5.1 The Directions a Matrix Cannot Turn

Take the $3 \times 3$ matrix $A$ whose rows are $(2, 1, 0)$ , $(1, 2, 0)$ , and $(0, 0, -1)$ . Fed a vector $\mathbf{v}$ , it hands back $A\mathbf{v}$ pointing somewhere else — for almost every $\mathbf{v}$ . This section is a hunt for the exceptions: the directions $A$ cannot turn.

The figure plots $A$ as a field of arrows. At each grid point $(x, y, z)$ , an arrow shows the direction of $A(x, y, z)$ , computed componentwise as $(2x + y, \; x + 2y, \; -z)$ . The arrows are all drawn at nearly the same short length, so their size is carried by color instead: dark purple where $A(x, y, z)$ is short, yellow where it is long. Threaded through the field are four straight lines through the origin: red along $(1, 1, 0)$ , blue along $(1, -1, 0)$ , green along the $z$ -axis, and a pale gray line along the $x$ -axis. (The coordinate axes are left out of this figure so that nothing is drawn on top of these lines.) Three of them are genuinely special. One is a decoy.

A cube-shaped grid of small arrows for the vector field that sends each point to A times that point, over a dim floor grid. The arrows are all drawn at nearly the same short length and are colored by the strength of the field, from dark purple (weak) to yellow (strong). The coordinate axes are not drawn, so four straight lines through the origin stand alone: a red line along the direction (1, 1, 0), a blue line along (1, -1, 0), a green line along the z-axis, and a pale gray line along the x-axis. Arrows sitting on the red, blue, and green lines point exactly along those lines; arrows on the pale gray line point noticeably off it. — The vector field $F(\mathbf{x}) = A\mathbf{x}$ for the matrix with rows $(2, 1, 0)$ , $(1, 2, 0)$ , $(0, 0, -1)$ , with arrows colored by the length of $A\mathbf{x}$ . Four candidate lines pass through the origin: red along $(1, 1, 0)$ , blue along $(1, -1, 0)$ , green along the $z$ -axis, and a pale gray decoy along the $x$ -axis.

Explore the figure

Orbit slowly. Most arrows point off whatever line they sit near — the matrix is turning those vectors. On which of the four marked lines do the arrows lie exactly along the line itself?
Zoom in on the pale gray line near $(1, 0, 0)$ and read the field formula there: $F(1, 0, 0) = (2, 1, 0)$ . The arrow visibly leaves the $x$ -axis — that is why this line is drawn in gray rather than a color. Looking special is not being special; the field is the test.
On each line that passes the test, note whether the arrows point outward (away from the origin, with the position vector) or inward (back toward it). One of the three colored lines disagrees with the other two.
Now use the colors, since the arrow lengths barely vary. At equal distances from the origin, the red line's arrows are much brighter than the others — teal at $(1, 1, 0)$ , yellow out at $(2, 2, 0)$ — because there $A$ triples the position vector. The blue line's arrows and the green line's stay dark, because $A$ keeps those position vectors at their own length — reversing the green ones — so their color depends only on the distance from the origin. Three lines, three stretch factors; the colors rank them, and the arithmetic below pins them down.

Naming what you found

A nonzero vector $\mathbf{v}$ with $A\mathbf{v} = \lambda\mathbf{v}$ is an eigenvector of $A$ , and the scalar $\lambda$ is its eigenvalue. The equation says exactly what you hunted for: the line through $\mathbf{v}$ is invariant — $A$ may stretch, shrink, or flip vectors on it, but cannot turn them off it. The arrows encode $\lambda$ : outward means $\lambda > 0$ , inward means $\lambda < 0$ , and the color, which tracks the length $|\lambda|\,|\mathbf{v}|$ of $A\mathbf{v}$ , brightens with $|\lambda|$ . Check the three catches by arithmetic: $A(1, 1, 0) = (3, 3, 0) = 3\,(1, 1, 0)$ , so $\lambda = 3$ on the red line; $A(1, -1, 0) = (1, -1, 0)$ , so $\lambda = 1$ on the blue line — invariant pointwise, the field there merely restating each position; and $A(0, 0, 1) = (0, 0, -1) = -1\,(0, 0, 1)$ , so $\lambda = -1$ on the green line. The inward green arrows are not a failure of invariance — they are the visual signature of a negative eigenvalue. The decoy fails outright: $A(1, 0, 0) = (2, 1, 0)$ is no multiple of $(1, 0, 0)$ .

Observation found these lines; algebra can find them blind. Rearranged, the eigenvector equation says $(A - \lambda I)\mathbf{v} = \mathbf{0}$ has a nonzero solution — which, by the last chapter, happens exactly when $A - \lambda I$ collapses: $\det(A - \lambda I) = 0$ . That determinant, as a function of $\lambda$ , is the characteristic polynomial.

Example 1 The characteristic polynomial confirms the hunt

Compute $\det(A - \lambda I)$ for the matrix above and verify its roots are the three eigenvalues found above.

Show solution

Subtracting $\lambda$ down the diagonal, $A - \lambda I$ has rows $(2 - \lambda, 1, 0)$ , $(1, 2 - \lambda, 0)$ , $(0, 0, -1 - \lambda)$ . The third row and column touch the rest of the matrix only through zeros, so the determinant factors into the upper-left $2 \times 2$ block times the corner entry: $\det(A - \lambda I) = \big((2 - \lambda)^2 - 1\big)(-1 - \lambda)$ . Expanding the block, $(2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)$ , so $\det(A - \lambda I) = -(\lambda - 3)(\lambda - 1)(\lambda + 1)$ . The roots are $\lambda = 3$ , $1$ , and $-1$ — precisely the outward triple stretch on red, the pointwise fix on blue, and the inward flip on green.

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

The characteristic polynomial, computed ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The characteristic polynomial of the same matrix the figure draws, expanded by a computer algebra system rather than by hand — and checked against the factored form worked out above. Its roots are the three stretch factors checked above: $3$ on red, $1$ on blue, $-1$ on green. ( $L$ stands for $\lambda$ .)

-1,\; 1,\; 3

The eigenvalues, certified ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The same three numbers again — but this time the computer algebra system starts from the matrix itself, extracts the complete set of eigenvalues, and certifies each one's algebraic multiplicity along the way. All three are simple, listed here in ascending order, and they agree with the roots worked out above: the inward flip on green, the pointwise fix on blue, the outward triple stretch on red.

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.