Many systems have several quantities evolving together, each rate depending on the others: two
chemicals reacting, two populations competing, the position and velocity of a spring. Stack
the unknowns into a vector \mathbf{x}(t) and the coupling into a
matrix A, and a linear system is simply
\mathbf{x}' = A\mathbf{x}, \qquad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.
For the scalar equation x' = \lambda x the answer is
x = e^{\lambda t}x_0. The whole game here is to find directions in
which the matrix system behaves like that scalar one — and those directions are the
eigenvectors of A.
Why e^{\lambda t}\mathbf{v} works
Step 1 — guess a single-mode solution. Try
\mathbf{x}(t) = e^{\lambda t}\mathbf{v} for a constant scalar
\lambda and a constant vector \mathbf{v}.
Differentiate:
\mathbf{x}'(t) = \lambda e^{\lambda t}\mathbf{v}.
Step 2 — substitute into \mathbf{x}' = A\mathbf{x}.
\lambda e^{\lambda t}\mathbf{v} = A\,e^{\lambda t}\mathbf{v}.
Step 3 — cancel the scalar. The factor
e^{\lambda t} is never zero, so divide it out:
A\mathbf{v} = \lambda \mathbf{v}.
This is precisely the eigenvalue equation:
e^{\lambda t}\mathbf{v} solves the system exactly when
\lambda is an eigenvalue of A and
\mathbf{v} a matching eigenvector. With two independent
eigenpairs the general solution is their superposition,
\mathbf{x} = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2.
Worked example: solve and classify
Take
A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} and solve
\mathbf{x}' = A\mathbf{x} step by step.
Step 1 — the characteristic equation. Eigenvalues satisfy
\det(A - \lambda I) = 0:
\det\!\begin{pmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{pmatrix} = (1-\lambda)^2 - 4 = 0.
Step 2 — solve for \lambda.
(1-\lambda)^2 = 4 gives
1 - \lambda = \pm 2, so
\lambda_1 = 3, \qquad \lambda_2 = -1.
Step 3 — eigenvector for \lambda_1 = 3. Solve
(A - 3I)\mathbf{v} = 0:
\begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix}\mathbf{v} = 0 \;\Rightarrow\; -2v_1 + 2v_2 = 0 \;\Rightarrow\; v_1 = v_2, \quad \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.
Step 4 — eigenvector for \lambda_2 = -1. Solve
(A + I)\mathbf{v} = 0:
\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}\mathbf{v} = 0 \;\Rightarrow\; v_1 + v_2 = 0 \;\Rightarrow\; \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.
Step 5 — assemble the general solution.
\mathbf{x}(t) = C_1 e^{3t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + C_2 e^{-t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}.
Step 6 — classify the equilibrium. The eigenvalues are
\lambda_1 = 3 > 0 and \lambda_2 = -1 < 0
— real, opposite signs. One direction grows, the other decays, so the origin is a
saddle: unstable, with trajectories sweeping in along
\mathbf{v}_2 and out along \mathbf{v}_1.
For \mathbf{x}' = A\mathbf{x} with a 2\times2 matrix
A, the eigenvalues of A determine both the solution
and the geometry of the origin:
-
Real distinct \lambda_1 \neq \lambda_2:
\mathbf{x} = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2.
Same sign → a node (stable if both negative); opposite signs → a
saddle (always unstable).
-
Complex \lambda = \alpha \pm i\beta: solutions
spiral, e^{\alpha t}(\cos\beta t,\ \sin\beta t) — a
spiral (decaying if \alpha < 0, growing if
\alpha > 0), or a center of closed orbits if
\alpha = 0.
-
Repeated \lambda_1 = \lambda_2: a degenerate
node; if there is only one eigenvector the second solution gains a factor
t, (\mathbf{v}\,t + \mathbf{w})e^{\lambda t}.
Rabbits x and foxes y obey the
Lotka–Volterra system
x' = x(\alpha - \beta y), \qquad y' = y(\delta x - \gamma).
It is nonlinear, so eigenvalues do not solve it outright — but linearising near the
coexistence equilibrium gives a matrix with purely imaginary eigenvalues, a
center. That is why predator and prey numbers cycle endlessly, the fox peak
lagging the rabbit peak, tracing closed loops in the phase plane — the
(x, y) picture where time is hidden and we read motion straight
off the curves. The phase plane is the right place to see a node, saddle, spiral,
or center, which is exactly what the figure below lets you do.