Complex Eigenvalues and Planar Phase Portraits

Definition Let $A$ be a real $n \times n$ matrix. A complex number $\lambda$ with $\det(A - \lambda I) = 0$ is an eigenvalue of $A$, and a nonzero vector $v$ with complex entries and $Av = \lambda v$ is a complex eigenvector for it. Taking complex conjugates in $Av = \lambda v$ gives $A\bar{v} = \bar{\lambda}\bar{v}$, because the entries of $A$ are real. So the nonreal eigenvalues of a real matrix come in conjugate pairs $\alpha \pm i\beta$ with real $\alpha$ and $\beta \ne 0$, and the conjugate of an eigenvector for one is an eigenvector for the other. Write $v = p + iq$ with real vectors $p$ and $q$. When $\beta \ne 0$, the vectors $p$ and $q$ are linearly independent. Otherwise $v$ would be a complex multiple of a single real vector $w \ne 0$, which would then satisfy $Aw = \lambda w$, and the real vector $Aw$ cannot equal the nonreal multiple $\lambda w$. For a real $2 \times 2$ matrix the eigenvalues are nonreal exactly when $(\operatorname{tr} A)^2 < 4\det A$, and then $\alpha = \tfrac{1}{2}\operatorname{tr} A$.

Definition Let $A$ be a real matrix with $Av = \lambda v$, where $\lambda = \alpha + i\beta$, $\beta \ne 0$ and $v = p + iq$. The complex function $Z(t) = e^{\lambda t}v$ solves $X' = AX$ by the same computation as for a real mode. Euler’s formula $e^{\lambda t} = e^{\alpha t}(\cos\beta t + i\sin\beta t)$ splits it as $Z = U + iW$ with the real vector functions

Both are solutions. Substituting $Z = U + iW$ into $Z' = AZ$ gives $U' + iW' = AU + iAW$, and since $A$, $U$ and $W$ are real, the real parts and the imaginary parts agree separately. At $t = 0$ they equal $p$ and $q$, which are independent. So for a $2 \times 2$ matrix every real solution is $X = c_1U + c_2W$ for exactly one pair of real constants, fixed by $X(0) = c_1p + c_2q$. The conjugate eigenpair gives $U - iW$ and nothing new. The factor $e^{\alpha t}$ is the envelope, and the trigonometric factors repeat with period $2\pi/\lvert\beta\rvert$.

Definition For a planar autonomous system $X' = F(X)$, the phase plane is the plane of states $X = (x_1, x_2)^{T}$. Each solution traces a trajectory, the curve of its points $X(t)$, marked with an arrow in the direction of increasing $t$. The vector $F(X)$ at a point is the velocity of the solution passing through it, so it is tangent to that trajectory. A phase portrait shows the equilibria, the points where $F(X) = 0$, and enough trajectories with arrows to show the behavior everywhere. When solutions are unique, two different trajectories never meet. For $X' = AX$ the origin is always an equilibrium, and it is the only one exactly when $\det A \ne 0$. A change of coordinates $Y = PX$ with a real invertible matrix $P$ gives $Y' = PAP^{-1}Y$, a matrix with the same eigenvalues, and it carries trajectories to trajectories. It can turn circles into ellipses, but it cannot change whether solutions grow, decay or stay on closed curves.

Definition Let $A$ be a real $2 \times 2$ matrix with real nonzero eigenvalues $\lambda_1$ and $\lambda_2$, and write $\tau = \operatorname{tr} A$ and $\Delta = \det A$. The origin of $X' = AX$ is a node when the eigenvalues have the same sign, which means $\Delta > 0$ and $\tau^2 \ge 4\Delta$. It is a stable node, asymptotically stable with every solution tending to $0$ as $t \to \infty$, when both are negative, so $\tau < 0$. It is an unstable node, with every nonzero solution unbounded as $t \to \infty$, when both are positive, so $\tau > 0$. With distinct eigenvalues the two eigenvector lines are invariant, and every nonzero trajectory not on the line of the eigenvalue farther from $0$ approaches the origin tangent to the line of the eigenvalue closer to $0$, as $t \to \infty$ for a stable node and as $t \to -\infty$ for an unstable one. A repeated eigenvalue gives a star node when $A = \lambda I$, with every line through the origin invariant, and otherwise a degenerate node with a single eigenvector line. The origin is a saddle when $\lambda_1 < 0 < \lambda_2$, which means $\Delta < 0$. Solutions on the line of $\lambda_1$ tend to $0$, solutions on the line of $\lambda_2$ move away, and every other solution is unbounded as $t \to \infty$, so a saddle is unstable. A zero eigenvalue, which occurs exactly when $\Delta = 0$, gives infinitely many equilibria and falls outside these types.

Definition Let $A$ be a real $2 \times 2$ matrix with eigenvalues $\alpha \pm i\beta$, where $\beta \ne 0$ and $\alpha \ne 0$. In terms of $\tau$ and $\Delta$ this means $\tau^2 < 4\Delta$ and $\tau \ne 0$, with $\alpha = \tau/2$. The origin of $X' = AX$ is then a spiral, also called a focus. Every nonzero solution has the form $X(t) = e^{\alpha t}Q(t)$, where $Q$ is a nonzero periodic vector function with period $T = 2\pi/\lvert\beta\rvert$ whose values trace an ellipse centered at the origin, so $X(t + T) = e^{\alpha T}X(t)$. When $\alpha < 0$ the solutions wind around the origin infinitely often while tending to it as $t \to \infty$, and the origin is a stable spiral, which is asymptotically stable. When $\alpha > 0$ they wind outward and are unbounded, an unstable spiral. All trajectories turn the same way, counterclockwise when the lower left entry $a_{21}$ of $A$ is positive and clockwise when it is negative. The reason is that the angular rate $x_1x_2' - x_2x_1'$ is a quadratic form in the state that keeps one sign away from the origin, and at $(1, 0)^{T}$ it equals $a_{21}$. That entry cannot be $0$ here, since $\tau^2 - 4\Delta = (a_{11} - a_{22})^2 + 4a_{12}a_{21}$ is negative.

Definition Let $A$ be a real $2 \times 2$ matrix with eigenvalues $\pm i\beta$, where $\beta \ne 0$. This means $\tau = 0$ and $\Delta > 0$, with $\beta^2 = \Delta$. The origin of the linear system $X' = AX$ is then a center. Every nonzero solution is periodic with period $2\pi/\lvert\beta\rvert$, and its trajectory is an ellipse centered at the origin, a circle only for special matrices. The origin is stable but not asymptotically stable, since solutions keep to their ellipses forever without approaching it. This conclusion holds for the exactly linear system. A nonlinear system whose linear part at an equilibrium has the eigenvalues $\pm i\beta$ may have a center, a stable spiral or an unstable spiral there. Such an equilibrium is not hyperbolic, so the linear part does not decide, and the nonlinear terms do.