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Defining e^(At)

The scalar equation x' = a x has solution x = e^(at) x0. Bottle that same miracle for a whole vector system x' = A x, and you get e^(At) — a matrix that is literally an exponential, born from the very same power series, and it turns out to be the flow that carries any starting state forward in time.

The one scalar fact worth copying

Go back to the simplest differential equation in the whole subject: x' = a x, a single number a and a single unknown x(t). Its solution is x(t) = e^(at) x0, where x0 = x(0). One scalar exponential carries the starting value forward forever — multiply x0 by e^(at) and you have the state at any time t. Everything in this guide is one stubborn act of imitation: we want that exact sentence, word for word, for a whole system x' = A x where x is now a vector and A a square matrix. The dream is a solution that reads x(t) = (something)(t) x0, with the same shape and the same magic.

In the previous guide you built a fundamental matrix for the homogeneous system: a matrix whose columns are independent solutions, so any solution is a combination of them. That already solves the system. But a generic fundamental matrix is awkward — its value at t = 0 is some arbitrary invertible matrix, so reading off the answer in terms of the starting state x0 takes an extra inverse. The matrix exponential is the fundamental matrix tuned to the cleanest possible normalization, and that normalization is exactly what makes it act like e^(at).

Defining e^(At) by the series that already worked

How do you even raise e to a matrix? You do not invent anything; you reuse the definition of e^x you already trust. Recall that for an ordinary number, e^x = 1 + x + x^2/2! + x^3/3! + ... . That infinite sum needs only three operations: adding, multiplying, and dividing by counting numbers. A square matrix can do all three — you can add matrices, multiply A by itself to get A^2, A^3, and so on, and divide by 2!, 3! — so the very same series is legal with A in place of x. That sum is the definition of the matrix exponential, written e^(At).

scalar :  e^(at)  =  I + (at) + (at)^2/2! + (at)^3/3! + ...      (a number)

matrix :  e^(At)  =  I + (At) + (At)^2/2! + (At)^3/3! + ...      (a matrix)

          where  I = identity matrix,  (At)^k = A^k t^k

properties :  e^(A*0) = I        (the t=0 term alone survives)
              d/dt e^(At) = A e^(At)   (differentiate the series term by term)
Same series, with the identity matrix I playing the role of the leading 1. The two boxed properties are everything.

Two facts fall straight out of that series and they are the whole point. First, at t = 0 every term with a t vanishes and only the identity I is left, so e^(A*0) = I — it starts as the do-nothing matrix. Second, differentiate the series term by term in t and you recover A times the same series, so d/dt e^(At) = A e^(At). Read those two lines together: the columns of e^(At) form a principal fundamental matrix — a fundamental matrix that equals I at t = 0. The first fact is the normalization the ordinary fundamental matrix lacked; the second says e^(At) genuinely solves x' = A x.

The solution formula we were chasing

Now put it together and claim the prize. Set x(t) = e^(At) x0 for a constant vector x0. Differentiate: x'(t) = (d/dt e^(At)) x0 = A e^(At) x0 = A x(t), so it solves the system. And at t = 0 it gives e^(A*0) x0 = I x0 = x0, so it hits the right starting state. That is the unique solution of the initial value problem x' = A x, x(0) = x0. Stand it next to the scalar line and the imitation is complete — same shape, same role, only the symbol e^(at) has grown into the matrix e^(At).

Why it deserves the name flow

The identity e^(At) e^(As) = e^(A(t+s)) reads better as a story than as algebra. Think of e^(At) as a machine that takes the system's state and advances it by t seconds. Run it for s seconds, then for t more seconds, and you have advanced by t + s seconds total — which is exactly what e^(A(t+s)) does in one go. Time-stepping composes by adding the times. That is why e^(At) is called the flow of the linear system: like a river, it carries every point steadily downstream, and where you end up depends only on how long you drift, not on whether you paused along the way.

This composition law has a formal name, the semigroup property, and the word semigroup is precise, not loose. You can always step forward and you can compose steps (that is what makes it a semigroup), and because A is a constant matrix you can also run time backward — e^(-At) is the genuine inverse of e^(At), undoing the drift. So here this flow is actually a full group, reversible in both directions. The reason it is usually called a semigroup is that the same construction for more general evolution (think heat spreading out, which cannot be reversed) only steps forward, and the shared name keeps the family together.

Same object, two more names

You will meet e^(At) wearing other hats, and it helps to recognize it through the disguise. In control engineering and signals it is called the state-transition matrix, often written Phi(t): the operator that transitions the system's state from time 0 to time t, the very same x(t) = Phi(t) x0. The name simply emphasizes the job rather than the formula. Whenever a book says state-transition matrix for a constant-coefficient linear system, mentally translate: that is e^(At), no more and no less.

And the connection back to any ordinary fundamental matrix is clean. If Psi(t) is any fundamental matrix you happened to build — columns independent solutions, but Psi(0) not necessarily the identity — then e^(At) = Psi(t) Psi(0)^(-1). The trailing factor Psi(0)^(-1) is just the rescaling that forces the value at t = 0 to become I. So the matrix exponential is not a new solution; it is the one fundamental matrix that has been normalized so that reading off x0 needs no extra inverse. Every fundamental matrix is a relabeled e^(At), and e^(At) is the tidiest member of the family.

What you now hold, and what comes next

Step back and see how much one object carries. e^(At) is defined by a series you already knew, it equals the identity at t = 0, it solves x' = A x, it gives the initial value problem its one-line answer x(t) = e^(At) x0, it composes by adding times, and it reverses cleanly. For the homogeneous constant-coefficient linear system, that genuinely is the entire story — the final guide of this rung makes that boast precise. What remains is mechanical: how to write e^(At) down for an actual matrix, and how to handle a push from outside.

  1. To define it: write e^(At) = I + At + (At)^2/2! + ... , the scalar exponential series with A in place of the number.
  2. To solve x' = A x with x(0) = x0: the answer is simply x(t) = e^(At) x0 — verify it by checking the two properties e^(A*0) = I and d/dt e^(At) = A e^(At).
  3. To combine flows: use e^(At) e^(As) = e^(A(t+s)) and e^(-At) = (e^(At))^(-1); never split e^(A+B) into a product unless AB = BA.
  4. To actually compute it: skip the series and turn to the next guide's methods (diagonalization, Jordan form, Cayley-Hamilton/Putzer, Laplace).