The Matrix Exponential

The matrix exponential $e^{At}$ is the fundamental solution operator for constant-coefficient linear systems. Its computation requires understanding the Jordan normal form of $A$ .

Definition and Computation

Definition5.3Matrix exponential

For an $n \times n$ matrix $A$ , the matrix exponential is

$e^{At} = \sum_{k=0}^{\infty}\frac{(At)^k}{k!} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!} + \cdots$

This series converges for all $t$ and all $A$ . The function $\Phi(t) = e^{At}$ satisfies $\Phi' = A\Phi$ , $\Phi(0) = I$ .

Theorem5.3Computation via diagonalization

If $A = PDP^{-1}$ where $D = \text{diag}(\lambda_1, \ldots, \lambda_n)$ , then

$e^{At} = Pe^{Dt}P^{-1} = P\,\text{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t})\,P^{-1}.$

More generally, if $A = PJP^{-1}$ where $J$ is the Jordan form, then $e^{At} = Pe^{Jt}P^{-1}$ .

ExampleMatrix exponential of a $2 \times 2$ matrix

For $A = \begin{pmatrix}2 & 1\\0 & 2\end{pmatrix}$ (Jordan block with $\lambda = 2$ ):

$A = 2I + N$ where $N = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $N^2 = 0$ .

$e^{At} = e^{2t}e^{Nt} = e^{2t}(I + Nt) = e^{2t}\begin{pmatrix}1 & t\\0 & 1\end{pmatrix}.$

Properties

RemarkKey properties

$e^{At}$ is always invertible: $(e^{At})^{-1} = e^{-At}$ .
$\frac{d}{dt}e^{At} = Ae^{At}$ .
If $A$ and $B$ commute ( $AB = BA$ ): $e^{(A+B)t} = e^{At}e^{Bt}$ .
$\det(e^{At}) = e^{\text{tr}(A)\cdot t}$ (always positive).
The eigenvalues of $e^{At}$ are $e^{\lambda_j t}$ where $\lambda_j$ are eigenvalues of $A$ .

Cayley-Hamilton Method

Theorem5.4Cayley-Hamilton approach

By the Cayley-Hamilton theorem, $e^{At}$ can be expressed as a polynomial in $A$ of degree at most $n-1$ :

$e^{At} = \alpha_0(t)I + \alpha_1(t)A + \cdots + \alpha_{n-1}(t)A^{n-1}$

where the coefficients $\alpha_k(t)$ are determined by the equations $e^{\lambda_j t} = \sum_{k=0}^{n-1}\alpha_k(t)\lambda_j^k$ for each eigenvalue $\lambda_j$ (with derivative conditions for repeated eigenvalues).

ExampleCayley-Hamilton computation

For $A = \begin{pmatrix}0&1\\-2&-3\end{pmatrix}$ with eigenvalues $\lambda = -1, -2$ :

$e^{At} = \alpha_0(t)I + \alpha_1(t)A$ where $e^{-t} = \alpha_0 - \alpha_1$ and $e^{-2t} = \alpha_0 - 2\alpha_1$ .

Solving: $\alpha_1 = e^{-t} - e^{-2t}$ , $\alpha_0 = 2e^{-t} - e^{-2t}$ .

$e^{At} = \begin{pmatrix}2e^{-t}-e^{-2t} & e^{-t}-e^{-2t}\\-2e^{-t}+2e^{-2t} & -e^{-t}+2e^{-2t}\end{pmatrix}.$