Matrix Exponentials and Repeated Modes

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The idea

The scalar exponential generalizes to $e^{At}=I+At+A^2t^2/2!+\cdots$ . For a constant matrix, differentiation gives $(e^{At})'=Ae^{At}$ and $e^{A0}=I$ . Thus $\mathbf x=e^{At}\mathbf x_0$ solves any homogeneous constant system.

Visual guide

VISUAL GUIDEA stable system can have temporary component growth

x₁ = te⁻ᵗ
x₂ = e⁻ᵗ

For the defective matrix [[−1, 1], [0, −1]] and initial state (0, 1), x₁ = te⁻ᵗ first rises, then decays; x₂ = e⁻ᵗ decreases throughout. The polynomial factor comes from the repeated mode, not an unstable eigenvalue.

Method and assumptions

If $A=PDP^{-1}$ , then $e^{At}=Pe^{Dt}P^{-1}$ . For a Jordan block $A=\lambda I+N$ with nilpotent $N$ , factor $e^{At}=e^{\lambda t}e^{Nt}$ ; the nilpotent series terminates. In general, $e^{A+B}=e^Ae^B$ requires commuting matrices.

Worked example: a defective matrix

For $A=\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$ , write $A=-I+N$ with $N^2=0$ . Then $e^{At}=e^{-t}\begin{pmatrix}1&t\\0&1\end{pmatrix}$ . Initial state $(0,1)$ yields $(te^{-t},e^{-t})$ , which cannot be produced by a single eigenvector alone.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

If N²=0, what is eᴺᵗ?

I+tN

I+N

Hint 1 · Find a starting point

Expand the matrix exponential as a power series.

Hint 2 · Take the next step

Every power N² and higher is zero.

Show the reasoning

Answer: I+tN

Only I and tN survive. The identity is necessary to give eᴺ⁰=I.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: rotation

For $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , $J^2=-I$ . Splitting the series into even and odd powers gives $e^{Jt}=I\cos t+J\sin t$ , the rotation matrix. State norm is preserved.

Interpreting the result

A negative repeated eigenvalue can produce a transient polynomial factor before eventual decay. A plot over a short interval may show growth in one component even though all states eventually approach zero.

Practice

What is $e^{A0}$ ?
Compute $e^{Nt}$ when $N^2=0$ .
Does $te^{-t}$ eventually decay?

Show worked solutions

The identity matrix.
$I+tN$ , since every higher power vanishes.
Yes. Its maximum for $t\ge0$ is at $t=1$ , after which it tends to zero.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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