Review / Part I — Explicit & Qualitative Methods / Ch 8

The Matrix Exponential

Ch 8 — Part I — Explicit & Qualitative Methods

For any square matrix \(A\), the matrix exponential \(e^{tA}\) generalizes the scalar exponential and gives the solution to \(\mathbf{x}' = A\mathbf{x}\).

Definition

$$e^{tA} = \sum_{k=0}^\infty \frac{(tA)^k}{k!} = I + tA + \frac{t^2 A^2}{2!} + \cdots$$ The series converges for every matrix \(A\).

Why it solves the system

\(\dfrac{d}{dt} e^{tA} = A e^{tA}\). Hence \(\mathbf{x}(t) = e^{(t-t_0)A} \mathbf{x}_0\) solves \(\mathbf{x}' = A \mathbf{x},\; \mathbf{x}(t_0) = \mathbf{x}_0\).

Computing it

If \(A = PDP^{-1}\) with \(D\) diagonal (or block-diagonal for complex/defective cases), then \(e^{tA} = P e^{tD} P^{-1}\), and \(e^{tD}\) is just exponentials of the diagonal entries.

For a \(2\times 2\) defective matrix with repeated eigenvalue \(\lambda\), use \(e^{tA} = e^{\lambda t}(I + t(A - \lambda I))\).

Practice Problems

Problem 1easy
Compute \(e^{tA}\) for \(A = \begin{pmatrix}2 & 0 \\ 0 & 5\end{pmatrix}.\)
Hint
Diagonal.
Solution

\(e^{tA} = \begin{pmatrix}e^{2t} & 0 \\ 0 & e^{5t}\end{pmatrix}\).

Answer: See solution.

Problem 2hard
Compute \(e^{tA}\) for \(A = \begin{pmatrix}3 & 1 \\ 0 & 3\end{pmatrix}.\)
Hint
Defective: \(\lambda = 3\) repeated. Use \(e^{tA} = e^{3t}(I + t(A-3I))\).
Solution

\(A - 3I = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\). So \(e^{tA} = e^{3t}\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}\).

Answer: See solution.

Problem 3medium
If \(A\mathbf{v} = \lambda \mathbf{v}\), what is \(e^{tA}\mathbf{v}\)?
Hint
Use the series.
Solution

\(e^{tA}\mathbf{v} = \sum_k \frac{t^k A^k \mathbf{v}}{k!} = \sum_k \frac{t^k \lambda^k \mathbf{v}}{k!} = e^{\lambda t}\mathbf{v}\).

Answer: \(e^{\lambda t}\mathbf{v}.\)