The directions a matrix leaves alone
Tonight's idea
A \(2 \times 2\) matrix \(A\) moves every vector \(\mathbf v\) to \(A\mathbf v\) (grade 10). A few directions do not turn: \(A\mathbf v = \lambda \mathbf v\) for some number \(\lambda\) (lambda). Such a \(\mathbf v \ne \mathbf 0\) is an eigenvector and \(\lambda\) its eigenvalue (eigen: German for "own"). Along that line the matrix is just a number.
To find them: \(A\mathbf v = \lambda\mathbf v\) means \((A - \lambda I)\mathbf v = \mathbf 0\), where \(I\) is the identity matrix. A nonzero vector sent to zero means \(A - \lambda I\) squashes the plane flat, so its determinant, the area factor, is zero:
\(\det(A - \lambda I) = (a - \lambda)(d - \lambda) - bc = 0 \quad\text{for } A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},\)
a quadratic in \(\lambda\): the characteristic equation. Solve it for \(\lambda\), then solve \((A - \lambda I)\mathbf v = \mathbf 0\) for \(\mathbf v\). For \(B\) in the figure, \((2 - \lambda)^2 - 1 = 0\) gives \(\lambda = 3, 1\).
Why care: \(A^n\mathbf v = \lambda^n\mathbf v\). Split a vector into eigenvectors and powers of \(A\) become powers of numbers.
Watch one: Fibonacci in closed form
The Fibonacci rule \(F_{n+1} = F_n + F_{n-1}\) is a matrix step:
\[\binom{F_{n+1}}{F_n} = M\binom{F_n}{F_{n-1}}, \quad M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}, \qquad\text{so}\qquad \binom{F_{n+1}}{F_n} = M^n\binom{1}{0}.\]
The characteristic equation \(\lambda^2 - \lambda - 1 = 0\) has roots \(\varphi = \tfrac{1 + \sqrt5}{2} \approx 1.618\) and \(\psi = \tfrac{1 - \sqrt5}{2} = -1/\varphi\). Since \(\varphi^2 = \varphi + 1\), \(M(\varphi, 1) = (\varphi + 1, \varphi) = \varphi(\varphi, 1)\): an eigenvector, and likewise \((\psi, 1)\). As \(\varphi - \psi = \sqrt5\), \((1, 0) = \frac{1}{\sqrt5}\big[(\varphi, 1) - (\psi, 1)\big]\), so \(M^n(1, 0) = \frac{1}{\sqrt5}\big[\varphi^n(\varphi, 1) - \psi^n(\psi, 1)\big]\). The second entry is \(F_n\):
\(F_n = \dfrac{\varphi^n - \psi^n}{\sqrt5}\) (Binet's formula). Check \(n = 10\): \((122.992 - 0.008)/\sqrt5 = 55\).
Drag a vector and watch where its image lines up with it: mathdetours.com/demos/eigen.html.
Your turn
- \(A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}\). Find both eigenvalues and an eigenvector for each.
- Show that the quarter turn \(R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) has no real eigenvalue. Why must it not? What would you expect for a reflection?
- Expand the characteristic equation in the callout to \(\lambda^2 - (a + d)\lambda + (ad - bc) = 0\). Deduce (Vieta, grade 10) that the eigenvalues multiply to \(\det A\) and add to \(a + d\). Check for \(M\) and \(B\).
- The sequence \(a_0 = 0\), \(a_1 = 1\), \(a_{n+1} = a_n + 2a_{n-1}\) runs 0, 1, 1, 3, 5, 11, 21, … Write the rule as a matrix \(J\), find its eigenvalues and eigenvectors, write \((1, 0)\) in terms of them, and hence find a closed formula for \(a_n\). Check \(n = 5\).
Stretch: the eigenvector that runs the web
Each column of \(P = \begin{pmatrix} 0.9 & 0.5 \\ 0.1 & 0.5 \end{pmatrix}\) sums to 1: it shifts probability between two states. Show \(\lambda = 1\) is an eigenvalue, find its eigenvector with entries summing to 1, and find the other eigenvalue. Why is that eigenvector the long-run distribution, whatever the start?
Answers
- \((3 - \lambda)(2 - \lambda) = 0\): \(\lambda = 3\) and \(\lambda = 2\). For \(\lambda = 3\), \(A - 3I = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}\) kills \((1, 0)\). For \(\lambda = 2\), \(A - 2I = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) kills \((1, -1)\). A triangular matrix has its eigenvalues on the diagonal.
- \(\lambda^2 + 1 = 0\), no real root. A quarter turn moves every direction, so no real eigenvector can exist. (The roots are \(\pm i\): complex numbers as rotations again, grade 10.) A reflection leaves its mirror line alone (\(\lambda = 1\)) and reverses the perpendicular direction (\(\lambda = -1\)); for the reflection in \(y = x\), \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), the eigenvectors are \((1, 1)\) and \((1, -1)\).
- \((a - \lambda)(d - \lambda) - bc = \lambda^2 - (a + d)\lambda + (ad - bc)\). By Vieta, the roots multiply to the constant term \(ad - bc = \det A\) and add to \(a + d\) (the trace). \(M\): \(\varphi\psi = -1 = \det M\), \(\varphi + \psi = 1\). \(B\): \(3 \cdot 1 = 4 - 1\), \(3 + 1 = 4\). Since \(\varphi\psi = -1\), \(|\psi| = 1/\varphi \lt 1\), which is why \(\psi^n/\sqrt5\) fades and \(F_n\) is the nearest integer to \(\varphi^n/\sqrt5\).
- \(J = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}\), \(\lambda^2 - \lambda - 2 = 0\), \(\lambda = 2\) and \(-1\), eigenvectors \((2, 1)\) and \((1, -1)\). \((1, 0) = \tfrac13(2, 1) + \tfrac13(1, -1)\), so \(J^n(1, 0) = \tfrac13\big[2^n(2, 1) + (-1)^n(1, -1)\big]\) and the second entry gives \(a_n = \dfrac{2^n - (-1)^n}{3}\). \(n = 5\): \((32 + 1)/3 = 11\).
Stretch. \(\det(P - I) = (-0.1)(-0.5) - (0.5)(0.1) = 0\), so \(\lambda = 1\) works. \((P - I)\mathbf v = \mathbf 0\) gives \(-0.1v_1 + 0.5v_2 = 0\), \(v_1 = 5v_2\): the eigenvector is \((\tfrac56, \tfrac16)\). The eigenvalues add to the trace 1.4, so the other is 0.4, with eigenvector \((1, -1)\). Any starting distribution is \((\tfrac56, \tfrac16) + c(1, -1)\) for some \(c\), and each step multiplies the second part by 0.4, so it dies away. This is the steady state of a Markov chain (grade 10) and PageRank is exactly this eigenvector for the matrix of web links: mathdetours.com/demos/pagerank.html.