Lesson 25 of 25

Classify a whole parameter family, including its boundaries

Use trace, determinant and discriminant together, then resolve the exceptional cases by structure.

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Try

Consider the family x′ = A(a)x with A(a) = [a,−3;1,1], where a is any real parameter and x is a two-component column vector. At a = −1, a trajectory turns around the origin without decaying. At a = −3, many nonzero states do not move at all. Neither case is asymptotically stable at the origin, but they have different geometries. A rule that labels both only as “not a stable node” misses useful information.

Predict

List the quantities that can change a classification: an eigenvalue crossing zero, a pair changing from real to complex, or a complex pair changing the sign of its real part. Predict why checking the trace alone cannot locate all of these transitions. Also predict whether a repeated eigenvalue necessarily gives a star-shaped portrait with two independent eigenvectors.

Experiment

Change the parameter and coordinate basis

Temporary analytic teaching values. Controls do not save evidence or award understanding. Record a prediction and explanation in the lesson practice.

Parameter case Coordinate basis Elapsed time

A(a) = [a, −3; 1, 1]. Center; stable, not attracting. Full real-parameter region: a = −1. Named radical cases use the exact repeated-root formula; their displayed decimals are approximations.

Trace 0.00000, determinant 2.00000. Fixed physical initial state x(0) = (1,−1). Time 0.25: x = (1.42780, -0.93815).

Physical solution matrix E(t), with E(0)=I: [ 0.69332, -0.73447; 0.24482, 1.18297 ]
Basis matrix S, with x = S z: [ 1.00000, 0.00000; 0.00000, 1.00000 ]
Transformed generator S⁻¹ A S: [ -1.00000, -3.00000; 1.00000, 1.00000 ]
Transformed solution matrix S⁻¹ E(t) S: [ 0.69332, -0.73447; 0.24482, 1.18297 ]

z(0) = (1.00000, -1.00000); z(t) = (1.42780, -0.93815). Reconstructing S z(t) gives (1.42780, -0.93815). Changing basis preserves the physical trajectory and changes its coordinates.

The displayed trajectory stays inside the fixed −4 to 4 window. It has 65 analytic samples and is not a time-integrator output or a proof of stability. The portrait always uses physical x coordinates; a reflected z portrait would reverse orientation.

Choose the exact zero-determinant case and then the center case. Compare the equations and classification rather than inferring infinite-time behavior from half a time unit. Next compare the lower repeated case with one parameter on either side. The solution matrix remains defined at the boundary. Changing coordinate basis preserves physical motion; the table below classifies the original physical system for every real a, not only the menu’s representatives.

Understand

Find and order every relevant boundary

The characteristic polynomial is λ² − Tλ + Δ, with trace T, determinant Δ and discriminant D:

T=a+1,\qquad\Delta=a+3,\qquad D=T^2-4\Delta=(a-1)^2-12.

The determinant vanishes at −3; repeated roots occur at L = 1 − 2√3 and U = 1 + 2√3; trace vanishes at −1. Their exact order is −3, L, −1, U. Between these boundaries the signs cannot change. For Δ negative there is a saddle. For Δ positive and real eigenvalues, the trace determines whether both are negative or positive. For complex eigenvalues, the real part is T/2. These sign arguments establish the open intervals; the boundary points need additional work.

Parameter	Eigenvalue structure	Origin and portrait
a below −3	Opposite real signs	Unstable saddle
a = −3	0 and −2	Equilibrium line; stable, not attracting
−3 < a < L	Distinct negative real roots	Asymptotically stable node
a = L	Repeated negative root, one eigenvector direction	Asymptotically stable defective node
L < a < −1	Complex roots, negative real part	Asymptotically stable spiral
a = −1	Pure imaginary roots	Center; stable, not attracting
−1 < a < U	Complex roots, positive real part	Unstable spiral
a = U	Repeated positive root, one eigenvector direction	Unstable defective node
a above U	Distinct positive real roots	Unstable node

Here “stable” means Lyapunov stable: sufficiently small initial perturbations stay small for all future time. “Asymptotically stable” additionally requires nearby trajectories to approach the origin. A short plotted trajectory cannot establish either universal statement.

Resolve the exceptional points by their structure

At a = −3, the kernel is the line spanned by (1,−1). The other eigenvalue is −2 with eigenvector (−3,1). Every solution is a constant kernel component plus a decaying component. This proves boundedness of the forward solution operator and hence stability. It does not imply convergence to the origin. For example initial state (1,0) decomposes as −(1/2)(1,−1) − (1/2)(−3,1), so its limit is (−1/2,1/2). Nonzero equilibria arbitrarily close to the origin are an even simpler obstruction to attraction.

At either repeated-root boundary, the off-diagonal entry −3 prevents A from being a scalar matrix. In dimension two, a matrix with a repeated eigenvalue and two independent eigenvectors would equal that scalar times the identity. Therefore these cases are defective, not stars. The flow has an exp(λt) factor times I + tN with nonzero N and N² = 0. At L, λ = 1 − √3 is negative, so the polynomial factor still decays; at U, λ = 1 + √3 is positive.

At a = −1, use a quadratic invariant rather than a guess from a sampled circle:

Q(x)=x_1^2+2x_1x_2+3x_2^2=(x_1+x_2)^2+2x_2^2,\qquad \frac{dQ}{dt}=0.

Q is positive definite, so its level curves are closed ellipses. Direct multiplication gives A² = −2I, and the solution matrix has sine/cosine frequency √2. Nonzero trajectories are periodic with period 2π/√2, remain bounded and never approach the origin. A nonlinear system with the same imaginary linearization need not preserve Q; this center conclusion is for the stated linear system.

Prove the rotation direction, not only one arrow

For nonzero x, the angular rate in the original oriented coordinates is:

\theta'=\frac{x_1x_2'-x_2x_1'}{x_1^2+x_2^2}=\frac{x_1^2+(1-a)x_1x_2+3x_2^2}{x_1^2+x_2^2}.

In the complex-eigenvalue interval, the numerator’s quadratic discriminant (1−a)² − 12 is negative and its leading coefficient is positive. The numerator is therefore positive for every nonzero x. All these spirals and the center turn counterclockwise in physical coordinates. A reflected coordinate basis reverses the orientation label without changing stability or the physical trajectory.

Changed family: use a new partition rather than copying the old one

For B(b) = [−1,2;b,−1], the trace is always −2, determinant is 1−2b and discriminant is 8b. Negative b gives stable spirals, now clockwise because the angular numerator is b x₁² − 2x₂². At b = 0 the matrix is defective with eigenvalue −1. Between 0 and 1/2 it has two negative real eigenvalues. At 1/2 it has a stable equilibrium line; above 1/2 it is a saddle. There is no center, unstable spiral or unstable node. The old parameter line and its exact boundaries do not transfer unchanged.

Apply

Draw the complete parameter line before checking selected values in the inspector. At each boundary explain which argument changes: determinant sign, discriminant sign, trace sign or eigenspace structure. Then solve the changed-family practice and justify every empty category as well as every nonempty one.

Continue

Use the graph, matrix and parameter transfer task for a new family and switched input. Return to nonlinear equilibria for the limits of linearization or fundamental matrices for direct construction at repeated roots.

Source scope: exhaustive parameter classification is an assessed action in MIT 18.03SC’s final materials, Fall 2011, consulted 28 October 2026. The family and changed-family tasks here are original.

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Practice and reasoning

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