Eigenvectors

Does a matrix turn every arrow? Guess first, then watch an arrow sweep round a circle while its image sweeps beside it.

Arrows that only stretch

Most arrows come out of a matrix turned. The picture at the top of the page shows it: as $\mathbf{v}$ goes round, its image $\mathbf{A}\mathbf{v}$ runs ahead of it or lags behind, at an angle to it.

But not always. Twice in each half turn, the image lies on the same line as the arrow. At 26.57°, which is 0.4636 radians, the image points the same way and is twice as long. At 135°, or $3\pi/4$ radians, it points the same way and is half as long. On those two lines the matrix does not turn anything. It only stretches or shrinks.

Did you guess that such lines exist? Many people guess “never”, because most arrows turn. After the clip, drag $\mathbf{v}$ round the circle yourself and try to line it up with its image. The angle readout falls to 0° only on the two drawn lines.

Name it

Let me check the first line with numbers. The arrow at 26.57° points along $(2,1)$, because its slope is $1/2$. The matrix sends it to

The image is the arrow itself, twice as long. The arrow at 135° points along $(1,-1)$, and $\mathbf{A}\,(1,-1)=(1.5-1,\ 0.5-1)=(0.5,-0.5)$, which is $0.5\,(1,-1)$: half as long, same direction.

An arrow that a matrix only stretches is called an eigenvector of the matrix, from the German eigen, “its own”. The stretch factor is its eigenvalue, written $\lambda$:

So $(2,1)$ is an eigenvector of $\mathbf{A}$ with eigenvalue $\lambda=2$, and $(1,-1)$ is one with $\lambda=0.5$.

An eigenvector is really a whole line, not one arrow. If $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$, then for any number $c$ the rule from A matrix moves the plane gives

So $(4,2)$ works too: $\mathbf{A}\,(4,2)=(8,4)=2\,(4,2)$. That is why the picture draws lines, not arrows. Every arrow on the line, long or short, pointing either way, is stretched by the same factor.

Why leave out the zero vector? Because $\mathbf{A}\mathbf{0}=\mathbf{0}=\lambda\mathbf{0}$ for every matrix and every $\lambda$. It would be an “eigenvector” of everything, with every factor, and tell us nothing.

Now an arrow that is not an eigenvector. Take $(1,0)$. The matrix sends it to $(1.5,0.5)$, which is not a multiple of $(1,0)$: its second entry is not zero. The image has turned from 0° to 18.43°.

The factor can be any number. A factor of 2 doubles the arrow; 0.5 halves it. A factor of 1 leaves the arrow exactly where it was. A factor of 0 crushes the whole line to the origin. A negative factor flips the arrow to point the other way along its line, so the angle readout would show 180°, not 0°. This matrix has no negative factor, which is why its readout reaches 0° on both lines.

Four pictures

Does every matrix have two such lines? Here are four matrices, each a simple map of the plane, to test the question on. For each, think about which arrows keep their line before you look.

• A projection drops every point straight onto a line, here the line through $(1,1)$: $\mathbf{A}_\text{proj}=\begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$.
• A reflection mirrors the plane in the same line: $\mathbf{A}_\text{refl}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$.
• A rotation by 90° turns every arrow a quarter turn: $\mathbf{A}_\text{rot}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$.
• A shear slides each row of the grid sideways, by its height: $\mathbf{A}_\text{shear}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$.

Projection. An arrow along $(1,1)$ is already on the line, so it stays: $\mathbf{A}_\text{proj}(1,1)=(1,1)$, factor 1. An arrow along $(1,-1)$ sits at a right angle to the line, so it drops onto the origin: $\mathbf{A}_\text{proj}(1,-1)=(0,0)$, factor 0.

Reflection. The mirror line keeps its arrows: $\mathbf{A}_\text{refl}(1,1)=(1,1)$, factor 1. An arrow at a right angle to the mirror flips: $\mathbf{A}_\text{refl}(1,-1)=(-1,1)$, factor $-1$. These are the projection’s two lines, with different factors.

Rotation. Every arrow turns by 90°, so no arrow stays on its own line. With numbers: $\mathbf{A}_\text{rot}$ sends $(a,b)$ to $(-b,a)$. Multiply matching entries and add, $a\cdot(-b)+b\cdot a=0$, the right-angle test of Length and the dot product. The image is always at a right angle to the arrow, so this matrix has no real eigenvector at all. (Complex eigenvalues returns to this.)

Shear. The shear sends $(x,y)$ to $(x+y,\,y)$. An arrow on the horizontal axis has $y=0$, so it stays: $(1,0)$ goes to $(1,0)$, factor 1. Any arrow with $y\neq0$ is pushed sideways off its line. So the shear has one line, not two.

So the answer is no. A 2×2 matrix can have two lines, one line, none, or, for a multiple of the identity, every line. And the two lines do not have to meet at a right angle. They do for the projection and the reflection, but the first matrix of this page has lines at 26.57° and 135°, which meet at 71.57°.

Worked example

Take $\mathbf{A}=\begin{bmatrix}1.5&1\\0.5&1\end{bmatrix}$, the matrix of the first picture.

1. The first line. $\mathbf{A}\,(2,1)=(3+1,\ 1+1)=(4,2)=2\,(2,1)$, so $\lambda=2$.
2. The second line. $\mathbf{A}\,(1,-1)=(1.5-1,\ 0.5-1)=(0.5,-0.5)=0.5\,(1,-1)$, so $\lambda=0.5$.
3. Not an eigenvector. $\mathbf{A}\,(1,0)=(1.5,0.5)$. It is not a multiple of $(1,0)$; it has turned by 18.43°.
4. The whole line. $\mathbf{A}\,(4,2)=(8,4)=2\,(4,2)$.
5. Where the sweep lights. Two arrows $(a,b)$ and $(c,d)$ lie on one line when one is a multiple of the other, which is when $ad-bc=0$. For $\mathbf{v}=(\cos\theta,\sin\theta)$ at angle $\theta$, the image is $(1.5\cos\theta+\sin\theta,\ 0.5\cos\theta+\sin\theta)$, and the test gives $0.5\cos^2\theta-0.5\sin\theta\cos\theta-\sin^2\theta=0$. Divide by $-\cos^2\theta$: $\tan^2\theta+0.5\tan\theta-0.5=0$, so $\tan\theta=0.5$ or $\tan\theta=-1$. Those are the angles 26.57° and 135°, and half a turn later, 206.57° and 315°.
6. Why the lines are worth finding. Write $(5,1)$ as steps along the two lines: $(5,1)=2\,(2,1)+1\,(1,-1)$. The matrix only stretches each piece, so

The direct way agrees: $(1.5\cdot5+1,\ 0.5\cdot5+1)=(8.5,3.5)$. Once an arrow is split along the eigenvectors, the matrix is only two stretches. The page Diagonalisation and matrix powers builds on this.

Try these before you read the answers.

• Is $(1,1)$ an eigenvector of $\begin{bmatrix}2&1\\1&2\end{bmatrix}$? Yes: it goes to $(3,3)=3\,(1,1)$.
• If $\mathbf{A}\mathbf{v}=3\mathbf{v}$, what is $\mathbf{A}(2\mathbf{v})$? It is $6\mathbf{v}$, because the whole line stretches by 3.
• Which directions does $\begin{bmatrix}3&0\\0&-2\end{bmatrix}$ keep, and with which factors? The horizontal axis, with 3, and the vertical axis, with $-2$: $(0,1)$ goes to $(0,-2)$, flipped.

Notice one more thing about the first matrix. Its diagonal entries add up to $1.5+1=2.5$, and its two factors add up to $2+0.5=2.5$ too. That is not luck. The page Finding eigenvalues explains it.

Where you’ll meet this

A guitar body or a bridge vibrates in a few special shapes, called modes. Each mode is an eigenvector of the matrix that describes how the parts push on one another, and its eigenvalue sets how fast that mode vibrates.

Whether a feedback system settles down or runs away is read from its eigenvalues. A web search engine once ranked pages by the eigenvector with eigenvalue 1 of a matrix built from the links between pages.

Reference card