the semigroup property
/ SEM-ee-group /
Common sense says that letting a system evolve for three seconds, and then for two more, should land you exactly where letting it evolve for five seconds straight would. The starting point should not matter how you slice the waiting. The semigroup property is the precise statement that the matrix exponential respects this: evolving by time t and then by time s composes into evolving by time t + s.
In symbols, e^(A(t+s)) = e^(At) e^(As) for all t and s. This is the exact echo of the scalar law e^(a(t+s)) = e^(at) e^(as) — the exponential turns addition of times into multiplication of operators. The proof rests on a quiet but essential fact: A commutes with itself, so the cross-terms in the product of the two series line up just as they do for numbers (the binomial theorem goes through), and the two sides agree term by term. Because A always commutes with A, e^(At) and e^(As) also commute with each other, and the family of maps for all t forms a smooth one-parameter group: e^(A0) = I is the do-nothing element, and e^(-At) is the exact inverse that runs the flow backward.
The reason it is called a semigroup and not just a group is a matter of habit from harder settings: for many evolution problems (like heat diffusion) you can only run time forward, never backward, so only nonnegative t makes sense and there is no inverse. For the matrix exponential you actually do have inverses, so it is a full group; the name 'semigroup' is inherited from that broader theory. The deeper warning lurks here too: e^(A) e^(B) equals e^(A+B) only when A and B commute. The clean law works for e^(At) and e^(As) precisely because both are built from the same A.
With the rotation A = [0, 1; -1, 0], e^(At) rotates the plane by angle t. The semigroup law e^(A(t+s)) = e^(At) e^(As) is just the everyday fact that rotating by t then by s is rotating by t + s. But for two different matrices it can fail: e^(A) e^(B) does not equal e^(A+B) in general, exactly because rotating about one axis then another is not the same as one combined rotation when the axes differ.
Addition of times becomes multiplication of operators — but only down a single matrix's own timeline, where everything commutes.
e^(A+B) = e^(A) e^(B) is false in general — it holds only when A and B commute (AB = BA); the semigroup law survives only because e^(At) and e^(As) share the same A.