Matrix Exponential — Mathematics Knowledge System

Overview

The matrix exponential e^A extends the scalar exponential to square matrices via a power series. It is the cornerstone of continuous-time linear dynamical systems: the solution to x'(t) = Ax with x(0) = x_0 is x(t) = e^{At} x_0. For diagonalizable matrices, e^A is computed efficiently using eigendecomposition; for general matrices, Jordan form is used.

Intuition

For a scalar a, e^a = 1 + a + a^2/2! + .... The same series defines e^A for matrices. The key property is that e^A 'exponentiates' the linear vector field Ax: just as e^{at} solves dx/dt = ax, e^{At} solves the matrix ODE. For diagonal A = diag(λ_1,...,λ_n), we get e^A = diag(e^{λ_1},...,e^{λ_n}) — each eigenvalue exponentiates independently.

Formal Definition

Definition

For a square matrix A ∈ M_n(C), the matrix exponential is:

e^A = \sum_{k=0}^{\infty} \frac{A^k}{k!} = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots

This series converges for every square matrix A.

Matrix exponential (power series)

\text{If } A = P D P^{-1}: \quad e^A = P\, e^D P^{-1}, \quad e^D = \operatorname{diag}(e^{\lambda_1},\ldots,e^{\lambda_n})

Diagonalizable case

\frac{d}{dt} e^{At} = A e^{At}, \quad e^{A \cdot 0} = I

ODE property

Notation

Notation	Meaning
$e^A$	Matrix exponential of A
$\exp(A)$	Alternative notation for e^A

Properties

Always invertible

(e^A)^{-1} = e^{-A}.

Determinant

\det(e^A) = e^{\operatorname{tr}(A)}.

Commutativity

\text{If } AB = BA, \text{ then } e^{A+B} = e^A e^B. \text{ (This fails if } AB \neq BA.)

Nilpotent case

\text{If } N^k = 0, \text{ then } e^N = I + N + N^2/2! + \cdots + N^{k-1}/(k-1)! \text{ (finite sum).}

Applications

Real-world · Engineering

Control systems: the state transition matrix e^{At} describes how a linear system evolves. Stability requires all eigenvalues of A to have negative real parts.

Worked Examples

1
A is diagonal, so e^A = diag(e^{λ_1}, e^{λ_2}).
$e^A = \begin{pmatrix}e^1 & 0 \\ 0 & e^2\end{pmatrix} = \begin{pmatrix}e & 0 \\ 0 & e^2\end{pmatrix}$

✓ Answer

e^A = diag(e, e^2).

Practice Problems

Mediumfree response

Compute e^A for A = [[2,1],[0,2]] using the Jordan block formula.

Mediumapplication

The system x'(t) = Ax(t) with A = [[-1,0],[0,-2]] and x(0) = (3, 1)^T. Find x(t).

Common Mistakes

Common Mistake

e^{A+B} = e^A e^B always.

This identity holds only when AB = BA. In general, e^{A+B} ≠ e^A e^B for non-commuting matrices (the Baker-Campbell-Hausdorff formula gives the correction).

Quiz

Summary

The matrix exponential e^A = Σ A^k/k! converges for every square matrix A.
For diagonal A = diag(λ_i): e^A = diag(e^{λ_i}). For diagonalizable A = PDP^{-1}: e^A = P e^D P^{-1}.
Key ODE: d/dt e^{At} = A e^{At}; solution to x' = Ax is x(t) = e^{At} x_0.
det(e^A) = e^{tr(A)}; e^A is always invertible with (e^A)^{-1} = e^{-A}.
e^{A+B} = e^A e^B only when AB = BA.

References

WebsiteWikipedia — Matrix exponential