Lecture · 16 min · advanced

Eigenvalues

The directions a linear map merely stretches

Eigenvalues and eigenvectors

Most vectors get rotated as well as stretched by a linear map $A$ . Eigenvectors are the special directions that refuse to rotate: $A$ sends $v$ to a scalar multiple of itself. That scalar is the eigenvalue. Diagonalization is the act of choosing a basis of such directions, so $A$ becomes a list of stretches.

On the board today

—Solve $A v = λ v$ as $(A - λ I) v = 0$ .
—Compute the characteristic polynomial of a $2 \times 2$ .
—Interpret $λ$ as a stretch factor along $v$ .

The eigen-equation

A v = λ v ⟺ (A - λ I) v = 0, v \neq = 0

$λ$ is an eigenvalue iff $A - λ I$ is singular iff $det (A - λ I) = 0$ .

Worked example · A symmetric $2 \times 2$

Find the eigenvalues of $A = (2112)$ .

1. $det (2 - λ 1 1 2 - λ) = (2 - λ)^{2} - 1 = λ^{2} - 4 λ + 3$ .
2. $(λ - 1) (λ - 3) = 0$ , so $λ = 1$ or $3$ .

Along $(1, - 1)$ the map stretches by $1$ ; along $(1, 1)$ it stretches by $3$ .

Pause the lecture

Can a real $2 \times 2$ matrix have no real eigenvectors?

Common mistake. Including $v = 0$ as an eigenvector. Zero is always a solution of $(A - λ I) v = 0$ ; it is excluded by definition. Eigenvectors are nonzero.

Practice this lecture

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