Eigenvalues and eigenvectors

Find the directions a matrix keeps on their original lines, calculate their scale factors, and use null spaces to describe every corresponding eigenvector.

Some vectors stay on the same line

A matrix usually changes both the length and direction of a vector. For a few special directions, it only scales. Finding those directions lets us understand part of the transformation through a single multiplier. Start with the chosen matrix A below. Multiplication takes a column vector $(x,y)^T$ to $(2x+y,x+2y)^T$; T means transpose, so the written pair is a column.

$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix},\qquad A\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix},\qquad A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\3\end{bmatrix}$$

The input $(1,0)^T$ leaves the horizontal line. The input $(1,1)^T$ stays on its line: its output is exactly three times the input. Such a nonzero input is called an eigenvector, and its multiplier is the corresponding eigenvalue. We use the Greek letter $\lambda$ (lambda) for that multiplier.

$$A\mathbf v=\lambda\mathbf v,\qquad\mathbf v\ne\mathbf0$$

The matrix sends v to a scalar multiple of itself. Here v=(1,1)ᵀ and λ=3.

An eigenvector need not point the same way afterward. If $A\mathbf v=-2\mathbf v$, the output reverses direction and doubles in length. More generally, $|\lambda|$ gives the length multiplier: values greater than 1 stretch, values between 0 and 1 shorten, and 1 preserves length. The absolute-value bars remove the sign. A negative sign reverses the ray, while the line stays the same. Two applications in this negative-eigenvalue case give $A^2\mathbf v=4\mathbf v$.

The eigenvalue may be zero: $\begin{bmatrix}1&0\\0&0\end{bmatrix}(0,1)^T=(0,0)^T=0(0,1)^T$. A nonzero vector has collapsed to the origin. But the zero vector itself is never an eigenvector: $A\mathbf0=\lambda\mathbf0$ holds for every multiplier, so it identifies none. Also, multiplying an eigenvector by any nonzero scalar keeps its eigenvalue: $A(c\mathbf v)=cA\mathbf v=\lambda(c\mathbf v)$.

Use the defining equation to find an unknown

  1. For our original A, suppose $(1,c)^T$ is an eigenvector whose eigenvalue is $2+c$. Multiplication gives $(2+c,1+2c)^T$; scaling gives $(2+c,c(2+c))^T$.
  2. The first coordinates agree automatically. The second require $1+2c=2c+c^2$, hence $c^2=1$ and $c=1$ or $c=-1$. Both satisfy the original vector equation, giving eigenvalues 3 and 1. A single coordinate equation would not have been enough.

We use square matrices so input and output live in the same coordinate space. Unless stated otherwise, vectors and multipliers are real. Always check the entire equation, especially when an input coordinate is zero; dividing corresponding coordinates could mean dividing by zero.