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 $(x, y) \mapsto (2x + y, \; x + 2y)$ . A thick 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 dashed answer lines are hidden in the scene; leave them hidden until step 4.

A cube of small arrows for the vector field F(x) = Ax, colored by length, with a thick 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 the arrows sitting on the line point along it instead of across it. Two dashed answer lines are hidden until revealed. — 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 arrows sitting on the line point along it instead of across it; dashed answer lines are hidden until you reveal them.

Explore the figure

Start at the default $a = 0.3$ and zoom toward the arrows sitting on the amber line. They cross it at a visible angle — this direction is being turned, so it is no eigenline. Notice that $a = 0$ would be the $x$ -axis: the dashed decoy from last section, now just one more failed angle.
Sweep $a$ slowly upward and watch only the arrows on the line. Somewhere near $a \approx 0.79$ they swing into perfect alignment, pointing outward along the line and noticeably long. Park there.
Keep sweeping. Alignment breaks, then returns near $a \approx 2.36$ — but differently: the arrows again lie along the line, yet they are shorter, matching the position vectors in length. Two alignments in the whole sweep, with two different strengths.
Reveal the two hidden dashed 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 $\mathbf{d}$ , and two plane vectors are parallel exactly when their cross-term vanishes:

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

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 squareDrag 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: $3 \times 1 = 3$ . A determinant is the product of the eigenvalues, which is the next section's fact, arriving early and for free.

Example 1 The eigenvalues at the two winning angles

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

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 long 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 modest 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 long 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. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.