Recipe

Solve \dfrac{d\mathbf x}{dt}=A\mathbf x by eigenvalues and eigenvectors

Use when: you have a constant-coefficient linear system \dfrac{d\mathbf x}{dt}=A\mathbf x (with or without an initial condition) and need its solution, or just the long-run behavior of its trajectories

Goal: the general solution \mathbf x(t), pinned to a specific trajectory if an initial condition is given, plus a stability classification of the origin

Taught in Linear systems of differential equations

1. 1

   Solve \det(A-\lambda I)=0 exactly as for any square matrix.

   Example

   For A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}: \det(A-\lambda I) = (2-\lambda)^2-1 = \lambda^2-4\lambda+3 = (\lambda-1)(\lambda-3)=0, so \lambda_1=1, \lambda_2=3.

   Why: eigenvalues are the growth or decay rates the whole solution is built out of — nothing else in this recipe can proceed without them.

2. 2

   Solve (A-\lambda I)\mathbf v = \mathbf 0 for each \lambda, the same way as in the eigenvalues chapter.

   Example

   \lambda_1=1: A-I=\begin{bmatrix}1&1\\1&1\end{bmatrix} \implies x+y=0 \implies \mathbf v_1=\begin{bmatrix}1\\-1\end{bmatrix}. \lambda_2=3: A-3I=\begin{bmatrix}-1&1\\1&-1\end{bmatrix} \implies -x+y=0 \implies \mathbf v_2=\begin{bmatrix}1\\1\end{bmatrix}.

   Why: each eigenvector is exactly the "compass heading" of one straight-line solution — the direction e^{\lambda t}\mathbf v never turns away from.

3. 3

   One term per independent eigenpair: \mathbf x(t) = c_1e^{\lambda_1t}\mathbf v_1 + c_2e^{\lambda_2t}\mathbf v_2.

   Example

   \mathbf x(t) = c_1e^{t}\begin{bmatrix}1\\-1\end{bmatrix} + c_2e^{3t}\begin{bmatrix}1\\1\end{bmatrix}

   Why: each e^{\lambda t}\mathbf v solves the system by direct substitution, and two independent eigenpairs span every solution of a 2\times2 system.

4. 4

   Set t=0 (every exponential becomes 1) and solve the resulting linear system for c_1,c_2.

   Example

   If \mathbf x(0) = \begin{bmatrix}4\\2\end{bmatrix}: c_1\begin{bmatrix}1\\-1\end{bmatrix}+c_2\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}4\\2\end{bmatrix} \implies c_1+c_2=4,\ -c_1+c_2=2 \implies c_1=1,\ c_2=3.

   • an initial condition \mathbf x(0) is given — solve for the specific c_1, c_2.
   • no initial condition is given — leave c_1, c_2 as arbitrary constants; the general solution from Step 3 is already the complete answer.

   Why: c_1, c_2 pin down exactly one trajectory out of the whole family the general solution describes, the one that starts at \mathbf x(0).

5. 5

   Negative real part decays toward the origin, positive real part grows away, and mixed signs (both real) give a saddle.

   ⚡ Shortcut

   For a 2\times2 system, skip straight here from Step 1 using only \operatorname{tr}(A) and \det(A) — no eigenvector, no constant-solving needed if only the long-run shape matters. Check the discriminant (\operatorname{tr}A)^2-4\det A: if it's negative, the eigenvalues are a complex-conjugate pair whose shared real part is exactly \tfrac12\operatorname{tr}(A), so the sign of \operatorname{tr}(A) alone (together with \det(A)>0, which guarantees a genuine conjugate pair rather than two negative reals) settles stable spiral vs. unstable spiral with no square root ever computed.

   Why: the sign of each eigenvalue's real part is exactly what e^{\lambda t} does as t\to\infty — grow, shrink, or (for a nonzero imaginary part) rotate while doing either — so it fully decides the shape of every trajectory without needing the exact numeric solution.