---
title: Homogeneous Constant-Coefficient Systems and Phase Portraits
module: Systems of First-Order Linear Equations
moduleNumber: 6
lessonNumber: 2
order: 602
summary: >
  For x' = Ax with A constant, the trial x = ξe^{rt} turns the differential
  equation into the eigenvalue problem Aξ = rξ. The eigenvalues fix the geometry
  of the phase plane: real opposite signs give a saddle, real same sign a node,
  complex a spiral, purely imaginary a center. Worked in the plane, these cases
  form the eigenvalue-type classification of equilibria.
topics: [Systems of First-Order Linear Equations]
sources:
  - book: Boyce
    ref: "Ch. 7 §7.5 Homogeneous Linear Systems with Constant Coefficients"
  - book: Boyce
    ref: "§7.6 Complex-Valued Eigenvalues"
  - book: Simmons
    ref: "Ch. 10 §56 Homogeneous Linear Systems with Constant Coefficients"
draft: false
---

The homogeneous system with a constant coefficient matrix,

$$
\mathbf{x}' = A\mathbf{x}, \qquad A \text{ a real } n \times n \text{ constant matrix},
$$

solves in closed form, and the qualitative theory of the subject grows from it.
The scalar analogue $x' = ax$ has solution $x = ce^{at}$, so a vector system
should have solutions built from an exponential times a fixed direction. Finding
those directions is the
[eigenvalue problem](/differential-equations/systems/matrices-eigenvalues-review)
$A\boldsymbol{\xi} = r\boldsymbol{\xi}$.

## The exponential ansatz

Try a solution of the form

$$
\mathbf{x} = \boldsymbol{\xi}\,e^{rt},
$$

with the scalar $r$ and the constant vector $\boldsymbol{\xi} \neq \mathbf{0}$ to
be found. Then $\mathbf{x}' = r\boldsymbol{\xi}e^{rt}$, and substituting into
$\mathbf{x}' = A\mathbf{x}$ gives $r\boldsymbol{\xi}e^{rt} = A\boldsymbol{\xi}
e^{rt}$. Cancelling the never-zero factor $e^{rt}$ leaves

$$
A\boldsymbol{\xi} = r\boldsymbol{\xi}, \qquad\text{equivalently}\qquad
(A - rI)\boldsymbol{\xi} = \mathbf{0}.
$$

The vector $\boldsymbol{\xi}e^{rt}$ solves the differential equation exactly when
$r$ is an eigenvalue of $A$ and $\boldsymbol{\xi}$ a belonging eigenvector. The
differential-equation problem has become the algebra problem
$\det(A - rI) = 0$.

> **Theorem (Solutions from eigenpairs).** If $(r_i, \boldsymbol{\xi}^{(i)})$ is
> an eigenvalue-eigenvector pair of the constant matrix $A$, then
> $\mathbf{x}^{(i)}(t) = \boldsymbol{\xi}^{(i)} e^{r_i t}$ solves
> $\mathbf{x}' = A\mathbf{x}$. If $A$ has $n$ linearly independent eigenvectors
> $\boldsymbol{\xi}^{(1)}, \dots, \boldsymbol{\xi}^{(n)}$ with eigenvalues
> $r_1, \dots, r_n$, the general solution is
> $$
> \mathbf{x} = c_1\boldsymbol{\xi}^{(1)} e^{r_1 t} + \cdots
>   + c_n\boldsymbol{\xi}^{(n)} e^{r_n t}.
> $$

That these $n$ solutions form a fundamental set is immediate from their
Wronskian. Factoring the exponentials out of each column,

$$
W\big[\mathbf{x}^{(1)}, \dots, \mathbf{x}^{(n)}\big](t)
= e^{(r_1 + \cdots + r_n)t}\,
\det\!\big[\boldsymbol{\xi}^{(1)}\ \cdots\ \boldsymbol{\xi}^{(n)}\big],
$$

which is never zero because the exponential is positive and the eigenvectors are
independent. In the $2 \times 2$ case the solution lives in a plane and can be
drawn as a phase portrait.

## The phase plane

For a planar system, a solution $\mathbf{x}(t) = \big(x_1(t), x_2(t)\big)$ traces
a curve in the $x_1x_2$-plane as $t$ increases. That plane, with a
representative sample of solution curves drawn on it, is the standard picture of
a two-dimensional system.

> **Definition (Phase plane, trajectory, phase portrait).** The **phase plane**
> is the $x_1x_2$-plane of solution values. A **trajectory** is the curve traced
> by one solution as $t$ varies. The **direction field** attaches to each point
> the tangent vector $A\mathbf{x}$; a **phase portrait** is a plot of several
> trajectories, giving the qualitative behavior of all solutions at once.

Since $\det A \neq 0$ is assumed, the only equilibrium (where $A\mathbf{x} =
\mathbf{0}$) is the origin. The central question is whether nearby trajectories
approach the origin or leave it as $t \to \infty$, i.e. whether $\mathbf{x} =
\mathbf{0}$ is stable or unstable. The eigenvalues settle it.

## Real eigenvalues of opposite sign: the saddle

When the two eigenvalues are real with $r_1 > 0 > r_2$, one exponential grows
and the other decays.

> **Worked example.** For $\mathbf{x}' = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}\mathbf{x}$,
> the characteristic equation $(1-r)^2 - 4 = r^2 - 2r - 3 = (r-3)(r+1) = 0$ gives
> $r_1 = 3$, $r_2 = -1$. For $r_1 = 3$, $(A - 3I)\boldsymbol{\xi} = \mathbf{0}$ is
> $-2\xi_1 + \xi_2 = 0$, so $\boldsymbol{\xi}^{(1)} = (1, 2)^\top$; for $r_2 = -1$,
> $(A + I)\boldsymbol{\xi} = \mathbf{0}$ is $2\xi_1 + \xi_2 = 0$, so
> $\boldsymbol{\xi}^{(2)} = (1, -2)^\top$. The general solution is
>
> $$
> \mathbf{x} = c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} e^{3t}
> + c_2 \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{-t}.
> $$

The two eigenvectors set two straight-line trajectories through the origin.
Along $x_2 = 2x_1$ (the direction of $\boldsymbol{\xi}^{(1)}$) solutions move
_out_, since $e^{3t}$ grows; along $x_2 = -2x_1$ (direction of
$\boldsymbol{\xi}^{(2)}$) solutions move _in_, since $e^{-t}$ decays. Every other
trajectory is a combination: for large $t$ the $e^{3t}$ term dominates, so
trajectories become asymptotic to the line $x_2 = 2x_1$; running $t \to -\infty$
they approach $x_2 = -2x_1$.

$$
% caption: Saddle point (eigenvalues 3 and -1): trajectories flow out along the
% unstable eigendirection and in along the stable one, bending like hyperbolas.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{hot}{HTML}{C0603A}
  \draw[black, ->] (-3,0) -- (3,0) node[right, black]{$x_1$};
  \draw[black, ->] (0,-2.6) -- (0,2.6) node[above, black]{$x_2$};
  % unstable eigenline x2 = 2 x1 (outflow)
  \draw[acc, thick] (-1.35,-2.7) -- (1.35,2.7);
  % stable eigenline x2 = -2 x1 (inflow)
  \draw[hot, thick] (-1.35,2.7) -- (1.35,-2.7);
  % arrows on eigenlines
  \draw[acc, thick, ->] (0.5,1.0) -- (0.75,1.5);
  \draw[acc, thick, ->] (-0.5,-1.0) -- (-0.75,-1.5);
  \draw[hot, thick, ->] (0.75,-1.5) -- (0.9,-1.8);
  \draw[hot, thick, ->] (-0.75,1.5) -- (-0.9,1.8);
  % hyperbolic trajectories (approx)
  \draw[black] (-2.7,-1.3) .. controls (-1.2,-0.2) and (-1.0,0.2) .. (-1.4,1.9);
  \draw[black] (2.7,1.3) .. controls (1.2,0.2) and (1.0,-0.2) .. (1.4,-1.9);
  \draw[black] (-2.7,0.9) .. controls (-1.4,0.5) and (-0.9,-0.4) .. (-0.3,-2.4);
  \draw[black] (2.7,-0.9) .. controls (1.4,-0.5) and (0.9,0.4) .. (0.3,2.4);
  \fill (0,0) circle (1.6pt);
  \node[acc, anchor=west, font=\scriptsize] at (1.4,2.5) {$x_2 = 2x_1$};
  \node[hot, anchor=east, font=\scriptsize] at (-1.4,2.5) {$2x_1 + x_2 = 0$};
  \node[anchor=north, font=\scriptsize, text=black] at (0,-2.75) {saddle (unstable)};
\end{tikzpicture}
$$

This picture — inflow along one eigendirection, outflow along the other, and
hyperbola-like curves between — is typical of every $2 \times 2$ system whose
eigenvalues are real and of opposite sign. The origin is a **saddle point**, and
it is always unstable, since almost every trajectory eventually departs.

## Real eigenvalues of the same sign: the node

When both eigenvalues are real and share a sign, both exponentials pull the same
way, and every trajectory approaches (or leaves) the origin.

> **Worked example.** For $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x}$,
> the characteristic equation $(-2-r)^2 - 1 = r^2 + 4r + 3 = (r+1)(r+3) = 0$ gives
> $r_1 = -1$, $r_2 = -3$. For $r_1 = -1$, $(A + I)\boldsymbol{\xi} = \mathbf{0}$ is
> $-\xi_1 + \xi_2 = 0$, so $\boldsymbol{\xi}^{(1)} = (1, 1)^\top$; for $r_2 = -3$,
> $(A + 3I)\boldsymbol{\xi} = \mathbf{0}$ is $\xi_1 + \xi_2 = 0$, so
> $\boldsymbol{\xi}^{(2)} = (1, -1)^\top$. The general solution is
>
> $$
> \mathbf{x} = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{-t}
> + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-3t}.
> $$

Both terms decay, so all trajectories approach the origin. The rates differ:
$e^{-3t}$ dies far faster than $e^{-t}$, so for large $t$ the slow term
$c_1\boldsymbol{\xi}^{(1)}e^{-t}$ dominates. Unless $c_1 = 0$, every trajectory
reaches the origin _tangent to the slow eigendirection_ $x_2 = x_1$; the only
trajectories coming in along the fast direction $x_2 = -x_1$ are the ones with
$c_1 = 0$.

$$
% caption: Nodal sink (eigenvalues -1 and -3): all trajectories decay to the
% origin, arriving tangent to the slow eigendirection x2 = x1.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{hot}{HTML}{C0603A}
  \draw[black, ->] (-3,0) -- (3,0) node[right, black]{$x_1$};
  \draw[black, ->] (0,-2.6) -- (0,2.6) node[above, black]{$x_2$};
  % slow eigenline x2 = x1
  \draw[acc, thick] (-2.4,-2.4) -- (2.4,2.4);
  % fast eigenline x2 = -x1
  \draw[hot, thick] (-2.4,2.4) -- (2.4,-2.4);
  % inflow arrows
  \draw[acc, thick, ->] (1.6,1.6) -- (1.2,1.2);
  \draw[acc, thick, ->] (-1.6,-1.6) -- (-1.2,-1.2);
  \draw[hot, thick, ->] (1.6,-1.6) -- (1.2,-1.2);
  \draw[hot, thick, ->] (-1.6,1.6) -- (-1.2,1.2);
  % curved trajectories bending toward slow line
  \draw[black, ->] (-0.4,2.5) .. controls (-0.6,1.2) and (-1.1,0.6) .. (-1.7,0.35);
  \draw[black, ->] (2.5,0.4) .. controls (1.2,0.6) and (0.6,1.1) .. (0.35,1.7);
  \draw[black, ->] (0.4,-2.5) .. controls (0.6,-1.2) and (1.1,-0.6) .. (1.7,-0.35);
  \draw[black, ->] (-2.5,-0.4) .. controls (-1.2,-0.6) and (-0.6,-1.1) .. (-0.35,-1.7);
  \fill (0,0) circle (1.6pt);
  \node[acc, anchor=west, font=\scriptsize] at (1.7,2.35) {slow: $x_2 = x_1$};
  \node[hot, anchor=east, font=\scriptsize] at (-1.7,2.35) {fast: $x_1 + x_2 = 0$};
  \node[anchor=north, font=\scriptsize, text=black] at (0,-2.75) {stable node};
\end{tikzpicture}
$$

The origin is a **node**. It is asymptotically stable when both eigenvalues are
negative (a sink, as drawn) and unstable when both are positive (a source, the
same picture with arrows reversed).

## Complex eigenvalues: spirals and centers

A real matrix can have complex eigenvalues, which must come in conjugate pairs
$r = \lambda \pm i\mu$ with real $\lambda, \mu$. The belonging eigenvectors are
also complex conjugates, and the solution $\boldsymbol{\xi}e^{rt}$ is
complex-valued. Real solutions are recovered by taking real and imaginary parts,
which is legitimate because for a real matrix the real and imaginary parts of any
complex solution are themselves solutions.

Write the eigenvector as $\boldsymbol{\xi} = \mathbf{a} + i\mathbf{b}$ with real
$\mathbf{a}, \mathbf{b}$, and expand $e^{(\lambda + i\mu)t} = e^{\lambda t}(\cos
\mu t + i \sin \mu t)$. Separating $\boldsymbol{\xi}e^{rt}$ into real and
imaginary parts gives two real solutions:

$$
\mathbf{u}(t) = e^{\lambda t}\big(\mathbf{a}\cos\mu t - \mathbf{b}\sin\mu t\big),
\qquad
\mathbf{v}(t) = e^{\lambda t}\big(\mathbf{a}\sin\mu t + \mathbf{b}\cos\mu t\big).
$$

The exponential $e^{\lambda t}$ sets the amplitude and the trigonometric factors
set the rotation. The sign of the real part $\lambda$ determines whether the
trajectory spirals in or out.

> **Worked example.** For
> $\mathbf{x}' = \begin{pmatrix} -\tfrac12 & 1 \\ -1 & -\tfrac12 \end{pmatrix}\mathbf{x}$,
> the characteristic equation is $r^2 + r + \tfrac54 = 0$, so $r = -\tfrac12 \pm i$.
> The eigenvector for $r_1 = -\tfrac12 + i$ is
> $(1, i)^\top = (1,0)^\top + i(0,1)^\top$, so $\mathbf{a} = (1,0)^\top$,
> $\mathbf{b} = (0,1)^\top$, and
>
> $$
> \mathbf{u}(t) = e^{-t/2}\begin{pmatrix} \cos t \\ -\sin t \end{pmatrix},
> \qquad
> \mathbf{v}(t) = e^{-t/2}\begin{pmatrix} \sin t \\ \cos t \end{pmatrix}.
> $$
>
> Their Wronskian is $W[\mathbf{u}, \mathbf{v}](t) = e^{-t} \neq 0$, so they are a
> fundamental set. Each trajectory winds around the origin while its distance
> shrinks like $e^{-t/2}$: a spiral inward, making infinitely many circuits.
>
> With $\lambda = 0$ the amplitude factor $e^{0} = 1$ is constant, so the
> trajectory retraces a closed curve instead. For
> $\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\mathbf{x}$ the
> eigenvalues are $\pm i$ and the solutions are $(\cos t, -\sin t)^\top$ and
> $(\sin t, \cos t)^\top$, whose trajectories are circles about the origin.

$$
% caption: Complex eigenvalues. Left: negative real part gives a spiral sink,
% winding inward. Right: purely imaginary eigenvalues give a center of closed
% orbits.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % --- spiral (left) ---
  \begin{scope}
    \draw[black, ->] (-2.4,0) -- (2.4,0) node[right, black]{$x_1$};
    \draw[black, ->] (0,-2.4) -- (0,2.4) node[above, black]{$x_2$};
    \draw[acc, thick, ->] (2.1,0)
      .. controls (2.1,1.35) and (1.15,2.15) .. (0,2.05)
      .. controls (-1.75,1.9) and (-1.85,0.0) .. (-1.35,-0.75)
      .. controls (-0.75,-1.55) and (0.55,-1.5) .. (1.05,-0.7)
      .. controls (1.45,-0.05) and (1.0,0.75) .. (0.35,0.9)
      .. controls (-0.35,1.05) and (-0.75,0.35) .. (-0.5,-0.2)
      .. controls (-0.3,-0.6) and (0.2,-0.5) .. (0.15,-0.05);
    \fill (0,0) circle (1.6pt);
    \node[anchor=north, font=\scriptsize, text=black] at (0,-2.6) {spiral sink};
  \end{scope}
  % --- center (right) ---
  \begin{scope}[xshift=6cm]
    \draw[black, ->] (-2.4,0) -- (2.4,0) node[right, black]{$x_1$};
    \draw[black, ->] (0,-2.4) -- (0,2.4) node[above, black]{$x_2$};
    \draw[acc, thick] (0,0) circle (0.8);
    \draw[acc, thick] (0,0) circle (1.5);
    \draw[acc, thick] (0,0) circle (2.1);
    \draw[acc, thick, ->] (1.5,0.05) -- (1.5,-0.05);
    \draw[acc, thick, ->] (0.8,0.05) -- (0.8,-0.05);
    \fill (0,0) circle (1.6pt);
    \node[anchor=north, font=\scriptsize, text=black] at (0,-2.6) {center (closed orbits)};
  \end{scope}
\end{tikzpicture}
$$

The origin is a **spiral point** when $\lambda \neq 0$ — asymptotically stable
for $\lambda < 0$, unstable for $\lambda > 0$ — and a **center** when $\lambda =
0$, which is stable but not asymptotically stable. The direction of winding
(clockwise or counterclockwise) is fixed by the matrix and found by checking the
tangent vector $A\mathbf{x}$ at one convenient point.

## Transitions: a one-parameter sweep

Because the portrait type is fixed by the eigenvalues, and eigenvalues move
continuously with the matrix, a system carrying a parameter passes through
several types as the parameter varies. The values where the type changes are
**bifurcation values**.

> **Worked example.** Consider
> $\mathbf{x}' = \begin{pmatrix} \alpha & 2 \\ -2 & 0 \end{pmatrix}\mathbf{x}$ with a
> real parameter $\alpha$. The characteristic equation is
> $r^2 - \alpha r + 4 = 0$, so
>
> $$
> r = \frac{\alpha \pm \sqrt{\alpha^2 - 16}}{2}.
> $$
>
> The discriminant $\alpha^2 - 16$ changes sign at $\alpha = \pm 4$, and the real
> part $\alpha/2$ changes sign at $\alpha = 0$. Tracking the eigenvalues across the
> range passes through every case of the classification.

| Range of $\alpha$ | Eigenvalues | Portrait |
| --- | --- | --- |
| $\alpha < -4$ | real, both negative | stable node |
| $-4 < \alpha < 0$ | complex, negative real part | spiral sink |
| $\alpha = 0$ | purely imaginary $\pm 2i$ | center |
| $0 < \alpha < 4$ | complex, positive real part | spiral source |
| $\alpha > 4$ | real, both positive | unstable node |

The bifurcation values are $\alpha = -4$ and $\alpha = 4$, where real eigenvalues
collide and split into a complex pair (node to spiral), and $\alpha = 0$, where
the pair crosses the imaginary axis (stable spiral to unstable, through a
center). The center at $\alpha = 0$ is the delicate transition: an arbitrarily
small change in $\alpha$ destroys the closed orbits, spiralling them in or out.

## The eigenvalue classification

Collecting the three main cases and the transitional ones, the type of the
equilibrium and its stability are determined by the eigenvalues.

| Eigenvalues of $A$ | Equilibrium type | Stability |
| --- | --- | --- |
| Real, opposite signs $(r_1 > 0 > r_2)$ | saddle point | unstable |
| Real, same sign, unequal | node | stable if both $< 0$, unstable if both $> 0$ |
| Complex, nonzero real part | spiral point | stable if $\operatorname{Re} < 0$, unstable if $> 0$ |
| Purely imaginary $(\pm i\mu)$ | center | stable, not asymptotically |
| Real, equal (defective) | improper node | stable if $< 0$, unstable if $> 0$ |

The last row, equal eigenvalues with too few eigenvectors, is the defective case;
its solution needs the
[generalized eigenvectors](/differential-equations/systems/repeated-eigenvalues-fundamental-matrices)
of the following lesson. Reading the table geometrically, plot the eigenvalues in
the complex plane: the
sign of the real part is the stability switch, and whether the pair is on the
real axis, off it, or on the imaginary axis picks node/saddle, spiral, or center.

$$
% caption: Eigenvalues in the complex plane. Left half-plane decays (stable),
% right half grows; real axis gives nodes and saddles, off-axis pairs spiral,
% the imaginary axis gives centers.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{hot}{HTML}{C0603A}
  % shaded half-planes
  \fill[acc!8] (-3,-2.2) rectangle (0,2.2);
  \fill[hot!8] (0,-2.2) rectangle (3,2.2);
  \draw[black, ->] (-3.2,0) -- (3.4,0) node[right, black]{Re};
  \draw[black, ->] (0,-2.4) -- (0,2.6) node[above, black]{Im};
  % stable spiral pair (left, off axis)
  \fill[acc] (-1.6,1.2) circle (2pt);
  \fill[acc] (-1.6,-1.2) circle (2pt);
  \node[acc, anchor=south, font=\scriptsize] at (-1.6,1.3) {spiral sink};
  % unstable spiral pair (right, off axis)
  \fill[hot] (1.7,0.9) circle (2pt);
  \fill[hot] (1.7,-0.9) circle (2pt);
  \node[hot, anchor=south, font=\scriptsize] at (1.7,1.0) {spiral source};
  % center pair (imaginary axis)
  \fill[black!70] (0,1.8) circle (2pt);
  \fill[black!70] (0,-1.8) circle (2pt);
  \node[black!70, anchor=west, font=\scriptsize] at (0.1,1.9) {center};
  % real eigenvalues (nodes/saddle) markers on axis
  \fill[acc] (-2.4,0) circle (2pt);
  \fill[acc] (-0.7,0) circle (2pt);
  \node[acc, anchor=north, font=\scriptsize] at (-1.5,-0.15) {node (both $< 0$)};
  \fill[hot] (2.5,0) circle (2pt);
  \node[hot, anchor=south, font=\scriptsize] at (2.5,0.1) {saddle: one each side};
  \node[anchor=north, font=\scriptsize, text=black] at (-1.5,-2.1) {stable half-plane};
  \node[anchor=north, font=\scriptsize, text=black] at (1.6,-2.1) {unstable half-plane};
\end{tikzpicture}
$$

The same classification reads off the trace and determinant of $A$. The
eigenvalues of a $2 \times 2$ matrix satisfy $r^2 - (\operatorname{tr} A)r + \det A
= 0$, so $r_1 + r_2 = \operatorname{tr} A$ and $r_1 r_2 = \det A$; the sign of
$\det A$, the sign of $\operatorname{tr} A$, and the sign of the discriminant
$(\operatorname{tr} A)^2 - 4\det A$ decide the type without solving for the roots.

$$
% caption: The trace-determinant plane. The parabola det = (tr)^2 / 4 separates
% nodes (below it) from spirals (above); below the horizontal axis lie saddles,
% and the vertical axis with positive determinant carries centers. Trajectories
% decay on the stable (left) half and grow on the unstable (right) half.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \definecolor{hot}{HTML}{C0603A}
  \fill[acc!8] (-3.2,0) rectangle (0,3.0);
  \fill[hot!8] (0,0) rectangle (3.2,3.0);
  \draw[black, ->] (-3.4,0) -- (3.5,0) node[right, black]{tr $A$};
  \draw[black, ->] (0,-1.6) -- (0,3.2) node[above, black]{det $A$};
  \draw[black!70, thick, domain=-3.3:3.3, samples=60] plot (\x, {(\x)*(\x)/4});
  \node[acc, font=\scriptsize, align=center] at (-1.35,2.4) {stable\\spiral};
  \node[hot, font=\scriptsize, align=center] at (1.35,2.4) {unstable\\spiral};
  \node[acc, font=\scriptsize, align=center] at (-2.4,0.55) {stable\\node};
  \node[hot, font=\scriptsize, align=center] at (2.4,0.55) {unstable\\node};
  \node[black!70, font=\scriptsize] at (0,-0.9) {saddle points};
  \node[black!70, font=\scriptsize, anchor=west] at (0.14,1.7) {center};
  \fill[black!70] (0,1.7) circle (1.6pt);
  \node[anchor=north, text=black, font=\scriptsize] at (-1.7,-1.4) {stable half};
  \node[anchor=north, text=black, font=\scriptsize] at (1.7,-1.4) {unstable half};
\end{tikzpicture}
$$

## Summary

Solving $\mathbf{x}' = A\mathbf{x}$ is finding eigenpairs, and reading the phase
portrait is reading eigenvalue signs.

- **Method.** Solve $\det(A - rI) = 0$ for eigenvalues, then
  $(A - rI)\boldsymbol{\xi} = \mathbf{0}$ for eigenvectors; combine
  $\boldsymbol{\xi}^{(i)}e^{r_i t}$. For a complex pair, take real and imaginary
  parts to get $\mathbf{u}, \mathbf{v}$.
- **Geometry.** Real opposite signs, saddle; real same sign, node; complex with
  nonzero real part, spiral; purely imaginary, center. The real part is the
  stability switch.

Near an equilibrium a nonlinear system resembles its
[linearization](/differential-equations/nonlinear/phase-plane-autonomous-stability),
so these five portraits classify its equilibria as well.