Eigenvectors: the directions a matrix only stretches

Which vectors does a matrix keep on their own line through 0?

For a nonzero vector \(v\), staying on its own line means \(Av = \lambda v\) for a single number \(\lambda\): on that line the matrix acts like multiplication by \(\lambda\). Then \(v\) is an eigenvector and \(\lambda\) is its eigenvalue.

Before you start · 3 quick checks on columns, vectors on one line and determinants, about 2 min

Setup: eigenvectors and eigenvalues

From the first page on matrices: a real 2×2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) sends a vector \(x = (x_1, x_2)\) to \(x_1Ae_1 + x_2Ae_2\), where \(Ae_1 = (a, c)\) and \(Ae_2 = (b, d)\) are its columns.

This page works in the real plane: the entries of \(A\) are real numbers, vectors are in \(\mathbb{R}^2\), and every number \(\lambda\) is real. So "eigenvalue" and "eigenvector" on this page mean real eigenvalues and real eigenvectors.

A nonzero vector \(w\) lies on the line through 0 and a nonzero vector \(v\) exactly when \(w\) is a multiple of \(v\): \(w = \lambda v\) for some number \(\lambda\).

Definition. A vector \(v \ne 0\) in \(\mathbb{R}^2\) is an eigenvector of \(A\) if \(Av = \lambda v\) for some real number \(\lambda\), called the eigenvalue of \(A\) for \(v\). In the plane: \(Av\) lies on the line through 0 and \(v\). The eigenvalue can be positive, negative (then \(Av\) points the opposite way from \(v\)) or zero (then \(Av = 0\)).

If \(Av = \lambda v\), then \(A(tv) = tAv = \lambda(tv)\) for every number \(t\). So every nonzero vector on the line through 0 and \(v\) is an eigenvector with the same eigenvalue, and that line is called an eigen-direction of \(A\).

Why the zero vector is left out

\(A0 = 0 = \lambda 0\) for every number \(\lambda\). If the zero vector counted as an eigenvector, every number would be an eigenvalue of every matrix, and the word would carry no information.

1 See

Find the directions that stay on their line

The dot is the tip of a vector \(v\) of length 1. Turn it around the circle: drag it, focus it and use the arrow keys, or type an angle. The figure draws \(Av\) for the matrix of the current goal. When \(Av\) lies on the dashed line through 0 and \(v\), press Record direction.

Angle of v, in degrees

How the angle, the tolerance and recording work

2 Try

Count the directions, then build matrices

Two questions first. The explorer below opens when you have answered both.

The explorer opens when you have answered the two questions above.

3 Prove

Why the eigenvalues solve det(A − λI) = 0

Theorem

Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) be any real 2×2 matrix and \(\lambda\) any real number. Then \(\lambda\) is an eigenvalue of \(A\) exactly when \(\det(A - \lambda I) = 0\), and

\[ \begin{aligned} \det(A - \lambda I) &= (a - \lambda)(d - \lambda) - bc \\ &= \lambda^2 - (\operatorname{tr} A)\lambda + \det A. \end{aligned} \]

So the eigenvalues of \(A\) are the real roots of its characteristic polynomial \(\lambda^2 - (\operatorname{tr} A)\lambda + \det A\), with eigenvectors in \(\mathbb{R}^2\). Over the complex numbers a polynomial with no real root still has complex roots, such as \(\pm i\) for the quarter turn of stage 1, where \(\lambda^2 + 1 = 0\); complex eigenvalues are outside this page.

Here \(I\) is the identity matrix, so \(\lambda I\) has \(\lambda\) on the diagonal and 0 elsewhere, and \(\lambda I v = \lambda v\) for every vector \(v\). Build the proof one line at a time; each line has one correct choice.

What follows

Finding the eigenvectors

For an eigenvalue \(\lambda\), the eigenvectors are the nonzero solutions of \((A - \lambda I)v = 0\). If \(A - \lambda I\) is the zero matrix, every nonzero vector is one. Otherwise its determinant is 0, so its two rows lie on one line through 0, and \((A - \lambda I)v = 0\) says the same as \(pv_1 + qv_2 = 0\) for one nonzero row \((p, q)\). Its solutions are the multiples of \((q, -p)\): one line.

How many eigen-directions

The characteristic polynomial has two, one or no real roots, depending on the sign of \((\operatorname{tr} A)^2 - 4\det A\).

Positive: two different eigenvalues. For each, \(A - \lambda I\) is not the zero matrix (otherwise \(A = \lambda I\), a multiple of \(I\), which has only one eigenvalue), so each gives exactly one line. The two lines differ, since a nonzero \(v\) with \(Av = \lambda_1 v = \lambda_2 v\) would force \(\lambda_1 = \lambda_2\). Two eigen-directions.
Zero: one eigenvalue, \(\lambda = \tfrac12 \operatorname{tr} A\). If \(A = \lambda I\), every direction is an eigen-direction; otherwise exactly one is.
Negative: no real eigenvalue and no eigen-direction in the plane.

Trace, determinant and the eigenvalues

When the characteristic polynomial has real roots \(\lambda_1\) and \(\lambda_2\) (equal when there is only one), it factors as \((\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2\). Comparing coefficients:

\[ \lambda_1 + \lambda_2 = \operatorname{tr} A, \qquad \lambda_1\lambda_2 = \det A. \]

In pictures, when \(A\) has two eigen-directions with eigenvectors \(v_1\) and \(v_2\): \(A\) sends the parallelogram with sides \(v_1\) and \(v_2\) to the one with sides \(\lambda_1v_1\) and \(\lambda_2v_2\), whose signed area is \(\lambda_1\lambda_2\) times as large. That factor is \(\det A\), as on the determinant page.

4 Apply

Problems

Each problem has hints and a full solution. Answers are graded exactly: type whole numbers, exact decimals such as 0.5, or fractions such as 1/2, without thousands separators. Where a problem asks for two numbers, separate them with a comma.

Key point To find eigenvalues, solve \(\lambda^2 - (\operatorname{tr} A)\lambda + \det A = 0\). For each root \(\lambda\), solve \((A - \lambda I)v = 0\) for a nonzero \(v\). An eigenvector for the eigenvalue 1 is a vector that \(A\) leaves unchanged, such as a split of a market that stays the same from month to month.

Level Intro Prerequisites A matrix is a function that moves the plane Time ~35 min

Related: Matrix multiplication is composition and What the determinant measures. Next in linear algebra: Solving Ax = b: one, none, or infinitely many solutions (in preparation).