From the recipe to a picture
The previous guide handed you a working recipe: for a linear system x' = A x, you guess x = v e^(lambda t), the guess collapses to the eigenvalue problem A v = lambda v, and each eigenpair (lambda, v) gives one solution. This guide zooms in on the friendliest case — when A has two distinct real eigenvalues — and asks a different question. Not 'what is the formula?' but 'what does the answer actually look like as motion in the plane?' That is the leap from solving to SEEING.
Set the scene with a concrete 2-by-2 example so the discussion has something to point at: x' = A x with A = [3, 0; 0, -1]. Reading off the matrix, A v = lambda v is solved by lambda1 = 3 with eigenvector v1 = (1, 0), and lambda2 = -1 with eigenvector v2 = (0, 1). Two different real numbers, two directions. The general solution the recipe produces is x(t) = C1 (1, 0) e^(3t) + C2 (0, 1) e^(-1t). Everything below is about what that single line of algebra MEANS geometrically.
Why an eigenvector traps the solution on a line
Start a trajectory exactly on an eigenvector — say at the point x(0) = c v1 for some scalar c. Then the solution is x(t) = c v1 e^(lambda1 t). Look hard at this: at every instant the position is some number, c e^(lambda1 t), times the FIXED vector v1. The number changes with time, but the direction never does. So the moving point stays forever on the line through the origin spanned by v1 — it can only slide in and out along that ray, never veering off it. That trapped trajectory is a straight-line solution.
The reason is exactly the eigenvector condition A v = lambda v. The velocity field of the system is x' = A x, so at a point x = c v1 the velocity is A(c v1) = c (A v1) = c lambda1 v1 — a vector pointing along v1 itself. The flow's arrow there is parallel to the line, so it pushes the point straight along the ray and never sideways. An eigenvector direction is precisely a direction the matrix does not rotate; that is what makes it invariant under the whole flow.
The sign of the eigenvalue then decides which way the point slides. If lambda > 0, e^(lambda t) grows, so the trajectory rushes OUT along the ray, away from the origin — the line is unstable. If lambda < 0, e^(lambda t) decays, so the trajectory creeps IN toward the origin — the line is stable. In our example v1 with lambda1 = 3 is an outward (unstable) line; v2 with lambda2 = -1 is an inward (stable) line. Same geometry, opposite traffic.
Two lines become the skeleton of the whole plane
A typical trajectory does not start on either eigenvector, so it is a genuine mixture x(t) = C1 v1 e^(lambda1 t) + C2 v2 e^(lambda2 t) with both constants nonzero. The two straight-line solutions still govern it, because as time runs the two exponentials grow at different rates, and the faster one wins. The eigenvector with the larger eigenvalue dominates as t increases; the one with the smaller eigenvalue dominates as t decreases. Every curve is therefore tangent to one eigenline at one end of time and parallel to the other eigenline at the opposite end.
This is why the phase portrait — the gallery of all trajectories drawn together in the phase plane — is so legible. Draw the two eigenlines first, mark inward or outward arrows on each from the eigenvalue signs, and the rest of the curves are forced to thread between them, hugging the lines near the origin and far away. The two straight-line solutions are the skeleton; every other trajectory is flesh draped over those bones.
Same signs make a node; opposite signs make a saddle
With two distinct real eigenvalues there are only two qualitative pictures, decided by the signs of lambda1 and lambda2. When BOTH are negative, every trajectory decays inward and the origin is a stable node (a sink): all curves swoop into the origin, entering tangent to the slow eigenline (the one with the smaller |lambda|). When both are positive it is an unstable node (a source) — the same picture with every arrow reversed. The slow direction always dictates the tangent because the fast term dies first on the way in.
When the eigenvalues have OPPOSITE signs — one positive, one negative, exactly our A = [3, 0; 0, -1] — the origin is a saddle point. One eigenline pulls trajectories in (the stable direction, lambda2 = -1) while the other flings them out (the unstable direction, lambda1 = 3). A nearby trajectory swings in along the stable line, slows, then peels away out along the unstable line, tracing a hyperbola-like sweep — the shape of water draining past a ridge. Only the two trajectories sitting exactly on the stable eigenline ever reach the origin; every other path eventually escapes.
A reusable reading of the matrix
You rarely need the full solution just to classify the picture. For a 2-by-2 matrix the two eigenvalues are the roots of lambda^2 - (trace) lambda + (det) = 0, where the trace is the sum of the diagonal and det is the determinant. Two short numbers — trace and determinant — already pin down the case, which is the whole point of the trace-determinant plane: a single map on which every linear system lands as one dot, and the dot's location tells you node, saddle, or (next guide) spiral at a glance.
Two distinct real eigenvalues lambda1, lambda2 (so b^2 - 4ac > 0 for lambda^2 - T*lambda + D)
det < 0 -> opposite signs -> SADDLE (always unstable)
det > 0, trace < 0 -> both negative -> stable NODE (sink)
det > 0, trace > 0 -> both positive -> unstable NODE (source)
tangent rule: trajectories enter/leave tangent to the SLOW eigenline
(the eigenvector whose |lambda| is smaller)- Find the eigenvalues from lambda^2 - (trace) lambda + (det) = 0; confirm they are real and distinct (discriminant positive).
- Find an eigenvector for each by solving (A - lambda I) v = 0; these are the two eigenlines through the origin.
- Mark each line with an arrow: outward if its lambda > 0, inward if its lambda < 0.
- Classify: opposite signs give a saddle, both negative a stable node, both positive an unstable node; sketch the other curves hugging the lines.
Two honest caveats before moving on. First, this entire clean picture needs the eigenvalues to be real AND distinct: a complex pair rotates instead of pointing along fixed rays (the next guide), and a repeated eigenvalue can fail to supply two independent eigenvectors, collapsing the two lines into one and demanding a generalized eigenvector (the guide after that). Second, for a NONLINEAR system this analysis describes only the linearization near an equilibrium, and that local picture is trustworthy only when the equilibrium is hyperbolic — no eigenvalue with zero real part. A node or saddle of the linearization is a reliable guide nearby; a borderline case is not.