01 / 21

Stability in
Linear Systems

Explore how eigenvalues determine whether a system converges, diverges, or oscillates: the foundation of long-term behavior in linear algebra.

Linear Algebra

What Is a Linear Dynamical System?

A linear dynamical system describes how a state vector evolves over discrete time steps. At each step, the state is multiplied by a fixed matrix:

Here \( A \) is an \( n \times n \) matrix and \( \mathbf{x}_0 \) is the initial state. After \( k \) steps the state is \( \mathbf{x}_k = A^k \mathbf{x}_0 \). The central question: what happens as \( k \to \infty \)?

Why Stability Matters

★ Population Models
Predict whether a species grows without bound or reaches a steady state.
★ Economics
Determine if prices, supply, or demand settle to equilibrium over time.
★ Engineering
Ensure a control system (robot arm, thermostat) doesn't oscillate wildly.

Eigenvalues: The Key Ingredient

Recall: \( \lambda \) is an eigenvalue of \( A \) if \( A\mathbf{v} = \lambda \mathbf{v} \) for some nonzero vector \( \mathbf{v} \). When we apply \( A \) repeatedly:

So the magnitude \( |\lambda| \) tells us whether that component of the state shrinks (\( |\lambda| < 1 \)), stays the same (\( |\lambda| = 1 \)), or grows (\( |\lambda| > 1 \)).

The Stability Theorem

Complex plane Stable region
\( |\lambda| < 1 \)
\( \lambda_1 \)
\( \lambda_2 \)
\( \lambda_3 \)
\( |\lambda| = 1 \)
 Unstable
\( |\lambda| > 1 \)
 Re Im

All eigenvalues must lie strictly inside the unit circle in the complex plane. Even one eigenvalue with \( |\lambda| \geq 1 \) prevents convergence to zero.

The Spectral Radius \( \rho(A) \)

\( \rho(A) < 1 \)
System is stable: \( A^k \to 0 \) as \( k \to \infty \).
\( \rho(A) = 1 \)
System is marginally stable: bounded but may not converge.
\( \rho(A) > 1 \)
System is unstable: \( \|A^k\| \to \infty \).

Stable vs Unstable vs Marginal

Property Stable Marginally Stable Unstable
Eigenvalue condition All \( |\lambda| < 1 \) \( |\lambda| \leq 1 \), at least one \( = 1 \) Some \( |\lambda| > 1 \)
Spectral radius \( \rho < 1 \) \( \rho = 1 \) \( \rho > 1 \)
Long-term behavior \( \mathbf{x}_k \to \mathbf{0} \) Bounded / periodic \( \|\mathbf{x}_k\| \to \infty \)
Example \( \lambda \) 0.5, −0.3 1, −1 2, 1.1

Visualizing the Three Cases

Stable k (time step)
\(x\)
 0 Marginal k
\(x\)
 Unstable k
\(x\)

How to Determine Stability

1
Write down the matrix \( A \) that governs the recurrence \( \mathbf{x}_{k+1} = A\mathbf{x}_k \).
2
Find the eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \).
3
Compute \( |\lambda| \) for each eigenvalue. If \( \lambda \) is complex, use \( |\lambda| = \sqrt{a^2 + b^2} \).
4
Find \( \rho(A) = \max|\lambda| \). If \( \rho(A) < 1 \): stable. If \( \rho(A) = 1 \): marginal. If \( \rho(A) > 1 \): unstable.

Example 1: A Stable System

1
Characteristic equation: Since \( A \) is upper triangular, eigenvalues are the diagonal entries: \( \lambda_1 = 0.5 \), \( \lambda_2 = 0.3 \).
2
Magnitudes: \( |\lambda_1| = 0.5 < 1 \) and \( |\lambda_2| = 0.3 < 1 \).
3
Spectral radius: \( \rho(A) = 0.5 < 1 \).

Example 2: An Unstable System

1
Eigenvalues: Diagonal entries give \( \lambda_1 = 1.2 \), \( \lambda_2 = 0.8 \).
2
Magnitudes: \( |\lambda_1| = 1.2 > 1 \) even though \( |\lambda_2| = 0.8 < 1 \).
3
Spectral radius: \( \rho(A) = 1.2 > 1 \).

Example 3: Complex Eigenvalues

1
Characteristic equation: \( \lambda^2 + 0.36 = 0 \), so \( \lambda = \pm 0.6i \).
2
Magnitudes: \( |\pm 0.6i| = \sqrt{0^2 + 0.6^2} = 0.6 < 1 \).
3
Spectral radius: \( \rho(A) = 0.6 < 1 \).

Spiraling Trajectories

\( x_1 \)
\( x_2 \)
 origin start
\( |\lambda| < 1 \): spiral inward (stable)

Complex eigenvalues with \( |\lambda| < 1 \) produce decaying spirals: the state rotates (oscillates) while shrinking toward zero. If \( |\lambda| > 1 \), the spiral moves outward.

The Dominant Eigenvalue

The eigenvalue with the largest magnitude is called the dominant eigenvalue. It controls the long-term behavior because all other components decay faster relative to it.

This is why the spectral radius \( \rho(A) = |\lambda_1| \) is the single number that determines stability: it is the growth rate of the dominant mode.

Common Mistakes

✕ Checking Only Real Parts
For discrete systems, stability depends on \( |\lambda| \), not \( \text{Re}(\lambda) \). A purely imaginary eigenvalue \( \lambda = 0.9i \) has \( |\lambda| = 0.9 < 1 \): stable!
✕ Ignoring One Eigenvalue
All eigenvalues must satisfy \( |\lambda| < 1 \). Even a single eigenvalue with \( |\lambda| \geq 1 \) makes the system unstable.
✕ Confusing Discrete and Continuous
In continuous systems (\( \dot{\mathbf{x}} = A\mathbf{x} \)), stability requires \( \text{Re}(\lambda) < 0 \). The unit circle criterion is for discrete systems only.

Summary: Discrete vs Continuous

Feature Discrete: \( \mathbf{x}_{k+1} = A\mathbf{x}_k \) Continuous: \( \dot{\mathbf{x}} = A\mathbf{x} \)
Stability criterion \( |\lambda| < 1 \) for all \( \lambda \) \( \text{Re}(\lambda) < 0 \) for all \( \lambda \)
Stability region Inside the unit circle Left half-plane
Key measure \( \rho(A) = \max|\lambda| \) \( \max \text{Re}(\lambda) \)
Solution form \( \mathbf{x}_k = A^k \mathbf{x}_0 \) \( \mathbf{x}(t) = e^{At}\mathbf{x}_0 \)

Question 1 of 5

A \( 2 \times 2 \) matrix \( A \) has eigenvalues \( \lambda_1 = 0.7 \) and \( \lambda_2 = -0.4 \). What is the long-term behavior of \( \mathbf{x}_{k+1} = A\mathbf{x}_k \)?

Question 2 of 5

For the discrete system \( \mathbf{x}_{k+1} = A\mathbf{x}_k \), which quantity must be less than 1 for stability?

Question 3 of 5

Matrix \( A \) has eigenvalues \( \lambda = 0.6 \pm 0.6i \). What is the spectral radius, and is the system stable?

Question 4 of 5

A matrix has eigenvalues \( \lambda_1 = 0.3 \), \( \lambda_2 = -1 \), and \( \lambda_3 = 0.5 \). The system is:

Question 5 of 5

In a discrete linear system, complex eigenvalues with \( |\lambda| < 1 \) cause the state to: