5.2 Hunting the Eigenline

The last section was a rigged hunt: four candidate lines were drawn in advance, and you merely had to judge them. Suppose nobody marks the candidates. The honest search is continuous — sweep a line through every direction and watch for the moment the field stops crossing it.

The figure below stages that search in the horizontal plane $z = 0$ , where our matrix from the last section acts by $(x, y) \mapsto (2x + y, \; x + 2y)$ . An amber test line runs through the origin at angle $a$ , controlled by a slider that rotates it from $0$ to $3.14$ — nearly a half-turn, which is all a line needs, since a line at angle $a + \pi$ is the same line. Two answer lines, one red and one blue, are hidden in the scene; leave them hidden until step 4.

A cube of small arrows for the vector field that sends each point to A times that point, colored by the strength of the field, with an amber test line through the origin in the horizontal plane. A slider named a rotates the test line from angle 0 to about pi. At exactly two angles, pi over 4 and 3 pi over 4, the field points along the line instead of across it: there the line runs through grid points whose arrows lie along it, brightly colored at pi over 4 and dark at 3 pi over 4. At the default angle of 0.3 the line passes through no arrow except at the origin. Two answer lines, red and blue, are hidden; the full XYZ 3D editor can reveal them. — The field $F(\mathbf{x}) = A\mathbf{x}$ with an amber test line $(t\cos a, \; t\sin a, \; 0)$ through the origin, rotated by the slider $a$ . At exactly two angles in the sweep, the field points along the line instead of across it, as the arrows the line passes through show; two answer lines are hidden, and the full XYZ 3D editor (↗) can reveal them.

Explore the figure

Start at the default $a = 0.3$ and zoom toward the amber line. Arrows are drawn only at whole-number grid points, and at this angle the line passes through none of them except the origin, so read the nearest ones: at $(1, 0, 0)$ and $(2, 1, 0)$ the arrows point more steeply than the line does — this direction is being turned, so it is no eigenline. Notice that $a = 0$ would be the $x$ -axis: the gray decoy from last section, now just one more failed angle.
Sweep $a$ slowly upward. Arrows sit exactly on the line only at the few angles where it passes through grid points. Near $a \approx 0.46$ it runs through $(2, 1, 0)$ , whose arrow still crosses it. Near $a \approx 0.79$ it runs through $(1, 1, 0)$ and $(2, 2, 0)$ , and now the arrows lie exactly along the line, pointing outward and brightly colored — teal and yellow. Park there.
Keep sweeping. The line meets more grid points near $a \approx 1.11$ , $1.57$ and $2.03$ , and each arrow there still crosses it. Alignment returns near $a \approx 2.36$ , where the line runs through $(-1, 1, 0)$ and $(-2, 2, 0)$ — but differently: the arrows again lie along the line, yet their colors are dark, because the field there is no longer than the position vector. Two alignments in the whole sweep, with two different strengths.
Choose Open in XYZ 3D ↗ under the figure — the full editor, where hidden objects can be shown — and reveal the two answer lines, answer: $(t, t, 0)$ and answer: $(t, -t, 0)$ . Drive the slider to each and confirm your parked angles were $\pi/4$ and $3\pi/4$ . Every other angle in the continuum — infinitely many candidates — lost the audition.

The alignment condition

The test direction at angle $a$ is $\mathbf{d} = (\cos a, \; \sin a)$ , and the field there is $A\mathbf{d} = (2\cos a + \sin a, \; \cos a + 2\sin a)$ . Alignment means $A\mathbf{d}$ is parallel to $\mathbf{d}$ — parallel in Section 0.4's sense, one vector a scalar multiple of the other — and for two vectors of the plane that happens exactly when their cross-term vanishes. Section 0.8 is where that test comes from: laid in the floor of space, two plane vectors have a cross product pointing straight up whose only nonzero component is this cross-term, and a cross product is zero precisely when the two vectors span no parallelogram, which is to say when they are parallel.

$\cos a \, (\cos a + 2\sin a) - \sin a \, (2\cos a + \sin a) = \cos^2 a - \sin^2 a = \cos 2a$

The mixed products cancel, and the entire misalignment of this field collapses to the function $\cos 2a$ . On the slider's range, $\cos 2a = 0$ exactly at $a = \pi/4 \approx 0.79$ and $a = 3\pi/4 \approx 2.36$ — the two places your sweep caught. Eigen-hunting is root-finding: a smooth misalignment function of the candidate direction, whose isolated roots are the eigenlines. That is why the winners are so rare, and it is the same rarity you met when a span collapsed at a single slider value in Chapter 2: special directions are knife-edge events, not neighborhoods.

One picture makes the two roots and their two eigenvalues visible at once. Feed the unit circle to $A$ instead of a single test direction: every direction is tested simultaneously, and the circle comes back as an ellipse.

Interactive figureThe two eigenlines of A, and what A does to the unit circleDrag inside the figure to pan; scroll to zoom.

XYZ Graph · viewer build 5edf91b

The unit circle (dashed) and its image under

A

. The two dashed lines through the origin are the eigenlines; the ellipse's long and short axes lie exactly along them, and their half-lengths are the eigenvalues

3

and

1

Find the ellipse's longest radius. It points along $y = x$ — the $a = \pi/4$ direction you parked at — and reaches $3$ units, which is the eigenvalue there.
Find the shortest radius. It points along $y = -x$ , the $a = 3\pi/4$ direction, and reaches only $1$ : that direction is held still, not stretched.
Every other point of the circle lands off the ray it started on — that is what "the direction was turned" looks like when all directions are tested at once. Only where the ellipse touches its own axes did a direction survive.
The readout gives $\det = 3$ . Check it against the two eigenvalues: $3 \times 1 = 3$ . That is no coincidence: a determinant is always the product of the eigenvalues. Write the characteristic polynomial in factored form, $\det(A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda)$ , and set $\lambda = 0$ : the left side becomes $\det A$ and the right side $\lambda_1 \lambda_2$ .

Example 1 The eigenvalues at the two winning angles

Find the eigenvalue of $A$ along each of the two alignment directions found above.

Show solution

At $a = \pi/4$ the direction has equal components, $\cos a = \sin a$ , so $A\mathbf{d} = (2\cos a + \cos a, \; \cos a + 2\cos a) = 3\,(\cos a, \; \cos a) = 3\mathbf{d}$ : eigenvalue $3$ , matching the bright outward arrows of step 2. At $a = 3\pi/4$ the components are opposite, $\cos a = -\sin a$ , so $A\mathbf{d} = (2\cos a - \cos a, \; \cos a - 2\cos a) = (\cos a, \; -\cos a) = \mathbf{d}$ : eigenvalue $1$ , matching the dark arrows of step 3. These are the red and blue lines of the previous section, recovered by search instead of by gift — and they agree with the roots $3$ and $1$ of the in-plane block of the characteristic polynomial. Blind algebra concurs: the in-plane block gives $\det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3$ , which vanishes at exactly $\lambda = 3$ and $\lambda = 1$ .

3 - 4 \cdot L + L^{2}

The in-plane characteristic polynomial ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The characteristic polynomial of the in-plane matrix with rows $(2, 1)$ and $(1, 2)$ , expanded by a computer algebra system: $\det(A - \lambda I) = \lambda^2 - 4\lambda + 3$ . ( $L$ stands for $\lambda$ .) The sweep found its roots with a slider; the next block finds them by algebra.

3,\; 1

The in-plane eigenvalues ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The two winning angles of the sweep, recovered as the roots of the characteristic polynomial by a computer algebra system: $\lambda = 3$ (the bright outward arrows at $\pi/4$ ) and $\lambda = 1$ (the pointwise fix at $3\pi/4$ ).

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.