Complex Eigenvalues
Differential Equations · Section 7.6 · generated practice set
Complex eigenvalues, real solutions from Euler's formula, and the spiral sink / source / centre split.
Practice this set → Fresh numbers on every attempt. No account needed.
Try one now
Find the eigenvalues of $$A = \begin{pmatrix} 5 & 4 \\ -5 & 1 \end{pmatrix}$$ and classify the origin for $\mathbf{x}' = A\mathbf{x}$.
Solution
- $\text{tr}A = 5 + 1 = 6$ and $\det A = (5)(1) - (4)(-5) = 25$.
- $\lambda^2 - 6\lambda + 25 = 0$ has discriminant $-64 < 0$, so the eigenvalues are complex.
- $\lambda = \dfrac{6 \pm \sqrt{-64}}{2} = 3 \pm 4i$.
- Complex eigenvalues always mean rotation. The real part decides what the rotation does: $\alpha > 0$, so trajectories spiral outward — an unstable spiral.
- The real solutions are built from $e^{3t}\cos(4t)$ and $e^{3t}\sin(4t)$; $\beta$ sets how fast the trajectory winds around, never whether it is stable.
Next problem, fresh numbers → No account needed. Your accuracy is tracked on this device.
Method
When $\det(A - \lambda I) = 0$ has complex roots $\lambda = \alpha \pm i\beta$, the algebra is the same but the picture rotates.
Building real solutions
Take the eigenvector $\mathbf{v} = \mathbf{a} + i\mathbf{b}$ for $\lambda = \alpha + i\beta$ (the conjugate pair gives nothing new). Then $$\mathbf{x} = e^{\alpha t}\Big[c_1\left(\mathbf{a}\cos\beta t - \mathbf{b}\sin\beta t\right) + c_2\left(\mathbf{a}\sin\beta t + \mathbf{b}\cos\beta t\right)\Big].$$ In practice: compute $\mathbf{v}e^{\lambda t}$, expand with Euler's formula, and take the real and imaginary parts as your two real solutions.
The phase portrait
- $\alpha < 0$ → spiral sink, asymptotically stable — inward spiral.
- $\alpha > 0$ → spiral source, unstable — outward spiral.
- $\alpha = 0$ → center — closed orbits. Stable, but not asymptotically stable: trajectories stay near the origin without approaching it.
$\beta$ sets how fast the trajectory winds around and never affects stability. The direction of rotation comes from the sign of the off-diagonal entries, most easily found by evaluating $A\mathbf{x}$ at a convenient point such as $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ and seeing which way the arrow points.
Definitions and theorems
Worked example
Classify the origin for $\mathbf{x}' = \begin{pmatrix} -1 & -4 \\ 1 & -1\end{pmatrix}\mathbf{x}$.
- $\text{tr}A = -2$, $\det A = 1 + 4 = 5$, so $\lambda^2 + 2\lambda + 5 = 0$.
- Discriminant $4 - 20 = -16 < 0$, so $\lambda = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i$.
- $\alpha = -1 < 0$ and $\beta = 2$.
- The origin is a spiral sink: trajectories wind around it while decaying like $e^{-t}$.
- They complete one full turn every $2\pi/\beta = \pi$ units of time, shrinking by a factor $e^{-\pi} \approx 0.043$ each turn.
Common mistakes
- Using both conjugate eigenvalues. One of the pair gives both real solutions; the other repeats them.
- Letting $\beta$ decide stability. Only the sign of $\alpha$ matters.
- Calling a center asymptotically stable. Orbits are closed — they stay nearby but never approach.
- Leaving $i$ in the final answer. The general solution of a real system is real; the complex arithmetic is scaffolding.
- Guessing the spiral direction from the eigenvalues. They do not encode it — test the vector field at a point.
Practice it
Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.