Eigenvectors

Take this transformation and try a few vectors by hand.

$$A = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$$

Start with $(1, 0)$. Reading the columns, the output is $(2, 1)$. It grew, but it also tilted: it is no longer pointing along the x-axis where it began.

Try $(0, 1)$. Output $(1, 2)$. Tilted as well.

Now try $(1, 1)$:

$$1\begin{bmatrix} 2 \ 1 \end{bmatrix} + 1\begin{bmatrix} 1 \ 2 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \end{bmatrix}$$

Look carefully at that answer. $(3,3)$ is exactly $3 \times (1,1)$. It got three times longer and it did not leave its line.

One more, $(1, -1)$: the output is $(1, -1)$. Completely unmoved.

Spin a vector around in the visual and watch. Almost everywhere, the input arrow and its output point in different directions. At two specific angles they snap into alignment.

Naming it

A vector that stays on its own line under a transformation is an eigenvector of it, and the factor it is stretched by is the eigenvalue. For our $A$:

• $(1, 1)$ is an eigenvector with eigenvalue $3$
• $(1, -1)$ is an eigenvector with eigenvalue $1$

Every eigenvector satisfies the same short sentence, written in symbols:

$$A v = \lambda v$$

Read it out loud: putting $v$ through the whole transformation does no more than stretching $v$ by the number $\lambda$. An entire matrix, on that one direction, collapses to a single multiplication.

Note that any multiple works too: $(2,2)$ is also stretched by 3. Eigenvectors come as whole lines, not single arrows.

Why they are worth hunting

In those two directions $A$ is not doing anything complicated. There is no rotation and no shear, only stretching. The complexity of a matrix lives entirely in the mismatch between the axes you happened to choose and the axes the transformation actually prefers.

Which raises an obvious question, and it is the next step: what if you stopped fighting it, and used the eigenvectors themselves as your axes?