Recipe

Classify a linear system's phase portrait from trace and determinant

Use when: you have $\mathbf x'=A\mathbf x$ for a $2\times2$ real matrix $A$ and want the portrait type and stability without solving for eigenvalues

Goal: the phase portrait type (node, saddle, spiral, or center) and whether the system is stable, computed from $\text{tr}(A)$ and $\det(A)$ alone

Taught in Linear systems x' = Ax, stability, and phase portraits

Quick recipe

Compute $\text{tr}(A)$ and $\det(A)$
Classify the eigenvalue type from $\det(A)$ and the discriminant
Determine stability from the sign of $\text{tr}(A)$

1
Add the two diagonal entries for the trace, and compute the ordinary $2\times2$ determinant $ad-bc$ .

Example

For $A=\begin{bmatrix}-1&2\\-2&-1\end{bmatrix}$ : $\text{tr}(A)=-1+(-1)=-2$ , and $\det(A)=(-1)(-1)-(2)(-2)=1+4=5$ .

Why: these two numbers are the only two coefficients of $\lambda^2-\text{tr}(A)\lambda+\det(A)=0$ , so they already carry everything the eigenvalues would tell you.
2
Compute the discriminant $\text{tr}(A)^2-4\det(A)$ , then use its sign together with $\det(A)$ 's own sign to pick exactly one case below.

Example

Here $\text{tr}(A)^2-4\det(A)=4-20=-16<0$ and $\det(A)=5>0$ , so this is the third case: a spiral.
- $\det(A)<0$ — real eigenvalues of opposite sign; the portrait is a saddle, always unstable regardless of the trace.
- $\det(A)>0$ and the discriminant is positive — real eigenvalues sharing one sign; the portrait is a node.
- $\det(A)>0$ and the discriminant is negative — a complex-conjugate eigenvalue pair; the portrait is a spiral, or a center if $\text{tr}(A)=0$ exactly.
Why: $\det(A)$ 's sign alone separates a saddle from everything else, and the discriminant then separates real eigenvalues from complex ones.
3
Once the portrait type is known from Step 2, read stability straight off $\text{tr}(A)$ 's sign, since every case except the saddle depends on it.

Example

Here $\text{tr}(A)=-2<0$ and the portrait is a spiral, so this is a stable spiral, decaying inward while orbiting the origin.
- portrait is a saddle — always unstable; the trace's sign plays no role here at all.
- portrait is a node or a spiral, and $\text{tr}(A)<0$ — stable, every trajectory decays to the origin.
- portrait is a node or a spiral, and $\text{tr}(A)>0$ — unstable, some trajectory grows without bound.
- portrait is a spiral candidate and $\text{tr}(A)=0$ exactly — a center, neither growing nor decaying.
⚡ Shortcut

If only stability is being asked, skip classifying the portrait shape entirely. The system is stable exactly when $\text{tr}(A)<0$ AND $\det(A)>0$ hold simultaneously, and unstable otherwise.

Why: stability is one fact true for every trajectory at once, and both conditions above together guarantee every eigenvalue has negative real part.

Common mistakes

Checking the discriminant before checking $\det(A)$ 's sign; a negative determinant is already a saddle, so the discriminant never needs computing.
Forgetting that a saddle is always unstable, regardless of what the trace happens to be.
Concluding a center whenever the discriminant is negative, without also checking that $\text{tr}(A)$ is exactly zero.
Mixing up which sign of $\text{tr}(A)$ means stable; a NEGATIVE trace is what gives decay toward the origin.