 1 > t^{2}/(n+1)(n+2) for n ≥ 5 (see?). This estimate has two consequences. Putting t = 1, we get η(1) > 5/6 > 1/\sqrt{2}, whence ξ(1) < 1/\sqrt{2} < η(1). Since P (2) = P (1)^{2} by the functional equation (‡), we now have ξ(2) = ξ(1)^{2} − η(1)^{2} < 0, and therefore ξ(λ) = 0 for some 0 < λ < 2. This is the first consequence of our estimate. The second one is that η(λ) — being positive — must equal 1, and hence P (λ) = J.

Since η(−t) = −η(−t), the range of η(t) for −λ ≤ t ≤ λ must therefore be the whole interval −1 ≤ s ≤ 1 — and the point P (t) = (ξ(t), η(t)) has no choice but to cover the first and fourth quadrants of the unit circle. The rule P (t + λ) = JP (t) does the rest. All that remains to do is to give the functions ξ(t) and η(t) their usual names cos(t) and sin(t). To compute λ alias π/2 — which played the role of κ in “Kreisbewegung” — remember the final exercise ibidem.

Differentiation. A special feature of the preceding arguments is that all differentiations occurring proir to (‡) were formal — based on dt^{n} = nt^{n−1}dt — as if we were dealing with polynomials (i.e., finite sums of monomials) instead of infinite series. For the latter it is not true that term-by-term differentiation always works. For instance, the series

∑_{n=1}^{∞} \frac{sin n^{2}t}{n^{2}}

converges uniformly on the whole real line (because | sin t| ≤ 1 always), hence defines a perfectly good continuous function f(t), but nobody seems to know whether f(t) is differentiable. One of the greatest virtues of power series is that their formal and functional derivatives coincide —

Slogan: power series can be differentiated term by term

— but it is not in the nature of mathematics to accept such declarations on faith. We have therefore avoided the use of this unproved slogan and worked with formal differentiation only. Its main property, the Product Rule, follows from the polynomial version by noting that the m-th term in the product of two power series does not know whether the higher terms of these series are zero (i.e., whether they are mere polynomials).

The proof of the “slogan” is not very difficult but somewhat circuitous. It depends on the uniform convergence of the formal derivative of a power series, which allows the (non-formal) integration of the latter (as a continuous function) to be performed term by term — the integral thereby becoming comparable to the original power series.

Generalizations. An interesting special case of (†) is e^{αt}e^{βAt} = e^{(αI+βA)t}, for scalars α and β. which is useful in the computation of e^{At}. Putting α = −\frac{1}{2} tr A, we get an A_{∘} = αI + A with zero trace, so that A_{∘}^{2} = δI for some δ, by Cayley-Hamilton. If δ ≠ 0, we divide A_{∘} by the square root of |δ| and get an A_{∗} whose square is = I, = 0, or = −I. In each of these three cases (assuming that A_{∗} is not itself equal to ±I or 0), it is not hard to find an invertible matrix M such that M^{−1}A_{∗}M has one of the standard forms

[ \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix} ] , [ \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ] , [ \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ] ,

and since M^{−1}e^{A_{∗}t}M = e^{M^{−1}A_{∗}Mt}, it suffices to solve the problem for each of these. The third of them is our J, the first two are even easier. Suitable matrices M = [V, W] can be found as follows.

a) If A^{2} = I, that is (A + I)(A − I) = 0, pick any vector U such that AU ≠ ±U and set V = (A − I)U, W = (A + I)U. Then A[V, W] = [−V, W] as desired.

b) If A^{2} = 0, pick any vector W such that AW ≠ 0 and put V = AW. Then A[V, W] = [0, V] as desired.

c) If A^{2} = −I, pick any vector V ≠ 0 and put W = AV. Then A[V, W] = [W, −V] as desired.