Linear Algebra

The Matrix Exponential

Solving a coupled linear system by exponentiating its matrix, computed through the eigenbasis.

Updated July 2026

A single linear equation $\overset{x}{˙} = a x$ has the solution everyone knows, $x (t) = e^{a t} x (0)$ . A coupled system of them,

\dot{x} = A x,

looks harder because the variables feed into each other. It is not, and the reason is that the same exponential works if you are willing to exponentiate a matrix.

Defining the exponential of a matrix

There is only one sensible definition, the power series, which converges for every square matrix:

e^{A} = n = 0 \sum \infty \frac{A ^{n}}{n !} = I + A + \frac{A ^{2}}{2 !} + \dots

With it, $e^{A t}$ has derivative $A e^{A t}$ , exactly as the scalar case does, so

x (t) = e^{A t} x (0)

solves the system for any initial condition. The whole difficulty has moved into one place: actually computing $e^{A t}$ .

Computing it through the eigenbasis

Summing an infinite series of matrix powers by hand is hopeless. The eigenbasis makes it trivial. If $A = P Λ P^{- 1}$ is diagonalisable, then $A^{n} = P Λ^{n} P^{- 1}$ , and the whole series collapses:

e^{A t} = P e^{Λ t} P^{- 1}, e^{Λ t} = diag (e^{λ_{1} t}, \dots, e^{λ_{n} t}) .

Exponentiating a diagonal matrix is just exponentiating its entries. So the recipe is: find the eigenvalues and eigenvectors, exponentiate the eigenvalues, transform back.

Consequence

The eigenvalues run the dynamics

In the eigenbasis the system uncouples into independent scalar equations $\overset{y}{˙}_{i} = λ_{i} y_{i}$ . Each eigenvalue is one decoupled mode: a negative real part decays, a positive one grows, an imaginary part oscillates. The long-term fate of the whole system is set by the eigenvalue with the largest real part, which is exactly the stability criterion for a linear system.

When there is no eigenbasis

Some matrices are defective: they do not have enough eigenvectors to diagonalise. The exponential still exists, since the series never cared, but you compute it through the Jordan form instead. A Jordan block splits as a multiple of the identity plus a nilpotent part $N$ , and a nilpotent's series terminates because $N^{k} = 0$ eventually:

e^{N t} = I + N t + \frac{N ^{2} t ^{2}}{2 !} + \dots + \frac{N ^{k - 1} t ^{k - 1}}{( k - 1 )!} .

The payoff is a genuine unification. Coupled oscillators, radioactive decay chains, the small-oscillation modes of the differential-equations notes, all of them are $\dot{x} = A x$ for some $A$ , and all of them are solved the same way: diagonalise, and the coupled problem falls apart into scalar exponentials you already understand.