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 $𝐯$, it hands back $A𝐯$ pointing somewhere else — for almost every $𝐯$. 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 $A(x,y,z)$, computed componentwise as $(2x+y,x+2y,-z)$, and the arrows are colored by length — dark purple for short, yellow for long. Threaded through the field are four straight lines through the origin: solid red along $(1,1,0)$, solid blue along $(1,-1,0)$, solid green along the $z$-axis, and a dashed gray line along the $x$-axis. Three of them are genuinely special. One is a decoy.

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Explore the figure

1. 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?
2. Zoom in on the dashed 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 dashed. Looking special is not being special; the field is the test.
3. 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 solid lines disagrees with the other two.
4. Now use the colors. At equal distances from the origin, the red line carries the long yellow arrows, roughly three times the position vector; the blue line's arrows match the position vector in length; the green line's match it too — but reversed. Three lines, three different stretch factors.

Naming what you found

A nonzero vector $𝐯$ with $A𝐯=\lambda 𝐯$ is an eigenvector of $A$, and the scalar $\lambda$ is its eigenvalue. The equation says exactly what you hunted for: the line through $𝐯$ 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 length ratio is $|\lambda |$. 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 $(A-\lambda I)𝐯=\mathbf{0}$ 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.

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.