The eigenvalues of a two-by-two matrix are decided by its trace τ and determinant Δ, and so is the picture. A negative determinant gives a saddle. A positive one gives a node when τ² − 4Δ is positive, a spiral when it is negative, and a centre when the trace is also zero. The borderline cases are degenerate.
Core ideas
Three ideas. The roots multiply to the determinant, the discriminant says whether they are real, and the sign of the real part says which way the orbits go.
The roots of λ² − τ·λ + Δ = 0 add to τ and multiply to Δ. A negative product means one positive root and one negative, so the system has a growing direction and a shrinking one: a saddle. A positive product means the roots share a sign, or are a complex pair.
The roots are real when τ² − 4Δ ≥ 0 and a complex pair when it is negative. For Δ > 0 that is the line between a node, where orbits bend but never wind round the origin, and a spiral, where they turn around the origin as they go.
A complex pair has the form α ± β·i, and α = τ/2. The orbit turns around the origin at a rate set by β and spirals out when α > 0, in when α < 0. With τ = 0 there is no growth or decay, and the orbit closes: a centre.
The material
The classification and its short proof, the table of regions, one example in full, and the borderline cases the table does not settle.
For x′ = A·x with trace τ and determinant Δ, the origin is a saddle if Δ < 0. If Δ > 0 it is a node when τ² − 4Δ > 0, a spiral when τ² − 4Δ < 0 and τ ≠ 0, and a centre when τ = 0.
The roots add to τ and multiply to Δ. If Δ < 0, the product is negative, so the roots are real and opposite in sign: a saddle.
If Δ > 0 and τ² − 4Δ > 0 the roots are real and have the same sign, which is the sign of τ: a node. If τ² − 4Δ < 0 the roots are τ/2 ± i·√(4Δ − τ²)/2, a complex pair whose real part is τ/2: a spiral if that is not zero. If τ = 0 and Δ > 0 the roots are the purely imaginary ±i·√Δ, and the real solutions are cosines and sines of the same angle: closed curves around the origin.
□
The classification is the whole of the story for a linear system in the plane. Two exact numbers, computed from four entries, sort every matrix with non-zero determinant. The dividing line between nodes and spirals is the parabola Δ = τ²/4 in the plane whose axes are τ and Δ, and the lab prints τ² − 4Δ as its own tile so that the side of the parabola is a number and not an impression.
Take the matrix with rows 1 2 and −2 1. Then τ = 2 and Δ = 1 + 4 = 5, and τ² − 4Δ = 4 − 20 = −16. The discriminant is negative, so the roots are complex: λ = 1 ± 2i.
The real part 1 is positive, so orbits turn around the origin and spiral out: an unstable spiral. The lab prints the roots as 1 ± 2i and calls the origin an unstable spiral.
For rows 1 2 and 3 0, Δ = −6 < 0, so the origin is a saddle with eigenvalues −2 and 3. For rows −1 0 and 0 −2, τ = −3 and Δ = 2, so τ² − 4Δ = 1 > 0 and the roots are real and negative: a stable node, where every orbit comes into the origin. For rows 0 2 and −2 0, τ = 0 and Δ = 4, so τ² − 4Δ = −16 and λ = ±2i: a centre, with closed orbits.
On the parabola itself, τ² − 4Δ = 0 and there is one repeated root. The matrix with rows −1 1 and 0 −1 has τ = −2, Δ = 1 and the repeated root −1. It has only one eigenline, and orbits bend round it as they come in: a degenerate node. A matrix that is a multiple of the identity also has a repeated root, with every direction an eigenvector, and is a star node instead.
The line Δ = 0 is not in the table. There one root is zero and the origin is not an isolated equilibrium, which the next lesson treats along with the stability criterion.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Determinant first, because it can finish the job at once; then the discriminant; then the trace.
τ = a + d and Δ = a·d − b·c, exactly. For rows 1 2 and −2 1 they are 2 and 5.
If Δ < 0 stop: it is a saddle. The roots have opposite signs, whatever else is true. If Δ = 0 the origin is not isolated.
This is the discriminant. Here 4 − 20 = −16. Positive means real roots, negative means a complex pair, zero means a repeated root.
For real roots the sign of τ is the sign of both: stable if negative. For a complex pair, τ/2 is the real part. With τ = 0 and Δ > 0 the origin is a centre.
Solve the quadratic and check the type against them: 1 ± 2i for this matrix has a positive real part, so the spiral is unstable.
The spiral preset. Four numbers, three of them computed, and then the roots.
The discriminant tile reads −16 and the eigenvalue tile reads 1 ± 2i. The determinant is positive, which rules out a saddle, and the trace is not zero, which rules out a centre.
Change the sign of the 1s on the diagonal and the real part becomes −1; the discriminant is still −16, and the same picture runs in the other direction, as a stable spiral.
You should be able to compute τ, Δ and τ² − 4Δ, name the type among saddle, node, spiral, centre and the borderline cases, and give the roots that justify it.
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