Tag: Phase Portraits

Eigenvalues & Eigenvectors Made Practical: From Matrices to Phase Portraits Eigenvalues and eigenvectors reveal the natural directions of a linear transformation. They power applications from vibration analysis to population models and are the engine behind solving linear systems of differential equations $latex \dot{\mathbf{x}} = M\mathbf{x}$. This article blends a freshman-friendly explanation with IB Math AIHL depth: steady states, coupled systems, phase portraits, stability, and complex-eigenvalue behavior. What are eigenvalues and eigenvectors? For a square matrix $latex A$, a nonzero vector $latex \mathbf{x}$ is an eigenvector with eigenvalue $latex \lambda$ if $latex A\mathbf{x}=\lambda \mathbf{x}$ Rearranging gives $latex (A-\lambda I)\mathbf{x}=0$. A nontrivial solution exists only when the characteristic equation holds: $latex \det(A-\lambda I)=0$ The roots of this polynomial are the eigenvalues. For an $latex n\times n$ matrix there are $latex n$ roots (counted with multiplicity), possibly complex. If $latex A$ is real, complex roots occur in conjugate pairs $latex a\pm bi$. Two core facts you’ll use constantly Trace–sum: $latex \displaystyle\sum_{i=1}^{n}\lambda_i=\text{tr}(A)$ (sum of diagonal entries). Determinant–product: $latex \lambda_1\lambda_2\cdots\lambda_n=\det(A)$. Why they matter Vibrations & stability: Natural frequencies of $latex M^{-1}K$ in $latex M\mathbf{Y}''=K\mathbf{Y}$ control how structures vibrate. Long-run behavior: Markov-like population models stabilize along eigenvectors of eigenvalue 1; other modes decay when $latex |\lambda|<1$. Data science: PCA finds eigenvectors of a covariance matrix to identify directions of maximum variance. From Linear Algebra to Differential Equations: $latex \dot{\mathbf{x}}=M\mathbf{x}$ General solution when $latex M$ has distinct real eigenvalues Let $latex M$ be $latex 2\times2$ with distinct real eigenvalues $latex \lambda_1,\lambda_2$ and corresponding eigenvectors $latex \mathbf{p}_1,\mathbf{p}_2$. Then every solution of the coupled system $latex \dot{\mathbf{x}}=M\mathbf{x}$ has the form $latex \mathbf{x}(t)=A e^{\lambda_1 t}\mathbf{p}_1+B e^{\lambda_2 t}\mathbf{p}_2$ where $latex A,B\in\mathbb{R}$ are determined by the initial condition $latex \mathbf{x}(0)$. Warm-up (uncoupled ➜ coupled) Uncoupled: $latex dx/dt=2x$, $latex x(0)=5$ gives $latex x(t)=5e^{2t}$. Coupled example: $latex \dot{x}=2x,\ \dot{y}=3y$ with $latex (x(0),y(0))=(3,5)$ yields $latex (x(t),y(t))=(3e^{2t},5e^{3t})$ — already in the eigenform above. Coupled system written as a matrix Any linear system like $latex \begin{cases}\dot{x}=a_{11}x+a_{12}y\\[4pt] \dot{y}=a_{21}x+a_{22}y\end{cases}$ is $latex \dot{\mathbf{x}}=M\mathbf{x}$ with $latex M=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}$ and $latex \mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}$. IB-style model: competing fungi Suppose areas (cm$latex ^2$) covered by two fungi $latex X$ and $latex Y$ evolve as $latex \dot{x}=0.4x-0.2y,\qquad \dot{y}=0.1x+0.1y$ Then $latex M=\begin{bmatrix}0.4&-0.2\\ 0.1&0.1\end{bmatrix}$. Write the system as $latex \dot{\mathbf{x}}=M\mathbf{x}$, find $latex \lambda_{1,2}$ from $latex \det(M-\lambda I)=0$, then construct $latex \mathbf{x}(t)=A e^{\lambda_1 t}\mathbf{p}_1+B e^{\lambda_2 t}\mathbf{p}_2$ and use $latex \mathbf{x}(0)=(10,6)^T$ to solve for $latex A,B$. Long-term ratios $latex \displaystyle\frac{x(t)}{y(t)}$ are governed by the eigenvector associated with the eigenvalue of largest real part. Worked example Given $latex \dot{x}=3x-4y,\qquad \dot{y}=x-2y,\qquad (x(0),y(0))=(4,-2)$ we have $latex M=\begin{bmatrix}3&-4\\1&-2\end{bmatrix}$. Solve $latex \det(M-\lambda I)=\left|\begin{array}{cc}3-\lambda & -4\\1 & -2-\lambda\end{array}\right|=(3-\lambda)(-2-\lambda)+4=\lambda^2-\lambda-2=0,$ so $latex \lambda = 2, -1.$ Eigenvectors: for $latex \lambda=2$, $latex (M-2I)\mathbf{p}_1=\mathbf{0}$ gives $latex \mathbf{p}_1=\begin{bmatrix}4\\1\end{bmatrix}$; for $latex \lambda=-1$, $latex \mathbf{p}_2=\begin{bmatrix}1\\1\end{bmatrix}$. Thus $latex \begin{bmatrix}x\\y\end{bmatrix}=A e^{2t}\begin{bmatrix}4\\1\end{bmatrix}+B e^{-t}\begin{bmatrix}1\\1\end{bmatrix}$ Apply $latex (x(0),y(0))=(4,-2)$ to get $latex 4=4A+B$ and $latex -2=A+B$, hence $latex A=2,\ B=-4$. Therefore $latex \begin{bmatrix}x\\y\end{bmatrix}=2e^{2t}\begin{bmatrix}4\\1\end{bmatrix}-4e^{-t}\begin{bmatrix}1\\1\end{bmatrix}$ Long-term ratio. As $latex t\to\infty$, the $latex e^{-t}$ term vanishes; solutions align with $latex \mathbf{p}_1$, so $latex \displaystyle\frac{x}{y}\to \frac{4}{1}=4\!:\!1$ Phase Portraits & Stability Equilibrium & classification For $latex \dot{\mathbf{x}}=M\mathbf{x}$ the origin $latex (0,0)$ is always an equilibrium. Its stability depends on eigenvalues: Both real and negative: stable node (sink). Trajectories move toward the origin as $latex t\to\infty$. Both real and positive: unstable node (source). Trajectories move away from the origin. Opposite signs: saddle point. Flow in along the eigenvector with negative eigenvalue; flow out along the positive one. Complex $latex a\pm bi$ with $latex a 0$: unstable spiral (outward spirals). Purely imaginary $latex \pm bi$: center. Closed orbits (ellipses/circles), neutrally stable. Reading the portrait quickly Eigenvectors draw the special straight-line trajectories through the origin. The sign of $latex \text{Re}(\lambda)$ decides in/out; the magnitude decides speed toward/away from the origin. To determine rotation direction for spirals, evaluate $latex dy/dt$ on the $latex x$-axis or $latex dx/dt$ on the $latex y$-axis at a test point. General solution with complex eigenvalues If $latex \lambda=a\pm bi$ and $latex \mathbf{v}_{1,2}$ are (possibly complex) eigenvectors, then solutions can be written as $latex \mathbf{x}(t)=e^{at}\!\left(A e^{ibt}\mathbf{v}_1+B e^{-ibt}\mathbf{v}_2\right)$ Using $latex e^{ibt}=\cos(bt)+i\sin(bt)$ shows oscillation with period $latex \displaystyle\frac{2\pi}{b}$ and exponential envelope $latex e^{at}$. Example: Population Mixing & Steady State Married–single model (Markov flavor) Let $latex A=\begin{bmatrix}0.7&0.2\\0.3&0.8\end{bmatrix},\quad \mathbf{w}_0=\begin{bmatrix}8000\\2000\end{bmatrix}$ Iterating $latex \mathbf{w}_{n+1}=A\mathbf{w}_n$ yields $latex \mathbf{w}_{12}=\begin{bmatrix}4000\\6000\end{bmatrix}$ and thereafter no change — a steady-state vector (eigenvalue 1). Any initial distribution converges to a multiple of that eigenvector; remaining components decay if their $latex |\lambda|<1$. Trace, Determinant, and Similarity — quick checks Check your work: If you compute eigenvalues numerically, verify that their sum equals $latex \text{tr}(A)$ and their product equals $latex \det(A)$. Coordinate change: If $latex B=S^{-1}AS$ (similar matrices), then $latex A$ and $latex B$ share the same eigenvalues. You’ve only changed basis. Further Reading MIT OCW 18.06: Linear Algebra — authoritative lectures and notes. Khan Academy: Eigenvalues & Eigenvectors — visual introductions and exercises. Wolfram MathWorld: Eigenvalue — concise reference and properties. Wikipedia: Phase portrait — phase-plane behavior and classification. Paul’s Online Notes: Systems of DEs — worked examples for $latex \dot{\mathbf{x}}=M\mathbf{x}$. 3Blue1Brown: Essence of Linear Algebra (Eigenvectors) — superb geometric intuition.