Answers · The mathematics underneath

Why do eigenvectors stay in the same direction after a transformation?

In shortAn eigenvector is defined as a vector that a matrix sends to a scaled copy of itself, so staying on its line is the condition that selects it, not a property of ordinary vectors.

An eigenvector stays on its line through the origin because the defining equation demands that the output be a scaled copy of the input. Ordinary vectors do not do this: a matrix usually turns them away from their original line. The definition is a filter, and eigenvectors are what survives it.

A\mathbf{v}=\lambda\mathbf{v},\qquad \mathbf{v}\neq\mathbf{0}.

The equals sign is the demand

Write the map as $A$ and the vector as $\mathbf{v}$ . The left side $A\mathbf{v}$ is what the map actually does. The right side $\lambda\mathbf{v}$ is a point on the line through $\mathbf{v}$ , because multiplying by a number cannot leave that line. The equals sign says those two things are the same. If $\lambda$ is positive the output is on the same ray, if $\lambda$ is negative it is on the opposite ray, and if $\lambda$ is zero the output is the zero vector and the line collapses. For a vector to qualify there must be one number $\lambda$ that works for both coordinates. If no such number exists, the vector is not an eigenvector. In two dimensions that means solving $\det(A-\lambda I)=0$ for $\lambda$ before any vector can be tested, which is why the factors are found first and the vectors second.

Take $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ . For $\mathbf{v}=(1,1)$ , $A\mathbf{v}=(2+1,1+2)=(3,3)=3(1,1)$ , so $\lambda=3$ . For $\mathbf{v}=(1,-1)$ , $A\mathbf{v}=(2-1,1-2)=(1,-1)=1(1,-1)$ , so $\lambda=1$ . For $\mathbf{v}=(2,0)$ , $A\mathbf{v}=(4,2)$ . If $(4,2)=\lambda(2,0)$ , the second coordinate gives $2=0$ , impossible, so $(2,0)$ is not an eigenvector and its line changes.

A\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}\neq \lambda\begin{pmatrix}2\\0\end{pmatrix}.

The natural guess is that a direction belongs to the vector alone, or that a map preserves a direction because it is linear. Neither holds. The same vector can be an eigenvector for one map and an ordinary vector for another. In the example above $(1,1)$ is an eigenvector of $A$ , but under a quarter turn $B=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ , $B(1,1)=(-1,1)$ . If $(-1,1)=\lambda(1,1)$ , the first coordinate gives $\lambda=-1$ and the second gives $\lambda=1$ , a contradiction. The quarter turn has no real eigenvector at all.

Why nothing else can happen

For a vector to keep its direction, $A\mathbf{v}$ must lie on the line spanned by $\mathbf{v}$ . Every point on that line is $\lambda\mathbf{v}$ for exactly one $\lambda$ , provided $\mathbf{v}$ is not zero. So 'keeps its direction' and ' $A\mathbf{v}=\lambda\mathbf{v}$ for some $\lambda$ ' are the same sentence written twice. There is no third possibility where the output is partly along the line and partly across it and still has the same direction: any across component changes the line. The equation is therefore not a theorem about all vectors. It is the condition that cuts the plane down to the special lines.

Along such a line the map has become arithmetic. If $\lambda=2$ and $\mathbf{v}$ starts with length $1$ , after one application its length is $2$ and after five it is $2^5=32$ . If $\lambda=1/2$ , after five applications it is $(1/2)^5=1/32$ . If $\lambda=-2$ , after five applications the factor is $(-2)^5=-32$ : the length is $32$ and the arrow points the opposite way. A factor of $1$ leaves the vector fixed, and a factor of $0$ sends it to zero in one step. That is why repeated maps forget their starting point: each surviving direction compounds at its own rate, and the largest rate dominates.

Where it stops being a direction

Two qualifications matter. A negative factor keeps the line but reverses the arrow, so 'same direction' is loose language for 'same line'. A zero factor sends every vector on that line to zero, so there is no surviving direction left. For a real matrix a complex factor means the line is not a real line at all: a quarter turn has no real eigenvector, and the honest description is a rotation. Non-square maps are outside the definition because input and output live in different spaces, so an output cannot be compared with the input line.

A shear such as $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ keeps the $x$ -axis: it sends $(1,0)$ to $(1,0)$ , so $\lambda=1$ . It sends $(0,1)$ to $(1,1)$ , which is not a multiple of $(0,1)$ , because a multiple of $(0,1)$ has first coordinate $0$ and this output has first coordinate $1$ . So the shear has only one real eigen-direction. That is a map with perfectly good behaviour but not enough surviving directions to write every vector as a mixture of them.

Worth keeping

An eigenvector is defined by the map sending it to a scalar multiple of itself, so staying on its line is the definition rather than a general property.
A negative factor reverses the arrow along the same line, and a zero factor collapses that line to the origin.
A quarter turn has no real eigenvector, so not every map has a full set of directions that are only scaled.

Where this goes next

This question is one step of Matrices as Maps, a course that works through its subject from the beginning. The page that teaches this part of it is Directions That Survive, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course