Eigenvalue stability
Eigenvalues tell us what happens to small disturbances near an equilibrium. They turn the Jacobian matrix into a direct description of local growth, decay and oscillation.
Where eigenvalues enter
After linearising a nonlinear model near an equilibrium \(\mathbf{x}^*\), we obtain
\[\frac{d\mathbf u}{dt}\approx J(\mathbf{x}^*)\mathbf u,\]where \(\mathbf u\) is a small displacement from equilibrium and \(J\) is the Jacobian evaluated there.
For local stability, we therefore need to understand solutions of
\[\frac{d\mathbf u}{dt}=J\mathbf u.\]First recall the one-dimensional case
For
\[\frac{du}{dt}=\lambda u,\]the solution is
\[\boxed{u(t)=u(0)e^{\lambda t}}.\]If \(\lambda\lt0\), the disturbance shrinks. If \(\lambda\gt0\), it grows.
Eigenvalues extend exactly this growth-or-decay idea to several dimensions.
What is an eigenvector?
An eigenvector \(\mathbf v\) of a matrix \(J\) is a nonzero direction satisfying
\[J\mathbf v=\lambda\mathbf v.\]The matrix does not turn this vector into a completely different direction. It returns the same direction, multiplied by the number \(\lambda\).
Why eigenvectors matter dynamically
Suppose a perturbation starts exactly along an eigenvector:
\[\mathbf u(0)=c\mathbf v.\]Then the linear system gives
\[\mathbf u(t)=c e^{\lambda t}\mathbf v.\]So the perturbation remains in that characteristic direction while its size changes according to \(e^{\lambda t}\).
Two modes in a two-dimensional system
When a two-dimensional matrix has two independent real eigenvectors, a general perturbation can be written as
\[\mathbf u(0)=c_1\mathbf v_1+c_2\mathbf v_2.\]Its evolution is
\[\boxed{\mathbf u(t)=c_1e^{\lambda_1t}\mathbf v_1+c_2e^{\lambda_2t}\mathbf v_2}.\]This equation is the main intuition behind eigenvalue stability. Each component evolves at the rate determined by its own eigenvalue.
Why all eigenvalues must decay
Suppose \(\lambda_1\lt0\) but \(\lambda_2\gt0\). The first mode shrinks, but the second grows.
A generic small disturbance contains some component in the growing direction. Eventually that component drives the state away from equilibrium.
The general stability rule
For a hyperbolic equilibrium:
\[\boxed{\operatorname{Re}(\lambda_i)\lt0\text{ for every }i\quad\Rightarrow\quad\text{locally asymptotically stable}},\]while
\[\boxed{\operatorname{Re}(\lambda_i)\gt0\text{ for at least one }i\quad\Rightarrow\quad\text{unstable}}.\]Why do we say real part?
Eigenvalues can be complex:
\[\lambda=a+bi.\]The associated exponential can be written as
\[e^{\lambda t}=e^{at}e^{ibt}.\]The factor \(e^{at}\) controls the amplitude. The imaginary part produces oscillation or rotation.
Therefore:
\[a\lt0\Rightarrow\text{oscillations shrink},\] \[a\gt0\Rightarrow\text{oscillations grow}.\]Real part and imaginary part have different jobs
| Part of \(\lambda=a+bi\) | Dynamical role |
|---|---|
| real part \(a\) | controls exponential growth or decay of the disturbance amplitude |
| imaginary part \(b\) | controls oscillation or rotation |
Stable node
If both eigenvalues are real and negative, both characteristic modes decay. Trajectories approach the equilibrium without sustained rotation.
For example,
\[\lambda_1=-1,\qquad\lambda_2=-3.\]The second mode decays faster because \(e^{-3t}\) disappears more rapidly than \(e^{-t}\).
Which eigenvalue dominates near a stable equilibrium?
At long times, the mode with real part closest to zero usually decays most slowly and therefore dominates the final approach.
For \(-1\) and \(-3\), the \(-1\) mode is the slow mode.
Unstable node
If both eigenvalues are real and positive, both modes grow. Nearby trajectories move away from the equilibrium.
Saddle point
If one eigenvalue is negative and the other positive, one direction attracts and the other repels.
For example,
\[\lambda_1=-1,\qquad\lambda_2=2.\]A disturbance exactly along the first eigenvector approaches the equilibrium, but almost any disturbance containing a component along the second eigenvector eventually moves away.
Therefore a saddle is unstable.
Stable spiral
If the eigenvalues are
\[\lambda_{1,2}=a\pm bi,\qquad a\lt0,\]trajectories rotate while their distance from equilibrium decreases.
In biological time series this often appears as damped oscillations.
Unstable spiral
If
\[\lambda_{1,2}=a\pm bi,\qquad a\gt0,\]the rotation remains, but the amplitude grows. Trajectories spiral away.
See the main phase-plane types
Centre and purely imaginary eigenvalues
For a linear system, a pair
\[\lambda_{1,2}=\pm bi\]can produce closed trajectories around the equilibrium. Disturbances neither decay nor grow.
For a nonlinear system, however, zero real parts make first-order linearisation inconclusive. Nonlinear terms may change the conclusion.
Repeated eigenvalues
If two eigenvalues are equal, the geometry depends partly on whether there are enough independent eigenvectors.
The equilibrium can still be stable when the repeated eigenvalue is negative, but trajectories may have a different geometric arrangement from a simple node with two distinct eigenvalues.
How are eigenvalues calculated?
Eigenvalues satisfy
\[\boxed{\det(J-\lambda I)=0}.\]For
\[J=\begin{pmatrix}a&b\\c&d\end{pmatrix},\]this becomes
\[\lambda^2-(a+d)\lambda+(ad-bc)=0.\]Since
\[\operatorname{tr}(J)=a+d,\qquad\det(J)=ad-bc,\]we can write
\[\boxed{\lambda^2-\operatorname{tr}(J)\lambda+\det(J)=0}.\]Trace and determinant intuition
The trace is the sum of the eigenvalues:
\[\lambda_1+\lambda_2=\operatorname{tr}(J),\]and the determinant is their product:
\[\lambda_1\lambda_2=\det(J).\]If the determinant is negative, the eigenvalues have opposite signs, so the equilibrium is a saddle.
If the determinant is positive and the trace is negative, both eigenvalues have negative real part, giving local asymptotic stability.
Node or spiral?
The discriminant
\[D=\operatorname{tr}(J)^2-4\det(J)\]distinguishes real from complex eigenvalues.
If \(D\gt0\), the eigenvalues are real. If \(D\lt0\), they form a complex conjugate pair.
Trace–determinant picture
Worked example
Suppose the Jacobian at an equilibrium is
\[J=\begin{pmatrix}-2&1\\1&-2\end{pmatrix}.\]The characteristic equation is
\[\det\begin{pmatrix}-2-\lambda&1\\1&-2-\lambda\end{pmatrix}=0.\]Thus
\[(-2-\lambda)^2-1=0,\]giving
\[\lambda_1=-1,\qquad\lambda_2=-3.\]Both are negative, so the equilibrium is locally asymptotically stable. Because they are real, the local phase portrait is a stable node rather than a spiral.
Biological meaning: recovery after perturbation
Suppose an ecological coexistence equilibrium is disturbed slightly. Negative-real-part eigenvalues mean the disturbance modes decay, so populations return toward coexistence.
Eigenvalues closer to zero correspond to slower recovery. This connects eigenvalues with the idea of resilience.
Biological meaning: invasion
At a disease-free or species-absence equilibrium, a positive eigenvalue associated with the invading population means a small introduction grows.
The equilibrium is then unstable to invasion.
Eigenvalues can change when parameters change
The Jacobian depends on model parameters and often on the equilibrium itself. As a parameter changes, an eigenvalue can cross the imaginary axis.
When the real part changes sign, stability can change. Such changes are central to bifurcation theory.
What eigenvalue stability does not tell us
Eigenvalues of the Jacobian describe local deterministic behaviour near an equilibrium. They do not by themselves describe global dynamics, large perturbations, stochastic extinction, or every possible attractor.
A stable equilibrium can coexist with other stable states elsewhere in the phase plane.
Borderline cases require care
If any eigenvalue has zero real part, the equilibrium is non-hyperbolic and linearisation may not decide its nonlinear stability.
A practical workflow
Find the equilibrium. Calculate the Jacobian and evaluate it there. Solve \(\det(J-\lambda I)=0\). Inspect the real parts of all eigenvalues. Use imaginary parts to identify oscillation. In two dimensions, trace, determinant and discriminant can simplify classification. Finally, interpret the result as a local statement about small perturbations.