Eigenvalues and Stability

The algebra behind every phase portrait.

12 problems · about 25 minutes · N-VM.C.11, A-REI.B.4

What this lesson teaches

The student classifies equilibria of linear systems from the eigenvalues of the matrix.

For x′ = Ax, the eigenvalues of A decide the entire phase portrait.
Both negative gives a stable node, both positive an unstable one, opposite signs a saddle.
Complex eigenvalues give spirals, and a zero real part gives closed orbits around a centre.

Straightforward practice. Get the method working first.

5 problems

Eigenvalues −5 and 2. Enter 1 stable node, 2 unstable node, 3 saddle.
Answer 3
Why A saddle.
What decides whether an equilibrium is stable?
Answer Whether the real parts of the eigenvalues are all negative.
Why The signs of the real parts.
Eigenvalues 3 and 5. Enter 1 stable node, 2 unstable node, 3 saddle.
Answer 2
Why Both positive.
Eigenvalues −6 and 1. Enter 1 stable node, 2 unstable node, 3 saddle.
Answer 3
Why Opposite signs.
Complex eigenvalues with real part 0. Enter 1 stable spiral, 2 unstable spiral, 3 centre.
Answer 3
Why No growth and no decay.

The same ideas with more to keep track of.

3 problems

Complex eigenvalues with real part 1. Enter 1 stable spiral, 2 unstable spiral, 3 centre.
Answer 2
Why A positive real part grows.
Trace 4 and determinant 3. What is the larger eigenvalue?
Answer 3
Why 3 and 1.
A trajectory starting exactly on an eigenvector direction. Does it leave that line? 1 yes, 0 no.
Answer 0
Why The system does not turn that direction.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Sort each eigenvalue situation by whether the origin is stable.
Answer Stable: Both eigenvalues negative, Complex with negative real part · Not stable: Both eigenvalues positive, Opposite signs
Why Everything hinges on whether the real parts are negative.
Eigenvalues −2 and −7. Enter 1 stable node, 2 unstable node, 3 saddle.
Answer 1
Why Both negative.
The Signs: Eigenvalues −4 and −1. Enter 1 stable node, 2 unstable node, 3 saddle.
Answer 1
Why A stable node.
The Spiral: Complex eigenvalues with real part −2. Does the spiral approach the origin? 1 yes, 0 no.
Answer 1
Why Yes.